Commutative Algebra and Algebraic Geometry

Robert Friedman

August 1, 2006 2 i

Disclaimer: These are rough notes for a course on commutative algebra and algebraic geometry. I would appreciate all suggestions concerning ty- pos, errors, unclear exposition or obscure exercises, and any other helpful feedback. Contents

1 Introduction to Commutative Rings 1 1.1 Introduction...... 1 1.2 Primeideals...... 5 1.3 Localrings...... 9 1.4 Factorizationinintegraldomains ...... 10 1.5 Motivatingquestions ...... 18 1.6 Theprimeandmaximalspectrum ...... 25 1.7 Gradedringsandprojectivespaces ...... 31 Exercises...... 37

2 Modules over Commutative Rings 43 2.1 Basicdeﬁnitions...... 43 2.2 Directandinverselimits ...... 48 2.3 Exactsequences...... 54 2.4 Chainconditions ...... 59 2.5 ExactnesspropertiesofHom...... 62 2.6 ModulesoveraPID...... 67 2.7 Nakayama’slemma ...... 70 2.8 Thetensorproduct ...... 73 2.9 Productsofaﬃnealgebraicsets ...... 81 2.10Flatness ...... 84 Exercises...... 89

3 Localization 98 3.1 Basicdeﬁnitions...... 98 3.2 Idealsinalocalization ...... 101 3.3 Localizationofmodules...... 103 3.4 Localpropertiesofringsandmodules ...... 104

ii CONTENTS iii

3.5 Regularfunctions ...... 107 3.6 Introductiontosheaves...... 112 3.7 Schemesandringedspaces ...... 118 3.8 Projasascheme ...... 123 3.9 Zariskilocalproperties ...... 126 Exercises...... 127

4 Integral Ring Homomorphisms 135 4.1 Deﬁnitionofanintegralhomomorphism ...... 135 4.2 The going up theorem and dimension ...... 140 4.3 Thegoingdowntheorem ...... 144 4.4 Moreondimension ...... 146 4.5 Noethernormalization,version1 ...... 148 4.6 Noethernormalization,version2 ...... 154 Exercises...... 158

5 Ideals in Noetherian Rings 166 5.1 Irreducible sets and radical ideals ...... 166 5.2 Associatedprimes...... 169 5.3 Primarydecomposition ...... 174 5.4 Artinianrings ...... 179 5.5 Fractional ideals and invertible modules ...... 184 5.6 Dedekinddomains ...... 192 5.7 Higherdimensions ...... 199 5.8 Extensions of Dedekind domains ...... 200 Exercises...... 208

6 Quasiprojective varieties 215 6.1 Projective and quasiprojective varieties ...... 215 6.2 Localringsandtangentspaces...... 216 6.3 Derivations, diﬀerentials, and tangent spaces ...... 225 6.4 Elementaryprojectivegeometry ...... 231 6.5 Products...... 238 6.6 Grassmannians ...... 244 Exercises...... 245 iv CONTENTS

7 Graded Rings 248 7.1 FiltrationsandtheArtin-Reeslemma ...... 248 7.2 Hilbertfunctions ...... 252 7.3 Thedegreeofaprojectivevariety ...... 256 7.4 Thedimensiontheorem...... 264 7.5 Applications of the dimension theorem ...... 268 7.6 Regularlocalrings ...... 273 7.7 Completions...... 275 7.8 Hensel’s lemma and the implicit function theorem ...... 285 Exercises...... 291 Chapter 1

Introduction to Commutative Rings

1.1 Introduction

Commutative algebra is primarily the study of those rings which most naturally arise in algebraic geometry and number theory.For example, let k be a field (typically algebraically closed, and often the field C of complex numbers). Then (affine) algebraic geometry is to a large extent the study of the ring R = k[x1,...,xn] and associated objects, for example ideals I in R or the corresponding quotients R/I. In number theory, one studies the ring Z 1+√ 3 as well as related rings, for example Z[i], where i = √ 1, or Z[ − ]. There − 2 are other kinds of associated rings, for example the field of quotients Q of Z or of k[x1,...,xn] (which is denoted by k(x1,...,xn), and called the field of rational functions in n variables). More complicated rings derived from the standard ones are for example the ring of p-adic integers Zp (defined for every prime number p), as well as its polynomial analogue, the ring of formal power series in n variables, denoted by k[[x1,...,xn]]. We will discuss the construction of these and other rings later. Although the rings above are very disparate, they have a great deal of structure in common, and one of the goals of commutative algebra is to elucidate this structure. One obvious property that all of the above rings share is that they are commutative, and as its name suggest, commutative algebra is almost exclu- sively concerned with such rings. In fact, from now on we shall always make the assumptions:

1 2 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS

Assumption: Every ring R is commutative, with a unity 1. (Note that the zero ring 0 is allowed.) Every ring homomorphism ϕ: R S is required to satisfy ϕ(1) = 1. In other words, ϕ takes the unique unity→ in R to the corresponding unity in S. In particular, if R′ is a subring of R, then we require that 1 R′. ∈ Thus, for example, if R1 and R2 are rings, then the Cartesian product R R is also a ring, with unity (1, 1), and the projections π : R R R 1 × 2 i 1 × 2 → i are ring homomorphisms in the above sense. However, in case R1 and R2 are nonzero, the subsets R1 0 and 0 R2 are not subrings in the sense described above. In more×{ technical} { terms,} × R R is a product in the 1 × 2 category of commutative rings (with unity), but not a coproduct. We shall describe the coproduct later. Quite often, rings come in pairs. Thus, implicit in the definition of k[x1,...,xn] is the ring homomorphism k k[x1,...,xn]; likewise, there are the obvious homomorphisms Z Z[i] or→Z Q or Z Z/nZ. Given → → → a ring homomorphism ϕ: R S, not necessarily injective, we call S an R-algebra, and refer (if there is→ any ambiguity) to the given homomorphism ϕ as the structural homomorphism. Morphisms, subalgebras, and quotients are required to be compatible with the given structural homomorphism. For example, if I is an ideal in R, then R/I is an R-algebra via the natural projection R R/I. → Another standard example of an R-algebra is the ring R[x1,...,xn] of polynomials in several variables with coefficients in R, with the obvious inclusion R R[x ,...,x ]. Here, R[x ,...,x ] is the set of all expressions → 1 n 1 n r xa1 xan , a1,...,an 1 ··· n a1,...,an 0 X≥ where r R and r = 0 for only finitely many indices a ,...,a . a1,...,an ∈ a1,...,an 6 1 n More formally, we can think of R[x1,...,xn] as the set of all “multi-sequences” in R, indexed by n-tuples of nonnegative integers, and such that only finitely many terms are nonzero. Addition of polynomials is defined componentwise:

r xa1 xan + s xa1 xan = (r +s )xa1 xan . a1,...,an 1 ··· n a1,...,an 1 ··· n a1,...,an a1,...,an 1 ··· n X X X Multiplication is deﬁned so as to insure that

(xa1 xan )(xb1 xbn )= xa1+b1 xan+bn . 1 ··· n 1 ··· n 1 ··· n 1.1. INTRODUCTION 3

This can be written eﬃciently with vector notation: if ~a =(a1,...,an) is an n-tuple of nonnegative integers, and we denote xa1 xan by x~a, then 1 ··· n

~a ~a ~b+~c r~ax s~ax = r~bs~cx . ! !   X~a X~a X~a ~b+X~c=~a   If, in the definition of R[x1,...,xn], we omit the requirement that only finitely many coefficients ra1,...,an are nonzero, we obtain a larger ring containing R and R[x1,...,xn], the ring of formal power series R[[x1,...,xn]]. The R-algebra R[x1,...,xn] has the following universal property with respect to R-algebras: if S is an R-algebra, to give an R-algebra homomorphism ϕ: R[x ,...,x ] S is equivalent to specifying n elements α ,...,α S. 1 n → 1 n ∈ In one direction, the αi are just given by ϕ(xi), so that ϕ determines n elements α ,...,α S. Conversely, given n elements α ,...,α S, the 1 n ∈ 1 n ∈ evaluation homomorphism ev : R[x ,...,x ] S defined by α1,...,αn 1 n →

evα1,...,αn (f(x1,...,xn)) = f(α1,...,αn) defines a homomorphism ϕ: R[x ,...,x ] S, and these constructions are 1 n → clearly inverse. In general, if an R-algebra S is the quotient of R[x1,...,xn] (in the sense of R-algebras), then we say that S is a finitely generated R- algebra. If in addition there exists a surjection R[x ,...,x ] S whose 1 n → kernel is a finitely generated ideal of R[x1,...,xn], then we say that S is a finitely presented R-algebra. Both algebraic geometry and algebraic number theory are very much relative theories, often concerned not just with a single ring R but rather with a ring R and an R-algebra S, and we shall try to emphasize this aspect a much as possible. The set of ideals I in a ring R, and in particular the sets of prime and maximal ideals, are among the most important objects of study in algebra. They can be generalized to modules: roughly speaking, a module over a ring R is the formal analogy of a vector space over a field k:

Deﬁnition 1.1.1. An R-module M consists of an abelian group M (with group operation denoted by addition), together with a function R M M, whose value at (r, m) is denoted by r m or rm, and is usually referred× → to as · multiplication, satisfying:

(i) For all r, s R and m M, r(sm)=(rs)m; ∈ ∈ 4 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS

(ii) For all r ,r R and m M, (r + r )m = r m + r m; 1 2 ∈ ∈ 1 2 1 2 (iii) For all r R and m , m M, r(m + m )= rm + rm ; ∈ 1 2 ∈ 1 2 1 2 (iv) For all m M, 1 m = m. ∈ · A submodule N of an R-module module M is deﬁned in the obvious way. If N is a submodule of an R-module module M, then there is a natural structure of an R-module on the abelian group M/N, and we will refer to this structure as the quotient R-module.

If M and N are R-modules, then a homomorphism f : M N is a → homomorphism of abelian groups satisfying f(rm) = rf(m) for all r R and m M. Isomorphisms of R-modules are deﬁned similarly. We write∈ ∈ M ∼= N if M and N are isomorphic. Example 1.1.2. The following are standard examples of modules:

(i) If k is a ﬁeld, then a k-module V is the same thing as a k-vector space.

(ii) A Z-module M is the same thing as an abelian group. Indeed, given a Z-module M, we associate to M its underlying abelian group (also denoted M). Conversely, if A is an abelian group, for all a A and ∈ n Z, deﬁning n a in the usual way gives A the structure of a Z- module.∈ ·

(iii) For R an arbitrary ring, an R-submodule of R is the same thing as an ideal I in R, and a quotient module of R is the same as a quotient R/I.

(iv) Again with R arbitrary, the n-fold Cartesian product Rn of all ordered n-tuples of elements of R is an R-module in the usual way, via componentwise addition and scalar multiplication:

(r1,...,rn)+(s1,...,sn)=(r1 + s1,...,rn + sn);

r(r1,...,rn)=(rr1,...,rrn).

The module Rn is called the free R-module of rank n.

(v) If M and M are two R-modules, we deﬁne the direct sum M M to 1 2 1 ⊕ 2 be the set (m1, m2) : mi Mi with addition and scalar multiplication deﬁned componentwise.{ Thus∈ } as a set M M is just the Cartesian 1 ⊕ 2 1.2. PRIME IDEALS 5

product of M and M . The direct sum M M of finitely many 1 2 1 ⊕···⊕ n R-modules is defined in a similar way. In particular, Rn = R R is the direct sum of n copies of R. ⊕···⊕ We shall consider more general kinds of sums and products of modules in the next chapter. However, at the end of this chapter, we shall need 0 the case of an infinite direct sum indexed by N (or by the set Z≥ of nonnegative integers). Given a collection M , n N of R-modules, n ∈ we define n N Mn to be the set of all sequences (m1, m2,... ), with ∈ mi Mi and mn = 0 for all sufficiently large n, or equivalently for all but∈ finitelyL many n. Addition and scalar multiplication are, as usual, defined pointwise.

Just as it is important to understand all ideals in a ring, we will also try to describe all R-modules up to isomorphism. While this is not in general possible, it is feasible for special classes of rings R and R-modules M, and we will frequently return to this question.

1.2 Prime ideals

Fix a ring R. The intersection of an arbitrary collection Iα α A of ideals { } ∈ in R is an ideal α A Iα. Given ideals I and J of R, we can also define the ideal product ∈ T I J = r s : r I,s J , · { i i i ∈ i ∈ } i X and it is an ideal in R contained in I J. For example, RI = I. Products of ∩ a finite or infinite number of ideals, and powers IN of an ideal I, are defined similarly. Ideal product is associative, i.e. (IJ)K = I(JK). We can also define the ideal sum

I + J = r + s : r I,s J . { ∈ ∈ } It is the smallest ideal of R containing both I and J. Moreover, the distribu- tive laws hold: I(J + K)= IJ + IK.

Inﬁnite sums α A Iα are deﬁned similarly. ∈ P 6 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS

Given r ,...,r R, let (r ,...,r )R =(r ,...,r ), the ideal generated 1 n ∈ 1 n 1 n by r1,...,rn, be the smallest ideal containing r1,...,rn. Clearly

(r ,...,r )R = t r : t R . 1 n { i i i ∈ } i X In particular rR =(r)= tr : t R is called the principal ideal generated by r. There is a similar{ notation∈ for} ideals generated by inﬁnitely many elements.

Definition 1.2.1. A zero divisor in R is an element r R,r = 0 such that ∈ 6 there exists an s R with s = 0 and rs = 0. The ring R is an integral domain if R = 0 and∈ R has no6 zero divisors, in other words if given r, s R 6 ∈ with rs = 0, then either r or s is zero. For example, Z is an integral domain, and Z/nZ is an integral domain if and only if n is prime. A field is an integral domain; in particular, a field always has at least two distinct elements 0 and 1. An ideal p in a commutative ring R is a prime ideal if p = R and, for all r, s R, if rs p then either r p or s p. Clearly, p is6 a prime ideal if ∈ ∈ ∈ ∈ and only if R/p is an integral domain.

A subring of an integral domain is again an integral domain. A field is an integral domain, and therefore every subring of a field is also an integral domain. Conversely, every integral domain R is contained in a field. In fact, one constructs the fraction field or field of quotients of an integral domain R by analogy with the construction of the rational numbers from the integers, and this field is in a sense the smallest field containing R. We shall describe this construction later. Note that if R is an integral domain, then so is R[x] (by considering terms of top degree) and therefore also R[x1,...,xn] by induction. In particular, if k is a field, then k[x1,...,xn] is an integral domain whose quotient field k(x1,...,xn) is called the field of rational functions in x1,...,xn. Lemma 1.2.2. Given a ring homomorphism f : R S and a prime ideal q 1 → in S, the ideal f − (q) is a prime ideal in R.

1 Proof. The ring R/f − (q) is a subring of S/q and hence is an integral domain.

Deﬁnition 1.2.3. A nilpotent element of R is an element r such that rn =0 for some n > 0. The ring R is reduced if the only nilpotent element is 0. 1.2. PRIME IDEALS 7

The radical of R (written √0) is the set of r R such that there exists an ∈ n > 0 such that rn = 0, i.e. such that r is nilpotent. Thus R is reduced if and only if √0 = 0. More generally, for an ideal I, we deﬁne the radical of I to be the set √I = r R : rn I for some n> 0 . { ∈ ∈ } Thus √I is the preimage in R of the set of nilpotent elements of R/I. An ideal I is a radical ideal if I = √I. The ideal I is a radical ideal if and only if R/I is reduced.

Lemma 1.2.4. For every ideal I, √I is an ideal of R.

Proof. First consider the case I = (0). If rn = 0 and sm = 0, then (r + s)n+m = 0, by applying the binomial theorem to (r + s)n+m. Thus √0 is closed under addition, and if rn = 0, then clearly (sr)n = 0 for all s R. So √0 is an ideal. For an arbitrary ideal I, √I is the inverse image of∈ the radical of R/I under the natural map R R/I, and so is an ideal in R. → Deﬁnition 1.2.5. A maximal ideal m of R is an ideal m such that m = R, and if I is an ideal such that m I, then I = m or I = R. 6 ⊆ Lemma 1.2.6. The ideal m is maximal if and only if R/m is a ﬁeld. Hence a maximal ideal is prime.

Proof. The ideal m is maximal if and only if R/m has no proper ideals. Thus we must show that a nonzero commutative ring k is a field if and only if it has no proper ideals. Clearly if k is a field, then it has no proper ideals. Conversely, if k is a commutative ring without proper ideals, then given a k, a = 0, the ideal (a) is a nonzero ideal and so (a)= R. Thus 1 (a), so∈ that there6 exists x R with ax = 1. Thus k is a field. ∈ ∈ The following will be used repeatedly:

Lemma 1.2.7. If R is a commutative ring and I is an ideal of R such that I = R, then I is contained in a maximal ideal m. 6 Proof. Consider the set of all ideals containing I and not equal to R, partially ordered by inclusion. We claim that we can apply Zorn’s lemma to this set to produce a maximal element, in other words a maximal ideal. It suﬃces to note that if J , α A is a totally ordered set of proper ideals containing α ∈ I, then α Jα is an ideal (because every two elements are contained in a Jβ S 8 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS for some β, so we can add), containing I, and = R (since if 1 J , then 6 ∈ α α 1 Jα for some α). Thus the hypotheses of Zorn’s lemma are satisﬁed, and a∈ maximal element of this set is a maximal ideal containing I. S Corollary 1.2.8. If R =0, then there exists a maximal ideal of R. 6 Corollary 1.2.9. An element u of R is a unit if and only if u does not lie in any maximal ideal. Proof. If u is a unit, then an ideal containing u contains 1 and so is equal to R, so that u is not contained in a maximal ideal. Conversely, if u is not a unit, then (u) is a proper ideal and thus it is contained in a maximal ideal m. So u m. ∈ Another application of Zorn’s lemma gives another description of the radical of an ideal: Lemma 1.2.10. Let I = R be an ideal. Then √I is the intersection of the prime ideals containing 6 I. Proof. First consider the case where I = (0) and √0 is the ideal of all nilpotent elements of R. Clearly, if r is nilpotent, then r p for every prime ideal ∈ p of R. Conversely, suppose that r is not nilpotent. Consider the set of all ideals J of R such that, for all n > 0, rn / J (note that such an idealI is automatically not equal to R). Clearly =∈, since (0) . A standard I 6 ∅ ∈ I application of Zorn’s lemma to , partially ordered by inclusion, shows that contains a maximal element pI. To see that p is prime, we must show that I for all a / p, b / p, the product ab / p. Since a / p, the ideal (a)+ p strictly ∈ ∈ ∈ ∈ contains p and hence does not lie in . Thus there exist s1 R, p1 p, and n1 I ∈ ∈ n1 > 0 such that r = as1 + p1. Likewise there exist s2 R, p2 p, and n2 ∈ ∈ n2 > 0 such that r = bs2 + p2. But then

n1+n2 r = abs1s2 + p for some p p, and so (ab)+ p strictly contains p as well. Thus ab / p. The case∈ of a general I now follows by applying the above to∈ the ring R/I and noting that the prime ideals of R containing I are in one-to-one correspondence with the prime ideals of R/I. Corollary 1.2.11. An ideal I is an intersection of prime ideals if and only if I is a radical ideal, i.e. I = √I. 1.3. LOCAL RINGS 9

Proof. Clearly, if I = √I, then I = √I is the intersection of all of the prime ideals containing I. Conversely, if I is an intersection of prime ideals, then an easy exercise shows that I = √I.

In fact, if R has stronger ﬁniteness hypotheses, for example if R = k[x1,...,xn], then every radical ideal is an intersection of ﬁnitely many prime ideals, and the prime ideals which appear are essentially unique if we impose an appropriate minimality condition. This is one analogue of unique factorization into primes. For ideals which are not radical ideals, there is a more subtle kind of “factorization,” called primary decomposition, which we shall discuss in Chapter 5.

1.3 Local rings

Deﬁnition 1.3.1. A ring R is local if there is a unique maximal ideal m of R. For example, a ﬁeld k is a local ring. If R is a local ring with maximal ideal m and S is a local ring with maximal ideal n, then a ring homomorphism 1 ϕ: R S is local if ϕ(m) n. Note that ϕ(m) n if and only if m ϕ− (n), → 1 ⊆ ⊆1 ⊆ if and only if m = ϕ− (n) since n and hence ϕ− (n) are proper ideals.

The following is immediate from Corollary 1.2.9

Lemma 1.3.2. R is local if and only if the set of all r in R such that R is not a unit is an ideal of R.

For example, the ring k[[x1,...,xn]] of formal power series with coefficients in a field k is a local ring with maximal ideal m = (x1,...,xn). It suffices to see that, if f(x ,...,x ) k[[x ,...,x ]] has constant term equal 1 n ∈ 1 n to 1, then f is invertible. This follows by formally taking the inverse: 1 =1 g + g2 , 1+ g − −··· and this infinite series makes sense if g m. Likewise the ring Z of p-adic ∈ p integers is a local ring. We shall discuss such rings in more detail later, and give more elementary methods for constructing local rings. Local rings are of fundamental importance in commutative algebra and many questions about rings and modules can be reduced to questions about local rings. More generally, there is the following definition: 10 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS

Deﬁnition 1.3.3. The Jacobson radical of R is the intersection of all the maximal ideals of R.

For example, if R is local with maximal ideal m, then its Jacobson radical is just m.

Lemma 1.3.4. The Jacobson radical of R is the set of all r R such that, ∈ for every s R, 1+ rs is a unit. ∈ Proof. Suppose that r lies in m for every maximal ideal m. Then so does rs for every s R. Thus 1 + rs / m, since otherwise 1 m. It follows that 1+ rs is contained∈ in no maximal∈ ideal, and is thus a unit.∈ Conversely, suppose that r is not contained in some maximal ideal m. Then the ideal (r, m) generated by r and m is an ideal properly containing m and so is R. Hence we can write 1 = t + ry where t m and y R. Thus 1+ r( y) m is not a unit. ∈ ∈ − ∈ 1.4 Factorization in integral domains

Throughout this section, R denotes an integral domain. Our goal is to collect some elementary results about factorization.

Deﬁnition 1.4.1. The element r R divides r′ (which we write as r r′) if ∈ | there exists x R such that r′ = xr. Two elements r, r′ are associates if ∈ r′ = ur, where u is a unit. Equivalently, r and r′ are associates if (r)=(r′). An element r R 0 is irreducible if r is not a unit and if whenever r = st for s, t R, then∈ s−{or }t is a unit. Thus r = 0 is irreducible if and only if the only elements∈ of R which divide r are either6 units or associates of r.

Lemma 1.4.2. If (r) is a prime ideal and r =0, then r is irreducible. 6 Proof. If r = st, then s or t lies in (r). We may suppose that s (r), say s = ru. Then r = rut, so that ut = 1 and u is a unit. ∈

However, the converse to the above need not hold: if r is irreducible, then (r) need not be a prime ideal. For example, in Z[√ 5], it is easy to check − that 2 is irreducible, but (2) is not prime since 6 = 2 3 = (1+√ 5)(1 √ 5), and neither 1 + √ 5 nor 1 √ 5 is in (2). One important· case− where− − this does hold is the case− where −R has− unique factorization: 1.4. FACTORIZATION IN INTEGRAL DOMAINS 11

Deﬁnition 1.4.3. An integral domain R is a unique factorization domain (UFD) if, given r not a unit in R, r = p1 pn, where the pi are irreducible elements, and moreover if r = q q ,··· where the q are also irreducible, 1 ··· m i then n = m and possibly after relabeling the qi, pi and qi are associates for all i.

In a UFD, two nonzero elements r and s are relatively prime if, whenever t r and t s, then t is a unit. We can define a greatest common divisor (gcd) | | of two nonzero elements r and s to be an element d such that d r, d s, and whenever e r and e s, then e d. Two gcds of r and s are associates.| Greatest| common divisors| for| a finite| collection of elements of R are defined similarly. An easy argument shows that gcds exist in a UFD. The gcd of tr1,...,trn is t times the gcd of r1,...,rn. If r is an irreducible element and s, t = 0, if r st, then by unique factorization r s or r t. 6 | | | Lemma 1.4.4. Let R be a UFD. Then r is irreducible if and only if (r) is a nonzero prime ideal. Moreover, in this case (r) is a minimal nonzero prime ideal, and every minimal nonzero prime ideal p is of the form (r) where r is irreducible.

Proof. We have seen that (r) nonzero and prime = r irreducible. Con- ⇒ versely, suppose that r is irreducible, and suppose that st (r). Then st = rx for some x R, and we claim that either s or t lies in∈ R. We may ∈ clearly assume that s and t are both nonzero. Then as r st, either r s or r t and so either s or t lies in (r). | | | Now suppose that p is any nonzero prime ideal. Then p contains a nonzero element x which is not a unit. At least one irreducible factor r of x must lie in p. Then (r) p, so that every nonzero prime ideal p contains a nonzero prime ideal of the⊆ form (r). Thus if p is a minimal nonzero prime ideal, then p = (r) for some irreducible element r. If now r is irreducible and p is a nonzero prime ideal contained in (r), then there exists an irreducible r′ such that (r′) p (r). Thus r′ = ru for some u R. Since r′ is irreducible and ⊆ ⊆ ∈ r is not a unit, u is a unit and thus (r)=(r′)= p. Thus (r) is minimal.

Basic examples of UFD’s come from the following deﬁnition:

Deﬁnition 1.4.5. An integral domain R is a principal ideal domain (PID) if every ideal in R is principal. 12 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS

For example, the ring Z is a PID. Likewise, if k is a field, then k[x] is a PID. In fact, for a field k we always have the following (long division with remainder for polynomials): Proposition 1.4.6. Let k be a field and let f(x),g(x) k[x] with g = 0. Then there exist unique q,r k[x] such that f = gq +∈ r with r =6 0 or deg r< deg g. ∈ Proof. Existence: induct on deg f, the cases f = 0 or deg f < deg g = d are clear. Write g = d a xi, with a = 0, and suppose that f = n b xi i=0 i d 6 i=0 i with n d and bn = 0. Then ≥ P6 P 1 n d h = f a− b x − g − d n has degree strictly less than n (or is zero). Thus h = gq′ + r with r = 0 or 1 n d deg r< deg g, and so f = g(q′ + ad− bnx − )+ r = gq + r as desired. To see uniqueness, if gq +r = gq′ +r′ with r′ = 0or deg r′ < deg g as well, then g (r r′) and deg(r r′) < deg g, which is only possible if r r′ = 0, | − − − so that r = r′. In this case gq = gq′, and so q = q′ as well. Remark 1.4.7. More generally, if R is any ring (not necessarily an integral domain), and g(x) is monic (i.e. the leading coefficient of g is 1), then the statement and proof of the proposition continue to hold. Corollary 1.4.8. If k is a field, then k[x] is a PID. Proof. If I is an ideal of k[x] and I = 0, choose a nonzero element g I of 6 ∈ minimal degree. If f I, write f = gq + r with deg r< deg g or r = 0. Since r = f gq I, we must∈ have r = 0 by the choice of g. Thus I =(g). − ∈ As usual, we define a root of f k[x] in a field K k to be an element ∈ ⊇ α K such that f(α)=0. ∈ Corollary 1.4.9. If α k, then α is a root of f if and only if (x α) divides ∈ − f. If f k[x] has degree n, then there are at most n elements α k such that f(α∈)=0. ∈ Proof. Given α k, we can write f(x)=(x α)q + r, where r k is a constant polynomial.∈ Thus f(α) = r, and so− f(α) = 0 if and only∈ if (x α) f. Clearly, if α = α , the polynomials (x α ) and (x α ) are − | 1 6 2 − 1 − 2 irreducibles which are not associates. Thus if α1,...,αk are roots of f, then (x α ) (x α ) f. It follows that k deg f. − 1 ··· − k | ≤ 1.4. FACTORIZATION IN INTEGRAL DOMAINS 13

Corollary 1.4.10. If k is an inﬁnite ﬁeld and f k[x] induces the zero ∈ function from k to itself, then f =0. We leave the corresponding statement about f k[x ,...,x ] as an ex- ∈ 1 n ercise. We now show that every principal ideal domain is in fact a unique factorization domain. Theorem 1.4.11. A PID is a UFD. Proof. The proof will be in several steps. Lemma 1.4.12. Let R be an integral domain with the property that, if

(a ) (a ) (a ) (a ) 1 ⊆ 2 ⊆···⊆ n ⊆ n+1 ⊆··· is an increasing sequence of principal ideals, then the sequence is eventually constant. Then every nonzero r R which is not a unit factors into a product of irreducibles. ∈ We can paraphrase the hypothesis of the lemma by saying that R satisfies the ascending chain condition (a.c.c) on principal ideals. Proof of Lemma 1.4.12. If not, let r R be an element, not zero or a unit, which does not factor into a product∈ of irreducibles. In particular, r itself is not irreducible, so that r = r1s1 where neither r1 nor s1 is a unit. Thus (r) is properly contained in (r1) and in (s1). Clearly, we can assume that at least one of r1, s1, say r1, does not factor into irreducibles (if both so factor, so does the product). By applying the above to r1, we see that (r1) is strictly contained in a principal ideal (r2), where r2 does not factor into a product of irreducibles. Continuing in this way, we can produce an infinite chain of principal ideals (r ) (r ) , each properly containing the previous one, 1 ⊂ 2 ⊂··· contradicting the hypothesis on R. Lemma 1.4.13. Suppose that R is an integral domain in which every nonzero element which is not a unit factors into a product of irreducibles. Moreover, suppose that r R is irreducible if and only if (r) is prime. Then R is a UFD. ∈ Proof. It suffices to show, in the usual way, that if an irreducible element r divides a product s1 sn, then r divides si for some i. This is an immediate consequence of the fact··· that (r) is prime. 14 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS

To complete the proof of Theorem 1.4.11, it suffices to show that a PID R satisfies the hypotheses of Lemma 1.4.12 and Lemma 1.4.13. First suppose that (r ) (r ) is an increasing sequence of ideals of R. It is easy 1 ⊆ 2 ⊆ ··· to check that I = i(ri) is again an ideal. Since R is a PID, I = (r) for some r R. Necessarily r (rN ) for some N. But then (r) (rN ) (r ) ∈ (r )=(S r). Thus∈ all inclusions are equalities, and (r⊆)=(r ⊆) N+1 ···⊆ i i n N for all n N. Thus the sequence is eventually constant. ≥ S Finally we must show that, if r R is irreducible, then (r) is a prime ideal. We begin by showing that gcds∈ exist in R and are of a very special form: Lemma 1.4.14. Let R be a PID, and let a, b R. Let d be a generator of the ideal (a, b), so that d = ar + bs for some r,∈ s R. Then d r, d s, and if ∈ | | e R is such that e r and e s, then e d. ∈ | | | Proof. Since a (a, b)=(d), d a, and by symmetry d b. Moreover, if e r and e s, then e ar +∈bs for all r, s | R, and hence e d. | | | | ∈ | For a, b R, let gcd(a, b)= d where d is any generator of (a, b). Clearly, d is uniquely∈ specified up to a unit. Moreover, d is uniquely specified by the property that d r, d s, and if e R is such that e r and e s, then e d. | | ∈ | | | Corollary 1.4.15. Let R be a PID, and let r R be an irreducible. Suppose that r st for some s, t R. Then either r s or∈r t. | ∈ | | Proof. Since gcd(r, s) r, either gcd(r, s) is an associate of r or gcd(r, s) is a unit. In the first case,| r s. In the second case, write 1 = rx + sy for some | x, y R. Then t = rxt + sty. Since r st, r t. ∈ | | Now suppose that r is irreducible, and that st (r), i.e. that r st. Since r is irreducible, it must divide one of s, t, and thus∈ either t or s lies| in (r). Thus (r) is a prime ideal. This then finishes the proof of Theorem 1.4.11.

In a PID, the gcd d of two elements r and s exists and is of the form ar + bs. This result fails for most UFD’s. For example, in k[x, y], the gcd of x and y is 1, but 1 cannot be written as ax + by. Related to the above is the following special property of a PID:

Lemma 1.4.16. If R is a PID, every nonzero prime ideal of R is maximal. 1.4. FACTORIZATION IN INTEGRAL DOMAINS 15

Proof. A nonzero prime ideal p in R is of the form (r) where r is irreducible. If p I, where I is an ideal and I = R, then I = (r′) for some r′. Since ⊆ 6 r′ r and r is irreducible, r and r′ must be associates. Thus I = p and so p is maximal.|

The ascending chain condition and the arguments of Theorem 1.4.11 are so fundamental that we generalize them as follows:

Proposition 1.4.17. For a ring R, the following two conditions are equivalent:

(i) Every ideal I of R is ﬁnitely generated: if I is an ideal of R, then I =(r ,...,r ) for some r R. 1 n i ∈ (ii) Every increasing sequence of ideals is eventually constant, in other words if I I I I , 1 ⊆ 2 ⊆···⊆ n ⊆ n+1 ⊆··· where the I are ideals of R, then there exists an N N such that for n ∈ all k N, I = I . ≥ k N If the ring R satisﬁes either of the equivalent conditions above, then R is called a Noetherian ring.

Proof. (i) = (ii): given an increasing sequence of ideals I I , let ⇒ 1 ⊆ 2 ⊆··· I = n In. Then as in the proof of Theorem 1.4.11, I is an ideal, and hence

I =(r1,...,rn) for some ri R. Thus ri Ini for some ni. If N = maxi ni, thenSr I for every i. Hence,∈ for all k ∈ N, I =(r ,...,r ) I I i ∈ N ≥ 1 n ⊆ N ⊆ k ⊆ I. It follows that I = I = I for all k N. k N ≥ (ii) = (i): Let I be an ideal of R and choose r I. Set I =(r ). If ⇒ 1 ∈ 1 1 I = I1, stop. Otherwise there exists an r2 I I1. Set I2 =(r1,r2). If I = I2, stop, otherwise there exists an r I I ∈. Inductively− suppose that we have 3 ∈ − 2 found Ik = (r1,...,rk) with Ik I. If I = Ik we are done, otherwise there exists r I I and we set⊆I = (r ,...,r ). So if I is not ﬁnitely k+1 ∈ − k k+1 1 k+1 generated, we have constructed a strictly increasing sequence I1 I2 , contradicting the assumption on R. Thus I is ﬁnitely generated.⊂ ⊂···

Clearly, the proof of Lemmas 1.4.12 and 1.4.13 together with Lemma 1.4.4 imply the following: 16 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS

Theorem 1.4.18. Suppose that R is a Noetherian integral domain. Then R is a UFD if and only if, for every nonzero r R, the element r is irreducible if and only if (r) is prime. ∈ Next we turn to the study of polynomials over a UFD. The main result here is: Theorem 1.4.19. If R is a UFD, then so is R[x]. By induction, we have the obvious corollary:

Corollary 1.4.20. The rings Z[x1,...,xn] and k[x1,...,xn] for k a ﬁeld are UFD’s.

We note that Z[x1,...,xn] is not a PID if n 1 and that, if k is a field, then k[x ,...,x ] is not a PID if n 2. ≥ 1 n ≥ Proof of Theorem 1.4.19. We begin with a preliminary definition. Given a nonzero polynomial f(x) = a xi R[x], define the content c(f) to be i i ∈ the gcd of the coefficients ai. Here c(f) is well-defined up to a unit, or as an P element of the multiplicative set R/R∗. In general f(x) = c(f)f0(x), where the gcd of the coefficients of f0 is 1. Define f to be primitive if c(f) = 1, in other words if the gcd of the coefficients of f is 1. If f = rf0 where f0 is primitive, then c(f)= r. More generally, if r R,r = 0, then c(rf)= rc(f). We will show: ∈ 6 Claim 1.4.21. Let R be a UFD with quotient field K. (i) If r R is irreducible, then r is an irreducible element of R[x]. ∈ (ii) If f (x) R[x] is a primitive polynomial which is irreducible in K[x], 0 ∈ then f0(x) is an irreducible element of R[x]. (iii) Every element of R[x] factors into a product of elements which are either irreducible elements in R or primitive polynomials which are irreducible in K[x]. Moreover such a factorization is unique up to order and associates. Clearly the claim implies Theorem 1.4.19, indeed somewhat more as it describes all of the irreducibles in R[x]. Proof of Claim 1.4.21. It is easy to check that all of the elements described in (i) and (ii) of the claim are indeed irreducible. Next we prove: 1.4. FACTORIZATION IN INTEGRAL DOMAINS 17

Lemma 1.4.22 (Gauss lemma). If f and g are primitive, then fg is primitive. Proof. Let r be an irreducible element of R such that r divides all coeﬃcients of fg. Given a polynomial in R[x], we use a bar to denote its reduction in R/rR, which is an integral domain since r is irreducible and R is a UFD. By hypothesis fg = 0, and so f¯g¯ = 0. Since (R/rR)[x] is an integral domain, either f¯ =0org ¯ = 0. In particular either f or g is not primitive, a contradiction. Corollary 1.4.23. For all f,g R[x], c(fg)= c(f)c(g). ∈ Proof. If f = c(f)f0 and g = c(g)g0 where f0 and g0 are primitive, then fg = c(f)c(g)f0g0, where, by the lemma, f0g0 is primitive. Hence c(fg) = c(f)c(g). Returning to the proof of Claim 1.4.21, given f R[x], let g ,...,g be ∈ 1 n the irreducible factors of f in K[x]. After clearing denominators, we can write g = λ h with h R[x], and by factoring out c(h ) we can assume i i i i ∈ i that hi is primitive. So

f =(λ λ )h h , 1 ··· k 1 ··· k where the λ K and the h R[x] are primitive. Thus h h is primitive i ∈ i ∈ 1 ··· k as well, by the Gauss lemma and induction. Now there is r R such that r(λ λ ) R. Thus ∈ 1 ··· k ∈ c(rf)= rc(f)= c(r(λ λ )h h )= r(λ λ ). 1 ··· k 1 ··· k 1 ··· k

It follows that c(f)= λ1 λk and in particular that λ1 λk R. We have now written f as ch ···h where the h are primitive··· polynomials∈ in R[x] 1 ··· k i which are irreducible in K[x]. If we further factor c into irreducibles in R, we have a factorization as claimed. We must ﬁnally show that this factorization is unique. Suppose that

f = r r h h = r′ r′ h′ h′ , 1 ··· n 1 ··· k 1 ··· m 1 ··· ℓ where the ri′ are irreducible in R and the hi′ are primitive polynomials in R[x] which are irreducible in K[x]. Up to a unit r r = r′ r′ since they 1 ··· n 1 ··· m are both equal to c(f), and so we can reorder until n = m and ri = ri′ up to associates by unique factorization in R. We also know by unique factorization 18 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS

in K[x] that k = ℓ and that after reordering we can assume that hi and hi′ are associates in K[x]. Thus h′ = th for some t K, t = 0. Choosing an i i ∈ 6 s such that st R, we have sh′ = sth , and taking contents we ﬁnd that s ∈ i i and st are associates. Thus t is a unit in R, and so hi and hi′ are associates in R[x]. As an easy corollary of Theorem 1.4.19, we can prove the famous Eisen- stein criterion for a polynomial to be irreducible:

Theorem 1.4.24 (Eisenstein criterion). Let R be a UFD with quotient ﬁeld K, let r R be an irreducible and let f = n a xi R[x] satisfy: ∈ i=0 i ∈ (i) For all i < n, r a ; P | i (ii) r ∤ a , so that a =0 and hence n = deg f; n n 6 2 (iii) r ∤ a0.

Then f is irreducible in K[x].

Proof. Since r ∤ an, the gcd of r and c(f) is a unit. Thus, if we write f = c(f)f0, then f0 is a primitive polynomial satisfying the same hypotheses as f. So we may as well assume that f is primitive in R[x]. In this case, f is irreducible in K[x] if and only if it is irreducible in R[x]. As usual, we shall let a bar stand for the image of a polynomial in (R/rR)[x]. If f = gh where g, h R[x] and both g and h have positive degree, then f¯ = gh =g ¯h¯. ∈ n d n d On the other hand, f¯ = cx , where c = 0. Thusg ¯ = c1x and h¯ = c2x − , and both have constant term equal to6 zero. It follows that r divides the 2 constant terms of both g and h, so that r a0. This contradicts (iii). Thus f is irreducible. |

1.5 Motivating questions in algebraic geometry and number theory

Now that we have some preliminary results concerning commutative rings, we describe some of the questions which arise naturally in the two main applications of commutative algebra. We begin with the case of algebraic geometry. Fix an algebraically closed field k (the classical case is k = C). In its crudest form, algebraic geometry 1.5. MOTIVATING QUESTIONS 19 seeks to study the common zeroes of polynomials in n variables. Tradition- n n ally, the vector space k is denoted by Ak in algebraic geometry, and is called affine n-space (over k). The allowable functions on this space are the functions which can be defined algebraically via the field operations, namely the polynomial functions, and these form the ring k[x1,...,xn]. (It is easy to see, using the fact that k is algebraically closed, that a rational function f/g is n well-defined on all of Ak if and only if it is a polynomial.) Given a polynomial f(x ,...,x ) k[x ,...,x ], its set of zeroes is 1 n ∈ 1 n V (f)= (x ,...,x ) An : f(x ,...,x )=0 . { 1 n ∈ k 1 n } More generally, given a finite collection f ,...f k[x ,...,x ], their set of 1 k ∈ 1 n common zeroes is defined in a similar way:

V (f ,...,f )= (x ,...,x ) An : f (x ,...,x )=0 for all i . 1 k { 1 n ∈ k i 1 n } k Thus V (f1,...,fk)= i=1 V (fi). Clearly, V (f1,...,fk) only depends on the ideal I =(f1,...,fk) generated T by the fi, and we shall also use the notation V (I) for V (f1,...,fk). A priori, it might seem that we could also consider inﬁnite collections of polynomials. However, a famous theorem of Hilbert says that it is always enough to look at the common zeroes of ﬁnitely many. In the language of ideals, the theorem is as follows:

Theorem 1.5.1 (Hilbert’s Basis Theorem). Every ideal in k[x1,...,xn] is ﬁnitely generated.

In fact, we shall prove the following:

Theorem 1.5.2 (Hilbert’s Basis Theorem II). If R is a Noetherian ring, then R[x] is Noetherian.

Corollary 1.5.3. If R is a Noetherian ring, then R[x1,...,xn] is Noethe- rian. In particular, the rings k[x1,...,xn] (k a ﬁeld) and Z[x1,...,xn] are Noetherian.

Proof of the basis theorem. Let I be an ideal in R[x]. Set

k I = r R : there exists f(x) I with f(x)= a xi and a = r . k { ∈ ∈ i k } i=0 X 20 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS

k i Clearly Ik is an ideal in R, and Ik Ik+1 since if f(x) = i=0 aix with k+1 i ⊆ k+1 ak = r, then xf(x)= ai 1x , so that the coefficient of x is r as well. i=1 − P Thus, since R is Noetherian, there exists an n0 such that Ik = In for all P 0 k n0, and In0 = (r1,...,rA) for ri R. By construction, for each i, 1 i ≥A, there exists an f (x) I of degree∈ n and leading coefficient r . For≤ ≤ i ∈ ≤ 0 i k n , the ideal I is finitely generated. Thus, after enlarging the collection ≤ 0 k fi(x) we can further assume that the ideals Ik,k n0, are generated by {leading} coefficients of suitable polynomials f (x) I≤of degree k. We claim i ∈ that I = (f1(x),...,fA(x)). Clearly (f1(x),...,fA(x)) I. Conversely, let f(x) I, and suppose that deg f(x) = k. By induction,⊆ starting with ∈ k = 0 n0, we can assume that all polynomials in I of degree less than k lie in≤ (f (x),...,f (x)). If k n , then the leading coefficient of f is in 1 A ≥ 0 the ideal generated by the leading coefficients of the fi, and since each fi has ai degree n0, it follows that f i x sifi has degree

(iii) V (I1+I2)= V (I1) V (I2). More generally, V ( α A Iα)= α A V (Iα). ∩ ∈ ∈ (iv) V (I I )= V (I I )= V (I ) V (I ). P T 1 · 2 1 ∩ 2 1 ∪ 2 Proof. The proofs of (i)—(iii) are easy and left to the reader. To see (iv), note that I I I I I . Thus V (I ) V (I I ) V (I I ). Since 1 · 2 ⊆ 1 ∩ 2 ⊆ 1 1 ⊆ 1 ∩ 2 ⊆ 1 · 2 the same holds for V (I2), we have V (I ) V (I ) V (I I ) V (I I ). 1 ∪ 2 ⊆ 1 ∩ 2 ⊆ 1 · 2 1.5. MOTIVATING QUESTIONS 21

On the other hand, if p / V (I ) V (I ), then p / V (I ) and p / V (I ). ∈ 1 ∪ 2 ∈ 2 ∈ 2 So there exist g I such that g (p) = 0, and likewise g I such that 1 ∈ 1 1 6 2 ∈ 2 g2(p) = 0. Thus g1g2(p) = 0, so that p / V (I1 I2). It follows that V (I1 I2) V (I )6 V (I ), so that equality6 holds throughout.∈ · This proves (iv). · ⊆ 1 ∪ 2 n Given a subset X of Ak , we can consider the ideal I(X) of all functions f k[x ,...,x ] such that f(p)=0 for all p X. Clearly, we have: ∈ 1 n ∈ Lemma 1.5.5. (i) If X X , then I(X ) I(X ). 1 ⊆ 2 2 ⊆ 1 (ii) I(X X )= I(X ) I(X ). 1 ∪ 2 1 ∩ 2 (iii) X V (I(X)). ⊆ (iv) I I(V (I)). ⊆ n The ideal I(X) is interesting for the following reason. Just as with Ak , we want to describe those functions on X which can be defined algebraically. It is natural to look at restrictions of polynomials in k[x1,...,xn] to the subset X. (We will see later that it makes no difference if we try to enlarge this to the set of rational functions which are everywhere defined on X.) Two polynomials have the same restriction to functions on X if and only if their difference lies in I(X). Thus the natural ring of algebraically defined functions on X is the quotient ring k[x1,...,xn]/I(X).

n Definition 1.5.6. Let X be a closed algebraic subset of Ak . We call X an affine algebraic set. The affine coordinate ring A(X) is the ring k[x1,...,xn]/I(X).

Given a closed algebraic set X = V (I), what is the relation between I and I(X)? On the one hand, we have the following easy result:

n Proposition 1.5.7. V (I(X)) is the smallest algebraic subset of Ak contain- n ing X. In particular, if X is an algebraic subset of Ak , then V (I(X)) = X.

n Proof. Clearly, V (I(X)) is an algebraic subset of Ak containing X. If Y = V (J) contains X, then J I(Y ) I(X), and hence V (I(X)) V (J)= Y . ⊆ ⊆ n ⊆ Thus V (I(X)) is the smallest algebraic subset of Ak containing X. The last statement is then clear.

In this context, Hilbert proved another fundamental result, which we shall prove later: 22 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS

Theorem 1.5.8 (Hilbert’s Nullstellensatz). In the above notation,

I(V (I)) = √I, where √I is the ideal of k[x1,...,xn] deﬁned by

√I = f k[x ,...,x ] : f N I for some positive integer N . { ∈ 1 n ∈ } Since I I(V (I)), √I I(V (I)), because if a power of a function f is ⊆ ⊆ zero at all points of a set, then the same must be true for the function f. So the point of the theorem is the opposite inclusion I(V (I)) √I. ⊆ Corollary 1.5.9. The function I V (I) is an order reversing bijection 7→ from the set of radical ideals of k[x1,...,xn] to the set of closed algebraic subsets of An, with inverse X I(X). k 7→ The proof we shall give for the Nullstellensatz will also be nonconstruc- tive. However, a great deal of interest has centered on finding effective versions of the Nullstellensatz. For example, if f vanishes on V (g1,...,gk), then m one wants to find an effective bound for an integer m such that f = i higi as well as bound the degrees of the hi. One very important special case of the Nullstellensatz, from whichP the full theorem can be derived by a formal trick, is the following:

Theorem 1.5.10. Let m be a maximal ideal of k[x1,...,xn]. Then there exists a point p =(c ,...,c ) An such that 1 n ∈ k m =(x c ,...,x c )= f k[x ,...,x ] : f(p)=0 = I( p ). 1 − 1 n − n { ∈ 1 n } { }

In particular, maximal ideals in k[x1,...,xn] are in one-to-one correspon- n dence with points of Ak . Note that the correspondence is as follows: the maximal ideal m corresponds to the point p = V (m), and the point p corresponds to the maximal ideal I( p ). { } n Corollary 1.5.11. If X is an algebraic subset of Ak , then the maximal ideals of A(X) are in one-to-one correspondence with the points of X. Proof. A maximal ideal of A(X) is the same thing as a maximal ideal m of k[x1,...,xn] containing I(X). But I(X) m if and only if V (m) V (I(X)) = X. ⊆ ⊆ 1.5. MOTIVATING QUESTIONS 23

The above corollary explains the signiﬁcance of the maximal ideals in k[x1,...,xn]. It is natural to characterize those algebraic sets corresponding more generally to prime ideals. This is the goal of the following deﬁnition:

n Deﬁnition 1.5.12. A nonempty closed algebraic subset X of Ak is irreducible if, whenever X = X1 X2, where the Xi are closed algebraic subsets n ∪ of Ak , then either X = X1 or X = X2.

n For example, a point is an irreducible closed subset of Ak .

n Proposition 1.5.13. An algebraic subset X of Ak is irreducible if and only if I(X) is a prime ideal.

Proof. First suppose that X = X X with neither X nor X equal to X. In 1∪ 2 1 2 particular Xi X and so I(X) I(Xi). We cannot have I(X)= I(Xi), for then X = V (I⊆(X)) = V (I(X )) =⊆ X . So there must exist f I(X ) I(X), i i i ∈ i − i =1, 2. But then f f I(X X )= I(X), and so I(X) is not prime. 1 2 ∈ 1 ∪ 2 Conversely, suppose that I(X) is not prime, and choose f / I(X), i = i ∈ 1, 2, such that f1f2 I(X). (Note that I(X) is a proper ideal since X = . Set X = V (f ) X∈. Note that X = X, for otherwise X V (f ) and6 so∅ i i ∩ i 6 ⊆ i fi I(X). Moreover, since f1f2 I(X), X V (f1f2)= V (f1) V (f2) and thus∈ X =(V (f ) X) (V (f ) X∈)= X X⊆. Thus X is not irreducible.∪ 1 ∩ ∪ 2 ∩ 1 ∪ 2 n Definition 1.5.14. An irreducible closed algebraic subset X of Ak is called an (affine) algebraic variety. Equivalently, X is an affine algebraic variety if and only if A(X) is an integral domain. In this case, the quotient field K(X) of A(X) is called the function field or field of rational functions on X.

We shall see that every aﬃne algebraic set X is a union of ﬁnitely many irreducible algebraic sets Xi, and the Xi are unique up to order if there are no containment relations among them, i.e. X is not contained in X if i = j. i j 6 The ideals which are of the form I(X) are not necessarily prime, but have the weaker property that they are radical ideals, in other words ideals I such that I = √I. The Nullstellensatz sets up a correspondence between algebraic n sets in Ak and radical ideals I in k[x1,...,xn]. Of course, we might wish to study all ideals in k[x1,...,xn]. For example, in case n = 1, every ideal is principal, and the study of ideals in k[x] is the study of monic polynomials in k[x]. But for n = 2, the study of ideals in k[x, y] is already considerably more complicated. We can say a little bit about prime ideals as follows: 24 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS

Proposition 1.5.15. In k[x1,...,xn], every nonzero minimal prime ideal is principal. Proof. This is a consequence of Lemma 1.4.4 and Corollary 1.4.20.

Thus, in k[x1, x2], there are three kinds of prime ideals: maximal ideals (x a , x a ), principal ideals (f(x , x )) which are prime (equivalently, 1 − 1 2 − 2 1 2 f(x1, x2) is irreducible), and the zero ideal (0). In fact, it is not hard to show (Exercise 1.15) that these are the only prime ideals. Note that, in this case, the maximal ideals are generated by two elements (and no fewer), and the minimal nonzero primes are generated by one element. We shall show later 2 that every proper closed subset of Ak is a union of ﬁnitely many points and closed subsets of the form V (f), where f is irreducble. However, for n > 2, the picture is much more complicated. While every maximal ideal in k[x1, x2, x3] is generated by three elements, and every minimal nonzero prime ideal is generated by one element, there exist intermediate prime ideals p. Given such a prime ideal p, i.e. a prime ideal which is neither maximal nor minimal, we shall show in Chapter 4 that every prime ideal which properly contains it is maximal, and every nonzero prime ideal properly contained in it is a minimal nonzero prime ideal. We say that the ring k[x1, x2, x3] has dimension three. However, while some of intermediate prime ideals can be generated by two elements, some can only be generated by three elements, and it is not easy to characterize those which can be generated by only two. Let us make some (much briefer) comments about some motivating questions in number theory. A great deal of elementary number theory is concerned with factorization in the PID Z: in particular, every natural number factors into a product of primes, in an essentially unique way. A similar statement holds in the ring Z[i] of Gaussian integers; this result is at the heart of the description of which natural numbers are sums of two squares, i.e. for which natural numbers n the Diophantine equation x2 + y2 = n has a solution. On the other hand, the ring Z[√ 5] is not a PID, and so the Diophantine equation x2 +5y2 = n is harder to− study. Of course, the most 2πi/k th famous ring of this sort is the ring Z[ζk], where ζk = e is a primitive k root of unity, connected to the Fermat equation xk + yk = n. However, the ring Z[√ 5], as well as many of the rings of algebraic numbers encountered − in number theory, does have a very remarkable property which is a weaker substitute for unique factorization: Every ideal in Z[√ 5] (and in much more general rings of algebraic numbers) has a unique factor− ization (in the 1.6. THE PRIME AND MAXIMAL SPECTRUM 25 appropriate sense) into prime ideals. Indeed, this fact is at the origin of the name “ideal,” a sort of ideal number where unique factorization would hold. In turn, this unique factorization of ideals into prime ideals has repercussions in Riemann surface theory and in the study of algebraic curves.

1.6 The prime and maximal spectrum

We begin by considering the case of algebraic geometry: fix an algebraically closed field k, and let R = k[x1,...,xn]. More generally, we consider a closed n algebraic set X in Ak corresponding to the radical ideal I = √I, with affine coordinate ring A(X) = k[x1,...,xn]/I. Our first goal is to show that the collection of all algebraic sets forms a category. In other words, given two affine algebraic sets X and Y , we must describe the possible morphisms from X to Y . 1 n 1 We first describe morphisms to Ak. A morphism from Ak to Ak should be the same thing as a function with values in k. Since the allowable functions are polynomials, these will just be polynomials and the ring of all morphisms n 1 from Ak to Ak is just k[x1,...,xn]. Hence, if X is an algebraic subset of n 1 Ak , then it is natural to define the set of morphisms from X to Ak to be n the affine coordinate ring A(X). More generally, a morphism from Ak to m n 1 Ak should be given by an m-tuple (f1,...,fm) of morphisms from Ak to Ak, m and a morphism from X to Ak should be given by an m-tuple (g1,...,gm) where the gi A(X) for all i, and hence gi is in the image of the natural homomorphism∈ A(An) A(X). This motivates: k → n Definition 1.6.1. Suppose that X is a closed algebraic set in Ak and Y is m a closed algebraic set in Ak . A function G: X Y is a morphism from X to Y if there exists a polynomial function F = (→f ,...,f ): An Am such 1 m k → k that F X = G. Two morphisms G1,G2 : X Y are equal if they coincide as functions.| → Note that, by definition, every morphism from X to Y lifts to a morphism between the two affine spaces containing them. It is easy to see that the identity function and constant functions are morphisms. The composition of two morphisms is again a morphism, since the composition of two polynomials is again a polynomial. Proposition 1.6.2. Let X An and Y An be two algebraic subsets, and ⊆ k ⊆ k let G: X Y be a morphism. For every f A(Y ), the function G∗f = → ∈ 26 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS f G is in A(X), and defines a k-algebra homomorphism ϕ: A(Y ) A(X). ◦ → Conversely, every k-algebra homomorphism ϕ: A(Y ) A(X) is of the form → G∗ for a unique morphism G: X Y . → Proof. Since the composition of two morphisms is again a morphism, it is easy to see that G∗ defines a k-algebra homomorphism A(Y ) A(X). Con- → versely, suppose that ϕ: A(Y ) A(X) is a k-algebra homomorphism. By → definition, A(X) = k[x1,...,xn]/I(X) and A(Y ) = k[y1,...,ym]/I(Y ) for the appropriate (radical) ideals I(X) and I(Y ). Lety ¯i be the image of yi in A(Y ), and let fi k[x1,...,xn] be a polynomial whose image in A(X) is ϕ(¯y ). Let F = (f ,...,f∈ ): An Am be the induced morphism. Clearly, i 1 m k → k there is a commutative diagram

∗ k[y ,...y ] F k[x ,...,x ] 1 m −−−→ 1 n

 ϕ  A(Y ) A(X). y −−−→ y We claim that F (X) Y . By the Nullstellensatz, it suﬃces to show that, if ⊆ h I(Y ), then h(F (X)) = 0. In other words, we must show that F ∗I(Y ) ∈ ⊆ I(X). By the commutativity of the above diagram, the image of F ∗I(Y ) in A(X) is zero, and hence F ∗I(Y ) I(X) as desired. Thus we have produced a morphism G: X Y . It is easy⊆ to see that the morphism G is independent of → the choice of the lifts fi of ϕ(¯yi) (though the morphism F does indeed depend on these choices). It is straightforward to check that these two constructions are inverses.

n The set Ak has a natural topology, the Zariski topology. n Deﬁnition 1.6.3. A subset of Ak is a Zariski closed subset if it is of the form V (I) for some ideal I k[x1,...,xn]. It is Zariski open if it is the complement of a Zariski closed⊆ subset. By Lemma 1.5.4, the Zariski closed n subsets of Ak deﬁne a topology for which they are the closed subsets, the Zariski topology.

1 For example, the Zariski topology on Ak is the finite complement topology: the proper closed sets are the finite sets. In particular, the topological 1 space Ak only depends on the cardinality of k and is thus somewhat uninter- esting: it is only when we consider the functions k[x] on this space that the link with the algebra becomes evident. It is easy to see that, for n,m > 0, 1.6. THE PRIME AND MAXIMAL SPECTRUM 27 the Zariski topology on An+m is never the product topology on An Am, and k k × k in fact it is always coarser in the sense of having fewer open (or equivalently 2 closed) sets. For example, on Ak, the closed sets in the product topology 2 1 1 Ak = Ak Ak would be finite unions of horizontal and vertical lines and × 2 points or all of Ak, and a subset of the form V (f), for f an irreducible polynomial in k[x1, x2], is not of this form. The following shows that the Zariski topology is always far from Haus- dorff: Lemma 1.6.4. (i) The open sets An V (f) are a basis for the Zariski k − topology. In fact, every open set is a union of finitely many sets of this type.

n (ii) If U1 and U2 are two nonempty Zariski open subsets of Ak , then U1 U = . In particular, every nonempty open set is dense. ∩ 2 6 ∅ Proof. (i) To see that every open set is a union of sets of the form An V (f), k − it suffices to note that V (I)= f I V (f). Moreover, if I =(f1,...,fk), then ∈ V (I) = k V (f ). The finiteness statement then follows from Hilbert’s i=1 i T basis theorem. T (ii) It suffices to show that two nonempty sets U1, U2 of the form Ui = n Ak V (fi) intersect. Now Ui is nonempty if and only if fi is not 0, and clearly− U U = An V (f f ). Thus if neither f nor f is zero, then f f 1 ∩ 2 k − 1 2 1 2 1 2 is nonzero as well, proving (ii). In terms of closed sets, the lemma can be rephrased as follows: Lemma 1.6.5. (i) Every closed set is an intersection of finitely many sets of the form V (f).

n (ii) Ak is not the union of two proper closed subsets. n For a closed algebraic subset X of Ak , we deﬁne the Zariski topology on n X to be the subspace topology. Since X is a closed subset of Ak , a closed n subset Y of X is a closed subset of Ak contained in X. We also have: Lemma 1.6.6. If G: X Y is a morphism, then G is continuous in the Zariski topology. →

n m Proof. It suﬃces to check the case of a morphism F : Ak Ak . In this case, m 1 → if Z = V (g1,...,gk) is a closed subset of Ak , then F − Z = V (F ∗g1,...,F ∗gk) n is a closed subset of Ak . 28 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS

For many reasons, one wants to be able to consider more general rings than those of the form A(X), i.e. finitely generated reduced k-algebras. As a first step, it is often natural to consider rings of the form k[x1,...,xn]/I, where I is an arbitrary ideal of k[x1,...,xn], not necessarily reduced. Fur- thermore, for many applications, we might want to allow k to be an arbitrary field, not necessarily algebraically closed, or even to consider quotients of Z[x1,...,xn]. At this level of generality, it is worth looking first at arbitrary commutative rings R. Our goal now will be to see how much of the above carries through. In general, we will try to define a contravariant functor from the category of commutative rings to the category of topological spaces and then decree that R is the ring of “functions” on the corresponding topological space, where we use the word function in a very general sense. n Note that, for the case of Ak , or more generally for an affine algebraic set X, the underlying set X is just the set of maximal ideals in A(X). For an arbitrary commutative ring R, we have the maximal spectrum of R, the set maxSpec R of maximal ideals in R, and as we shall shortly see, this set has a natural topology. Unfortunately, this construction is not functorial in a natural way: if f : R S is a homomorphism of commutative rings, and n → 1 is a maximal ideal in S, then the ideal f − (n) need not be a maximal ideal in R; a priori, it is only prime. Perhaps the simplest example is the inclusion 1 homomorphism i: Z Q: (0) is a maximal ideal in Q, but i− ((0)) = (0) is only a prime ideal in→Z. (If R and S are both finitely generated k-algebras, where k is an algebraically closed field, then we leave it as an exercise to 1 show that, for every maximal ideal n in S, f − (n) is a maximal ideal in R.) One way to remedy this problem is to consider the set of all prime ideals of R. While this solution has some unnatural features, it also has many advantages, some of which we shall try to describe below. Definition 1.6.7. If R is a ring, we let Spec R, the spectrum or prime spectrum of R, denote the set of prime ideals of R. Given a ring homomorphism 1 f : R S, we let f ∗ : Spec S Spec R denote the function q f − (q). → → 7→ Note that (f g)∗ = g∗ f ∗ (exercise). ◦ ◦ We will discuss the problem of making the set of all spectra into a category in 3.7. For the moment, we will view the category of all spectra as opposite to§ the category of all commutative rings, i.e. given two commutative rings R and S, the set of morphisms from Spec S to Spec R will be identified with the set of homomorphisms from R to S, via f f ∗. We can also define a topology on Spec R as7→ follows. 1.6. THE PRIME AND MAXIMAL SPECTRUM 29

Deﬁnition 1.6.8. If I is an ideal of R, deﬁne

V (I)= p Spec R : I p . { ∈ ⊆ }

If I = (f1,...,fk), then we denote V (I) simply by V (f1,...,fk). More generally, if the ideal I is generated by the set X, we set V (X)= V (I). Lemma 1.6.9. (i) If I I , then V (I ) V (I ). 1 ⊆ 2 2 ⊆ 1 (ii) V (0) = Spec R and V (1) = . ∅

(iii) V (I1+I2)= V (I1) V (I2). More generally, V ( α A Iα)= α A V (Iα). ∩ ∈ ∈ (iv) V (I I )= V (I I )= V (I ) V (I ). P T 1 · 2 1 ∩ 2 1 ∪ 2 Proof. The proof is the same as the proof of Lemma 1.5.4, once we have the following lemma.

Lemma 1.6.10. Let p be a prime ideal of R, and let I1 and I2 be two ideals. If p I I , then p I or p I . ⊇ 1 · 2 ⊇ 1 ⊇ 2 Proof. If p does not contain either I or I , then there exists r I p and 1 2 1 ∈ 1 − r2 I2 p. But then r1r2 I1 I2 and r1r2 / p, since p is prime. This is a contradiction.∈ − ∈ · ∈ Corollary 1.6.11. The subsets V (I) of Spec R satisfy the axioms for the closed subsets of a topology on Spec R, the Zariski topology. We also call the induced topology on maxSpec R the Zariski topology. It is then easy to check: Proposition 1.6.12. Let X be an aﬃne algebraic set. Then the two deﬁni- tions of the Zariski topology on maxSpec A(X)= X agree. We have the following, which is left as an exercise: Lemma 1.6.13. Let f : R S be a ring homomorphism, i.e. S is an R- → algebra. Then the induced function f ∗ : Spec S Spec R is continuous. →

For example, for a ring R, we have a morphism R R[x1,...,xn]. De- n → ﬁne Spec R[x1,...,xn] = AR, aﬃne n-space over Spec R. Then there is a n continuous function AR Spec R. We now give some examples→ to illustrate what Spec R might look like. 30 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS

Example 1.6.14. R = Z: Spec Z = (0) (p) : p is a prime number . { }∪{ } Since every ideal in Z is principal, the closed sets in Spec Z are all of the form V (n), where n 0. If n = 0, V (n) = Spec Z. Otherwise, V (n) = (p ),..., (p ) , where≥ the p are the prime factors of n (and in particular { 1 k } i V (1) = ). In particular, the point (0) of Spec Z is contained in no proper closed subset∅ of Spec Z, and hence its closure (0) is all of Spec Z. Note { } that maxSpec Z is neither open nor closed in Spec Z. A similar picture holds for every principal ideal domain R, in particular for R = k[x]: the closed subsets of Spec R are either Spec R itself, or ﬁnite subsets of Spec R not containing (0), and (0) = Spec R. { } More generally, if R is an integral domain, then (0) is a prime ideal of R and hence (0) Spec R. If (0) V (I), then I (0) and hence I = (0). In this case V (I)∈ = Spec R. It follows∈ as in the above⊆ example that (0) is not contained in any proper closed subset of Spec R, and hence (0) = Spec R. We call (0) the generic point of Spec R. Despite the fact that{ generic} points are far from being closed, they are quite useful in algebraic geometry. On the other hand, if m is a maximal ideal in R, then it is easy to see that m is a closed point of Spec R. In fact, we leave it as an exercise to show that the closed points of Spec R exactly correspond to the maximal ideals of R.

Example 1.6.15. Let R = k[x1, x2], where k is algebraically closed. As we have seen, there are three kinds of prime ideals in R, i.e. elements of Spec R: maximal ideals m =(x a , x a ), corresponding to points of A2; principal 1 − 1 2 − 2 k prime ideals (f), corresponding to irreducible polynomials f(x1, x2); and (0). From this, one can show that the proper closed sets in the Zariski topology on Spec R are all finite unions of closed sets of the form V (m) = p for a closed point p Spec R or of the form V (f), where f is irreducible.{ } ∈ A similar picture holds for Spec Z[x], using Exercise 1.15. The maximal ideals of Z[x], i.e. the closed points of Spec Z[x], are of the form (p, f(x)), where p is a prime number and f(x) Z[x] is a polynomial whose image in ∈ (Z/pZ)[x] is irreducible. The other nonzero prime ideals are principal, and are thus of the form (p), where p is a prime number, or (f), where f(x) Z[x] is a polynomial whose coefficients generate the unit ideal and whose∈ image in Q[x] is irreducible. Using the structure homomorphism Z Z[x], we have the corresponding function Spec Z[x] Spec Z. It is natural→ to think of Spec Z[x] Spec Z as a fibration. The fiber→ over a prime ideal (p) is the → 1.7. GRADED RINGS AND PROJECTIVE SPACES 31

1 affine line Spec(Z/pZ)[x]= AZ/pZ and the fiber over the generic point (0) is 1 the affine line Spec Q[x]= AQ.

Generalizing the example of Spec k[x1, x2], let X be any aﬃne algebraic set. Then the closed sets on X are in one-to-one correspondence with the closed sets on Spec A(X). Of course, a similar statement holds for the open sets. However, Spec A(X) has many more points than X. In fact, the points of Spec A(X) are exactly the nonempty irreducible closed subsets of X. These extra points are something of a defect in the deﬁnition of Spec; geometrically, one tends to visualize X or equivalently the closed points in Spec A(X). However, the language of non-closed points can also be very useful, for example in dealing with questions about “generic behavior,” and to give a more precise meaning to the terminology of “generic points.”

1.7 Graded rings and projective spaces

Definition 1.7.1. A graded ring R = n 0 Rn is a ring such that Ri Rj ≥ · ⊆ Ri+j for all i, j. A homomorphism f : R = n 0 Rn S = n 0 Sn is L ≥ → ≥ graded if f(Rn) Sn for all n 0. ⊆ ≥ L L For example, S = R[x1,...,xn] is graded by degree: Let Sd be the submodule of S generated by all monomials rxk1 xkn with k + + k = d. 1 ··· n 1 ··· n More generally, given positive integers wi (the weights), we can defined a k1 kn grading on S by declaring that a monomial rx1 xn has weighted degree d = k w + + k w . ··· 1 1 ··· n n It is possible to use different kinds of grading in defining rings and mod- 1 ules. For example, the ring k[x, x− ] of finite Laurent series is naturally graded by Z. One can also consider “bigraded rings” R = n,m 0 Rn,m, and we shall see an example in Chapter 6. In general, one can consi≥ der rings graded by any abelian semigroup. However, here we shallL only consider the case of rings graded by the nonnegative integers. If R is a graded ring, R = R is an ideal of R, 1 R , and R is + n>0 n ∈ 0 0 a subring of R. An element r R is homogeneous of degree n if r Rn, and ∈ L ∈ in general the projection of r into rn is the homogeneous component of r of degree n. An ideal I is a homogeneous ideal if I = n 0(I Rn), if and only ≥ ∩ if I can be generated by homogeneous elements. Clearly, I is homogeneous if and only if, for every r I, if r = r is the decompositionL of r into its ∈ n n homogeneous pieces, then rn I for every n. If f is homogeneous of degree ∈ P 32 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS

n, f = i gifi where the fi are homogeneous of degree di, then f = i hifi where the hi are the homogeneous components of gi of degree n di. P − P Most natural operations on ideals preserve homogeneous ideals. For example, it is easy to see that the sum, product, and intersection of homogeneous ideals is homogeneous. The radical of a homogeneous ideal is homogeneous. A homogeneous ideal p is a prime ideal if and only if, for all homogeneous r, s R, if rs p, then either r or s lies in p. We leave the proofs of these statements∈ as∈ exercises.

Let R = n 0 Rn be a graded ring. A graded R-module M = n 0 Mn ≥ ≥ is an R-module satisfying Rn Mk Mn+k for all n, k 0. L · ⊆ ≥ L n Geometrically, graded rings arise as follows. For R = k a ﬁeld, let Pk be the n-dimensional projective space over k: Pn is the quotient of kn+1 0 k −{ } by the equivalence relation , where v1 v2 if and only if there exists a ∼ ∼ n+1 λ k∗ such that λv1 = v2. Traditionally, we label the coordinates on k by∈x ,...,x . A point of Pn is an equivalence class for the relation , and the 0 n k ∼ equivalence class containing (a0,...,an) is often denoted by (a0 :a1 : :an), and can be viewed as a generalized ratio. However, we shall just denote··· this class by (a0,...,an) as well. We can clearly identify the equivalence class n+1 containing (a0,...,an) with the unique line in k through the origin and containing (a ,...,a ), and two points (a ,...,a ), (b ,...,b ) kn+1 lie in 0 n 0 n 0 n ∈ the same equivalence class if and only if they lie on the same line through n+1 n the origin in k . Thus the points of Pk are in one-to-one correspondence with the lines in kn+1 through the origin. Let S = k[x0,...,xn] with the usual grading. The ring S is some- n times called is called the homogeneous coordinate ring of Pk . A nonconstant polynomial f(x0,...,xn) S does not have a well-deﬁned value at a n ∈ point of Pk . However, it makes sense to say that a homogeneous polynomial f(x0,...,xn) S is zero at a point (a0,...,an), since if f is homogeneous of ∈ d degree d, then f(λa0,...,λan)= λ f(a0,...,an).

Deﬁnition 1.7.2. If f S , let ∈ d V (f)= (a ,...,a ) Pn : f(a ,...,a )=0 . + { 0 n ∈ k 0 n }

More generally, if f1,...,fk S are homogeneous (not necessarily of the same ∈ n degrees), then we deﬁne V+(f1,...,fk) to be the subset of Pk consisting of the (a0,...,an) such that fi(a0,...,an) = 0 for all i. For a homogeneous n ideal I, we deﬁne V+(I) similarly. Finally, a subset X of Pk is a (closed) 1.7. GRADED RINGS AND PROJECTIVE SPACES 33

n algebraic subset of Pk if it is of the form V+(I) for some homogeneous ideal I. Of course, given a homogeneous ideal I in S, we can always consider the n+1 corresponding subset V (I) of Ak . We will discuss the relationship between V+(I) and V (I) shortly. The sets V+(I) satisfy the same formal properties as those given in Lemma 1.5.4, and hence are the closed subsets for a topology n on Pk , the Zariski topology. Example 1.7.3. Let V be a vector subspace of kn+1. Then

(V 0 )/ (kn+1 0 )/ = Pn. −{ } ∼ ⊆ −{ } ∼ k We let P(V )=(V 0 )/ . It is a closed algebraic subset of Pn since −{ } ∼ k V itself is defined by linear (i.e. homogeneous degree one) equations. We call such a subset a linear subspace of Pn. For example, if a = 0, then k 0 6 (a0,...,an) = V+(a0x1 a1x0,...,a0xn anx0). In the other direction, if { } − n+1 − ℓ is a nonzero linear function on k , in other words an element of S1, then n V+(ℓ) is a a linear space in Pk , called a hyperplane. In particular, taking the functions xi, we obtain the standard coordinate hyperplanes Hi = V+(xi). n The geometry of linear subspaces of Pk exactly parallels the geometry of vector subspaces of kn+1, and we will describe this in more detail later. Given a subset X of Pn, we let I(X) S be the homogeneous ideal k ⊆ generated by the homogeneous polynomials vanishing on X. Clearly I(X) is a radical ideal of S, and X V+(I(X)) as in the affine case, with equality if ⊆ n and only if X is an algebraic subset of Pk . One minor but annoying difference with the affine case is the following: By definition, V+(x0,...,xn) = , or equivalently V (S ) = , where S is the homogeneous maximal ideal∅ of S + + ∅ + generated by x0,...,xn. In particular, V+(S+)= V+(S). Theorem 1.7.4 (Homogeneous Nullstellensatz). Let I be a homogeneous ideal in S and let f S be a nonconstant polynomial such that f vanishes on V (I). Then f ∈√I. + ∈ We shall prove this later (Proposition 1.7.10); it follows easily from the usual Nullstellensatz. As a consequence, we have: Corollary 1.7.5. The function I V (I) sets up a one-to-one correspon- 7→ + dence from homogeneous radical ideals in S not equal to S+ to algebraic subsets of Pn. Its inverse is X I(X). k 7→ 34 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS

n As in the affine case, if X is an algebraic subset of Pk , we can define its homogeneous coordinate ring R(X) = S/I(X). It is a graded ring, but it is not an abstract invariant of X (unlike the affine coordinate ring of an n algebraic subset of Ak ). As in the affine case, there is the following:

n Deﬁnition 1.7.6. A nonempty closed algebraic subset X of Pk is irreducible if, whenever X = X X , where the X are closed algebraic subsets of Pn, 1 ∪ 2 i k then either X = X1 or X = X2. The set X is closed if and only if the ideal I(X) is a homogeneous prime ideal. An irreducible closed algebraic subset n of Pk is also called a projective variety.

n The importance of Pk is that it is in a very natural sense a compactifica- n n tion of Ak . Of course, strictly speaking Ak is already compact, in the sense that every open cover has a finite subcover (exercise). But this is really just a reflection of the fact that the Zariski topology is very badly non-separated, n and Ak is noncompact in the sense that there are sequences in it which have no convergent subsequences and so there are “points at infinity.” On the n other hand, Pk has many of the strong properties we want compact sets to have, and which we shall describe later. Since compact spaces are usually n much better behaved than noncompact ones, Pk is a much better place to n 2 do geometry than Ak . For example, in Ak, two different lines will intersect 2 unless they are parallel. But in Pk, every pair of distinct lines intersect in 2 2 exactly one point. Indeed, classically Pk is obtained from Ak by adding a 2 “line at infinity” to Ak, which represents all of the possible directions of lines, and two parallel lines meet each other on the line at infinity at the point corresponding to their common direction. n Let us explain how this works for Pk . We have the hyperplane H0 = n n 1 V+(x0) Pk . As a set, H0 ∼= Pk− with coordinates x1,...,xn, and it is easy to see that⊆ in fact this is a homeomorphism in the Zariski topology (use for example the inclusion k[x ,...,x ] k[x ,...,x ] as well as the surjection 1 n ⊆ 0 n k[x0,...,xn] k[x1,...,xn] defined by x0 0 to identify the Zariski closed → 7→ n 1 subsets of H0 with the usual closed subsets of Pk− ). The complement U0 = Pn H is thus an open subset of Pn. Clearly, every equivalence class lying k − 0 k in U0 has a unique representative of the form (1,y1,...,yn), and hence there n is a bijection U0 ∼= Ak .

Proposition 1.7.7. Via the above bijection, the open set U0 is homeomorphic n to Ak . 1.7. GRADED RINGS AND PROJECTIVE SPACES 35

n Proof. We must show that, if X is a closed subset of Pk , then X U0 corre- n ∩ n sponds to a closed subset of Ak , and moreover that every closed subset of Ak arises in this way. Given f k[x0,...,xn] homogeneous of degree d, define inh ∈ f (y1,...,yn) = f(1,y1,...,yn), a polynomial of degree at most d. Con- d versely, given g k[y1,...,yn], let g(x0,...,xn) = x0g(x1/x0,...,xn/x0), where d = deg g.∈ It is easy to see that g is a homogeneous polynomial of de- inh d e inh inh gree d, that (g) = g, and that f = xe0− f for d = deg f . It is then easy n to check that, if X = V+(f1,...,fk) ise a closed algebraic subset of Pk , then inh inh n X U0 correspondse to the algebraic subsetgV (f1 ,...,fk ) of Ak . Moreover, ∩ n if Y = V (g1,...,gj) is an algebraic subset of Ak , then Y corresponds to the intersection V (g ,..., g ) U . This completes the proof. + 1 j ∩ 0 Of course, there is nothing special about the index 0. For every j, 0 e e n n 1 n ≤ j n, Hj is a closed subset of Pk homeomorphic to Pk− and Uj = Pk Hj ≤ n n − is an open set of Pk homeomorphic to Ak . It is clear that U0,...,Un is an n { } open cover of Pk , usually called the standard affine open cover. Let I be a homogeneous ideal in S = k[x0,...,xn]. We have defined the n algebraic set V+(I) Pk . Of course, we can also define the algebraic set n+1 ⊆ V (I) A . The following is then clear from the definitions: ⊆ k

Lemma 1.7.8. Let I be a homogeneous ideal in S with V+(I) = . Then n+1 6 ∅ V (I) is the union of the lines in Ak corresponding to points of V+(I). Equivalently, if V (I) = , and π : An+1 0 Pn is the projection, then + 6 ∅ k −{ }→ k 1 V (I)= π− (V (I)) 0 . + ∪{ }

In particular, V (I) is closed under multiplication by k∗. Note however that if V (I)= , then V (I) is either 0 or , depending + ∅ { } ∅ on whether √I =(x0,...,xn) or √I = S.

n Definition 1.7.9. Let X be an algebraic subset of Pk , and suppose that X = V+(I) for some homogeneous ideal I S+. Then the algebraic subset n+1 ⊆ V (I) of Ak is independent of the choice of I. It is called the affine cone over the projective algebraic set X, and is denoted C(X). In particular, C( )= 0 . ∅ { } Proposition 1.7.10. The correspondence X C(X) defines a bijection n 7→ from the set of all algebraic subsets of Pk to the set of all algebraic subsets n+1 of A which are closed under multiplication by k∗ and which contain 0 . k { } 36 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS

Under this correspondence, I(X)= I(C(X)). Finally, if X = V (I), f S + ∈ + is homogeneous, and V+(f) contains X, then there exists an N > 0 such that f N I. ∈

Proof. In one direction, we associate to X = V+(I) the algebraic set C(X). n+1 In the other direction, let Z be an algebraic subset of Ak , closed under multiplication by k∗ and containing 0 . If f I(Z) and λ k∗, then f(λx ,...,λx ) I(Z) as well. Hence{ } the homogeneous∈ components∈ of 0 n ∈ f are in I(Z), and so I(Z) is a homogeneous ideal (Exercise 1.29). Let n X be the algebraic set in Pk defined by I(Z). Then clearly Z = C(X). Conversely I(C(X)) is a homogeneous radical ideal in S, containing the homogeneous radical ideal I(X) and such that V (I(C(X))) = V (I(X). Thus I(C(X)) = I(X), and so the projective variety defined by this construction by C(X) is just X. Thus the two constructions are inverse to each other. Finally suppose that X = V (I), f S is homogeneous of positive + ∈ + degree, and V+(f) contains X. Then V (f) contains C(X). It follows from the Nullstellensatz that f I(C(X)) = I(X). On the other hand, V (I)= C(X), so that √I = I(C(X))∈ = I(X). Hence f √I, so that there exists an N > 0 such that f N I. ∈ ∈ For general rings R, we have defined the topological space Spec R, which n generalizes (in a certain sense) algebraic subsets of Ak . In the graded case, there is a corresponding construction:

Deﬁnition 1.7.11. Let R = n 0 Rn be a graded ring. Let Proj R be ≥ the set of homogeneous prime ideals in R which do not contain R+. For a L homogeneous ideal I, deﬁne V+(I)= p Proj R : I p . Then the V+(I) satisfy the axioms for the closed subsets{ of∈ a topology⊆ on} Proj R, the Zariski topology.

For example, for a ring R and S = R[x0,...,xn] with its usual grading, n we denote Proj R[x0,...,xn] by PR. Warning: A graded homomorphism f : R S of graded rings does not → necessarily induce a function f ∗ : Proj S Proj R. The problem is that a → homogeneous prime ideal q in S which does not contain S+ may pull back to an ideal of R containing R+. For example, for the ring homomorphism k[x ,...,x ] k[x ,...,x ] (m < n) given by inclusion, the homogeneous 0 m → 0 n prime ideal (x0,...,xm) in k[x0,...,xn] pulls back to k[x0,...,xm]+. Geo- metrically, this corresponds to the fact that linear projection kn+1 km+1 → 1.7. GRADED RINGS AND PROJECTIVE SPACES 37 does not induce a function on the corresponding projective spaces, since it is not deﬁned at points (0,..., 0, a1,...,an m) where the ﬁrst a+1 coordinates − vanish. We will discuss the geometric meaning of projection later. We will discuss the relationship between Proj and Spec in more detail in n the chapter on localizations, and generalize the picture of Pk as a union of n open subsets homeomorphic to Ak .

Exercises

Exercise 1.1. In the ring Z, what are (n)+(m)? (n) (m)? (n) (m)? (The ∩ · argument should work for any PID.) In k[x, y], describe (x)+(y); (x) (y); (xy) (y); (x) (y); (xy) (y). ∩ ∩ · · Exercise 1.2. (i) In a ring R, show that, if u is a unit and r is nilpotent, then u + r is a unit. (ii) Let R[x] be the ring of polynomials with coeﬃcients in R. If R is an integral domain, show that R[x] is an integral domain. (In fact, show that deg(fg) = deg f deg g.) · (iii) With R[x] as above, show that, if a0,...,an R are nilpotent, then n i ∈ so is i=0 aix . If a0 R is a unit in R and a1,...,an R are nilpotent, n i ∈ ∈ then i=0 aix is a unit in R[x]. P n i (iv) With R[x] as above, show that, if ai R and i=0 aix is nilpotent, P n∈ i then a0,...,an are nilpotent, and that, if aix is a unit in R[x], then a0 i=0 P is a unit in R and a1,...,an are nilpotent. (To prove the ﬁrst statement, for P example, it is enough to prove that ai lies in every prime ideal p for every i. Reduce mod p and use (ii).)

Exercise 1.3. Let R be a ring, I1,...,In ideals of R such that I1 + +In = N ··· R. Show that, if N ,...,N are positive integers, then I 1 + + INn = R. 1 n 1 ··· n Exercise 1.4. Let R be a ring, I and J ideals of R. Show: √I = √I; √I J = √I J = √I √J; √I + J = √I + √J. If p is ap prime ideal · ∩ ∩ pN and N is a positive integer, what is ?p Exercise 1.5. (i) Generalizing thep case of two ideals, show that, if p is a prime ideal of R, I ,...,I are ideals of R and p contains I I , then 1 n 1 ∩···∩ n p contains Ij for some j. 38 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS

(ii) Suppose that p1,... pn are prime ideals of R and that I is an ideal of R with I p . Then I p for some i. (Show that, if I p for every i, ⊆ i i ⊆ i 6⊆ i then I pi by induction on n, the case n = 1 being trivial. Thus we may 6⊆ iS assume that, for all i, there exists an ri I such that ri / pj for all i = j. S ∈ ∈ 6 If ri / pi for some i, we are done, so we can assume that ri pi for all i. In this case,∈ consider the element ∈

s = rj, i j=i X Y6 and argue that s I but s / p for every i.) ∈ ∈ i Exercise 1.6. Let R be a ring. Show that the set of prime ideals of R, ordered by inclusion, contains minimal elements, and in fact that every prime ideal of R contains a minimal prime ideal. (Apply Zorn’s lemma to the set of all prime ideals contained in a ﬁxed prime ideal, ordered by reverse inclusion.)

Exercise 1.7. (Chinese remainder theorem.) Suppose that I1 and I2 are two ideals of R such that I1 + I2 = R (we say that I1 and I2 are coprime or comaximal). Show that I I = I I and that the natural homomorphism 1 · 2 1 ∩ 2 R R/I R/I → 1 × 2 deﬁned by a (a mod I , a mod I ) induces an isomorphism R/I I = 7→ 1 2 1 · 2 ∼ R/I1 R/I2. Generalize by induction to the case of k ideals I1,...,Ik such that I×+ I = R for i = j. i j 6 Exercise 1.8. Let R be a Noetherian ring and let I be an ideal in R. Then there exists an m > 0 such that (√I)m I, in other words every ideal in a ⊆ Noetherian ring contains a power of its radical.

Exercise 1.9. Let R be a Noetherian integral domain, and suppose that, for all a, b R, the ideal (a) (b) is principal (this essentially says that least common multiples,∈ and hence∩ greatest common divisors, exist). Show that R is a UFD. (Show that r irreducible = (r) prime as follows: suppose that ⇒ r ab and that r a. First show that (r) (a)=(ra). Since r ab, ab (r) and hence| ab (r) 6| (a)=(ra). But then∩ab = ars for some s| R,∈ so b = rs ∈ ∩ ∈ and r b.) | 1.7. GRADED RINGS AND PROJECTIVE SPACES 39

Exercise 1.10. (i) Let k be an infinite field, and suppose that f(x1,...,xn) n ∈ k[x1,...,xn] be such that f(a1,...,an) = 0 for all (a1,...,an) k . Show that f = 0. ∈ (ii) Let k be algebraically closed, and let f,g k[x ,...,x ] be relatively ∈ 1 n prime. Using the Nullstellensatz, show that there exists p kn such that ∈ f(p) = 0, g(p) = 0. Conclude that, if a rational function f/g k(x1,...,xn) is defined6 at all points of kn, then f/g k[x ,...,x ]. ∈ ∈ 1 n Exercise 1.11. Let k be algebraically closed, and let X be an affine algebraic variety. Suppose that U is a nonempty Zariski open subset of X and f ∈ A(X) satisfies: f U = 0. Show that f = 0. Hence if U is a nonempty open subset of X and f| , f A(X) are such that f U = f U, then f = f . (If 1 2 ∈ 1| 2| 1 2 Y = X U, then Y is closed, so that there exists a g A(X) with g Y = 0. But then− fg = 0.) Show that, if X is an affine algebraic∈ variety and U,| V are two nonempty Zariski open subsets, then U V = 0. ∩ 6 Do the same results hold if X is not an affine variety? Exercise 1.12. Let R be a finitely generated k-algebra, where k is an algebraically closed field, and let I be an ideal in R such that I = √I. Show that I is the intersection of all of the maximal ideals containing it. (In fact, this statement is true even if k is not algebraically closed.)

Exercise 1.13. Argue carefully that Z[x] and k[x1,...,xn], n > 1, are not PIDs. Exercise 1.14. Let p(x) be a polynomial in k[x] which is not a perfect square. Show that f(x, y)= y2 p(x) is an irreducible element of k[x, y]. − Exercise 1.15. Let R be a PID, and let p be a nonzero prime ideal of R[x]. Show that p is either of the form (f), where f is an irreducible element of R[x], or of the form (r, f), where r is an irreducible element of R and f(x) R[x] is a polynomial whose image in the PID (R/rR)[x] is irreducible. (Let∈ K be the quotient ﬁeld of R. If p contains a nonzero element of R, which we may assume to be irreducible, show that p = (r) or (r, f) as described. If p contains f,g such that f and g are relatively prime in K[x], show that p contains a nonzero element of R. In the remaining case, show that p = (f) for some primitive and irreducible polynomial f.)

Exercise 1.16. Let f : R S be a ring homomomorphism and let f ∗ be the induced map Spec S →Spec R. → 40 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS

1 (i) Show that, for every ideal I of R, (f ∗)− V (I) = V (IS), where IS is the ideal of S generated by f(I) and hence that f ∗ is continuous in the Zariski topology.

(ii) If S = R/I, show that f ∗ is a homeomorphism from Spec S to the closed subset V (I) of Spec R.

(iii) If g : S T is another ring homomorphism, show that (g f)∗ = f ∗ g∗. → ◦ ◦ Exercise 1.17. Let R be a ring and let p Spec R. Let k(p) be the quotient field of the integral domain R/pR, and∈ let f : R k(p) be the induced → homomorphism. Show that the image of the one point space Spec k(p) under f ∗ is the point p. Exercise 1.18. Let R and S be two finitely generated k-algebras, where k is an algebraically closed field, and let f : R S be a homomorphism of → 1 k-algebras. Show that, if n is a maximal ideal in S, then f − (n) is a maximal ideal in R. Is this necessarily true if R and S are not finitely generated as k-algebras? (Again, this statement is true even if k is not algebraically closed.) Exercise 1.19. Let p be a prime ideal in R. Viewing p as a point of Spec R, show that p = V (p). Conclude that p is a closed point of Spec R if and only if it is{ maximal.} Exercise 1.20. Let S be a ring and R an S-algebra. Let I be an ideal of S[x1,...,xn] and set X = Spec S[x1,...,xn]/I. An R-valued point of X over Spec S is a morphism Spec R X commuting with the natural morphisms to Spec S, in other words an →S-algebra homomorphism S[x ,...,x ] R. 1 n → Show that an R-valued point of X over Spec S is the same thing as an n-tuple n (r1,...,rn) R such that f(r1,...,rn) = 0 for all f I. Interpret when R = R = k,∈ where k is a field. ∈ Exercise 1.21. Let R be a ring and I an ideal contained in √0. Show that the natural map Spec(R/I) Spec R is a homeomorphism. → Exercise 1.22. Let X be a topological space. A collection of subsets Yα of X has the finite intersection property if every finite intersection Y Y is α1 ∩···∩ αn nonempty. A space X is compact if every collection of closed subsets with the finite intersection property has a nonempty intersection. (You should verify that this is equivalent to the usual definition in terms of open covers.) Show 1.7. GRADED RINGS AND PROJECTIVE SPACES 41

n that Spec R is compact. Similarly, Ak , and more generally every algebraic n set in Ak , is compact in the Zariski topology.

Exercise 1.23. Let R be a ring. Let X1 = V (I1) and X2 = V (I2) be two closed subsets of Spec R. Show that X X if and only if √I √I . 1 ⊇ 2 1 ⊆ 2 Exercise 1.24. Deﬁne a subset X of Spec R to be irreducible by the analogue of Deﬁnition 1.5.12. Show that X is irreducible if and only if X = V (p), where p is a prime ideal of R.

Exercise 1.25. Let R be a ring. For f R, let (Spec R) be the set of prime ∈ f ideals p such that f / p. Show that (Spec R)f is an open subset of Spec R. Moreover, the collection∈ of open sets (Spec R) : f R is a basis for the { f ∈ } topology of Spec R: for each p Spec R and each open subset U of Spec R containing p, there exists an f ∈R such that p (Spec R) U. Show also: ∈ ∈ f ⊆ (i) (Spec R) (Spec R) = (Spec R) ; f ∩ g fg (ii) (Spec R) = f is nilpotent; f ∅ ⇐⇒ (iii) (Spec R) = Spec R f is a unit; f ⇐⇒ (iv) More generally, (Spec R) = (Spec R) (f) = (g), and in f g ⇐⇒ fact (Spec R) (Spec R) (f) (g). f ⊆ g ⇐⇒ ⊆ p p p p (v) Spec R = i(Spec R)fi if and only the the fi generate the unit ideal. S Exercise 1.26. Let R be a graded ring. If I1 and I2 are two homogeneous ideals in R, show that I + I , I I , and I I are homogeneous. If I is 1 2 1 · 2 1 ∩ 2 homogeneous, show that √I is homogeneous.

Exercise 1.27. Let R be a graded ring and let I be a homogeneous ideal of R. Show that I is a prime ideal if and only if, for every two homogeneous elements r, s R, rs I if and only if either r I or s I. ∈ ∈ ∈ ∈ Exercise 1.28. Let R be a graded ring. If u is a unit of R, show that u R and in particular u is homogeneous of degree zero. Now suppose ∈ 0 that, in addition, R is a UFD. Show that, if f R is homogeneous, then every irreducible factor of f is homogeneous. ∈ 42 CHAPTER 1. INTRODUCTION TO COMMUTATIVE RINGS

Exercise 1.29. (i) Let k be an inﬁnite ﬁeld and let S = k[x1,...,xn] with n the usual grading. Show that f Sd if and only if, for all λ k and a k , f(λa)= λdf(a). ∈ ∈ ∈ (ii) With k and S as above, show that an ideal I of S is a homogeneous ideal if and only if, for all f(x) I and λ k∗, f(λx) I. ∈ ∈ ∈ Exercise 1.30. Let R be a ring, and let R[x] be given its usual grading, so 0 0 that Proj R[x]= PR. Show that PR = Spec R.

Exercise 1.31. (i) In the notation of Proposition 1.7.7, for all f1, f2 inh inh inh inh ∈ k[x0,...,xn], show that (f1 + f2) = f1 + f2 and that (f1f2) = (f inh)(f inh). Likewise, for g k[y ,...,y ], show that (^g g ) = g g , but 1 2 ∈ 1 n 1 2 1 2 ^ that (g1 + g2) is not necessarily equal to g1 + g2. n n e e (ii) Let Y = V (g1,...,gr) be an algebraic set in Ak . Viewing Ak as a subset n of Pk , show that, again in the notation ofe ofe Proposition 1.7.7, the closure n Y of Y in Pk is contained in V+(g1,..., gr). Show that, if I = I(Y ) and I = g : g I(Y ) , then I(Y ) = I. However, show that I(Y ) need not { ∈ } be equal to (g1,..., gr), by lookinge at thee following example: Y = V (y2 2 3 2 3 3 3 − ye1,y3 e y1) = (t, t , t ) : t k Ae k and Y Pk. It is easy to see that − 2 {2 3 ∈ } ⊆ ⊆3 V+(x0x2 x1,e x0x3 ex1) contains all points of Pk of the form (0, 0,r,s). But − 2 3− 2 2 since (y2 y1,y3 y1) contains y1y2 y3 and y1y3 y2 , Y V+(x0x2 x1, x0x3 − 2 − − 2 − ⊆ − 2 − x1x2, x1x3 x2) (and in fact I(Y )=(x0x2 x1, x0x3 x1x2, x1x3 x2)). Thus − − − 3− (0, 0,r,s) Y if and only if r = 0. The algebraic subset Y of Pk is called the twisted cubic∈ curve. Chapter 2

Modules over Commutative Rings

2.1 Basic deﬁnitions

We have deﬁned R-modules in Chapter 1 and described some of the basic ones: ideals I, quotients R/I, submodules and quotient modules more generally, the free rank n R-module Rn, and ﬁnite direct sums. We begin this chapter by giving more sophisticated constructions along these lines.

Deﬁnition 2.1.1. Let Ma a A be a collection of R-modules indexed by the ∈ set A. Then there is a natural{ } R-module structure on the Cartesian or direct product

M = ϕ: A M : ϕ(a) M for all a A , a { → a ∈ a ∈ } a A a A Y∈ [∈ given as follows: addition is the usual addition of functions ((ϕ1 + ϕ2)(a)= ϕ (a)+ ϕ (a)), and (rϕ)(a)= r ϕ(a) using the R-module structure on M . 1 2 · a The direct sum a A Ma is the R-submodule of a A Ma given by the ϕ such that ϕ(a) = 0 for∈ all but finitely many a. ∈ L Q Of course, if A is finite, there is no difference between the Cartesian product and the direct sum. If Ma = R for all a, we denote the Cartesian product by RA. It is the set of all functions f : A R. The corresponding → direct sum F = FR(A) is the set of all functions f : A R which are zero for all but finitely many A. It is called the free R-module→ on A. By logic

43 44 CHAPTER 2. MODULES OVER COMMUTATIVE RINGS or convention, F ( ) = 0. An R-module which a free module on some set R ∅ is called a free R-module. We often write the element f F (A) as a ∈ R finite formal sum a A ra a, where ra = f(a), and the addition and R- module structure are∈ defined· in the obvious way. Suppose that M is an P R-module and that there is an isomorphism ϕ: M FR(A) for some A, 1 → and let B = ϕ− (A). We say that B is an R-basis of M or that M is freely generated by B over R. The intersection of arbitrarily many submodules of an R-module M is a submodule. If N1, N2 are submodules of M, then so is N1 + N2. The sum of an arbitrary collection of submodules of M is defined similarly. In general, we cannot define the product of two submodules of M. However, if N is a submodule of M and I is an ideal in R, we define

k IN = r m : r I, m N { i i i ∈ i ∈ } i=1 X to be the subset of M given by finite sums (of any length) of products of elements of I times elements of N. It is again a submodule of M. Given m M, there is the cyclic submodule Rm =(m) M, which by definition is ∈ ⊆ rm : r R . { ∈ } For m1,...,mn M, we define the submodule generated by m1,...,mn to be ∈ n (m ,...,m )= r m : r R . 1 n { i i i ∈ } i=1 X Note that (m1,...,mn) = Rm1 + + Rmn. The submodule generated by an arbitrary collection r , a A is··· defined similarly. a ∈ Definition 2.1.2. The module M is finitely generated or a finite R-module if M =(m ,...,m ) for some m ,...,m M. 1 n 1 n ∈ For an R-module M, the annihilator Ann M is by definition

Ann M = r R : rm = 0 for all m M . { { ∈ ∈ } It is an ideal of R. For example, Ann(R/I) = I. If m M, then Ann m is by deﬁnition the annihilator of the submodule Rm of M∈. 2.1. BASIC DEFINITIONS 45

A nonzero R-module M is simple if it has no proper submodules, i.e. the only submodules of M are 0 and M. Every simple module M is cyclic; in fact, M = Rm for every m M, m = 0. A quotient R/I is simple if and only if I is a maximal ideal.∈ It is easy6 to see that every simple R-module is in fact isomorphic to R/I, where I is a maximal ideal of R. A composition series for an R-module M is an increasing sequence 0= M M M = M 0 ⊂ 1 ⊂···⊂ ℓ of submodules Mi, such that Mi/Mi 1 is simple for 1 i ℓ. Of course, it − is a very special property for R-modules to have a composition≤ ≤ series. For example, if R = k is a field, then an R-module V is a k-vector space, a simple k-module is a one-dimensional vector space, and a composition series is an increasing sequence of subspaces 0= V V V = V 0 ⊂ 1 ⊂···⊂ ℓ such that dim(Vi/Vi 1) = 1. In this case, ℓ = dim V and V is finite dimen- − sional; conversely, every finite dimensional k-vector space has a composition series. We have the following standard result: Theorem 2.1.3 (Jordan-Hölder). Let M be an R-module. Suppose that

0= M M M = M and 0= N N N ′ = M 0 ⊂ 1 ⊂···⊂ ℓ 0 ⊂ 1 ⊂···⊂ ℓ are two composition series for M. Then ℓ = ℓ′, and there exists a permutation σ of the set 1,...,ℓ such that Ni/Ni 1 = Mσ(i)/Mσ(i) 1 for all i. { } − ∼ − If M has a composition series 0 = M0 M1 Mℓ = M, then by the theorem the integer ℓ does not depend⊂ on the⊂···⊂ choice of a composition series. It is called the length of M and written ℓ(M). For example, for a field k, the length of a k-module V is dim V . Let f : M N be a homomorphism of R-modules. Then the kernel of → f (viewed as a homomorphism of abelian groups) is an R-submodule Ker f of M, and the image Im f is an R-submodule of N. By definition, the cokernel Coker f of f is the quotient module N/ Im f. There is a natural isomorphism of R-modules M/ Ker f ∼= Im f. For example, if M is an R- module and m M, then the function f : R M defined by f(r) = rm ∈ → induces a homomorphism R M with kernel Ann m and image Rm, and hence an isomorphism from R/→Ann m to Rm. More generally, we have the variations for modules of the isomorphism theorems for groups: 46 CHAPTER 2. MODULES OVER COMMUTATIVE RINGS

Proposition 2.1.4. (i) If f : M N is an R-module homomorphism and → K is a submodule of M contained in Ker f, then there is a unique homomorphism j : M/K N which is onto Im f, such that f = j π, where π : M M/K is the→ natural map. Moreover Ker j = Ker f/K◦ → ⊆ M/K.

(ii) Let M be an R-module and let N1 and N2 be two submodules of M, with N N . Then N /N is naturally isomorphic to a submodule of 1 ⊆ 2 2 1 M/N1, and (M/N1)/(N2/N1) is isomorphic to M/N2. More precisely, the obvious map ψ : M/N M/N is a well-deﬁned homomorphism 1 → 2 with kernel N2/N1 and image M/N2.

(iii) Let M be an R-module and let N1 and N2 be two submodules of M. Then there is an isomorphism N /N N (N + N )/N . 1 1 ∩ 2 → 1 2 2 As usual, an R-module homomorphism f : M N is injective if it is one-to-one, if and only if Ker f = 0 and f is surjective→ if it is onto, if and only if Im f = N, if and only if Coker f = 0. If f is surjective, then N is isomorphic to M/ Ker f and in particular is isomorphic to a quotient of M. One very important way to construct new modules from old is by taking the set of all R-module homomorphisms from one module to another:

Deﬁnition 2.1.5. Let M and N be R-modules. We deﬁne HomR(M, N) to be the set of all R-module homomorphisms f : M N. It is itself an R-module via pointwise addition and multiplication by→ elements of R.

Let M , a A be a collection of R-modules. Given b A, there is a ∈ ∈ an R-module homomorphism πb : a A Ma Mb deﬁned by ϕ ϕ(b), called projection onto the bth factor (here∈ we→ view elements of the7→ direct product as functions ϕ). It is surjective.Q Likewise, there is a homomorphism ib : Mb a A Ma deﬁned by ib(m)(a)=0, a = b, and ib(m)(b) = b. It is an injection,→ called∈ inclusion into the bth factor.6 We then have the following: L

Proposition 2.1.6. Let Ma, a A be a collection of R-modules, and let N be an R-module. ∈

(i) If f : N a A Ma is a homomorphism, then for each a A, there → ∈ ∈ are induced homomorphisms fa = πa f : N Ma. Conversely, given homomorphismsQ f : N M for every◦ a →A, there is a unique ho- a → a ∈ momorphism f : N a A Ma such that fa = πa f for every a. → ∈ ◦ Q 2.1. BASIC DEFINITIONS 47

(ii) If g : a A Ma N is a homomorphism, then for each a A, there ∈ → ∈ are induced homomorphisms ga = g ia : Ma N. Conversely, given homomorphismsL g : M N for every◦ a →A, there is a unique ho- a a → ∈ momorphism g : a A Ma N such that ga = g ia for every a. ∈ → ◦

Proof. The first statementL in (i) is clear. For the second, given the fa, define the homomorphism f by the formula f(n)(a) = fa(n). The equality fa = π f and the uniqueness are clear. As for (ii), again the main point is a ◦ how to define g. Given ϕ a A Ma, let g(ϕ) = a A ga(ϕ(a)). The sum makes sense since only finitely∈ ∈ many of the ϕ(a) are∈ nonzero. It is then straightforward to see that gLhas the desired properties.P

In technical terms, a Ma is a product and a Ma is a coproduct in the category of R-modules. Q L Corollary 2.1.7. With hypotheses as in Proposition 2.1.6,

(i) HomR(N, a Ma)= a HomR(N, Ma); Q Q (ii) HomR( a Ma, N)= a HomR(Ma, N).

Corollary 2.1.8.L Let A beQ a set, and let F = FR(A) be the free module on A. Then, for every R-module N, there is a bijection from the set of R-module homomorphisms f : F N and the set of functions ψ : A N. → → Proof. We begin with the case where A has one element (leaving to the reader the case where A = ). In this case, we claim that there is a natural ∅ isomorphism Ψ: HomR(R, M) ∼= M. In fact, given f : R M, let Ψ(f) = f(1). Conversely, given m M, deﬁne f (r)= r m. It is→ easy to see that ∈ m · m f is an inverse to Ψ. Given the statement for R, the statement for 7→ m FR(A) follows from part (ii) of the above corollary.

In particular, we see that HomR(R, R) ∼= R via the map ϕ ϕ(1); the inverse is given by t R (r tr). More generally, we7→ can identify n m ∈ 7→ 7→ HomR(R , R ) with the free R-module of m n matrices Mn,m(R) in the usual way. ×

Lemma 2.1.9. An R-module M is ﬁnitely generated if and only if there exists a surjection Rn M. For every R-module M, there exists a free module F and a surjection→ F M. → 48 CHAPTER 2. MODULES OVER COMMUTATIVE RINGS

Proof. To see the first statement, for any R-module M and any finite collection m ,...,m M, there is an R-module homomorphism f : Rn M 1 n ∈ → by the previous corollary, defined by f(r1,...,rn) = i rimi, and its im- n age is Rm1 + + Rmn. Conversely, given a homomorphism f : R M, ··· P → the image of f is of the form Rm1 + + Rmn for some m1,...,mn M. The first statement of the lemma is··· then clear. To see the second,∈ note that the identity defines a homomorphism FR(M) M, which is clearly a surjection. → Definition 2.1.10. An R-module M is finitely presented if there exists a surjection f : Rn M such that the kernel of f is finitely generated. → In particular, a finitely presented R-module is finitely generated. In the exercises, we show that, if M is a finitely presented R-module and g : Rk M → is an arbitrary surjection, then Ker g is finitely generated.

Deﬁnition 2.1.11. If M is an R-module, then the dual module M ∨ is the R-module HomR(M, R).

Note that R∨ ∼= R by the proof of Corollary 2.1.8. Taking duals commutes with finite direct sums, by Corollary 2.1.7: (M1 M2)∨ ∼= M1∨ M2∨. In n n ∨⊕ ⊕ particular (R )∨ = R . However, a A Ma = a A Ma∨ in general. ∼ ∈ ∼ ∈ There is also the evaluation homomorphism ev: M M ∨∨, defined by m L → Q 7→ (ϕ ϕ(m)). It is easy to see that ev: R R∨∨ is an isomorphism and 7→ n n → hence that ev: R (R )∨∨ is an isomorphism. However, unlike the case of finite dimensional→ vector spaces, ev need not be injective or surjective in general.

2.2 Direct and inverse limits

Direct and inverse limits are a technical but very useful generalization of direct sums and products. We begin with the following: Deﬁnition 2.2.1. A directed set is a partially ordered set with the property that, for all a, b , there exists a c such that a c andA b c. ∈A ≤ ≤ Deﬁnition 2.2.2. Let be a directed set. A direct system M , f of A { a ab} R-modules indexed by consists of the following: a collection of R-modules A Ma indexed by and, for each pair (a, b) of elements a, b with a b, a homomorphismAf : M M satisfying: ∈A ≤ ab a → b 2.2. DIRECT AND INVERSE LIMITS 49

(i) For all a b c, f f = f ; ≤ ≤ bc ◦ ab ac

(ii) For all a, faa = Id.

Given two direct systems M , f , N , h of R-modules indexed by ,a { a ab} { a ab} A homomorphism of direct systems from Ma, fab to Na, hab is by deﬁnition a collection of homomorphisms h : M { N making} { the following} diagram a a → a commute: M ha N a −−−→ a

fab gab   Mb Nb. y −−−→hb y The idea of the direct limit is that we want to construct a module M which is in a sense the union of the Ma, glued together by the homomorphisms fab:

Definition 2.2.3. Let M , f be a direct system of R-modules indexed { a ab} by . We define the direct or injective limit lim Ma to be the R-module A a −→∈A a Ma M ′, where M ′ is the submodule of a Ma generated by ele- ∈A ∈A ments of the. form ia(m) ib fab(m) for all a b, a, b , and m Ma, L − ◦ ≤L ∈ A ∈ where ia : Ma a Ma is the natural inclusion. Note that the homomor- → ∈A phism ia : Ma a Ma induces a homomorphism fa : Ma lim Ma, and → ∈A → L a−→ it follows from theL construction that f = f f . ∈A a b ◦ ab Despite this somewhat confusing definition, direct limits are quite natural and arise in many situations. In working with them, it is best to use the following, which we leave as an exercise:

Lemma 2.2.4. If M = lim Ma, then every element of M is of the form a −→∈A fa(m) for some a and some m Ma. Moreover, if for some a and some m M f (m∈A)=0, then there∈ exists an element b with a ∈Ab such ∈ a a ∈A ≤ that fab(m)=0.

Remark 2.2.5. One can form direct limits with any partially ordered set , not necessarily directed. However, in this generality, the above lemma mayA not hold. 50 CHAPTER 2. MODULES OVER COMMUTATIVE RINGS

The next result says that the direct limit is characterized by a certain universal property:

Proposition 2.2.6. Suppose that Ma, fab is a direct system of R-modules indexed by . { } A (i) Let N be an R-module, and suppose that, for every a , we are given a homomorphism ϕ : M N such that, for all a, b∈A , a b, ϕ = a a → ∈A ≤ a ϕ f . Then there exists a unique homomorphism ϕ: lim M N b ◦ ab a → a−→ such that, for all a , ϕ = ϕ f . ∈A ∈A a ◦ a (ii) Let M be an R-module and h : M M a collection of homomor- a a → phisms such that h = h f for all a, b , a b. Suppose that, a b ◦ ab ∈ A ≤ given an R-module N and homomorphisms ga : Ma N for every a such that, for all a, b , a b, g = g f →, there exists a ∈ A ∈ A ≤ a b ◦ ab unique homomorphism g : M N such that, for all a , g = g h . → ∈A a ◦ a Then there is a natural isomorphism M ∼= lim Ma. a −→∈A Proof. For (i), by the universal property of the direct sum, there exists

ϕ: a Ma N such that, for all a , ϕa = ϕ ia, where ia : Ma ∈A → ∈ A ◦ → a Ma is the inclusion. Clearly ϕ(ia(m) ib fab(m)) = 0 for all m Ma. ∈AL − ◦ ∈ Thus, in the notation of Deﬁnition 2.2.3, ϕ(M ′) = 0, and hence ϕ induces a Le e homomorphism ϕ: lim Ma N withe the desired property. a → −→∈A e e Given M as in (ii), by the universal properties of lim Ma and of M, the ho- a −→∈A momorphisms fa : Ma lim Ma and ha : Ma M induce homomorphisms → a → −→∈A F : lim Ma M and H : M lim Ma. It is easy to see that the uniqueness a → → a −→∈A −→∈A properties force F H = Id, H F = Id. Thus M ∼= lim Ma. ◦ ◦ a −→∈A Corollary 2.2.7. If Ma, fab , Na, hab are two direct systems of R-modules indexed by and h {: M }N{ is a homomorphism} of direct systems from A a a → a Ma, fab to Na, hab , then there is an induced homomorphism h: lim Ma { } { } a → −→∈A lim Na. a −→∈A Proof. Apply (i) of the previous proposition to N = lim N and ϕ the a a a composition M N N. −→ ∈A a → a → 2.2. DIRECT AND INVERSE LIMITS 51

Example 2.2.8. Let = N, with the following partial ordering: n m if A and only if n m, and hence m = nd for some d N. Let Mn = Z for every n, and define,| for n m, ∈ f : M M by the formula f (a)= ma/n. nm n → m nm Clearly, if n m k, f f = f . We claim that lim M = Q (and mk ◦ nm nk n ∼ −→n Z in fact a slight variant of the construction works to give the∈ quotient field of every integral domain). Define ϕ : Z Q via ϕ (a) = a/n. Then one n → n checks that, if n m, then ϕ (a)= a/n =(ma/n)/m = ϕ f (a). n m ◦ nm Thus there is a homomorphism ϕ: lim Q, and it is straightforward to → −→n Z check that ϕ is an isomorphism. ∈ Example 2.2.9. Let M be an R-module, and let be the set of all sub- A modules of M, ordered as follows: N1 N2 if and only if N1 N2. We consider the collection of all submodules N of N as indexed by⊆ itself. For N N , let f beA the natural inclusion of N in N . Then it is easy to 1 2 N1,N2 1 2 see that lim N = M. More generally, if is a collection of submodules of A N−→ M with the∈A property that, if N , N , then there exists N such that 1 2 ∈A ∈A N + N N, then is a directed set under and lim N = N. 1 2 ⊆ A N−→ N ∈A [∈A Although an infinite direct sum of rings is not a ring in our sense, since it does not have the element 1, the direct limit of rings lim Ra is still a ring, a −→∈A since the image fa(1) of 1 Ra is easily checked to be independent of a and to be a unity in the direct∈ limit. One important example of this is the following:

Example 2.2.10. Let X be a topological space, let x X, and let x be the set of all open sets containing x. For U, V , define∈ U V if UU V . ∈ Ux ⊇ Let C(U) be the ring of continuous real-valued functions on U, viewed as a real vector space, and, if U V , let f : C(U) C(V ) be the restriction UV → homomorphism. Then CX,x = lim C(U) is a ring, the ring of germs of U x −→∈U continuous functions on X at x. An element of CX,x is an equivalence class 52 CHAPTER 2. MODULES OVER COMMUTATIVE RINGS of pairs (U, f), where U is an open set containing x and f is a continuous function defined on U, and two pairs (U, f) and (V,g) are equivalent if there is an open set W U V such that f W = g W . Note that CX,x is a local ring: if f is a continuous⊆ ∩ function defined| on| an open set U containing x, then the value f(x) is well-defined, and ev : f f(x) is a homomorphism x 7→ from CX,x to R. Let mx = Ker evx. Then mx is a maximal ideal of CX,x, and if f / mx, i.e. f(x) = 0, then 1/f is defined and continuous in some open set V ∈U with x V6. Clearly the equivalence class of the pair (V, 1/f) is an ⊆ ∈ inverse for (U, f). So mx is the unique maximal ideal of CX,x. There are many similar examples. For example, if X = Rn, or more generally X is a C∞ manifold, then we can define CX,x∞ = lim C∞(U), where U−→x n ∈U C∞(U) is the ring of C∞ functions on U. If X = C or C , or more generally a complex manifold, then we can define HX,x = lim H(U), where H(U) is U−→x the ring of holomorphic functions on U. In this case,∈U if X has dimension n as a complex manifold, one can also describe HX,x as the ring C z1,...,zn as the ring of convergent power series in n variables. All of the above{ rings} are also local; the content of this is that, if f is C∞ (resp. analytic) and nowhere zero, then 1/f is also C∞ (resp. analytic). One can ask about the analogous construction for the Zariski topology. We are not yet in a position to answer this question, since we have defined rings of functions on an algebraic set only in the affine case. We will return to this question in the next chapter.

There is also the dual concept of an inverse or projective limit:

Deﬁnition 2.2.11. Let be a partially ordered set. An inverse system A Ma, fab of R-modules indexed by consists of the following: a collection {of R-modules} M indexed by and,A for each pair (a, b) of elements a, b a A ∈A with a b, a homomorphism f : M M satisfying: ≤ ab b → a (i) For all a b c, f f = f ; ≤ ≤ ab ◦ bc ac

(ii) For all a, faa = Id.

Homomorphisms of inverse systems are deﬁned in the obvious way, by analogy with Deﬁnition 2.2.3. 2.2. DIRECT AND INVERSE LIMITS 53

Let M , f be an inverse system of R-modules indexed by . We define { a ab} A the inverse or projective limit lim Ma to be the R-module a ←−∈A lim M = ϕ M : f (ϕ(b)) = ϕ(a) for all a b, a, b a { ∈ a ab ≤ ∈ A} a←− a ∈A Y∈A Thus an element of the inverse limit is an element in the direct product satisfying obvious compatibility conditions. The projections πb : a Ma ∈A → Mb induce homomorphisms pb : lim Ma Mb. a → Q ←−∈A It is clear from the definitions that, if R is a ring for every a and a ∈ A fab is a ring homomorphism, then lim Ra is again a ring. a←− There is a universal property for∈A the inverse limit: Proposition 2.2.12. Suppose that M , f is an inverse system of R- { a ab} modules indexed by . A (i) Let N be an R-module, and suppose that, for every a , we are given ∈A a homomorphism ψa : N Ma such that, for all a, b , a b, ψa = f ψ . Then there exists→ a unique homomorphism∈Aψ : N ≤ lim M ab ◦ b → a a←− such that, for all a , ψ = p ψ. ∈A ∈A a a ◦ (ii) Suppose that M is an R-module and η : M M are homomorphisms a → a such that, for all a, b , a b, η = f η . Suppose further that, ∈ A ≤ a ab ◦ b given an R-module N and homomorphisms ρa : N Ma for every a such that, for all a, b , a b, ρ = f →ρ , there exists a ∈ A ∈ A ≤ a ab ◦ b unique homomorphism ρ: M N such that, for all a , ρ = η ρ. → ∈A a a ◦ Then there is a natural isomorphism M ∼= lim Ma. a ←−∈A (iii) If h : M N is a homomorphism of inverse systems, then there is a a → a an induced homomorphism lim Ma lim Na. a → a ←−∈A ←−∈A Proof. (i): The collection ψ defines a homomorphism ψ : N M , and a → a a the condition ψ = f ψ implies that the image of ψ is contained in the a ab b Q inverse limit. The uniqueness◦ is clear. e (ii), (iii): The arguments are the same as that for (ii)e of Proposition 2.2.6 and Corollary 2.2.7. 54 CHAPTER 2. MODULES OVER COMMUTATIVE RINGS

Example 2.2.13. With R = Z, let = N, with the usual ordering, and let n m A n Mn = Z/p Z, with fnm : Z/p Z Z/p Z the usual surjection for m n. n → ≥ The ring Zp = lim Z/p Z is then the ring of p-adic numbers. ←−n This example may be extended in two different directions. Replacing Z with an arbitrary ring R and the subgroups pnZ with the powers J n of some fixed ideal J (with the usual surjections R/J m R/J n for m n), we obtain the ring lim R/J n, which we will study in the→ chapter on completions.≥ ←−n Even more generally, given an abelian group A and a decreasing filtration of A by subgroups An, we can define lim A/An. ←−n On the other hand, if as in Example 2.2.8, we let = N with the partial A ordering n m if n m and Mn = Z/nZ, then for n m we have the canonical surjection f : Z/m| Z Z/nZ. We may thus define| Zˆ = lim Z/nZ. An nm → ←−n element of Zˆ consists of a sequence a Z/nZ for every n 1, such that, if n ∈ ≥ n m, then the image of a in Z/nZ is a . | m n As an application of the Chinese remainder theorem and unique factorization into primes, one can in fact show that

ˆ Z ∼= Zp, p Y where the product is over all prime numbers.

2.3 Exact sequences

One of the most basic constructions for modules is the formation of a quotient module M/N. We want to write this abstractly as follows:

Deﬁnition 2.3.1. Let M ′,M,M ′′ be R-modules. We say that the diagram

i π 0 M ′ M M ′′ 0 → −→ −→ → is a short exact sequence if i is injective, π is surjective, and Ker π = Im i. In this case, i is an isomorphism from M ′ to the submodule i(M ′) of M, and π induces an isomorphism from M/i(M ′) to M ′′. We shall also refer to M as an extension of M ′′ by M ′. 2.3. EXACT SEQUENCES 55

In general, given a (ﬁnite or inﬁnite) sequence of R-modules Mi and homomorphisms f : M M , which we write as i i → i+1 f − f f i 1 M i M i+1 , ··· −−→ i −→ i+1 −−→· · · we say that the sequence is exact if, for every i, Im fi = Ker fi+1. For example, given an R-module homomorphism f : M N, the following sequence → is always exact:

f 0 Ker f i M N π Coker f 0, → −→ −→ −→ → where i is the natural inclusion of Ker f in M and π is the natural projection f N N/ Im f = Coker f. To say that 0 M N is exact is to say that f → f → −→ is injective, to say that M N 0 is exact is to say that f is surjective, −→ →f and hence to say that 0 M N 0 is exact is to say that f is an isomorphism. Exact sequences→ are−→ a fundamental→ pattern in mathematics, f − f f and we shall use them freely. Every exact sequence i 1 M i M i+1 ··· −−→ i −→ i+1 −−→ can be broken into compatible short exact sequences via ··· f 0 Ker f M i Im f 0, → i → i −→→ i → where compatibility means that, via the equality Im fi = Ker fi+1, the right- most nonzero term of this sequence agrees with the leftmost nonzero term in the next. Here is a simple way to construct new exact sequences from old (proof left to the reader):

Lemma 2.3.2. Suppose that Ma′ , Ma and Ma′′ are collections of R-modules indexed by a , and that f : M ′ M and g : M M ′′ are homomor- ∈A a a → a a a → a phism. Then M ′ M M ′′ a → a → a a a a M∈A M∈A M∈A is exact if and only if, for every a , the sequence M ′ M M ′′ is ∈ A a → a → a exact.

i π Returning to the case of a short exact sequence 0 M ′ M M ′′ 0, → −→ −→ → one way that such a sequence arises for given M ′ and M ′′ is to take M = M ′ M ′′, i = i the natural inclusion of M ′ in the direct sum, and π = π , ⊕ 1 2 56 CHAPTER 2. MODULES OVER COMMUTATIVE RINGS

the projection of M ′ M ′′ = M ′ M ′′ to the second factor. In this case we ⊕ × say that the extension of M ′′ by M ′ (or the short exact sequence) is split. It is often important to know when an extension is essentially the same as a split extension. That is the purpose of the following: i π Lemma 2.3.3. If 0 M ′ M M ′′ 0 is a short exact sequence of → −→ −→ → R-modules, then the following statements are equivalent:

(i) There exists a homomorphism s: M ′′ M such that π s = Id. → ◦ (ii) There exists a homomorphism r : M M ′ such that r M ′ = Id. → | 1 (iii) There exists an isomorphism f : M = M ′ M ′′ such that i = f − i , ∼ ⊕ ◦ 1 where i : M ′ M ′ M ′′ is the inclusion into the ﬁrst factor, and 1 → ⊕ π = π f, where π : M ′ M ′′ is the projection onto the second factor. 2 ◦ 2 ⊕ Proof. We outline the construction of the relevant maps and leave it to the reader to check that they have the desired properties. (i) = (ii): let ⇒ 1 r = Id s π. (ii) = (iii): let f =(r, π). (iii) = (i): let s = f − i . − ◦ ⇒ ⇒ ◦ 2 Deﬁnition 2.3.4. A homomorphism s as in (i) above is called a section and a homomorphism r as in (ii) is called a retraction. If any of the equivalent conditions of Lemma 2.3.3 holds, we say that the exact sequence is split.

Corollary 2.3.5. Suppose that M ′′ is a free R-module. Then every short i π exact sequence 0 M ′ M M ′′ 0 is split. → −→ −→ → Proof. Suppose that M ′′ = F (A) is free on the set A. Since M M ′′ R → is surjective, the natural inclusion j : A FR(A) can be lifted to some functions ¯: A M (i.e. π s¯ = j). By Corollary→ 2.1.8, there is an induced homomorphism→s: F (A) ◦ M such that s(a)=s ¯(a) for all a A. It R → ∈ follows that π s = Id since this holds on the generating set A. Thus (i) of Lemma 2.3.3 is◦ satisﬁed, and so the sequence is split. We next give two basic results about exact sequences. The proofs are an example of the method of proof “by a diagram chase.” Lemma 2.3.6 (Snake lemma). Suppose that the following is a commutative diagram of R-modules, with exact rows:

α′ α′′ 0 M ′ M M ′′ 0 −−−→ −−−→ −−−→ −−−→ f g h

 β′  β′′  0 N′ N N′′ 0. −−−→ y −−−→ y −−−→ y −−−→ 2.3. EXACT SEQUENCES 57

Then there is an exact sequence

α′ α′′ β′ β′′ 0 Ker f 0 Ker g 0 Ker h ∂ Coker f 1 Coker g 1 Coker h 0, → −→ −→ −→ −→ −→ → where the above homomorphisms will be described in the course of the proof.

Proof. The homomorphisms α0′ , α0′′, β1′ , β1′′ are those induced from α′, α′′, β′, β′′ in the obvious way. It is then obvious that α0′ is injective and β1′′ is surjective and that Ker α′′ = Ker α′′ Ker g = Im α′ Ker g = Im α′ . If 0 ∩ ∩ 0 x Ker β′′ and n N projects onto x Coker g, then by definition β′′(n)= ∈ 1 ∈ ∈ h(m′′) for some m′′ M ′′. Since α′′ is surjective, m′′ = α′′(m) for some ∈ m M. But then β′′(n)= h α′′(m)= β′′(g(m)). It follows that n g(m) ∈ ◦ − ∈ Ker β′′ = Im β′. Thus x, the projection of n into Coker g, lies in Im β1′ . We must now construct the homomorphism ∂ and check exactness at the remaining two stages. Let x Ker h, and choose m M such that ∈ ∈ α′′(m) = x, which is possible since α′′ is surjective. Since 0 = h α′′(m) = ◦ β′′ g(m), g(m) Ker β′′ = Im β′. Thus g(m)= β′(n) for some n N, which ◦ ∈ ∈ is well-defined since β′ is injective. The only choice was that of m, which is uniquely determined mod Ker α′′ = Im α′. Replacing m by m + α′(m′), m′ M ′, replaces g(m) by g(m)+ g α′(m′) = g(m)+ β′(f(m′)), and so ∈ ◦ g(m)= β′(n+f(m′)). Thus the image of n in Coker f is well-defined, and we let ∂(x) be this image. It is straightforward if tedious to check that ∂ is an R-module homomorphism. We leave this, as well as the remaining exactness proofs, to the reader.

Lemma 2.3.7 (Five lemma). Suppose that the following is a commutative diagram of R-modules, with exact rows:

M α1 M α2 M α3 M α4 M 1 −−−→ 2 −−−→ 3 −−−→ 4 −−−→ 5

f1 f2 f3 f4 f5

 β1  β2  β3  β4  N N N N N. y1 −−−→ y2 −−−→ y3 −−−→ y4 −−−→ y5 If f1 is surjective and f2, f4 are injective, then f3 is injective. If f5 is injective and f2, f4 are surjective, then f3 is surjective.

Proof. We shall just prove the ﬁrst statement and leave the second as an exercise. If f3(x)=0, then 0= β3 f3(x)= f4 α3(x). Since f4 is injective, α (x) = 0, hence there exists y ◦ M such that◦ x = α (y). Since 0 = 3 ∈ 2 2 58 CHAPTER 2. MODULES OVER COMMUTATIVE RINGS f (x) = f α (y) = β f (y), β (f (y)) = 0 and thus f (y) = β (n) for 3 3 ◦ 2 2 ◦ 2 2 2 2 1 some n N . Since f is surjective, n = f (z) for some z M , and then ∈ 1 1 1 ∈ 1 f2(y) = β1(f1(z)) = f2(α1(z)). Since f2 is injective, y = α1(z). But then x = α2(α1(z)) = 0, so that f3 is injective. Finally, for future reference, we discuss exactness properties of direct and inverse limits.

Deﬁnition 2.3.8. Suppose that is a directed set and that M , f , A { a ab} Ma′ , fab′ , Ma′′, fab′′ are direct systems of R-modules indexed by . An exact {sequence} of{ direct} systems consists of homomorphisms of directedA systems g : M ′ M and h : M M ′′ such that, for all a , the sequence a a → a a a → a ∈A

ga ha M ′ M M ′′ a −→ a −→ a is exact. An exact sequence of inverse systems is deﬁned similarly.

ga ha Proposition 2.3.9. If Ma′ Ma Ma′′ is an exact sequence of direct systems, then the induced sequence−→ −→

g h lim Ma′ lim Ma lim Ma′′ a −→ a −→ a −→∈A −→∈A −→∈A is also exact.

Proof. Clearly h g = 0. If h(m) = 0, then, by Lemma 2.2.4, we can write ◦ m = f (n) for some n M . The image h (n) is then 0 in lim M ′′. a ∈ a a a a Thus, again by Lemma 2.2.4, there exists a b with a b−→such∈A that ∈ A ≤ f ′′ (h (n)) = 0. But f ′′ (h (n)) = h (f (n)), so that f (n) Ker h = Im g . ab a ab a b ab ab ∈ b b Writing fab(n) = x with x Mb′, it is clear that m = g(fb′(x)). Thus m Im g. ∈ ∈ For inverse systems, there is only the following weaker result:

Proposition 2.3.10. If 0 M ′ M M ′′ 0 is an exact sequence of → a → a → a → inverse systems, then the induced sequence

0 lim Ma′ lim Ma lim Ma′′ → a → a → a ←−∈A ←−∈A ←−∈A is also exact. 2.4. CHAIN CONDITIONS 59

Proof. The proof is straightforward and is left to the reader. For the following, we assume for simplicity that the inverse system is indexed by the natural numbers N, although one can formulate more general versions:

gn hn Proposition 2.3.11. If 0 M ′ M M ′′ 0 is an exact sequence → n −→ n −→ n → of inverse systems indexed by N, and the homomorphisms Mn′ +1 Mn′ are surjective for all n N, then the induced sequence → ∈ 0 lim Mn′ lim Mn lim Mn′′ 0 → n N → n N → n N → ←−∈ ←−∈ ←−∈ is also exact.

Proof. It suﬃces to prove the surjectivity of lim M lim M ′′. Let n N n n N n ∈ → ∈ (x ) M ′′ be such that f ′′ (x )= x for←− all n. By the←− surjectivity of n ∈ n n n+1,n n+1 n the homomorphism Mn M ′′, we can ﬁnd, for every n, an element yn Mn Q → n ∈ such that hn(yn) = xn. Suppose inductively that we have done so in such a way that fm,m+1(ym+1) = ym for all m < n. We claim that we can then choose y so that f (y ) = y . In any case, f (y ) y maps n+1 n+1,n n+1 n n+1,n n+1 − n to 0 via h , and so f (y ) y = g (z) for some z M ′ . We can then n n+1,n n+1 − n n ∈ n replace y by y g (w), where w M ′ is an element such that n+1 n+1 − n+1 ∈ n+1 fn′ +1,n(w)= z. It then follows that fn+1,n(yn+1)= yn as desired.

2.4 Chain conditions

We have defined Noetherian rings in Proposition 1.4.17 in Chapter 1. We now recall and generalize this definition. Definition 2.4.1. Let R be a ring. R is Noetherian if every increasing sequence of ideals is eventually constant, in other words if I I I I , 1 ⊆ 2 ⊆···⊆ n ⊆ n+1 ⊆··· where the I are ideals of R, then there exists an N N such that for all n ∈ k N, Ik = IN . More generally, for an arbitrary ring R, an R-module M≥is Noetherian if every increasing sequence of submodules of M is eventually constant. An R-module M is Artinian if every decreasing sequence of submodules of M is eventually constant, and the ring R is Artinian or an Artin ring if it is Artinian viewed as an R-module, in other words if every decreasing sequence of ideals in R is eventually constant. 60 CHAPTER 2. MODULES OVER COMMUTATIVE RINGS

We have essentially seen the following:

Lemma 2.4.2. Let R be a ring and let M be an R-module. Then the following are equivalent:

(i) M is Noetherian.

(ii) Every submodule of M is ﬁnitely generated.

(iii) Every nonempty collection of submodules of M has a maximal element.

Proof. The argument of Proposition 1.4.17 in Chapter 1 shows that (i) (ii). We now show that (i) (iii). ⇐⇒ ⇐⇒ (i) = (iii): If is a nonempty collection of submodules of M with ⇒ A no maximal element, choose M1 , which is possible since = . Since M cannot be a maximal element∈ of A , there exists M withA 6 M∅ M , 1 A 2 ∈A 1 ⊂ 2 i.e. M2 properly contains M1. Since M2 cannot be maximal, there exists M3 with M2 M3. Continuing in this way, we produce an inﬁnite strictly∈ A increasing sequence⊂ of submodules of M, contradicting (i). (iii) = (i): if M1 M2 is an increasing sequence of submodules of M, then⇒ the collection⊆ M ⊆···has a maximal element M . Since M M { i} n0 n0 ⊆ k for all k n0, it follows that Mn0 = Mk for all k n0. Thus the sequence is eventually≥ constant. ≥

Similar arguments show:

Lemma 2.4.3. Let R be a ring and let M be an R-module. Then the following are equivalent:

(i) M is Artinian.

(ii) Every nonempty collection of submodules of M has a minimal element.

The following is straightforward and left to the reader:

Lemma 2.4.4. Let R be a Noetherian ring and let I be an ideal of R. Then R/I is a Noetherian ring. Likewise, if R is an Artinian ring and I is an ideal of R, then R/I is Artinian. 2.4. CHAIN CONDITIONS 61

Lemma 2.4.5. Let R be a ring and let

0 M ′ M M ′′ 0 → → → → be an exact sequence of R-modules. Then M is Noetherian if and only if M ′ and M ′′ are Noetherian, and M is Artinian if and only if M ′ and M ′′ are Artinian. Proof. We shall just give the proof in the Noetherian case; the Artinian case is formally the same. If M is Noetherian, then every increasing sequence of submodules of M ′ is in particular an increasing sequence of submodules of M and thus is eventually constant. Hence M ′ is Noetherian. Likewise, an increasing sequence of submodules of M ′′ defines an increasing sequence of submodules of M, by inverse image and thus is eventually constant as well. Thus M ′′ is Noetherian. Conversely, suppose that M ′ and M ′′ are Noetherian. Let M M 0 ⊆ 1 ⊆··· be an increasing sequence of submodules of M. Since M ′′ is Noetherian, we can assume that for all large k, Mk has a fixed image N in M ′′. Fixing one such k , define, for j k , 0 ≥ 0

N = n M ′ : n + M M . j { ∈ k0 ⊆ j}

It is easy to see that the Nj are an increasing sequence of submodules of M ′ and that M = M +N for all j k . Since the N are eventually constant, j k0 j ≥ 0 j the Mj are as well.

Corollary 2.4.6. (i) If M1,...,Mn are Noetherian R-modules, then so is n i=1 Mi. (ii) IfLR is a Noetherian ring, then Rn is a Noetherian R-module and more generally every ﬁnitely generated R-module is Noetherian. Similar statements hold with Noetherian replaced by Artinian throughout.

In particular, if R is Noetherian, then an R-module M is ﬁnitely generated if and only if it is ﬁnitely presented. Lemma 2.4.7. Let R be a ring and let M be a nonzero R-module. Then the following are equivalent: (i) M is both Noetherian and Artinian. 62 CHAPTER 2. MODULES OVER COMMUTATIVE RINGS

(ii) There exists a composition series for M.

Proof. Suppose that M is both Noetherian and Artinian. Since M is Ar- tinian, there exists a minimal element M1 in the set of nonzero submodules of M. Clearly M is simple. Next, if M = M, there exists a minimal element 1 1 6 M2 of the set of submodules containing M1; again, M2/M1 is clearly simple. Continuing in this way, we ﬁnd

0 M M { } ⊂ 1 ⊂ 2 ⊂··· with Mi+1/Mi simple. Since M is Noetherian, this chain must terminate, and (up to reindexing) we have produced a composition series for M. Conversely, suppose that there exists a composition series for M. We argue by induction on the length ℓ(M). Clearly a simple R-module is both Noetherian and Artinian, and more generally if M has a composition series of length n then there is an exact sequence

0 M ′ M M ′′ 0, → → → → where M ′ is simple and ℓ(M ′′)= n 1. By induction and Lemma 2.4.5, we are done. −

2.5 Exactness properties of Hom

We can view Hom ( , ) as a functor of two variables (a bifunctor) from the R · · category of R-modules to itself. It is contravariant in the ﬁrst variable and covariant in the second. Holding one of the variables ﬁxed, we obtain either a contravariant or a covariant functor from the category of R-modules to itself. We turn now to the exactness properties of Hom.

Proposition 2.5.1. Suppose that

0 M ′ M M ′′ 0 → → → → is an exact sequence of R-modules, and that N is an R-module. Then the following are exact:

0 Hom (M ′′, N) Hom (M, N) Hom (M ′, N); → R → R → R 0 Hom (N, M ′) Hom (M, N) Hom (N, M ′′). → R → R → R 2.5. EXACTNESS PROPERTIES OF HOM 63

For example, if we take R = Z and the exact sequence

n 0 Z × Z Z/nZ 0, → −→ → → for N = Z, the sequence n 0 0 Z × Z → → −→ obtained by applying HomZ( , Z) to the above sequence is exact, but the · right hand homomorphism is not surjective. Similarly, taking N = Z/nZ, the sequence 0 0 0 Z/nZ → → → obtained by applying HomZ(Z/nZ, ) to the above sequence is exact, but · again the right hand homomorphism is not surjective. To find R-modules P such that Hom (P, ) is a functor with better ex- R · actness properties, we make the following definition: Definition 2.5.2. An R-module P is projective if, for every surjection π : → M ′′ of R-modules and homomorphism f : P M ′′, there exists an R-module → homomorphism f˜: P M such that f = π f˜. Equivalently, for every → ◦ surjection M M ′′, the corresponding map → Hom (P, M) Hom (P, M ′′) R → R is surjective. Another way to say this is that the functor M Hom (P, M) 7→ R is a covariant exact functor from the category of R-modules to itself. For example, the R-module Rn is projective. More generally a free R- module is projective. In fact, for R = Z, a Z-module is projective if and only if it is free. However, for most rings R, there exist projective R-modules which are not free. One important fact is the following: Lemma 2.5.3. Every R-module is a quotient of a projective R-module, in fact it is a quotient of a free R-module. Proof. This is immediate from Lemma 2.1.9 and the fact that a free module is projective. Corollary 2.5.4. Let M be an R-module. Then there exists an exact sequence P P M 0, ···→ 1 → 0 → → where each Pi is projective. In fact, we can even assume that each Pi is free. 64 CHAPTER 2. MODULES OVER COMMUTATIVE RINGS

Proof. Choose a surjection σ : P M, where P is a free R-module. Then 0 0 → 0 choose a surjection σ1 from a free R-module P1 to Ker σ0, and continue in the obvious way.

Such an exact sequence is called a projective resolution of M. In case each Pi is in fact free, such a resolution is called a free resolution of M. We can rephrase Corollary 2.5.4 by saying that the category of R-modules has enough projectives.

Proposition 2.5.5. Let P be an R-module. Then the following are equivalent:

(i) P is projective.

(ii) Every exact sequence

π 0 M ′ M P 0 → → → −→ splits, i.e. there exists a homomorphism s: P M such that π s = Id. → ◦ (iii) P is a direct summand of a free R-module.

Proof. (i) = (ii): Given an exact sequence as in (ii), the identity Id Hom (P,P )⇒ lifts to s Hom (P, M), which clearly satisﬁes π s = Id. ∈ R ∈ R ◦ (ii) = (iii): For every exact sequence as in (ii), M = P M ′. Now ⇒ ∼ ⊕ choose a surjection F P with kernel K, where F is free. Then F = P K, → ∼ ⊕ so that P is a direct summand of F . (iii) = (i): In general, the R-module N N is projective if and only ⇒ 1 ⊕ 2 if both N1 and N2 are projective, using the commutative diagram

Hom (N N , M) Hom (N N , M ′′) R 1 ⊕ 2 −−−→ R 1 ⊕ 2

  HomR(N1, M)  HomR(N2, M) HomR(N1, M ′′)  HomR N2, M ′′). ⊕y −−−→ ⊕y Thus if P is a direct summand of a free module, it is projective since a free module is projective.

To deﬁne an injective R-module, we reverse the arrows: 2.5. EXACTNESS PROPERTIES OF HOM 65

Deﬁnition 2.5.6. Let I be an R-module. Then I is injective if, for every injective R-module homomorphism i: M ′ M, if f : M ′ I is a homomorphism, then there exists a homomorphism→ g : M I →such that f = g i. Equivalently, I is injective if and only if, given→ an injection ◦ i: M ′ M, the induced homomorphism HomR(M, I) HomR(M ′, I) de- ﬁned by→g g i is surjective. We can also rephrase this→ by saying that the 7→ ◦ functor M HomR(M, I) is a contravariant exact functor from the category of R-modules7→ to itself. As a partial analogue of Proposition 2.5.5, we have: Proposition 2.5.7. For an R-module I, the following are equivalent: (i) I is injective. (ii) Every exact sequence

i 0 I M M ′ 0 → −→ → → splits, i.e. there exists a homomorphism r : M I such that r i = Id. → ◦ (iii) I is a direct summand of every R-module which contains it.

Proof. Clearly (i) = (ii) (take M ′ = I in the deﬁnition of injective) and (ii) = (iii) by the proof⇒ of Lemma 2.3.3. To see that (iii) = (i), suppose ⇒ ⇒ that i: M ′ M is an injective R-module homomorphism and f : M ′ I is → → a homomorphism. Deﬁne

I ′ M = I M/ Im(f, i). ⊕M ⊕ − Then the composite homomorphism ϕ: I I M I ′ M is injec- → ⊕ → ⊕M tive (where the first map is inclusion on the first factor), since if (x, 0) = (f(y), i(y)) for some y M ′, then y = 0 since i is injective and hence − ∈ x = f(y) = 0 as well. Thus I is a summand of I ′ M, and hence there ⊕M exists a retraction r : I M ′ M I. If j : M I M ′ M is the homomorphism induced by inclusion⊕ onto→ the second factor,→ ⊕ followed by projection, the composition g = r j : M I is easily seen to satisfy: g i = f. Hence ◦ → ◦ I is injective. Injective R-modules are not as easy to come by as projective R-modules, and in particular they are almost never finitely generated. In the exercises, we shall show the following: 66 CHAPTER 2. MODULES OVER COMMUTATIVE RINGS

Proposition 2.5.8. Let M be an R-module. Then there exists an injective R-module I and an injective homomorphism M I. Thus, there exists an exact sequence → 0 M I I .... → → 0 → 1 → Such an exact sequence is called an injective resolution of M, and we say that the category of R-modules has enough injectives. However, in case R is a PID we can characterize injective modules directly. Definition 2.5.9. Let R be an integral domain and M an R-module. Then r M is divisible if, for all r = 0, the homomorphism M × M given by 6 −→ multiplication by r is surjective. Equivalently, for all r R, r = 0, and all m M, there exists an n M such that rn = m. ∈ 6 ∈ ∈ Proposition 2.5.10. If R is an integral domain, then every injective R- module is divisible. If moreover R is a PID, then an R-module I is injective if and only if it is divisible. Proof. Let I be an injective R-module. Suppose that x I and that r R, r = 0. Consider f : R I defined by f(t) = tx and i:∈R R defined∈ by i(t6 ) = rt. Since I is injective,→ there exists a homomorphism→g : R I such → that g i = f. Then y = g(1) satisfies ry = g(r)= g(i(1)) = f(1) = x. Thus I is divisible.◦ Now suppose that R is a PID. It suffices to show that every divisible R- module I is injective. So suppose we are given i: M ′ M and f : M ′ I as in the definition. → →

Case I: M/M ′ is a cyclic R-module, i.e. of the form R/rR for some ideal of R, necessarily of the form rR. We begin by showing that we can give a very explicit description of M in terms of M ′ and R. Choose x M which ∈ maps onto the generator 1 R/rR. Thus rx = y M ′. Consider the ∈ ∈ homomorphism ϕ: M ′ R M defined by (a, t) i(a) tx. Then ϕ ⊕ → 7→ − is surjective since, given m m, there exists an s R such that m sx ∈ ∈ ≡ mod i(M ′), i.e. m = i(a)+sx = ϕ(a, s). We claim that Ker ϕ = R (y,r). − − · For ϕ(a, t) = 0 if and only if i(a)= tx M ′. It then follows that t (r), say t = rs for some s R, and then a = rsx∈ = sy, so that (a, t)= s(y,r∈ ). Thus ∈ M = (M ′ R)/R (y,r), and the injection i corresponds to the inclusion of ∼ ⊕ · M ′ into the first factor of M ′ R, followed by projection. ⊕ Since I is divisible, f(y) = rz for some z I. Defineg ˜: M ′ R I byg ˜(a, t) = f(a) tz. Theng ˜(y,r) = f(y) ∈rz = 0, so thatg ˜ ⊕induces→ a − − 2.6. MODULES OVER A PID 67 homomorphism (M ′ R)/R (y,r) = M I. Clearly the restriction to M ′ ⊕ · ∼ → is f. Case II: The general case. This will be a routine application of Zorn’s lemma. Consider pairs (N, ψ) consisting of a submodule N of M containing i(M) and a homomorphism ψ : N I such that ψ i = f. Order the set → ◦ of all such pairs via (N, ψ) (N ′, ψ′) if N N ′ and ψ′ N = ψ. Given Pa chain (N , ψ ) : i in , there exists an⊆ upper bound| (N, ψ) for { i i ∈ I} P I as follows: let N = i Ni and let ψ(n) be the common value of the ψi on n (where defined). Thus,∈I by Zorn’s lemma, there is a maximal element (M ,g ) for . If M S= M, then there is a submodule N, M N M, 0 0 P 0 6 0 ⊂ ⊆ such that N/M is cyclic. By Case I, there exists a ψ : N I extending i, 0 → contradicting the maximality of M0. Thus M0 = N and we have found the desired extension g0 of i.

For example, if R is a PID and K is its quotient ﬁeld, then K is an injective R-module. However, it is often better to look instead at the quotient K/R, viewed as an R-module, which is still divisible and hence is injective. We will use the injectivity of K/R in the next section.

2.6 Modules over a PID

Throughout this section, R is a PID. Our goal is to prove the following theorem:

Theorem 2.6.1. Every ﬁnitely generated R-module M is isomorphic to a direct sum of cyclic R-modules, i.e. modules of the form R/I for some ideal I, necessarily of the form (r). More precisely, M = RN (R/r R)ni , where ∼ ⊕ i i the ri are irreducible elements of R. Moreover, the integer N is unique, as L are the ri and ni up to associates and permutation.

Proof. The proof is not difficult, but it is long and will be broken up into steps. Before we begin, however, let us make a preliminary definition: Definition 2.6.2. Let R be an integral domain and M an R-module. A torsion element m M is an element m such that there exists an x R with ∈ ∈ xm = 0 and x = 0. The set of all torsion elements is a submodule τ(M), 6 the torsion submodule of M (sometimes written as Mtors). The module M is torsion free if τ(M)=0. If M = τ(M), then M is a torsion module. 68 CHAPTER 2. MODULES OVER COMMUTATIVE RINGS

Clearly τ(M M )= τ(M ) τ(M ), τ(R)=0, τ(R/I)= R/I for every 1 ⊕ 2 2 ⊕ 2 nonzero ideal I in R. We have an exact sequence 0 τ(M) M M/τ(M) 0, → → → → and it is a simple exercise to check that M/τ(M) is torsion free. For example, N ni if M ∼= R i(R/riR) , where the ri are irreducible elements of R, and in ⊕ ni N particular are nonzero, then τ(M) ∼= i(R/riR) and M/τ(M) ∼= R . We shall first analyzeL finitely generated torsion free R-modules, and conclude by studying finitely generated torsion R-modules.L Step I: A finitely generated torsion free R-module M is isomorphic to a submodule of Rk for some k. We may clearly assume that M = 0. Suppose that M is generated by 6 m1,...,mn and choose a maximal subset of m1,...,mn , say m1,...,mk, { k } which is linearly independent over R (i.e. if i=1 rimi = 0, then ri = 0 for all i). Since M is torsion free and nonzero, k 1. Let M ′ M be the P ≥ ⊆ k submodule generated by m1,...,mk. Then the homomorphism i: R M k → defined by the mi (i(r1,...,rk)= i=1 rimi) is injective by the definition of k linear independence, and its image is M ′. Thus M ′ = R . P ∼ For j>k, there exists an rj R, rj = 0, and si R, 1 i k, such k ∈ 6 ∈ ≤ ≤ that r m + s m = 0. Hence r m M ′ for j = k +1,...,n. If we set j j i=1 i i j j ∈ r = rk+1 rn, then rmj M ′ for j>k, and rmi M ′ for i k, since ·····P ∈ ∈ ≤ the m for i k generate M ′. Thus rM M ′. On the other hand, since i ≤ ⊆ M is torsion free, multiplication by r induces an isomorphism r : M rM. k · → Thus, M is isomorphic to the submodule rM of M ′ ∼= R , proving Step I. Step II: A submodule M of Rn is isomorphic to Rm for some m n. ≤ The proof goes by induction on n. In case n = 1, a submodule of R is an ideal in R, which is either (0) or of the form rR for some nonzero r R. ∈ Since R is an integral domain, rR ∼= R as R-modules. Thus Step II holds if n = 1. For the inductive step, if π : Rn+1 Rn is the projection onto the first n factors, with kernel R, then we have→ a commutative diagram with exact rows: 0 R Rn+1 Rn 0 −−−→ −−−→ −−−→ −−−→

x x x 0 M′ M M′′ 0, −−−→  −−−→  −−−→  −−−→ where the vertical arrows are inclusions; here M ′ = Ker π M and M ′′ = ∩ n π(M). Then M ′ is a submodule of R, and hence M ′ = 0 or R, and M ′′ R , ∼ ⊆ 2.6. MODULES OVER A PID 69

m so by the inductive hypothesis M ′′ = R with m n. In particular, M ′′ is ∼ ≤ free, so that the bottom exact sequence is split: M = M ′ M ′′ and so either ∼ ⊕ M = Rm with m n < n +1, or M = Rm+1 with m +1 n + 1. ∼ ≤ ∼ ≤ Step III: Every finitely generated R-module M is isomorphic to τ(M) Rn ⊕ for some n. This n is uniquely specified and is called the rank of M. Indeed, by Step I and Step II, M/τ(M) is finitely generated (since M is) n and torsion free, and so M/τ(M) ∼= R for some n. In particular, M/τ(M) is free, so that the exact sequence

0 τ(M) M M/τ(M) 0 → → → → is split: M = τ(M) M/τ(M) = τ(M) Rn for some n. ∼ ⊕ ∼ ⊕ To see the uniqueness, if M = τ(M) Rn = τ(M) Rm, then M/τ(M) = ∼ ⊕ ∼ ⊕ ∼ Rn = Rm. But it is an easy exercise (Exercise 2.26) to show (for any com- ∼ n m mutative ring R, not necessarily an integral domain), that R ∼= R if and only if n = m. If M is finitely generated, it follows from Step III that τ(M) is a quotient of M and is therefore finitely generated as well. To conclude the proof of Theorem 2.6.1, it will suffice to analyze finitely generated torsion modules. Clearly, if M is a finitely generated torsion module, then there exists x R, x = 0, such that xM = 0. ∈ 6 Step IV: Suppose that M is an R-module such that xM = 0, where x R ∈ and x = 0. Suppose further that x = r r where gcd(r ,r ) = 1. Then 6 1 2 1 2 M = M M , where ∼ 1 ⊕ 2 M = m M : r m =0 . i { ∈ i }

To see this, use the fact that R is a PID to write 1 = ar1 + br2 for some a, b R. We have the sum map M1 M2 M defined by (m1, m2) m +∈m . If m + m = 0, then m = m⊕ = →m satisfies: r m = r m =7→ 0, 1 2 1 2 1 − 2 1 2 and so m = 1 m = ar1m + br2m = 0. Thus the sum map is injective. Finally, if m M· , then m = ar m + br m. As xm = r r m = 0, we see that ∈ 1 2 1 2 r (br m)= r (ar m) = 0. Thus m = m + m , where m = br m M and 1 2 2 1 1 2 1 2 ∈ 1 m2 = ar1m M2. So the map M1 M2 M is an isomorphism. By taking∈ the factorization of x ⊕into a→ product of powers of non-associated irreducible elements ri, we see that it is enough to show: Step V: Let M be a finitely generated R-module, and let r be an irreducible element of R such that raM = 0. Then M is isomorphic to a direct sum of 70 CHAPTER 2. MODULES OVER COMMUTATIVE RINGS cyclic R-modules; in fact, we shall show that

k ni M ∼= R/r R, i=1 M for some nonnegative integers ni. (We will not check the uniqueness part of Theorem 2.6.1 for τ(M).) Assume that a > 0 is the smallest positive integer such that raM = 0. The proof goes by induction on the number of generators. Clearly the result holds if M is generated by one element. Assume inductively that the result has been shown for all modules generated by at most k elements. If M is generated by m1,...,mk+1, then after reordering we may suppose that a a 1 a r m = 0, that r − m = 0, and that r m = 0 for all i > 1. Thus there is 1 1 6 i an exact sequence

0 R/raR M M/Rm 0, → → → 1 → and, by induction, since M/Rm1 can be generated by k generators, M/Rm1 a is a direct sum of cyclic R-modules. Note that r (M/Rm1) = 0, since the same is true for M. It suffices to show that the above exact sequence splits. In order to split the sequence, let K be the quotient field of R and consider the homomorphism ψ : R/raR K/R defined by ψ(1) = 1/ra. Note that ψ is well-defined since ψ(ra) =→ 0 in K/R. The image of ψ is the submodule R (1/ra) of K/R. Clearly t (1/ra) R if and only if ra t, if and only if · · ∈ | t raR. So the submodule R (1/ra) of K/R is isomorphic to R/raR via ψ. ∈ Now we use the fact that K/R· is divisible, and hence by Proposition 2.5.10 is an injective R-module. Since R/raR is included in M, by the definition of injective we can lift the homomorphism ψ to a homomorphism ψ : M K/R. → Since raM = 0, if t is in the image of ψ, then rat = 0 in K/R. It follows as in the previous paragraph that t R (1/ra). But then composinge ψ : M a 1 a ∈ a· a→ R (1/r ) with ψ− : R (1/r ) R/r Re defines a retraction ρ: M R/r R · · → → which splits the above exact sequence. This concludes the proof ofe Step V and hence of Theorem 2.6.1.

2.7 Nakayama’s lemma

This section discusses the ubiquitous Nakayama’s lemma and some consequences: 2.7. NAKAYAMA’S LEMMA 71

Lemma 2.7.1 (Nakayama’s lemma). Let R be a commutative ring and let J be an ideal contained in the Jacobson radical of R. Let M be an R-module, and suppose either that M is ﬁnitely generated or that J is nilpotent, in other words that J n =0 for some n> 0. If JM = M, then M =0.

Proof. First assume that J is nilpotent. Then M = JM = J 2M = = ··· J nM = 0. In the case that M is ﬁnitely generated, suppose that M is generated by the k elements m1,...,mk. The proof is by induction on the number k. If k = 1, then M ∼= R/I for some ideal I, and JM = M implies that 1 r mod I for some r J. Thus 1 r I. But since r J and J is contained≡ ∈ − ∈ ∈ in the Jacobson radical, 1 r is a unit. Thus I = R and M = 0. − Suppose that k > 1. Since M = JM, there exist r ,...,r J such 1 k ∈ that mk = r1m1 + + rkmk. Thus (1 rk)mk = r1m1 + + rk 1mk 1. − − Since 1 r is again··· a unit, it follows that− m lies in the submodule··· of M − k k generated by m1,...,mk 1. Thus M can be generated by k 1 elements, so − by induction M = 0. −

Corollary 2.7.2. Let R be a commutative ring and let J be an ideal contained in the Jacobson radical of R. Let M be an R-module, and suppose either that M is ﬁnitely generated or that J is nilpotent. If N is a submodule of M such that M = N + JM, then N = M.

Proof. Apply Nakayama’s lemma to the module M/N.

Corollary 2.7.3. Let R be a local ring with maximal ideal m, and let k = R/m. Let M be a ﬁnitely generated R-module. Then m1,...,mk generate M if and only if their images in M/m M span the k-vector space M/m M. · · Proof. Since R is local, its Jacobson radical is just m. Now apply the above corollary to the submodule N of M generated by m1,...,mk.

Corollary 2.7.4. Let R be a local ring with maximal ideal m, and let M and N be two ﬁnitely generated R-modules. Then a homomorphism f : M N is surjective if and only if the induced homomorphism f¯: M/mM N/→mN → is surjective.

Corollary 2.7.5. Let R be a local ring. Then a ﬁnitely generated projective R-module M is free. 72 CHAPTER 2. MODULES OVER COMMUTATIVE RINGS

Proof. As usual, we let m be the maximal ideal of R and let k = R/m. Choose m1,...,mk M such that the images of the mi are a k-basis for M/mM. There is thus∈ a map Rn M and the induced map kn M/mM is an isomorphism. By Corollary 2.7.3,→ the map Rn M is surjective.→ If K → is the kernel, then 0 K Rn M 0 → → → → is exact. Since M is projective, this exact sequence splits: Rn = M K. Now ∼ ⊕ the second projection Rn K is surjective, and so K is finitely generated. → Moreover if we reduce mod m, we find that kn = (M/mM) (K/mK), in ∼ ⊕ such a way that the projection kn M/mM is an isomorphism of k-vector spaces. Thus K/mK = 0. Applying→ Nakayama’s lemma to the finitely generated R-module K, we see that K = 0 and thus that Rn M is an → isomorphism.

We conclude this section with another approach to Nakayama’s lemma, via the following result, which we shall also need later. For a homomorphism f : M M from an R-module to itself, we let f n = f f be the → ◦···◦ composition of f with itself n times, i.e. f 1 = f, and inductively f n+1 = f f n. Thus, given a polynomial p(x) R[x], we can deﬁne p(f) as well. ◦ ∈ Theorem 2.7.6 (Cayley-Hamilton). Let I be an ideal in R and let M be a ﬁnitely generated R-module, generated by n elements. Suppose that f : M M is a homomorphism such that f(M) I M. Then there exists → n n 1⊆ · i a polynomial p(x) R[x] of the form x + a x − + + a , with a I for ∈ 1 ··· n i ∈ every i, such that p(f): M M is the zero homomorphism. →

Proof. Suppose that M is generated by m1,...,mn. Let R[x] act on M as follows: R acts in the usual way, and x m = f(m). It is straightforward to verify that this action turns M into· an R[x]-module. By hypothesis, we can write x m = f(m ) = b m for some b I. Equivalently, · i i j ij j ij ∈ (δijx bij)mj = 0. If we let A be the n n matrix xI B with coeﬃcients i − P × − m P 1 m2 in R, and m be the column vector m =  . , then A m = 0, where 0 . ·   mn   is the zero column vector. Now, by Cramer’s  rule, there exists an n n × matrix A′ with coeﬃcients in R such that A′A = (det A)I. It follows that (det A)I m = A′A m = 0, and hence that det A m = 0 for every i. Since · · · i 2.8. THE TENSOR PRODUCT 73

the mi generate M, det A M = 0. But it is clear that det A = p(x) = n n 1 · i x + a1x − + + an, with ai I for every i, and that det A acts on M as p(f). Thus p(f···)(M) = 0 as claimed.∈

Corollary 2.7.7 (Second proof of Nakayama’s lemma). Let R be a commutative ring and let J be an ideal contained in the Jacobson radical of R. Let M be a ﬁnitely generated R-module. If JM = M, then M =0.

Proof. Applying the Cayley-Hamilton theorem to f = Id = multiplication by 1 and I = J, we see that (1+ a1 + + an)M = 0 for some elements a J. Thus (1 + a)M = 0 for some a ···J. But 1 + a is then a unit, so that i ∈ ∈ M = 0. Here is another corollary of the Cayley-Hamilton theorem:

Corollary 2.7.8. Let M be a ﬁnitely generated R-module. If f : M M → is surjective, then f is an isomorphism. Moreover, the inverse of f is of the form q(f) for some polynomial q(x) R[x]. ∈ Proof. As in the proof of the Cayley-Hamilton theorem, we can view M as an R[x]-module via x m = f(m). Let I = xR[x]=(x), viewed as an ideal of R[x]. Since f is surjective,· x M = M and hence IM = M. Applying the · Cayley-Hamilton theorem to Id: M M, viewed as an R[x]-module, there → n n 1 exists a polynomial p(t) R[x][t] of the form t + a1t − + + an, with i ∈ n n ··· i ai (x ), such that p(Id) = 0. Thus Id = (Id) = i=1 ai. Since ai (x ), ∈ n − ∈ we can write i=1 ai = xq(x) for some polynomial r(x). In terms of f, this says that −f q(f) = Id, and hence f has an inverseP as claimed. P◦ 2.8 The tensor product

Deﬁnition 2.8.1. Suppose that M, M ′ and N are R-modules. An R-bilinear function f : M M ′ N is a function satsifying: × →

(i) For all m , m M and m M ′, f(m + m , m) = f(m , m) + 1 2 ∈ ∈ 1 2 1 f(m2, m).

(ii) For all m , m M ′ and m M, f(m, m + m ) = f(m, m ) + 1 2 ∈ ∈ 1 2 1 f(m, m2).

(iii) For all r M and m M, n M ′, f(rm, n)= f(m, rn)= rf(m, n). ∈ ∈ ∈ 74 CHAPTER 2. MODULES OVER COMMUTATIVE RINGS

In other words, the function m f(m, a) is R-linear (i.e. an R-module 7→ homomorphism) for each ﬁxed a M ′ and conversely. Multilinear functions are deﬁned similarly. ∈ The tensor product is a universal object for bilinear functions:

Deﬁnition 2.8.2. Let M1, M2 be R-modules. A tensor product for M1 and M is an R-module M M , together with an R-bilinear function 2 1 ⊗R 2 ϕ: M1 M2 M1 R M2, such that, for every R-module N and R-bilinear function× f : M→ M⊗ N, there exists a unique R-module homomorphism 1 × 2 → f˜: M M N such that f = f˜ ϕ. 1 ⊗R 2 → ◦ As with all universal objects, tensor products, if they exist, are unique up to a unique isomorphism. Next we show that tensor products do indeed exist:

Proposition 2.8.3. For all R-modules M1, M2, the tensor product of M1 and M2 exists. Proof. Set M M = F (M M )/Q, the free module on the product 1 ⊗R 2 R 1 × 2 M M , modulo the submodule Q generated by the relations 1 × 2 (m + m , m) (m , m) (m , m), (m, m + m ) (m, m ) (m, m ), 1 2 − 1 − 2 1 2 − 1 − 2 (rm, n) r(m, n), (m, rn) r(m, n). − − Clearly the natural map M M M M is R-bilinear. We denote the 1 × 2 → 1 ⊗R 2 image of (m , m ) by m m . Every R-bilinear function f : M M 1 2 1 ⊗ 2 1 × 2 → N defines an R-module homomorphism FR(M1 M2) N which factors through the quotient by Q, and thus defines a× homomorphism→ f˜: M 1 ⊗R M N for which f(m , m )= f˜(m m ). 2 → 1 2 1 ⊗ 2 A useful consequence of the construction is that M1 R M2 is generated as an R-module by elements of the form m m . ⊗ 1 ⊗ 2 Warning: To define a homomorphism f : M1 R M2 N, it is not enough to specify the values of f on elements of the form⊗ m →m , because they may 1 ⊗ 2 satisfy somewhat complicated relations. Instead, it is typically necessary to construct f be defining a bilinear map on M1 M2 and then invoking the universal property. ×

More generally, given M1,...,Mk, we can deﬁne a universal object for multilinear maps M1 Mk N, which we denote by M1 R R Mk. The construction above×···× also shows→ that this universal object exists.⊗ ···⊗ Here are some easily established properties of : ⊗ 2.8. THE TENSOR PRODUCT 75

Lemma 2.8.4. (i) For all R-modules M and N, M N = N M. ⊗R ∼ ⊗R

(ii) Given R-modules M1, M2 and N, let Bil(M1, M2; N) denote the R- module of R-bilinear maps from M1 M2 to N. Then there are natural isomorphisms ×

Bil(M , M ; N) = Hom (M M , N) = 1 2 ∼ R 1 ⊗ 2 ∼ ∼= HomR(M1, HomR(M2, N)) ∼= HomR(M2, HomR(M1, N)).

(iii) For all R-modules M, the bilinear map R M M deﬁned by (r, m) × → 7→ rm induces an isomorphism R M = M. ⊗R ∼ (iv) If M , i I is a collection of R-modules and N is an R-module, then i ∈ ( M ) N = (M N). Thus in particular Rn M = M n. i i ⊗ ∼ i i ⊗ ⊗ ∼

(v) GivenL R-modulesLM1, M2, N1, N2 and homomorphisms f : M1 N1 and g : M N , there is a unique induced R-module homomorphism→ 2 → 2 f g : M M N N which satisfies m m f(m ) g(m ). ⊗ 1 ⊗R 2 → 1 ⊗R 2 1 ⊗ 2 7→ 1 ⊗ 2 (vi) There is a natural isomorphism M (M M ) = (M M ) M , 1⊗R 2⊗R 3 ∼ 1⊗R 2 ⊗R 3 and both of these R-modules are isomorphic to M M M . 1 ⊗R 2 ⊗R 3 Proof. (i) is clear since Bil(M1, M2; N) ∼= Bil(M2, M1; N). Next we prove (ii). The isomorphism Bil(M , M ; N) = Hom (M 1 2 ∼ R 1 ⊗ M2, N) is the definition of tensor product. Given f Bil(M1, M2; N), the function m f(m, ) defines a homomorphism of R-modules∈ 7→ · Bil(M , M ; N) Hom (M , Hom (M , N)). 1 2 → R 1 R 2 Conversely, given Φ: M Hom (M , N), the function 1 → R 2

f(m1, m2)=Φ(m1)(m2) is R-bilinear and thus there is an R-module homomorphism

Hom (M , Hom (M , N)) Bil(M , M ; N), R 1 R 2 → 1 2 which is the inverse function to the map

Bil(M , M ; N) Hom (M , Hom (M , N)) 1 2 → R 1 R 2 76 CHAPTER 2. MODULES OVER COMMUTATIVE RINGS described above. Thus Bil(M1, M2; N) ∼= HomR(M1, HomR(M2, N)), and the isomorphism Bil(M1, M2; N) ∼= HomR(M2, HomR(M1, N)) is similar. For an R-module N, we have Bil(R, M; N) ∼= HomR(R, HomR(M, N)) ∼= HomR(M, N) under the natural map f f(1, ). Thus the bilinear map R M M given by (r, m) rm is7→ universal· for bilinear maps from R ×M to→R-modules N, and so by7→ the universal properties of tensor products × R M = M under the map r m rm. This proves (iii). ⊗R ∼ ⊗ 7→ To see (iv), we again argue by the universal property, noting that for all R-modules P ,

Hom (( M ) N,P ) = Hom ( M , Hom (N,P )) R i ⊗R ∼ R i R i i M M = Hom (M , Hom (N,P )) = Hom (M N,P ) = ∼ R i R ∼ R i ⊗ ∼ i i Y Y = Hom ( (M N),P ). ∼ R i ⊗R i M

Thus i(Mi R N) has the universal property of ( i Mi) R N and so they are isomorphic.⊗ ⊗ L L To see (v), the map (m1, m2) f(m1) g(m2) is an R-bilinear map M M N N , and thus there7→ is an induced⊗ homomorphism M 1 × 2 → 1 ⊗R 2 1 ⊗R M N N which satisﬁes m m f(m ) g(m ). 2 → 1 ⊗R 2 1 ⊗ 2 7→ 1 ⊗ 2 Finally, to see (vi), if we deﬁne Trilin(M1, M2, M3; N) to be the R-module of all R-trilinear maps from M M M to N, then arguments as in the 1 × 2 × 3 proof of (i) show that

Trilin(M1, M2, M3; N) ∼= HomR(M1, Bil(M2, M3; N)) ∼= = Hom (M , Hom (M M , N)) = Hom (M (M M ), N). ∼ R 1 R 2 ⊗R 3 ∼ R 1 ⊗R 2 ⊗R 3 This sets up an isomorphism M M M = M (M M ), and 1 ⊗R 2 ⊗R 3 ∼ 1 ⊗R 2 ⊗R 3 there is a similar isomorphism M M M = (M M ) M . 1 ⊗R 2 ⊗R 3 ∼ 1 ⊗R 2 ⊗R 3 Next we discuss the operation of change of rings, or (depending on the context) extension of scalars. Let S be an R-algebra and let M be an R- module.

Lemma 2.8.5. There is a unique structure of an S-module on M S for ⊗R which s (m s )= m (s s ). 1 ⊗ 2 ⊗ 1 2 2.8. THE TENSOR PRODUCT 77

Proof. Use the map S M S M S deﬁned by (s ,m,s ) m × × → ⊗R 1 2 7→ ⊗ (s1s2) M R S. For s1 ﬁxed, this is R-bilinear and so induces a function S (M∈ S⊗) M S, which is R-bilinear. Thus there is a multiplication × ⊗R → ⊗R by S on M S, extending the R-module action and satisfying (st)(m s′)= ⊗R ⊗ s(t(m s′)), thus giving an S-module structure. ⊗

Example 2.8.6. 1) Let k K be fields. For a k-vector space V , we can ⊆ form V k K. Then V k K is a K-vector space and it follows from (iv) of Lemma⊗ 2.8.4 that dim ⊗V K = dim V . This is an intrinsic way to extend K ⊗k k scalars. 2) Let I be an ideal in R; we will show that M (R/I) = M/IM. ⊗R ∼ (However it is not in general true that M I = IM.) ⊗R One checks easily that this construction is transitive in the following sense: if S T is a map of R-algebras, then the T -modules M R T and (M S) →T are isomorphic. ⊗ ⊗R ⊗S Another related construction is the tensor product of two R-algebras: let S and T be two (not necessarily commutative) R-algebras. Define a multiplication on S T induced from the 4-linear map ⊗R S T S T S T × × × → ⊗R given by (s1, t1,s2, t2) (s1s2) (t1t2). So there is an induced (S R T ) R (S T ) S T . Here7→ 1 1⊗ is the identity and R is in the center.⊗ ⊗ ⊗R → ⊗R ⊗ Example 2.8.7. 1) R[x] S = S[x]. ⊗R 2) Mn(R) R S = Mn(S). where Mn(R) is the noncommutative R-algebra of n n matrices⊗ with coefficients in R. × 3) R[x] R[y] = R[x, y]. ⊗R ∼ This last follows from 1): R[x] R[y] = R[y][x] = R[x, y]. It is also ⊗R ∼ ∼ a categorical fact. More generally, if S and T are R-algebras, there are homomorphisms S S T and T S T , and the images of S and → ⊗R → ⊗R T commute. In particular, if S and T are commutative, then S R T is commutative also. ⊗

Proposition 2.8.8. For commutative R-algebras S and T , S R T is a coproduct in the category of commutative R-algebras. ⊗ 78 CHAPTER 2. MODULES OVER COMMUTATIVE RINGS

Proof. For S and T arbitrary, given R-algebra homomorphisms f : S A → and g : T A, the map S T A deﬁned by (s, t) f(s)g(t) is R-bilinear → × → 7→ and so deﬁnes a (unique) module homomorphism f g : S R T A, with the property that the composition S S T A⊗ is f, and⊗ similarly→ for → ⊗R → T . If A is commutative, then it is straightforward to check that f g is an algebra homomorphism. Conversely, a homomorphism ϕ: S T⊗ A ⊗R → induces homomorphisms f : S A and g : T A such that ϕ = f g. Thus S T is a coproduct. → → ⊗ ⊗R

In particular, if R and S are commutative rings, then R Z S is their coproduct in the category of commutative rings. ⊗ We have defined the topological space Spec R and associated to each ring homomorphism R S a continuous function Spec S Spec R. Thus it is natural to expect→ that tensor product has a “geometric”→ interpretation, related to the usual Cartesian product. If f : Y X and g : Z X are two → → functions, we define the fiber product

Y Z = (y, z) Y Z : f(y)= g(z) . ×X { ∈ × }

Note that there is an induced function h: Y X Z X defined by h(y, z)= f(y) = g(z). The fiber product is a product× in→ the category of sets over X (sets A, together with a function A X). The fiber product of two → topological spaces is similarly defined. If S and T are R-algebras, then we define the fiber product Spec S Spec R Spec T to be Spec(S R T ). Since the category of spectra is the opposite× to the category of commutative⊗ rings, it follows formally that Spec(S R T ) is a fiber product in the category of spectra. However, its underlying⊗ topological space is not the same as the fiber product in the sense of topological spaces of Spec S and Spec R (just as the product topology on An Am is not the product topology). For ex- k × k ample, k[x , x ] = k[x ] k[x ], but the natural function Spec k[x , x ] 1 2 ∼ 1 ⊗k 2 1 2 → Spec k[x1] Spec k[x2] is not a homeomorphism, in fact, it is not even injective. On× the other hand, let S be an R-algebra corresponding to the morphism f : Spec S Spec R. If x = m is a closed point of R, so that we have the inclusion i: →m Spec R, corresponding to the homomorphism { } → 1 R R/m = k, then we have the fiber f − (x), which may be viewed as → the (set-theoretic) fiber product Spec S Spec k. Then the fiber can ×Spec R be identified with the fiber product Spec(S R R/m), because a prime ideal of S R/m = S/mS is the same thing as⊗ a prime ideal q of S containing ⊗R 2.8. THE TENSOR PRODUCT 79

1 m, and, since m is maximal, q contains m if and only if f − (q) = m. For a generalization, see Exercise 3.10 in the next chapter. In case R = k is an algebraically closed field and S and T are finitely generated, reduced k-algebras, then we have the classical picture of S and T as the affine coordinate rings A(Y ), A(Z) for algebraic sets Y and Z. It is then natural to compare the k-algebras A(Y ) A(Z) and A(Y Z). We ⊗k × will return to this question shortly. We turn now to the exactness properties of the tensor product. Here (M , M ) M M 1 2 7→ 1 ⊗R 2 is a bifunctor from R-modules to R-modules which is covariant in both variables and symmetric in the obvious sense. It is not however exact. Instead we have the following: Lemma 2.8.9. For every exact sequence of R-modules

f g M ′ M M ′′ 0, −→ −→ → and for every R-module N, the induced sequence

M ′ N M N M ′′ N 0 ⊗R → ⊗R → ⊗R → is exact.

Proof. Since M ′′ N is generated by elements of the form m n, m M ′′ ⊗R ⊗ ∈ and the map M M ′′ is surjective, the induced map M R N M ′′ R N is surjective. Clearly→ (g Id) f Id = 0, and so it suffices⊗ to→ show⊗ that Ker(g Id) = Im(f Id).⊗ Let◦I be⊗ the submodule Im(f Id) of M N. ⊗ ⊗ ⊗ ⊗R The claim Ker(g Id) = Im(f Id) is equivalent to the statement that the ⊗ ⊗ induced map from (M N)/I to M ′′ N is an isomorphism (we already ⊗R ⊗R know that it is well-defined and surjective). To find an inverse map, define h: M ′′ R N (M R N)/I by first defining a bilinear map M ′′ N (M ⊗N)/I →as follows:⊗ (m, n) m˜ n mod I, wherem ˜ M×is any→ ⊗R 7→ ⊗ ∈ element satisfying g(˜m) = m. This is well-defined, since ifm ˜ 1 andm ˜ 2 are two such, thenm ˜ m˜ Im f, and thus 1 − 2 ∈ (˜m n) (˜m n)=(˜m m˜ ) n 1 ⊗ − 2 ⊗ 1 − 2 ⊗ which lies in Im(f Id) = I. Thus there is an induced map h: M ′′ R N (M N)/I, which⊗ is easily checked to be an inverse to the natural⊗ map→ ⊗R (M N)/I M ′′ N. ⊗R → ⊗R 80 CHAPTER 2. MODULES OVER COMMUTATIVE RINGS

Corollary 2.8.10. If I is an ideal in R, M (R/I) = M/IM. ⊗R ∼ Proof. Apply the previous lemma to the exact sequence

I R R/I 0. → → → Thus the sequence

I M R M (R/I) M 0 ⊗R → ⊗R → ⊗R → is exact. We know that R M = M, and under this identiﬁcation the ⊗R ∼ image of I R M is IM. Thus (R/I) R M is isomorphic to the quotient M/IM. ⊗ ⊗

For example, if n and m are relatively prime, (Z/nZ) Z (Z/mZ) = 0,and ⊗ more generally (Z/nZ) Z (Z/mZ) = Z/dZ, where d = gcd(m, n). Likewise ⊗ Q Z (Z/nZ)=0 for n> 0. ⊗ Corollary 2.8.11. If M and N are ﬁnitely generated R-modules, then so is M N. ⊗R The following corollary is just a restatement of Nakayama’s lemma.

Corollary 2.8.12. Let R be a ring and M a ﬁnitely generated R-module. If J is an ideal of R contained in the Jacobson radical of R and M R (R/J)=0, then M =0. ⊗

Corollary 2.8.13. Suppose that R is a local ring and that M and N are ﬁnitely generated R-modules such that M R N =0. Then either M or N is zero. ⊗

Proof. Suppose for example that N = 0. Then by Nakayama’s lemma, 6 N/mN = 0. In particular, there is a surjection N R/m 0. If we tensor with M6, we see that there is a surjection M N→ M → (R/m). Since ⊗R → ⊗R M N = 0, M (R/m) = 0 as well, Thus M = 0 by Corollary 2.8.12. ⊗R ⊗R 2.9. PRODUCTS OF AFFINE ALGEBRAIC SETS 81 2.9 Products of aﬃne algebraic sets

In this section, we shall we discuss products of affine algebraic sets and in particular of affine algebraic varieties. Note that the Zariski topology on n m n+m Ak Ak = Ak is not the same as the product topology. For example, × 1 in the Zariski topology on Ak the only proper closed sets are finitely many 2 points, and it is easy to check that an irreducible algebraic curve in Ak is not a closed set in the product topology on A1 A1. However, we do have the k × k following:

n+m n m Lemma 2.9.1. The projection maps Ak Ak and Ak are continuous. n m →n+m For all p Ak , the inclusion p Ak Ak defines a homeomorphism m ∈ n{+m} × → from Ak onto its image in Ak in the subspace topology, and likewise for An q . k ×{ } Proof. Since projection is a polynomial map, the first statement is clear. As m n+m for the second, the map Ak Ak defined by x (p, x) is continuous and its image is the closed set →p Am. The inverse7→ map p Am Am is { } × k { } × k → k defined by taking the restriction of projection, which is also continuous, and so both maps are homeomorphisms.

n m If X1 and X2 are closed subsets of Ak and Ak respectively, then X1 1 1 n+m × X = π− (X ) π− (X ) is a closed subset of A . Explicitly, if X = 2 1 1 ∩ 2 2 k i V (Ji), then we have X1 X2 = V (J1,J2), where J1 k[x1,...,xn] k[x ,...,x ,y ,...,y ] and× J k[y ,...,y ] k[x ,...,x⊆ ,y ,...,y ]. ⊆ 1 n 1 m 2 ⊆ 1 m ⊆ 1 n 1 m As for aﬃne coordinate rings, we have the following:

n m Proposition 2.9.2. Let X1 and X2 be closed algebraic subsets in Ak and Ak respectively. Then the aﬃne coordinate ring A(X X ) of X X An+m 1 × 2 1 × 2 ⊆ k is A(X ) A(X ). 1 ⊗k 2 Proof. The isomorphism

k[x ,...,x ] k[y ,...,y ] k[x ,...,x ,y ,...,y ] 1 n ⊗k 1 m → 1 n 1 m induces a surjection A(X ) A(X ) A(X X ). Here the map is deﬁned 1 ⊗k 2 → 1 × 2 as follows: given f g A(X ) A(X ), map it to the function i i ⊗ i ∈ 1 ⊗k 2 P (p ,p ) f (p )g (p ). 1 2 7→ i 1 i 2 i X 82 CHAPTER 2. MODULES OVER COMMUTATIVE RINGS

To see that this map is injective, suppose that it is not, and choose a nonzero ℓ f g in the kernel with ℓ minimal among all such expressions. Clearly i=1 i ⊗ i the gi must then be linearly independent and by symmetry the same is true forP the f . In particular, the f are all nonzero. Choose p X such that i i 1 ∈ 1 f (p ) = α = 0 for some i. By hypothesis f (p )g = α g is the zero i 1 i 6 i i 1 i i i i function on X2. Hence the gi are not linearly independent, contradiction. P P n m Corollary 2.9.3. Let X1 and X2 be closed algebraic subsets in Ak and Ak respectively, and let Ii = I(Xi) be the ideal of all functions vanishing on Xi. Then

I(X X )= I(X )k[x ,...,x ,y ,...,y ]+ I(X )k[x ,...,x ,y ,...,y ]. 1 × 2 1 1 n 1 m 2 1 n 1 m

Proof. Abbreviate k[x1,...,xn] by k[x], and similarly for k[y] and k[x, y]. Using the isomorphism k[x] k[y] = k[x, y], it is clearly enough to show that ⊗k ∼ the kernel of the composite homomorphism k[x] k k[y] A(X1) k A(X2) is I(X ) k[y]+ k[x] I(X ). Using the exactness⊗ of→ tensor product⊗ over 1 ⊗k ⊗k 2 k, there is a commutative diagram with exact rows and columns 0 0 0

   0 I(X1) k I(X2) k[x] k I(X2) A(X1) k I(X2) 0 −−−→ ⊗y −−−→ ⊗y −−−→ ⊗y −−−→

   0 I(X1) k k[y] k[x] k k[y] A(X1) k k[y] 0 −−−→ y⊗ −−−→ ⊗y −−−→ y⊗ −−−→

   0 I(X1) k A(X2) k[x] kA(X2) A(X1) k A(X2) 0 −−−→ ⊗y −−−→ ⊗y −−−→ ⊗y −−−→

   0 0 0 y y y A diagram chase then shows that the kernel of the natural homomorphism from k[x] k[y] to A(X ) A(X ) is I(X ) k[y]+ k[x] I(X ), as ⊗k 1 ⊗k 2 1 ⊗k ⊗k 2 claimed.

Corollary 2.9.4. If k is an algebraically closed ﬁeld, and R and S are two reduced k-algebras, then R S is also reduced. ⊗k 2.9. PRODUCTS OF AFFINE ALGEBRAIC SETS 83

Proof. If R and S are finitely generated over k, the result follows from Propo- sition 2.9.2 and the fact that affine coordinate rings are reduced. If R and S are arbitrary, then two elements of R S are contained in the image in ⊗k R S of R′ S′, where R′ and S′ are finitely generated subalgebras of R, ⊗k ⊗k S respectively, and the result then follows from the case where R and S are finitely generated over k. Remark 2.9.5. If k is not algebraically closed, then the tensor product of two reduced k-algebras need not be reduced (Exercise 2.30). However, one does have the following result: We say that a field K is a separably generated extension of k if K is a separable algebraic extension of a pure transcendental extension of k. For example, if k is algebraically closed, or more generally perfect, and hence for all k such that char k = 0, then one can show that every extension field of k is separably generated over k. There is then the following fact: if K is a separably generated extension of k, and L is an arbitrary extension of k, the k-algebra K L is reduced. ⊗k We turn now to affine algebraic varieties. There is the following topological result: n m Lemma 2.9.6. If X1 and X2 are irreducible algebraic subsets of Ak and Ak respectively, then X X is an irreducible algebraic subset of An+m. 1 × 2 k Proof. Suppose that X1 X2 = Z W , where Z and W are closed subsets of An+m. Then, for all x× X , ∪ k ∈ 1 x X = (( x X ) Z) (( x X ) W ). { } × 2 { } × 2 ∩ ∪ { } × 2 ∩ Since X is irreducible, x X is irreducible as well by Lemma 2.9.1, and 2 { } × 2 so either x X2 = (( x X2) Z) or x X2 = (( x X2) W ). Thus either{ }x × X {Z }or × x ∩X W{.} Define × { } × ∩ { } × 2 ⊆ { } × 2 ⊆ Z′ = x X : x X Z ; { ∈ 1 { } × 2 ⊆ } W ′ = x X : x X W . { ∈ 1 { } × 2 ⊆ } Then X = Z′ W ′. Moreover 1 ∪ Z′ = x X :(x, y) Z = π ((X y ) Z) , { ∈ 1 ∈ } 1 1 ×{ } ∩ y X2 y X2 \∈ \∈ and so Z′ is a closed subset of X1. Likewise W ′ is a closed subset of X1. Since X is irreducible, either X1 = Z′ or X1 = W ′. If say X1 = Z′, then X X = Z. It follows that X X is irreducible. 1 × 2 1 × 2 84 CHAPTER 2. MODULES OVER COMMUTATIVE RINGS

It follows that the affine coordinate ring of X X , which is A(X ) 1 × 2 1 ⊗k A(X2), must be an integral domain. Thus: Corollary 2.9.7. Let k be an algebraically closed field and let R, S be two k-algebras which are integral domains. Then R S is an integral domain. ⊗k Proof. In case R and S are finitely generated over k, this follows from Lemma 2.9.6 and Proposition 2.9.2. If R and S are arbitrary, then two elements of R S are contained in the image in R S of R′ S′, where R′ ⊗k ⊗k ⊗k and S′ are finitely generated subalgebras of R, S respectively, and the result then follows from this special case. Of course, the above corollary fails in general if k is not algebraically closed. For example, if k[α] = k[x]/(p(x)) is a simple extension of k, where p(x) k[x] is an irreducible polynomial, then k[α] K = K[α]= K[x]/(p(x)), ∈ ⊗k and so k[α] k K fails to be an integral domain for a field K where p(x) factors in a nontrivial⊗ way.

Corollary 2.9.8. Let k be an algebraically closed ﬁeld and let p be a prime ideal in k[x1,...,xn]. Let K be an extension ﬁeld of k. Then the induced ideal p K of K[x ,...,x ] is a prime ideal. ⊗k 1 n Proof. Using the exact sequence

0 p K k[x ,...,x ] K (k[x ,...,x ]/p) K 0, → ⊗k → 1 n ⊗k → 1 n ⊗k → we see that the quotient of K[x1,...,xn] by p k K is the integral domain (k[x ,...,x ]/p) K. Thus p K is a prime⊗ ideal. 1 n ⊗k ⊗k In terms of algebraic geometry, the above corollary says that a geometrically irreducible variety is absolutely irreducible.

2.10 Flatness

As with the bifunctor Hom , the modules N such that M M N is an R 7→ ⊗R exact functor play a special role. It is enough to require that, if M ′ M is → an injection, then M ′ N M N is an injection. ⊗R → ⊗R Deﬁnition 2.10.1. The R-module N is ﬂat (over R) if, for every injection M ′ M of R-modules, the induced map M ′ N M N is an injection. → ⊗R → ⊗R 2.10. FLATNESS 85

Lemma 2.10.2. (i) R is a ﬂat R-module.

(ii) If M , i I are a collection of R-modules, then M is flat if and i ∈ i i only if Mi is flat for every i. L (iii) A projective R-module N is flat. In particular, a free R-module N is flat.

Proof. (i) is clear from the isomorphism R R M ∼= M. (ii) is easy. As for (iii), the case where N is free follows from⊗ (i) and (ii), and then the case where N is projective follows from (ii), the statement for N free, and the fact that every projective module is a summand of a free R-module.

If M is finitely generated, then the property that M is flat is almost the same as that M is projective. In fact, in the next chapter, we shall show that M is finitely generated and projective if and only if it is flat and finitely presented. But in general there are many flat modules which are not projective. For example, Q is a flat Z-module. More generally, if R is a PID and M is an R-module, then M is flat if and only if it is torsion free (exercise). Thus flatness is a somewhat mysterious technical definition. It is however of fundamental importance in algebraic geometry, where it arises in the following setting. Given a morphism of algebraic sets f : X Y , corresponding in the affine case to a ring homomorphism R S, → → the statement that S is a flat R-module means that the morphism f is well behaved, in particular that the fibers of f fit together nicely in some sense.

Lemma 2.10.3. Let N be an R-module. Then the following are equivalent:

(i) N is ﬂat.

(ii) For every exact sequence of R-modules

0 M ′ M M ′′ 0, → → → → the induced sequence

0 M ′ N M N M ′′ N 0 → ⊗R → ⊗R → ⊗R → is exact. 86 CHAPTER 2. MODULES OVER COMMUTATIVE RINGS

(iii) For every exact sequence

M M , ···→ i → i+1 →··· the induced sequence

M N M N ···→ i ⊗R → i+1 ⊗R →··· is exact.

(iv) For all finitely generated R-modules M ′ and M and injections M ′ → M, the induced homomorphism M ′ N M N is injective. ⊗R → ⊗R (v) For all ideals I in R, the map I N N is injective. ⊗R → (vi) For all finitely generated ideals I in R, the map I N N is injective. ⊗R → Proof. Clearly (i), (ii), and (iii) are equivalent, and (i) = (iv) = (vi), (i) = (v) = (vi). ⇒ ⇒ ⇒ ⇒ We show first that (iv) = (i). Let M and M ′ be arbitrary R-modules ⇒ and suppose that f : M ′ M is an injection. We must show that f Id → k ⊗ is an injection. Suppose that m n Ker f Id. Let M ′ be the i=1 i ⊗ i ∈ ⊗ 0 submodule of M ′ generated by m1,...,mk. Since f(mi) ni = 0, the sum P i ⊗ (f(mi), ni) FR(M N) is a finite sum gj of the standard generators i ∈ × j P (x1 + x2, n) (x1, n) (x2, n), . . . , of the submodule Q which we factor out P − − P by to form the tensor product. Let M0 be the submodule of M generated by the f(mi) and all of the first coordinates of the elements gj. Thus there is a well-defined injective map f : M ′ M and m n Ker(f Id). By 0 0 → 0 i i ⊗ i ∈ 0 ⊗ assumption i mi ni = 0 M0′ R N, and thus its image in M ′ R N is zero as well. Thus ⊗f Id is injective,∈ ⊗ so thatP by definition N is flat.⊗ ⊗ The proofP of (vi) = (v) is similar and we shall leave it to the reader. ⇒ Thus the only thing left to prove is that (v) = (iv). First we claim that if S is a submodule of Rk for some k, then the map⇒ S N Rk N is ⊗R → ⊗R injective. The proof is by induction on k; the case k = 1 is just the statement for ideals. In general, if S is a submodule of Rk, we have the diagram

0 S′ S S′′ 0 −−−→ −−−→ −−−→ −−−→

 k k 1 0 R R R− 0. −−−→ y −−−→ y −−−→ y −−−→ 2.10. FLATNESS 87

k k 1 Here S′′ = Im(S) under the projection R R − . Tensoring with N, we → ﬁnd the commutative diagram

K S′ N S N S′′ N 0 −−−→ ⊗R −−−→ ⊗R −−−→ ⊗R −−−→

 k k 1 0 N N N− 0, −−−→ y −−−→ y −−−→ y −−−→ with exact rows, where K is the kernel of the homomorphism S′ N ⊗R → S R N. The second and fourth vertical arrows are injective by hypothesis or⊗ by induction. So the map S N N k is injective by the Five lemma. ⊗R → Finally, let M ′ and M be arbitrary R-modules, with M ﬁnitely generated, k and suppose that M ′ is a submodule of M. Choose a surjection R M, → and let M ′ be the preimage of S under the surjection. If S0 is the kernel of the surjection Rk M, then we have a commutative diagram →

0 S S M ′ 0 −−−→ 0 −−−→ −−−→ −−−→

k  0 S 0 R M 0. −−−→ −−−→ y −−−→ y −−−→ Taking the tensor product with N gives

S N S N M ′ N 0 0 ⊗R −−−→ ⊗R −−−→ ⊗R −−−→ f g

k   S0 R N R R N M R N 0. ⊗ −−−→ ⊗y −−−→ ⊗y −−−→ Here f is an injection. Extending the diagram on the right by adding another 0 to each row enables us to use the Five lemma to conclude that g is an injection. Thus (iv) holds.

The following corollary says that the ﬂatness of N is equivalent to saying that every relation between elements of N is essentially a formal consequence of relations in R:

Proposition 2.10.4. Let N be an R-module. Then N is ﬂat if and only if, for every collection r R, i = 1,...,n and m N, if r m = 0, then i ∈ i ∈ i i i there exists a positive integer ℓ and elements sij R, nij N, 1 i n and 1 j ℓ, such that r s =0 for every j and,∈ for all∈iP, m =≤ ≤s n . ≤ ≤ i i ij i j ij j P P 88 CHAPTER 2. MODULES OVER COMMUTATIVE RINGS

Proof. Note that if mi = j sijnj with i risij = 0 as above, then i rimi = 0. P P P First suppose that N is flat, and suppose that i rimi = 0. We have the homomorphism Rn R with (t ,...,t ) r t . Let K be the kernel. → 1 n 7→ i i i Thus the sequence K Rn R is exact. SincePN is flat, the sequence K N N n N is→ also exact.→ As (m ,...,mP ) is in the kernel of the ⊗R → → 1 n homomorphism N n N, there exist σ K and n N, 1 j ℓ, such → j ∈ j ∈ ≤ ≤ that (m1,...,mn) is the image of j σj nj. Writing σj = (sij,...,snj) n ⊗ ∈ K R , we have i risij = 0 for all j, since σj K, and tracing through ⊆ n n P ∈ the isomorphism R R N ∼= N , we find that mi = j sijnj. Conversely, supposeP⊗ that the second condition of the proposition holds. We shall show that, for every ideal I of R, the map IP N N is injective. ⊗ → Suppose that r I and that r m is in the kernel of the map I N i ∈ i i ⊗ i ⊗R → N. Thus rimi = 0, and so there exist sij R and nj N such that i P ∈ ∈ risij = 0 and mi = sijnj. Thus i P j P r m =P r s n = ( (r s ) n =0. i ⊗ i i ⊗ ij j i ij ⊗ j i i j j i X X X X X Thus I N N is injective. ⊗R → The main application is the following: Proposition 2.10.5. Let R be a local ring and M a finitely generated R- module. Then M is flat if and only if it is projective if and only if it is free. Proof. We know that M free = M projective = M flat, so it is enough to show that M flat = M free.⇒ Let m be the maximal⇒ ideal of R and let ⇒ k = R/m. Choose m1,...,mn M which map onto a k-basis for M/mM. Thus there is an induced homomorphism∈ Rn M, which is surjective by Nakayama’s lemma. We must show that it is→ injective. By induction on k, we show that, if m1,...,mk is any collection such that the images in M/mM k are linearly independent and i=1 rimi = 0, then ri = 0. The case k = n then shows that Rk M is injective and hence an isomorphism. → P For k = 1, if rm1 = 0, then there exist sj R and nj M such that rs = 0 for all j and m = s n . Since m /∈ mM, there∈ exists a j such j 1 j j j 1 ∈ that sj / m. But then sj is a unit, and rsj = 0, so that r = 0. For∈ the inductive step, supposeP that the result has been established for i k. If k+1 r m = 0, then there exist s R and n M such ≤ i=1 i i ij ∈ j ∈ P 2.10. FLATNESS 89

that mi = j sijnj and i risij = 0. Since the images of the mi are linearly independent in M/mM, mk+1 / mM. As before, there exists a j such that sP / m, andP thus r ∈= k ( s /s )r = k c r , k+1,j ∈ k+1 i=1 − ij k+1,j i i=1 i i say. Now 0 = k+1 r m = k r (m + c m ). Since the images of i=1 i i i=1 i i P i k+1 P m1,...,mk+1 are linearly independent in M/mM, the same is true for the im- P P ages of m1 +c1mk+1,...,mk +ckmk+1. Thus by induction ri =0, i =1,...,k, and rk+1mk+1 = 0. Using the case k = 1, we see that rk+1 = 0 as well. Thus, taking k = n, the map Rn M is an isomorphism and we are done. → Exercises

Exercise 2.1. (Schur’s lemma) Suppose that M1 and M2 are two simple R-modules. Let ϕ: M1 M2 be a homomorphism (of R-modules). Show that ϕ is either 0 or an isomorphism.→

Exercise 2.2. Suppose that R is a ring and that 0 M ′ M M ′′ 0 is an exact sequence of R-modules. → → → →

(i) Show that, if M ′ and M ′′ are ﬁnitely generated R modules, then so is M.

(ii) Show that, if M is ﬁnitely generated, then so is M ′′.

(iii) Suppose that M is ﬁnitely generated and that M ′′ is projective. Show that M ′′ and M ′ are ﬁnitely generated.

Exercise 2.3. Is Q a projective Z-module? Why or why not?

Exercise 2.4. Let I be a directed set, let (Mi, fij) be a directed system of R-modules and let f : M lim M = M be the natural homomorphisms. i i → i I i −→ ∈ (i) Show that every element of M is of the form fi(mi) for some i and some m M . i ∈ i

(ii) Suppose that, for some i and some mi Mi fi(mi) = 0. Show that there exists a j with i j and f (m )=0.∈ ≤ ij i Exercise 2.5. (i) Let R be a ring, I an ideal of R, and m a maximal ideal of R with quotient ﬁeld k. Show that the number of generators of I is at least as great as the dimension of the k-vector space I/m I. · 90 CHAPTER 2. MODULES OVER COMMUTATIVE RINGS

(ii) Let k be a field. Show that the minimum number of generators of the ideal (x2,xy,y2) of k[x, y] is 3. (iii) Let F : R3 I be the surjective homomorphism defined by → (f , f , f ) f x2 + f xy + f y2. 1 2 3 7→ 1 2 3 Show that the homomorphism G: R2 R3 defined by → G(g ,g )= g (y, x, 0) + g (0, y, x) 1 2 1 − 2 − defines an isomorphism from R2 to Ker F .

Exercise 2.6. Let Mi, i I be a collection of R-modules, and let N be an R-module. Show that there∈ is a natural homomorphism Hom (N, M ) Hom (N, M ). R i → R i i I i I M∈ M∈ Show that this homomorphism need not be an isomorphism. However, show that it is injective and is an isomorphism if N is ﬁnitely generated.

Exercise 2.7. If I is a directed set, (Mi, fij) a direct system of R-modules, and N an R-module, give a description of HomR(lim Mi, N). i I −→∈ Exercise 2.8. Let I be a directed set and (Mi, fij) a direct system of R- modules. A subset J I is coﬁnal if, for every i I, there exists a j J such that i j. Show⊆ that, in this case, J is also∈ directed and construct∈ a ≤ natural isomorphism lim Mj lim Mi. j J → i I −→∈ −→∈ Exercise 2.9. Let M be an R-module, and suppose that there are two exact sequences f 0 K N 1 M 0; → 1 → 1 −→ → f 0 K N 2 M 0. → 2 → 2 −→ → (i) Let f : N1 N2 M be deﬁned by f(n1, n2) = f1(n1)+ f2(n2), and let K = Ker⊕f. Show→ that there is an exact sequence 0 K K N 0, → 2 → → 1 → and thus by symmetry an exact sequence 0 K K N 0. → 1 → → 2 → 2.10. FLATNESS 91

(ii) (Schanuel’s lemma) Suppose in addition that N1 and N2 are projective. Show that K N = K N . 1 ⊕ 2 ∼ 2 ⊕ 1 (iii) Let M be a ﬁnitely presented R-module, and let g : Rm M be any → surjection. Show that Ker g is ﬁnitely generated. Exercise 2.10. Show that one can generalize Proposition 2.3.11 as follows: gn hn let 0 M ′ M M ′′ 0 be an exact sequence of inverse systems → n −→ n −→ n → indexed by N, and, for m n, let f ′ : M ′ M ′ be the homomorphisms ≥ n,m m → n of the inverse system for M ′ . Suppose that, for all n N, there exists n ∈ N n + 1 such that, for m N, the image Im f ′ = Im f ′ . In other ≥ ≥ m,n N,n words, the images of the fm,n′ eventually stabilize. Then the induced sequence

0 lim Mn′ lim Mn lim Mn′′ 0 → n N → n N → n N → ←−∈ ←−∈ ←−∈ is exact. (Identify Mn′ with its image in Mn. First, one can assume that N = n + 1 by passing to a subsequence of integers. Let (x ) lim M ′′. n ∈ n N n Suppose inductively that we have found y M , i n, such←− that∈ y i ∈ i ≤ i 7→ xi and fi,i 1(yi) = yi 1. Choose yn+1 Mn+1 such that yn+1 xn+1. − − ∈ 7→ Then fn+1,n(yn+1) yn = zn Mn′ . If zn 1 = fn,n′ 1(zn), then zn 1 = − ∈ − − − fn′ +1,n 1(wn+1) for some wn+1 Mn′ +1. Now we can replace yn by yn + zn f −(w ) and y by y ∈ w , and leave the other y unchanged.)− n+1,n n+1 n+1 n+1 − n+1 i The condition that the images of the fm,n′ eventually stabilize is called the Mittag-Leffler condition. It is satisfied if Mn′ is an Artinian module for every n. Exercise 2.11. Let R be a ring, M a Noetherian R-module, and f : M → M an R-module homomorphism. Show that, if f is surjective, then it is an isomorphism. (Consider the increasing sequence of submodules Ker f n.) State and prove the corresponding result for Artinian R-modules. Exercise 2.12. (Cohen) Let R be a ring such that every prime ideal in R is finitely generated. Then R is Noetherian. (If nonempty, the collection of all non-finitely generated ideals of R has a maximal element. It suffices to show that such a maximal element is prime. If I is such a maximal element, and I is not prime, then there exist r, s R I such that rs I. Then I +(r) is finitely generated, by f ,...,f , say.∈ Write− f = g + a r∈with g I, and 1 n i i i i ∈ let I =(g ,...,g ). Then I +(r)= I +(r). If I = t R : rt I , then 0 1 n 0 1 { ∈ ∈ } I1 is an ideal containing I and s, hence I1 is finitely generated. Finally show that I = I0 + rI1 and thus that I is finitely generated, a contradiction.) 92 CHAPTER 2. MODULES OVER COMMUTATIVE RINGS

Exercise 2.13. Let M be an R-module. Show that there is an injection from M into an injective R-module I as follows:

(i) Let A be an abelian group and N an R-module. Show that HomZ(N, A) is an R-module under the action r ϕ(n)= ϕ(rn) and that there is a nat- · ural isomorphism of R-modules HomZ(N, A) ∼= HomR(N, HomZ(R, A)). (Given ϕ HomZ(N, A), deﬁneϕ ˜(n)(r) = ϕ(rn), and ﬁnd an inverse ∈ map.)

(ii) Show, in the notation of (i), that if A is an injective Z-module, then the R-module HomZ(R, A) is an injective R-module. Thus, if F is a free R-module, then HomZ(F, A) is an injective R-module.

(iii) For an abelian group A, let Aˆ = HomZ(A, Q/Z). Show that the natural ˆ homomorphism A Aˆ is injective. (First show that, if A A is → 1 → 2 an injective homomorphism of abelian groups, then the induced homo- ˆ ˆ morphism Aˆ1 Aˆ2 is injective, and then verify the statement directly for cyclic groups.)→

(iv) Now start with an arbitrary R-module M. Taking A = Q/Z, (i) shows that Mˆ = HomZ(M, Q/Z) is an R-module. Thus it is the quotient of a free R-module F . Dualizing the surjection F Mˆ 0, and using → ˆ → the exactness of HomZ( , Q/Z), there is an injection Mˆ Fˆ, and since ˆ · → M Mˆ is injective, there is an injection M Fˆ. Show that the → ˆ ˆ → injections Mˆ Fˆ and M Mˆ are R-module homomorphisms and conclude then→ there is an injection→ from M to an injective R-module.

ˆ Exercise 2.14. Show that the homomorphism Z Zˆ = HomZ(Q/Z, Q/Z) ˆ → is injective but not surjective. Describe Zˆ by identifying it with a group we have seen in the text. (If f : Q/Z Q/Z is a homomorphism, then by 1 → 1 restriction f deﬁnes f : Z/Z Q/Z, whose image lies in Z/Z and is thus n n → n given by an integer mod n. What is the necessary compatibility between fn and f if n m?) m | Exercise 2.15. Show that an arbitrary direct product of injective R-modules is injective. 2.10. FLATNESS 93

Exercise 2.16. Let R be a PID, and r R a nonzero element. Show that ∈ the ring R/rR, viewed as an R/rR-module, is an injective R/rR-module. (It suffices to show that, if M is an R/rR-module and R/rR M is an inclusion, then there is a retraction M R/rR. Use the fact that,→ if K is → the quotient field of R, then K/R is an injective R-module and argue as in the last part of the classification of modules over a PID.)

Exercise 2.17. For n, m N, identify the abelian group (Z/nZ) Z (Z/mZ) ∈ ⊗ and show that (Z/nZ) Z Q = 0. What is Q Z Q? ⊗ ⊗ Exercise 2.18. Let M be an R-module. Show that the natural homomorphism ev: R Hom (R, M) M ⊗R R → induced by the bilinear map (r, ϕ) ϕ(r) is an isomorphism. (Reminder: it does not suffice to show that the two7→ sides are isomorphic.) Exercise 2.19. Let k be a field, let R = k[x, y], and let m be the maximal ideal (x, y) of R. Describe the structure of m R m as follows: beginning with he exact sequence ⊗ 0 R R R m 0, → → ⊕ → → where the map R R m is defined by (f,g) fx + gy and the map R R R is defined⊕ → by r (ry, rx), argue7→ that there is an exact → ⊕ 7→ − sequence ψ m m m m m 0, −→ ⊕ → ⊗R → where ψ is defined by m m (my, mx). Next show that the map ϕ ∈ 7→ − m m m2 = m m defined by (a, b) ax + by is surjective and that its ⊕ −→ · 7→ kernel contains the image of ψ. Finally show that Ker ϕ/ Im ψ is generated by the image e of (x, y), with xe = ye = 0 in Ker ϕ/ Im ψ, but e = 0. Thus 6 there is an exact sequence of R-modules

0 R/m m m m2 0, → → ⊗R → → and in particular m m has torsion. ⊗R Exercise 2.20. Let f : R S be a ring homomorphism, R, S commutative, and let M and N be S-modules.→ Thus we can also view M and N as R- modules via f. Show that M S N is always a quotient of M R N, and that the natural homomorphism⊗ is an isomorphism if f is surjective.⊗ Give an example to show that they do not always agree if f is not surjective. 94 CHAPTER 2. MODULES OVER COMMUTATIVE RINGS

Exercise 2.21. Let R be a ring, S an R-algebra, and M and R-module. Show that the S-module S R M has the following universal property: Let N be an S-module and f : M⊗ N be a homomorphism of R-modules. Then there is a unique homomorphism→ of S-modules fˆ: S M N, such that ⊗R → fˆ(1 m)= f(m) for all m M (and hence fˆ(s m)= sf(m) for all m M and⊗s S). ∈ ⊗ ∈ ∈ The above shows that, for every R-module M and S-module N, there is an S-module homomorphism HomR(M, N) HomS(S R M, N). Show that this homomorphism is an isomorphism, with→ inverse is⊗g Hom (S ∈ S ⊗R M, N) (m g(1 m)). Compare also Chapter 3, Exercises 3.18 and 3.19. 7→ 7→ ⊗ Exercise 2.22. Let R be a local ring with maximal ideal m and residue ﬁeld k. Suppose that 0 N Rn M 0 is an exact sequence of R-modules. Show that N and M→ are→ free R→-modules→ if and only if the induced sequence 0 N/mN kn M/mM 0 → → → → is exact. (If the sequence is exact, choose m1,...,md M whih map onto a k-basis, and consider the induced homomorphism Rd ∈ M, together with a lift of this to a homomorphism Rd Rk. The direct sum→ homomorphism N → ⊕ Rd Rn is surjective, and by the arguments in the proof of Corollary 2.7.5 it is→ an isomorphism.) Exercise 2.23. Let R be a local ring with maximal ideal m and residue ﬁeld k. Suppose that Rk Rn M 0 is an exact sequence of R-modules, → → → where the homomorphism Rk Rn corresponds to the k n matrix A. Suppose that, for some d n, the→ (d +1) (d +1) minors of×A are all zero, and that there exists some≤ d d minor of×A which is a unit in R, i.e. does × not lie in m. Show that M is free. (Show that there exist v1,...,vd Im A which are an R-basis for Im A and which can be completed to an R-basis∈ of Rn.) Is the converse statement true?

Exercise 2.24. (Graded Nakayama) Let R = n 0 Rn be a graded ring. A ≥ graded R-module M = n Z Mn is an R-module satisfying Rn Mk Mn+k ∈ L · ⊆ for all n 0 and k Z. Let M be a graded R-module such that Mn = 0 for at most≥ ﬁnitely many∈ L negative n and let I be a homogeneous ideal of6 R contained in the ideal n>0 Rn. If M = IM, show that M = 0. Exercise 2.25. Let fL: Rn Rn be an R-module homomorphism, corresponding to the n n matrix→A with coeﬃcients in R. × 2.10. FLATNESS 95

(i) Let p (x) be the characteristic polynomial det(x Id A). Show that A − pA(A)=0. (ii) If R = k is a ﬁeld, show that there is a unique monic polynomial qA(x), the minimal polynomial of A, such that, if P (x) k[x] is any polynomial with P (A) = 0, then q (x) P (x) in k[x]. ∈ A | Exercise 2.26. Let f : Rn Rm be an R-module homomorphism. → (i) If f is surjective, then n m. (The proof is elementary: there exists ≥ a maximal ideal m of R. Now consider the induced homomorphism (R/m)n (R/m)n and use the corresponding result for vector spaces.) → (ii) If f is an isomorphism, then n = m. (Same remark as in (i).)

(iii) If f is injective, then n m. (If n > m, let ι: Rm Rn be the ≤ → inclusion by zero on the last n m factors and consider the injection − ι f. If A is the corresponding matrix, then pA(A) = 0. By canceling ◦ n oﬀ a power of A, show that Im A contains a0R for some a0 = 0 and reach a contradiction.) 6

Exercise 2.27. (Jordan canonical form) Let k be an algebraically closed ﬁeld, and let V be a ﬁnite-dimensional k-vector space. Let F : V V be → a linear map. Show that there is a direct sum decomposition V = i Vi with A(Vi) Vi, elements λi k, and a basis e1,...,ek of Vi, such that ⊆ ∈ i L A(ej)= λiej + ej 1 for j > 1, and A(e1)= λie1. − Exercise 2.28. (Some results and examples concerning projective and injective modules.)

(i) Let R = Z/6Z, and view Z/3Z as an R-module via the quotient homomorphism. Show that Z/3Z is a projective R-module. The proof should work for R = Z/pqZ where p and q are relatively prime. Even more generally if R = R R is a product ring, show that the ideals 1 × 2 Ri of R are projective R-modules. (ii) If R = Z/8Z and M is the quotient Z/4Z, viewed as an R-module, is M a projective R-module?

(iii) The ring Z/6Z, or more generally a ring of the form R/xR, where R is a PID and x is not zero, is both projective and injective as a module 96 CHAPTER 2. MODULES OVER COMMUTATIVE RINGS

over itself. Show however that if R is an integral domain which is not a ﬁeld, then the only R-module which is both a projective and an injective R-module is the zero module.

(iv) Suppose that R is an integral domain and that every R-module is projective. Show that R is a ﬁeld. Does this hold if R is not an integral domain?

(v) Suppose that R is an integral domain such that R is an injective R- module. Show that R is a ﬁeld.

Exercise 2.29. (i) Let R = k[x] be a polynomial ring in one variable. Let S1 = k[z1] and S2 = k[z2] be polynomial rings, and view Si as an R-algebra via x = z2. Show that S is a free R-module of rank 2. Thus S S is a i i 1 ⊗R 2 free rank four R-module, with basis 1 = 1 1, e = z 1, e =1 z , and ⊗ 1 1 ⊗ 2 ⊗ 2 e3 = z1 z2. Write down the multiplication table for ei ej, and show that S S⊗is not an integral domain. · 1 ⊗R 2 (ii) For a general ring R, let S1 and S2 be two R-algebras and J an ideal of S2. Show that the R-algebra homomorphism S1 R S2 S1 R (S2/J) is surjective, and that its kernel is the ideal of S ⊗S which→ is⊗ the image of 1 ⊗R 2 S J under the natural map. 1 ⊗R (iii) With R and S1 as in (i), let T = k, viewed as an R-module via the 2 homomorphism R = k[x] k[x]/(x) ∼= k. Show that S1 R T ∼= k[z1]/(z1 ) as k-algebras. Hence, although→ S and T are integral domains,⊗ S T 1 1 ⊗R contains a nilpotent element.

Exercise 2.30. Let k be a field of characteristic p > 0, and let F = k(x) be the field of rational functions in one variable with coefficients in k. Let K = F (α), where αp = x, so that K is a finite inseparable extension of F . Show that K K has a nilpotent element. ⊗F Exercise 2.31. If R is a PID, show that an R-module M, not necessarily finitely generated, is flat if and only if it is torsion free.

Exercise 2.32. (i) Show that, if M and N are flat over R, then so is M N. ⊗R (ii) If S is a flat R-algebra and M is a flat S-module, then M is flat as an R-module. 2.10. FLATNESS 97

(iii) If S is an arbitrary R-algebra and M is a ﬂat R-module, then M S ⊗R is a ﬂat S-module.

Exercise 2.33. Let 0 M ′ M M ′′ 0 be an exact sequence of → → → → R-modules, and identify M ′ with its image in M. Let I be an ideal of R. Show that, if M ′′ is a flat R-module, then IM ′ = IM M ′. Is this true in ∩ general if M ′′ is not flat? Exercise 2.34. An R-module M is faithfully flat if M is a flat R-module, and, for all R-modules N, N = 0 if and only if N R M = 0. Show that the following are equivalent: ⊗ (i) M is faithfully flat.

(ii) M is flat and, for all R-modules N ′, N and homomorphisms f : N ′ → N, f = 0 if and only if the induced homomorphism f 1: N ′ M ⊗ ⊗R → N M is zero. ⊗R (iii) For all R-modules N ′,N,N ′′ and homomorphisms f : N ′ N and f g → g : N N ′′, the sequence N ′ N N ′′ is exact if and only if the → f 1 −→ g 1−→ sequence N ′ M ⊗ N M ⊗ N ′′ M is exact. ⊗R −−→ ⊗R −−→ ⊗R (iv) M is flat and, for every maximal ideal m of R, mM = M. 6 (It may help to remark that, if M is a flat R-module and f : N ′ N is an R-module homomorphism, then, for the induced homomorphism→f ⊗ 1: N ′ M N M, Ker(f 1) = (Ker f) M and Coker(f 1) = ⊗R → ⊗R ⊗ ∼ ⊗R ⊗ ∼ (Coker f) M.) ⊗R Exercise 2.35. With the terminology of the preceding exercise, suppose that f : R S is a local homomorphism of local rings and that S is flat over R. Show→ that S is faithfully flat over R. Exercise 2.36. For a Noetherian local ring R with maximal ideal m and residue field k, give a quick proof that a finitely generated R-module M is flat if and only if it is free, as follows: if M is flat and m1,...,mn M map onto a k-basis of M/mM, consider the homomorphism ϕ: Rn M∈ defined → by (r1,...,rn) i rimi. Nakayama’s lemma implies that ϕ is surjective. If K = Ker ϕ,7→ then K is finitely generated. Argue, again by Nakayama’s lemma, that K = 0.P Chapter 3

Localization

Localization, a generalization of the construction of the quotient ﬁeld of an integral domain, is a powerful method both for constructing new rings from old and for studying a given ring or module by “local” methods. In geometry, localization enlarges the category of regular functions to include rational functions deﬁned on a given open set, and gives a means for making sense of this procedure for a general commutative ring.

3.1 Basic deﬁnitions

Definition 3.1.1. A multiplicative subset S of a ring R is a subset S of R, closed under multiplication and such that 1 S. ∈ For example, R is a multiplicative subset of R, as is 1 . The set of all units in R is a multiplicative subset. If R is an integral domain,{ } then R 0 −{ } is a multiplicative subset of R. For a general ring R, the set of all r R which are not zero divisors is a multiplicative subset. ∈ For a multiplicative subset S of R, we would like to construct fractions r/s with denominators in S. Such a fraction is specified by the ordered pair (r, s), and the main point is to decide when two such expressions r/s and r′/s′ are equal. Motivated by the definition of the rational numbers, we could define r/s and r′/s′ to be equal if rs′ = sr′. This gives an equivalence relation if R is an integral domain, or more generally if S does not contain a divisor of zero. However, if S contains divisors of zero, it turns out that this is not necessarily an equivalence relation, i.e. we might have r/s = r′/s′ and r′/s′ = r′′/s′′ without having r/s = r′′/s′′. (If rt = 0 with neither r

98 3.1. BASIC DEFINITIONS 99 nor t equal to zero and t S, then the definition says that r/1=0/t and ∈ 0/t = 0/1, but r/1 = 0/1.) Instead we define an equivalence relation on 6 ≡ R S by: (r, s) (r′,s′) if there exists a t S such that t(rs′ sr′) = 0. It is clear× that is reflexive≡ and symmetric. To∈ see that it is transitive,− suppose ≡ that (r, s) (r′,s′) and that (r′,s′) (r′′,s′′). Thus there exist t, t′ S such ≡ ≡ ∈ that t(rs′ sr′) = 0 and t′(r′s′′ s′r′′) = 0. Then − − 0=(s′′t′)t(rs′ sr′)+(st)t′(r′s′′ s′r′′) − − =(s′tt′)(rs′′) ss′′r′tt′ + ss′′r′tt′ (s′tt′)(sr′′) − − =(s′tt′)(rs′′ sr′′). −

Thus (r, s) (r′′,s′′), and so is transitive. 1≡ ≡ Let S− R be the set of equivalence classes for . We denote the equiv- ≡ alence class containing (r, s) by r/s. A routine calculation shows that the operations, deﬁned on representatives of equivalence classes by

(r, s)+(r′,s′)=(rs′ + sr′,ss′);

(r, s)(r′,s′)=(rr′,ss′),

1 are well-defined and give S− R the structure of a commutative ring with 1 unity 1/1. There is a ring homomorphism R S− R defined by r r/1, 1 → 7→ which makes S− R into an R-algebra. However, this homomorphism need 1 not be injective. In fact, r/1 = 0 in S− R if and only if there exists an s S ∈ such that sr = 0. Nonetheless, we shall often denote r/1 simply by r. We 1 call S− R the ring of quotients of R with respect to S. Note that, for all s S, s/1 is a unit with inverse 1/s. ∈ We give some examples of this construction (some verifications left as exercises):

1 1. If S = 1 , S− R = R under the obvious map. Likewise, if S is the set { } ∼ 1 of all units in R, then S− R ∼= R. 1 2. S− R is the zero ring if and only if 0 S. ∈ 1 3. If R is an integral domain and S = R 0 , then S− R is the quotient ﬁeld of R. −{ }

1 4. The homomorphism R S− R is injective if and only if S does not contain zero or a divisor→ of zero. More generally, r/1 = 0 if and only if there exists s S with sr = 0. ∈ 100 CHAPTER 3. LOCALIZATION

2 1 5. Given r R, the set 1,r,r ,... is multiplicative. The ring S− R is ∈ { } denoted Rr. 6. If p is an ideal of R, then S = R p is multiplicative if and only if p − 1 is a prime ideal. In this case, we denote S− R by Rp and call it the localization of R at p.

For example, in 6) above, let R = k[x1,...,xn] for some ﬁeld k, and let m =(x1 a1,...,xn an) be the maximal ideal of all polynomial functions vanishing− at a = (a ,...,a− ) kn. Then R m is the set of all functions 1 n ∈ − which do not vanish at a, and Rm is the set of all quotients f/g, where g(a) = 0, which we can think of as the ring of all rational functions “deﬁned” at a.6 1 The ring S− R has the following universal property:

Proposition 3.1.2. Let S be a multiplicative subset of R.

(i) Let f : R R′ be an R-algebra such that, for all s S, f(s) is a unit → ∈1 in R′. Then there is a unique homomorphism fˆ: S− R R′ such that → fˆ(r/1) = f(r).

(ii) Suppose that g : R R is an R-algebra such that, for all s S, g(s) → ∈ is a unit in R, and suppose that R has the same universal property as 1 S− R: for all R-algebrase f : R R′ such that, for all s S, f(s) is a → 1 ∈ unit in R′, theree is a unique homomorphisme f˜: S− R R′ such that ˜ → 1 f(g(r)) = f(r). Then there is a unique isomorphism R ∼= S− R of R-algebras. e 1 Proof. To see (i), deﬁne fˆ(r/s)= f(r)f(s)− . It is easy to check that this is well-deﬁned and gives a ring homomorphism. The uniqueness of fˆ is clear. The proof of (ii) is straightforward.

2 For S = 1,r,r ,... as in (5) above, we can also construct Rr as the quotient ring{R[x]/(rx }1) = R[1/r]. (As a side comment, it then follows that r is nilpotent if and− only if R = 0 if and only if rx 1 is a unit in R[x], r − and, given a R, arn = 0 for some n > 0 if and only if a (rx 1). Of course, it is easy∈ to check these statements directly.) ∈ − Given r ,r R, the ring R contains inverses for r and r . For 1 2 ∈ r1r2 1 2 example, r1 [r2/(r1r2)] = 1. Thus, using Proposition 3.1.2, it is easy to check that R · (R ) (R ) , where we have identiﬁed r with its image in r1r2 ∼= r1 r2 ∼= r2 r1 2 3.2. IDEALS IN A LOCALIZATION 101

Rr1 and similarly for the image of r1 in Rr2 . Moreover, all of these rings are 1 n1 n2 identified with S− R, where S is the multiplicative set r1 r2 : n1, n2 0 . More generally, a multiplicative subset S of R is finitely{ generated if≥ there} exist r ,...,r R such that S = rn1 rn2 rnk : n 0 for all i . In this 1 k ∈ { 1 2 ··· k i ≥ } case S is generated by the ri. Clearly, if S is finitely generated, by r1,...,rk, 1 then S− R = Rr = R[x]/(rx 1), where r = r1 rk. Finally, if S is an ∼ − 1 ··· arbitrary multiplicative set, then S− R can be obtained as the direct limit of localizations over finitely generated subsets of S.

3.2 Ideals in a localization

1 Next we compare ideals in R with ideals in S− R: Proposition 3.2.1. Let R be a ring, let S be a multiplicative subset of R, 1 and let f : R S− R be the natural map. → 1 1 (i) Every ideal J in S− R is of the form IS− R for some ideal I of R.

1 1 1 (ii) If J is an ideal of S− R, f − (J)S− R = J.

1 1 (iii) If p is a prime ideal of R such that p S = , then f − (pS− R)= p. ∩ ∅ 1 (iv) The function q f − q sets up a one-to-one correspondence between 7→1 prime ideals of S− R and prime ideals p of R such that p S = , with 1 ∩ ∅ inverse function p pS− R. 7→ 1 Proof. (i), (ii): Let J be an ideal of S− R, and let

1 I = f − (J)= r R : r/1 J . { ∈ ∈ } 1 If r/s J, then r/1=(r/s)(s/1) J, so that r/1 IS− R. But then r/s = ∈ 1 ∈ 1 ∈ (1/s)(r/1) IS− R as well, so that J IS− R. On the other hand, clearly 1 ∈ ⊆ 1 1 1 IS− R J, so they are equal. Thus we see that J = f − (J)S− R = IS− R. ⊆ (iii): If r/1 = r′/s with r′ p, then there exists a t S such that ∈ ∈ tsr = tr′ p. As(ts) S and S p = , we must have r p. ∈ ∈ ∩1 ∅ ∈ 1 (iv): If q is a prime ideal in S− R, then we know that p = f − q is a prime 1 ideal of R, and clearly p S = , since otherwise f(f − (p)) would contain a unit. Conversely, suppose∩ that ∅p is a prime ideal of R such that p S = . 1 ∩ ∅ Let q be the ideal p(S− R); in other words, q = r/s : r p . { ∈ } 102 CHAPTER 3. LOCALIZATION

If (r /s )(r /s ) q, then (r /s )(r /s ) = r/s for some r p. Thus there 1 1 2 2 ∈ 1 1 2 2 ∈ exists a t S with (st)r r p. But as st S and S p = , we must ∈ 1 2 ∈ ∈ ∩ ∅ have r1r2 p and so either r1 or r2 p. Hence either r1/s1 or r2/s2 q, so that q ∈is prime. To see that these∈ are inverse constructions, note∈ that 1 1 1 1 f − (q)S− R = q, by (i), and f − (pS− R)= p by (iii). The following is then immediate from (i): Corollary 3.2.2. Let R be a Noetherian ring and let S be a multiplicative 1 subset of R. Then S− R is a Noetherian ring. Likewise, if R is an Artinian 1 ring and S is a multiplicative subset of R, then S− R is Artinian.

Corollary 3.2.3. If p is a prime ideal of R, then Rp is a local ring and its unique maximal ideal is pRp. The field Rp/pRp is the quotient field k(p) of R/p. 1 Proof. If q is a prime ideal of Rp, then f − (q) R (R p) = p. Thus 1 ⊆ − − q = f − (q)R pR . It follows that every prime ideal of R is contained in p ⊆ p p pRp, and thus that Rp is a local ring with maximal ideal pRp. To see the last statement, the composition R R/p k(p) induces a → → homomorphism Rp k(p), by Proposition 3.1.2, which is easily seen to be surjective. Thus the→ kernel of the homomorphism R k(p) is a maximal p → ideal, and hence is equal to pRp. It follows that k(p) ∼= Rp/pRp. We can now give a more efficient proof of Lemma 1.2.10 in Chapter 1, describing the intersection of all of the prime ideals in a given ring R. First a preliminary lemma: Lemma 3.2.4. Let S be a multiplicative subset of R with 0 / S and let m 1 1 ∈ be a maximal ideal of S− R. Then f − m is a prime ideal p of R such that p S = . ∩ ∅ Proof. Immediate from (iii) of Corollary 3.2.3. Corollary 3.2.5. Let R be a ring. Then the intersection of all of the prime ideals of R is the radical √0 of R. Proof. Clearly √0 is contained in every prime ideal. Conversely, suppose that x is not nilpotent. We must show that there exists a prime ideal p with 2 1 x / p. Let S = 1, x, x ,... . By assumption 0 / S. Thus S− R is not the ∈ { } ∈ 1 zero ring, and so there exists a maximal ideal m of S− R. By the previous 1 lemma p = f − (m) is a prime ideal such that p S = . In particular ∩ ∅ x / p. ∈ 3.3. LOCALIZATION OF MODULES 103 3.3 Localization of modules

Our goal now is to define a similar construction for R-modules. Let R be a ring, let S be a multiplicative subset of R, and let M be an R-module. We 1 can again define S− M to be the quotient of M S by the equivalence relation × , where (m, s) (m′,s′) if there exists a t S with t(s′m sm′) = 0 in M. ≡ ≡ ∈ − We denote the equivalence class of (m, s) by m/s. It is routine to check that 1 1 S− M is an S− R-module, via the multiplication (r/s) (m/t)=(rm)/(st). · The function f(m) = m/1 is a homomorphism of R-modules from M to 1 S− M, whose kernel is the set of m M for which there exists s S 2 ∈ 1 ∈ with sm = 0. If S = 1,r,r ,... , then as with rings we let S− M = M . { } r Likewise, if p is a prime ideal in R and S = R p, then we write Mp for 1 − S− M. 1 1 Note that M generates S− M as an S− R-module. In particular if M is 1 1 finitely generated over R, then S− M is a finitely generated S− R-module. 1 1 The map M S− M is a functor from R-modules to S− R-modules; given 7→ 1 1 f : M N, there is an induced map f¯: S− M S− N, obtained by checking that→ the function f¯(m/s)= f(m)/s is well-defined,→ and such that Id = Id and (g f)=¯g f¯. ◦ ◦ To relate this to tensor products, we have: Lemma 3.3.1. If R is a ring, S a multiplicative subset of R, and M and 1 1 1 R-module, then S− M = S− R M as S− R-modules. Under this isomor- ∼ ⊗R phism, if f : M N is an R-module homomorphism, then f¯ corresponds to Id f. → ⊗ 1 1 Proof. The natural R-bilinear map S− R M S− M given by (r/s,m) × → 7→ rm/s is well-defined (independent of the choice of representative for r/s) and 1 bilinear, and so induces a homomorphism of R-modules ρ: S− R R M 1 1 ⊗ → S− M. The homomorphism ρ is in fact a homomorphism of S− R-modules, by checking on generators. (Compare Exercise 2.21.) To go the other way, 1 define a function M S S− R M via (m, s) (1/s) m. If(m, s) × → ⊗R 7→ ⊗ ≡ (m′,s′), then there exists a t S such that ts′m = tsm′. But then ∈ 1 1 1 1 m = s′tm = tsm′ = m′. s ⊗ ss′t ⊗ ss′t ⊗ s′ ⊗ 1 1 Thus there is an induced function ψ : S− M S− R R M. It is easy to check that ρ ψ and ψ ρ are the identity. The→ statement⊗ about f¯ is left to the reader. ◦ ◦ 104 CHAPTER 3. LOCALIZATION

Next we turn to the exactness properties of localization:

f g Proposition 3.3.2. If M ′ M M ′′ is an exact sequence of R-modules, ¯ −→ −→ 1 f 1 g¯ 1 then S− M ′ S− M S− M ′′ is exact. In other words, localization is an −→ −→ exact functor.

Proof. Clearlyg ¯ f¯ = 0, so that Im f¯ Kerg ¯. Conversely, suppose that ◦ 1⊆ g¯(m/s) = 0, so that g(m)/s = 0 in S− M ′′. Then there exists t S such ∈ that tg(m) = 0 in M ′′. Thus g(tm) = 0, so that tm Ker g = Im f. Write ∈ tm = f(m′) for some m′ M ′. Then m/1 = f¯(m′/t) and m/s = f¯(m′/st). Thus m/s Im f¯. ∈ ∈ 1 Corollary 3.3.3. The ring S− R is ﬂat over R.

Corollary 3.3.4. Let N be an R-module and M, M ′ two submodules of N. Let S be a multiplicative subset of R. Then:

1 1 1 (i) If I is an ideal of R, then the S− R-submodule S− I of S− R is equal 1 to IS− R.

1 1 1 1 (ii) S− (M + M ′)= S− M + S− M ′ S− N. ⊆ 1 1 1 (iii) S− (M M ′)= S− M S− M ′. ∩ ∩ 1 1 1 (iv) S− N/S− M ∼= S− (N/M). 1 Proof. (i), (ii), and (iii) are exercises. To see (iv), apply S− to the exact sequence 0 M N N/M 0. → → → →

3.4 Local properties of rings and modules

Let M be an R-module. In many ways, the modules Mp or Mm, for p (resp. m) a prime (resp. maximal) ideal of R, are much simpler to study, and it is natural to see how much information about M can be obtained from the knowledge of these simpler modules. Of course, one of the simplest questions we can ask about a module is whether or not it is trivial. The following relates the triviality of M to that of its various localizations: 3.4. LOCAL PROPERTIES OF RINGS AND MODULES 105

Lemma 3.4.1. Let M be an R-module and let S be a multiplicative subset of R. Then the following are equivalent:

(i) M =0.

(ii) For all prime ideals p of R, Mp =0.

(iii) For all maximal ideals m of R, Mm =0.

Proof. Clearly (i) = (ii) = (iii). Conversely suppose that (iii) holds. ⇒ ⇒ Suppose that m M with m = 0. Let I = r R : rm =0 . By hypothesis I = R, so that I∈is contained6 in a maximal{ ideal∈ m. If the image} of m in M 6 m is zero, then there exists an r R m with rm = 0. But then r I m, a contradiction. ∈ − ∈ ⊆

Corollary 3.4.2. Let f : M N be an R-module homomorphism and let S be a multiplicative subset of R→.

(i) f is injective if and only the induced map f¯: M N is injective for p → p all prime ideals p of R if and only the induced map f¯: Mm Nm is injective for all maximal ideals m of R. →

(ii) f is surjective if and only the induced map f¯: Mp Np is surjective for all prime ideals p of R if and only the induced→ map f¯: M N m → m is surjective for all maximal ideals m of R.

Proof. We will just write down the case of maximal ideals. To see (i), let K = Ker f, so that 0 K M N is exact. For every maximal ideal m, → → → the sequence 0 K M N → m → m → m is also exact. Thus f is injective if and only if K = 0 if and only if Km =0 for all maximal ideals m of R if and only if f¯: Mm Nm is injective for all maximal ideals m of R. The proof of (ii) is similar. →

For an important application of these ideas, we have

Corollary 3.4.3. Let M be an R-module. Then M is flat if and only if, for all prime ideals p of R, the Rp-module Mp is flat over Rp, if and only if, for all maximal ideals m of R, the Rm-module Mm is flat over Rm. 106 CHAPTER 3. LOCALIZATION

1 Proof. First assume that M is flat over R. By Exercise 2.32, S− M = 1 1 S− R R M is a flat S− R-module. In particular Mp is flat over Rp for all prime⊗ ideals p. Conversely assume that Mm is flat over Rm for all maximal ideals m. Let N ′ N be an injection of R-modules. Then N ′ N is injective for all → m → m m, and thus N ′ M N M is injective. On the other hand, by m ⊗Rm m → m ⊗Rm m Exercise 3.17 N ′ M = (N ′ M) and N M = (N M) . It m ⊗Rm m ∼ ⊗R m m ⊗Rm m ∼ ⊗R m follows that the map N ′ M N M is injective for all N, and so M ⊗R → ⊗R is flat over R.

In general, a property P for rings R or R-modules M is called local if the following is true: the property P holds for an R-module M if and only if, for every maximal ideal m, P holds for the Rm-module Mm. As we have seen, the property of ﬂatness is a local property, as is the basic property of being nonzero. However neither projective nor ﬁnitely generated are local properties in this sense. An example is given in Exercise 3.22. We do however have the following basic result about projectivity:

Proposition 3.4.4. Let P be an R-module. Then the following are equivalent:

(i) P is projective and ﬁnitely generated;

(ii) P is ﬂat and ﬁnitely presented;

(iii) P is ﬁnitely presented, and, for all maximal ideals m, Pm is a free Rm-module.

Proof. (i) = (ii): we have seen in Chapter 2 that projective implies flat (Lemma 2.10.2),⇒ and that a finitely generated projective module is finitely presented (Exercise 2.2). (ii) = (iii): This follows from Corollary 3.4.3 and Proposition 2.10.5. ⇒ (iii) = (i): Since P is finitely presented, and hence finitely generated, ⇒ it suffices to show that P is projective. Suppose that M M ′′ is a sur- → jection of R-modules. We must show that HomR(P, M) HomR(P, M ′′) is surjective. It suffices to show that, for all maximal ideals m→of R, the induced homomorphism

(Hom (P, M)) (Hom (P, M ′′)) R m → R m 3.5. REGULAR FUNCTIONS 107 is surjective. By Exercise 3.19 below, since P is ﬁnitely presented, the natural homomorphism

(Hom (P, M)) Hom (P , M ) R m → Rm m m is an isomorphism, and similarly for

(Hom (P, M ′′)) Hom (P , M ′′ ). R m → Rm m m As P is a free R -module, the natural homomorphism Hom (P , M ) m m Rm m m → HomRm (Pm, Mm′′ ) is surjective. Thus, the same is true for the homomorphism (Hom (P, M)) (Hom (P, M ′′)) , for every maximal ideal m. Thus R m → R m Hom (P, M) Hom (P, M ′′) is surjective, so that P is projective. R → R 3.5 Regular functions

k Fix an algebraically closed field k. Let X = An, or more generally let X be an affine algebraic set. We want to look at the difference between the ring of functions on X, namely A(X), and functions which can be defined everywhere “locally” on X, in a manner that we will explain. It will be simpler to begin with the assumption that X is an affine algebraic variety, i.e. that A(X) is an integral domain, with quotient field K(X), the field of rational functions on X. We will define regular functions in two different ways. Definition 3.5.1. An element f K(X) is regular at x X if f = g/h for some g, h A(X) such that h(∈x) = 0. It is regular if it∈ is regular at all points of X. ∈ 6 This definition has the following easy consequences:

Proposition 3.5.2. (i) The ring of all regular functions at x is A(X)mx , where mx is the maximal ideal of all functions vanishing at x. (ii) The ring of all regular functions on X is A(X). Proof. (i) is just the deﬁnition. As for (ii), again by the deﬁnition, the ring of all regular functions is x X A(X)mx . By Exercise 3.7 below and the Nullstellensatz, this ring is just∈ A(X). (Compare also Exercise 1.10 for the n T case of Ak .) 108 CHAPTER 3. LOCALIZATION

The above definition is not easy to generalize directly to the case where A(X) has zero divisors. Here is an equivalent definition which lends itself more easily to generalization: Definition 3.5.3. Let F : X k be a function. Then F is regular if, for all → x X, there exists a Zariski open set Ux containing x and functions gx, hx A(∈X) such that h (y) = 0 for all y U , and such that F U = g /h . ∈ x 6 ∈ x | x x x For example, every f A(X) is regular (take Ux = X, gx = f, hx = 1). In fact, the next result says∈ that these are the only regular functions: Proposition 3.5.4. The ring of all regular functions on X is A(X). Proof. Since X is compact in the Zariski topology, there exists an open cover of X by finitely many nonempty open sets U1,...,UN and functions gi, hi A(X), with h = 0 in U , such that F U = g /h . Since A(X) is an integral∈ i 6 i | i i i domain, by Exercise 1.11, Ui Uj = . In Ui Uj, F Ui Uj = gi/hi Ui Uj = g /h U U , so that g h U∩ U6 =∅ g h U∩ U .| Again∩ by Exercise| ∩ 1.11, j j| i ∩ j i j| i ∩ j j i| i ∩ j g h = g h on X. Note that V (h ,...,h ) = , so by the Nullstellensatz, i j j i 1 N ∅ h1,...,hN generate the unit ideal, i.e. there exist ϕ1,...,ϕN A(X) such that ϕ h = 1. Set f = ϕ g A(X). Then h f = ∈ ϕ h g = i i i i i i ∈ j i i j i ϕihigj = ( ϕihi)gj = gj, so that f = gj/hj on each open set Uj. It i P i P P follows that F Uj is the restriction of the regular function f, for all j, and so FP= f. |P Note the “partition of unity” argument at the heart of the proof. In order to generalize the above, to affine coordinate rings A(X) which are not integral domains, or more generally finitely generated k-algebras R, and ultimately to arbitrary commutative rings, we must deal with the following two issues: (1) if R has divisors of zero, the homomorphism R Rf is not injective, and (2) if R has nilpotent elements, then we cannot→ think of the elements of R as functions in the usual sense. Once we begin to consider general finitely generated k-algebras, we might as well consider a general ring R. We have defined the topological space X = Spec R. However, there is no longer any clearly defined target space for functions on X, as there was for an affine algebraic set, namely the field k. Instead, for each p Spec R, we have the residue field k(p), the field of quotients of R/p. An element∈ r R defines a function ∈ f : Spec R k(p), r → p Spec R ∈a 3.5. REGULAR FUNCTIONS 109 by setting f (p) to be the image of r in R/p k(p). In case R is a finitely r ⊆ generated k-algebra, where k is an algebraically closed field, and m is a maximal ideal, then all of the k(m) are identified with k, and the restriction of fr to the maximal spectrum is the same as the function defined by r in the affine coordinate ring.

In general, the function fr is zero if and only if r p Spec R p = (0), ∈ ∈ and so R is reduced if and only if the function fr determines R. One could try T p to remedy this by considering instead functions from Spec R to p Spec R Rp, ∈ using the injectivity of the natural function R p Spec R Rp to recover r → ∈ ∈ R from the corresponding function. But in general this is rather` unnatural, and we will not try to view the elements of R as functions` in the usual sense. Instead, we proceed by trying to define functions locally in the Zariski topology and then to see what this definition really means. First recall that, given f R, we have defined (in Exercise 1.25) the open sets ∈ X = (Spec R) = p Spec R : f / p . f f { ∈ ∈ } They satisfy: (1) the X , f R, are a basis for the Zariski topology on X; f ∈ (2) Xf Xg = Xfg; (3) Xf = Xg (f)= (g); (4) X = α Xfα the f ∩generate the unit ideal. On⇐⇒ the other hand, we have the R-algebra⇐⇒R α p p f and the induced function ϕ : Spec R Spec R. S f f → Lemma 3.5.5. The image of ϕ : Spec R Spec R is X , and ϕ induces f f → f f a homeomorphism from Spec Rf to Xf . Moreover, Xf = Xg if and only if Rf and Rg are isomorphic as R-algebras.

Proof. By Proposition 3.2.1, the function ϕf identiﬁes the image of Spec Rf in X with p Spec R : f n / p for all n 0 . Since p is prime, f n / p for some n 1{ if∈ and only if f∈ / p. Thus the≥ image} of ϕ is exactly X∈, and ≥ ∈ f f ϕf is one-to-one, again by Proposition 3.2.1. We leave the homeomorphism statement to the reader. Clearly, if Rf and Rg are isomorphic R-algebras, then their images Xf and Xg are equal. Conversely, if Xf = Xg, then (f) = (g). Since g (g) = (f), there exists h R such that gn = hf. In particu- p p lar,∈f is invertible in R , so by the universal∈ property of localizations there p p g exists a homomorphism R R of R-algebras. Symmetrically, there ex- f → g ists an R-algebra homomorphism R R . By Exercise 3.1 below, these g → f two homomorphisms are inverses, and hence the R-algebras Rf and Rg are isomorphic. 110 CHAPTER 3. LOCALIZATION

We continue with the program of trying to find more functions on Spec R by allowing denominators. By analogy with the case of affine algebraic sets, since f “does not vanish” on the open set X = X V (f), it is reasonable to f − allow elements of Rf to be functions on Xf . By Lemma 3.5.5, if in addition Xf = Xg, then Rf ∼= Rg, so that up to isomorphism, Rf only depends on X and not on f. To define a function on X, given an open cover X f { fα } of X, we should look for functions defined on Xfα for every α which satisfy the natural compatibility condition: they agree on intersections Xfα Xfβ for all α, β. However, it turns out that we do not get any new functions∩ this way! The following theorem, whose proof again relies on a partition of unity argument, is an algebraic restatement of this fact.

Theorem 3.5.6. Let X : α A be an open cover of X = Spec R. { fα ∈ } Suppose, for every α A, that we are given rα Rfα such that, for every α, β A, ∈ ∈ ∈ r /1= r /1 in (R ) (R ) R . α β fα fβ ∼= fβ fα ∼= fαfβ

Then there exists a unique r R such that the image r/1 Rfα is equal to r for every α A. ∈ ∈ α ∈ More generally, let f R, and suppose that Xfα Xf : α A is an open cover of X . Given∈r R such that r {/1 = ∩r /1 in R∈ }, then f α ∈ ffα α β ffαfβ there exists a unique r Rf such that the image r/1 Rffα is equal to rα for every α A. ∈ ∈ ∈

Proof. First note that the statement for X implies that for Xf . So we will just consider the first statement in the theorem. Next let us prove uniqueness. It suffices to show that, if r R is such ∈ that r/1 = 0 in Rfα for every α A, then r = 0. Since r/1 = 0 in Rfα , there nα ∈ exists nα > 0 such that fα r = 0. Since Xfα : α A is an open cover { ∈ } nα of X, the fα, α A, generate the unit ideal. By Exercise 1.3, fα , α A ∈ nα ∈ also generate the unit ideal. Thus there is some finite sum α gαfα =1. It follows that r = r 1= g f nα r =0. P · α α α X To prove the existence of r, first suppose that the cover Xfα : α A is { ∈ } n finite; in this case, we can take A = 1,...,k for some k. Let r = h /f i { } i i i for some hi R and positive integers ni. If n ni for every i, we can write ni ∈n ni n n ≥ni n hi/fi = hifi − /fi . Replacing hi by hifi − , we now have ri = hi/fi . Thus we may assume that all of the ni are equal to n. The statement that 3.5. REGULAR FUNCTIONS 111

ri and rj have the same image in Rfifj means that, for every i and j, there Nij n Nij n exists a positive integer Nij such that (fifj) fj hi = (fifj) fi hj. Write Nij +n Nij Nij +n Nij this as fj (fi hi)= fi (fj hj). Choose an integer M Nij + n for M Nij n ≥ all i, j. Multiplying the above by (fifj) − − gives

M M n M M n fj (fi − hi)= fi (fj − hj).

M n M Replace hi by fi − hi for every i, so that ri now becomes hi/fi . The above M M expression then reads fj hi = fi hj for all i, j.

Since Xf1 ,...,Xfk is an open cover of X, f1,...,fk generate the unit { } M M ideal in R, and hence f1 ,...,fk also generate the unit ideal. Let 1 = g f M for some g R. Set r = g h . Then, for all j, i i i i ∈ i i i P M MP M fj r = gifj hi = gifi hj = hj. i i X X M It follows that r/1= hj/fj = rj for all j. This proves the existence of r in the case where the cover is ﬁnite.

If A is not necessarily ﬁnite, the cover Xfα has a ﬁnite subcover, by say X ,...,X , and so there exists r R such{ that} r/1= r for i =1,...,k. fα1 fαk αi For γ A arbitrary, we claim that r/∈ 1 = r as elements of R . Note that ∈ γ fγ Xf Xf is an open cover of Xf . By the compatibility condition for the γ ∩ αi γ rα, rγ and r/1 agree on the rings Rfγ fαi for every i, and so, by the uniqueness part of the proof applied to the ring Rfγ , r/1= rγ. Another way to restate the theorem is as follows:

Corollary 3.5.7. Suppose that Xfα : α A is an open cover of X = Spec R. Then the following sequence{ of R-modules∈ } is exact:

0 R R R . → → fα → fαfβ α A α,β A Y∈ Y∈

Here the homomorphism R α A Rfα sends r to the element whose pro- th → ∈ jection onto the α factor is r/1, and the homomorphism α A Rfα Q ∈ → α,β A Rfαfβ sends (rα) to the element whose projection onto the (α, β) fac- ∈ Q tor is rα/1 rβ/1, using the canonical identiﬁcations (Rf ) = (Rf )f = Q − α fβ ∼ β α ∼ Rfαfβ . Identical arguments prove the following version of the above results for R-modules M: 112 CHAPTER 3. LOCALIZATION

Theorem 3.5.8. Let M be an R-module. Suppose that X : α A is { fα ∈ } an open cover of X = Spec R, and that, for every α A, that we are given m M such that, for every α, β A, ∈ α ∈ fα ∈ m /1= m /1 in (M ) (M ) M . α β fα fβ ∼= fβ fα ∼= fαfβ Then there exists a unique m M such that the image m/1 M is equal ∈ ∈ fα to m for every α A. In other words, the following sequence is exact: α ∈ 0 M M M . → → fα → fαfβ α A α,β A Y∈ Y∈ A similar statement holds with M replaced by M and where X : α A f { fα ∈ } is only assumed to be an open cover of Xf .

3.6 Introduction to sheaves

Sheaves play a fundamental role in algebraic geometry and many other areas of mathematics, and examples of them arise constantly. They encode the idea that various classes of functions are characterized by local conditions, and provide a very efficient computational formalism for understanding global obstructions to problems which have local solutions. Like category theory, or prose to Molière’s Le bourgeois gentilhomme, readers should find that, without knowing it, they have been using sheaves for all of their mathematical lives. The main interest in this chapter will be to formalize the idea of defining functions on open subsets of an affine algebraic set or scheme.

Deﬁnition 3.6.1. Let X be a topological space. A presheaf on X (of abelian groups, rings, R-modules, . . . ) consists of the following:F for each open subset U of X, an abelian group, ring, R-module, . . . (U) called the group of sections of over U, and for each pair of openF sets U and V F with V U, a homomorphism ρUV : (U) (V ), called the restriction homomorphism⊆ , satisfying: F → F

1. ρUU = Id;

2. If W V U, then ρ = ρ ρ ; ⊆ ⊆ UW VW ◦ UV 3. ( )=0. F ∅ 3.6. INTRODUCTION TO SHEAVES 113

Condition (3) is often omitted in the definition of a presheaf. However, it is satisfied in all the examples we shall consider and follows formally from the definitions for sheaves, so we will assume it so as to avoid having to worry about the value of ( ). Sometimes we abbreviate ρ (s) by s V . F ∅ UV | One can paraphrase the definition of a presheaf as follows: the open subsets of X form a category, whose morphisms are inclusions, i.e. Hom(V, U)= , if V is not a subset of U, and Hom(V, U) consists of a single element if ∅V U. A presheaf on X is then a contravariant functor from this category to⊆ the category of abelian groups, rings, R-modules, . . . , satisfying (3) above. One can also define presheaves (and sheaves) of sets, but we shall not do so here. If is a presheaf on X, and U is an open subset of X, then by restricting to openF subsets of U we obtain a presheaf on U, called the restriction of F to U, and denoted U. F F| Let be a presheaf of rings on X and let be a presheaf of abelian groups.R Suppose that, for each open set U of X,F (U) is an (U)-module, F R compatibly with the restriction maps. In this case, we say that is a presheaf of -modules. F R Definition 3.6.2. Let X be a topological space and let be a presheaf on X. Then is a sheaf if the following two properties hold:F F 1. Let U be an open subset of X and let U : α A be an open cover { α ∈ } of U. If s (U) and ρ (s)=0 for all α, then s = 0. ∈F U,Uα 2. With U and U as in (1), if for all α we are given s (U ) such that, α α ∈F α for all pairs (α, β), ρUα,Uα U (sα)= ρU ,Uα U (sβ), then there exists an ∩ β β ∩ β s (U) (unique by (1)), such that ρ (s)= s for all α. ∈F U,Uα α

We can restate the conditions (1) and (2) by saying that, with U and Uα as in (1), the sequence

0 (U) (U ) (U U ) →F → F α → F α ∩ β α A (α,β) A A Y∈ Y∈ × is exact, where the homomorphism (U) α A (Uα) is the product of F → ∈ F the ρU,Uα and the homomorphism α A (Uα) (α,β) A A (Uα Uβ) is ∈ F Q→ ∈ × F ∩ given by the diﬀerence ρUα,Uα U ρU ,Uα U . ∩ β − β ∩ β Sheaves abound in nature. HereQ are just a fewQ of many examples. 114 CHAPTER 3. LOCALIZATION

Example 3.6.3. Let X be a topological space, and fix an abelian group A. Define a presheaf by: for each open nonempty subset U of X, (U)= A, F F and ρUV = Id. We call the constant presheaf with values in A. However, is not a sheaf in general.F For example, if A = 0 and there exist open sets F 6 U1 and U2 with U1 U2 = , then choosing a1 = a2 in A and for the open cover U , U of U ∩= U U∅ , the sections a 6 (U ) and a (U ) agree { 1 2} 1 ∪ 2 1 ∈F 1 2 ∈F 2 on U1 U2, but do not come from a section of (U). We will see how to remedy∩ this situation below. F Example 3.6.4. (Compare Example 2.2.10.) Let X be a topological space. We define a sheaf of R-algebras X as follows: for each open set U of X, (U) is the ring of continuous functionsC f : U R, with the restrictions CX → the usual ones. Likewise, if X is a C∞ manifold, we can define the sheaf R- algebras ∞ by setting ∞(U) equal to the ring of C∞ functions f : U R. CX CX → Clearly ∞ is a subsheaf of in the natural sense: for every U, ∞(U)is a CX CX CX subalgebra of X (U) and the restriction maps for X∞ are the ones induced by those for .C Likewise, if X is a complex manifold,C we can define the sheaf CX of holomorphic functions X , and it is a subsheaf of the ring of C∞ complex valued functions on X. H Virtually any of the standard constructions on C∞ manifolds fit into this framework. For example, there are sheaves of R-vector spaces defined by the set of all vector fields on an open set, or by the set of differential forms of degree k. Both of these example are in fact sheaves of ∞-modules. CX Example 3.6.5. For a construction which (partially) generalizes both of the preceding examples, suppose that G is a topological abelian group. For example, we could take G to be an abelian group A with the discrete topology, or G = R. Then there is an associated sheaf defined by: (U) is GX GX the group of continuous functions from U to G, with the usual restriction homomorphisms. In case G = A with the discrete topology, we denote this sheaf by AX and call it the constant sheaf with values in A. Note that AX (U) = A if U is connected, but not in general. There are the usual variations on this idea. For example, if X is a C∞ manifold and G is an abelian Lie group, then we could consider the group of all C∞ functions from U X to G. ⊆This idea can be generalized further still. Suppose that π : Y X is a → continuous map of topological spaces, and that the fibers of π have an abelian group structure in an appropriate sense. We would like this structure to vary continuously in some sense: in particular, if s1 and s2 are two sections of π, 3.6. INTRODUCTION TO SHEAVES 115

and we define (s1 + s2)(x) in the usual way, as s1(x)+ s2(x), then we want s1 + s2 to be continuous if s1 and s2 are continuous. One way to do this is to require that the function F : Y X Y Y on the fiber product defined by the addition be continuous, and similarly× → for the inverse Y Y , and lastly that → the identity section X Y be continuous. For example, if Y = X G and → × π = π1, where G is a topological group and the fiber group operations are those of G, then Y X Y = X (G G), and the function F : Y X Y Y is the function X ×(G G) ×X ×G defined by the multiplication× in→G. × × → × Under this assumption, if we define (U) to be the set of continuous 1 F 1 sections of the restriction of π to π− (U) U, i.e. functions s: U π− (U) such that π s = Id, then pointwise addition→ of two sections s ,s→of (U) ◦ 1 2 F is given by the composition

(s1,s2) 1 1 F 1 U π− (U) π− (U) π− (U) −−−→ ×U −→ and this is again continuous. Similar comments apply to the identity section and inverses, so that (U) is an abelian group. In this way, we deﬁne the F sheaf of sections of π : Y X. As above, if there is extra structure, for → example if X and Y are C∞ manifolds and all functions are C∞ functions, then we could instead look at the sheaf of C∞ sections of π.

Following the ideas of Example 2.2.10, we introduce the following:

Deﬁnition 3.6.6. Let x X, and let be a presheaf or sheaf on X. We deﬁne the stalk of to∈ be the directF limit lim (U), where the direct Fx F x U F limit is taken over the directed set of all open−→ sets∈ containing x, with the Ux ordering U V if U V . Note that, for every open set U containing x, there is an induced homomorphism⊇ (U) . We also call the group F →Fx Fx of germs of sections of at x. F We have already used the idea of a subsheaf or a subpresheaf of a sheaf G or presheaf : for all U an open subset of X, (U) is a subgroup of (U), and the restrictionF homomorphisms for are theG restrictions of thoseF for . More generally, a homomorphism F : G of presheaves or sheaves consistsF G→F of the following: for each U, a homomorphism F : (U) (U), such that, U G →F for all U, V with V U, the following diagram commutes (where the ρUV are the restriction homomorphisms⊆ for and the σ are the restriction F UV 116 CHAPTER 3. LOCALIZATION homomorphisms for ): G F (U) U (U) G −−−→ F σUV ρUV

 FV  (V ) (V ). G y −−−→ F y (This is the same thing as a morphism of functors.) For every x X, a homomorphism F : of presheaves induces homomorphisms F ∈: G→F x Gx → of the stalks at x. The homomorphism F is injective if F is injective for Fx U every U, and it is an isomorphism if FU is an isomorphism for every U. Warning: We do not define F to be surjective if FU is surjective for every U. Closely connected with this is the fact that, if is a subsheaf of a sheaf , then the presheaf which assigns to each open setGU the group (U)/ (U)F is F G not always a sheaf. Likewise, given a homomorphism F : of sheaves, G→F if we define Ker F (U) = Ker FU , then Ker F is again a sheaf. However, with the analogous definitions, Im F and Coker F are only presheaves and not in general sheaves. We now show that, to every presheaf , there is a canonically associated F sheaf ˆ, which is constructed essentially by forcing the sheaf axioms. F Theorem 3.6.7. Let be a presheaf on X. Then there exists a sheaf ˆ on X and a homomorphismF Φ: ˆ with the following universal property.F F → F Given a sheaf on X and a homomorphism F : , there exists a unique homomorphismG F˜ : ˆ such that F = F˜ ΦF. In → particular, G this property determines ˆ up toF a → unique G isomorphism. ◦ F Proof. Define a presheaf ˆ as follows. Given an open set U of X, ˆ(U) F F consists of all functions f : U x U x satisfying: For all u U, → ∈ F ∈ 1. f(u) ; ` ∈Fu 2. There exists an open set V U, with u V , and an s (V ), such that f(v) is the image of s in⊆ for all v∈ V . ∈ F Fv ∈ Note that there is a homomorphism Φ = Φ: ˆ defined as follows: for F F → F s (U), let Φ (s) be the function which sends u to the image of s in ∈ F U Fu for all u U. Clearly ΦU (s) satisfies (1) and (2) above. We claim∈ the following:

1. ˆ is a sheaf; F 3.6. INTRODUCTION TO SHEAVES 117

2. This construction is functorial, i.e. given a homomorphism F : , F → G then there is an induced homomorphism Fˆ : ˆ ˆ, such that the following diagram commutes: F → G

F F −−−→ G ΦF ΦG

 ˆ  ˆ F ˆ. Fy −−−→ Gy 3. If is a sheaf, then Φ : ˆ is an isomorphism. In particular ˆ G G G → G ˆ = ˆ, and Φ = Φ ˆ. F ∼ F F F Given the above, the rest of the proof is as follows. If F : is a c 1 F → G homomorphism, where is a sheaf, deﬁne F˜ = (Φ )− Fˆ : ˆ . The G G ◦ F → G commutativity of the diagram in (2) says that Φ F = Fˆ Φ , and hence 1 G ◦ ◦ F that F = (Φ )− Fˆ Φ = F˜ Φ . To see the uniqueness, if G: ˆ is G ◦ ◦ F ◦ F F → G another homomorphism satisfying F = G Φ , then Gˆ Φ = Fˆ, and thus ◦ F ◦ F 1 ˆ 1 ˆ ˜ G = (Φ )− G Φ ˆ = (Φ )− G Φ =cF. G ◦ ◦ F G ◦ ◦ F So we must prove statements (1)–(3). These arec straightforward if tedious exercises in the deﬁnition of a sheaf and properties of direct limits (Exercise 2.4), and are left to the reader.

Definition 3.6.8. We call the sheaf ˆ defined above the sheaf associated to F the presheaf or the sheafification of . F F Remark 3.6.9. If is a subpresheaf of the presheaf , then it is clear from the construction thatG ˆ is a subsheaf of the sheaf ˆ.F In particular, if is a G F G subpresheaf of a sheaf , then the sheafification of may be identified with a subsheaf of . F G F Using the above theorem, if ′ is a subsheaf of the sheaf , then we can F F define the sheaf quotient / ′ as the sheaf associated to the presheaf which F F assigns to every open set U the group (U)/ ′(U). Likewise we can define images and cokernels of sheaf homomorphisms,F F exact sequences, and so on. The meaning of these constructions, and their connections with problems which can be solved locally but not necessarily globally, will be discussed elsewhere. 118 CHAPTER 3. LOCALIZATION 3.7 Schemes and ringed spaces

We return now to the question of defining functions on open subsets of an affine algebraic set X or a spectrum X = Spec R. The basic idea comes from Theorem 3.5.6, which says the following: if we assign to the open subset Xf of X the ring Rf , then this behaves very much like a sheaf. The only problem is that not every open subset of X is of the form Xf . Instead, we can only say that the Xf form a basis for the topology of X and are closed under finite intersections. However, it is easy to surmount this technical difficulty:

Theorem 3.7.1. Let X be a topological space and let be a basis for the topology of X which is closed under finite intersections.U Suppose that we are given, for each U , an abelian group, ring, R-module, ..., (U), and U for each pair of open∈ U sets U and V in with V U, a homomorphismF U ⊆ ρUV : (U) (V ) satisfying the conditions of Definition 3.6.1 and Defi- FU →FU nition 3.6.2. Then there is a unique sheaf on X such that (U)= (U) U for all U . F F F ∈ U Proof. The proof is a minor variation on that of Theorem 3.6.7. First note that, for x X, we can still define the stalk ,x = lim (U). For an ∈ FU x U FU arbitrary open subset U of X, define (U) as follows:−→ ∈ (∈UU) consists of all F F functions f : U x U ,x satisfying: For all u U, → ∈ FU ∈

1. f(u) ,u; ` ∈FU 2. There exists an open set V U, with u V and V , and an ⊆ ∈ ∈ U s (V ), such that f(v) is the image of s in ,v for all v V . ∈FU FU ∈ Then the methods of the proof of Theorem 3.6.7 show that is a sheaf and that (U)= (U) for all U . The uniqueness is clear. F F FU ∈ U Definition 3.7.2. Let X be an affine algebraic set, or let X = Spec R. Then the sheaf of rings X on X defined by Theorem 3.7.1, starting from the basis = X : f RO and the requirement that (X ) = R , is called the U { f ∈ } OX f f structure sheaf of X. Note that, if R = A(X) is the affine coordinate ring of the affine algebraic set X, then Spec R and X have the same collection of open sets and X (U) does not depend on whether we view X as an affine algebraic set orO a spectrum.

We write for the stalk of at x. The following describes : OX,x OX OX,x 3.7. SCHEMES AND RINGED SPACES 119

Lemma 3.7.3. If x X corresponds to the prime ideal p of R, then = ∈ OX,x Rp.

Proof. Since the collection of Xf containing x is a coﬁnal subset of the collection of all open neighborhoods of x, = lim R . Note that x X X,x x X f f O ∈ f ∈ if and only if f / p. Given an R-algebra ϕ−→: R R′ such that, for all f R p, ϕ(f)∈ is a unit, there exists a unique R→-algebra homomorphism ∈ − from R to R′, for every f R p, and, by uniqueness, if X X , so f ∈ − f2 ⊆ f1 that there is an induced homomorphism R R , the homomorphisms f1 → f2 R R′ are compatible with the homomorphism R R . Hence there fi → f1 → f2 is an induced R-algebra homomorphism R′. Thus has the same OX,x → OX,x universal property as Rp, and so they are isomorphic as R-algebras by (ii) of Proposition 3.1.2.

In a very similar way, we have:

Definition 3.7.4. Let X = Spec R, and let M be an R-module. Then there is a sheaf of -modules M on X defined by Theorem 3.7.1, starting from OX M(Xf ) = Mf , called the sheaf on X associated to M. Its stalk at a point x X corresponding to p fis M (exercise). ∈ p f Definition 3.7.5. A ringed space is a pair (X, X ), where X is a topological space and is a sheaf of rings on X. Often weO omit from the notation. OX OX The ringed space (X, X ) is a locally ringed space if, for every x X, the stalk is a local ring.O ∈ OX,x For X = Spec R, the locally ringed space (X, X ) is called an affine scheme. O

Examples of ringed spaces are: a topological space X with = , OX CX the sheaf of continuous functions; a C∞ manifold X with = ∞; an OX CX affine scheme X = Spec R with X the sheaf constructed above. Note they are all locally ringed spaces. InO a locally ringed space X, a function s ∈ X (U) has a “value” s(x) for all x U, namely the image of s in the residueO field /m . In many cases, ∈s is determined by its values, i.e. the OX,x x homomorphism X (U) x U X,x/mx is injective. For example, this is O → ∈ O the case for affine schemes X = Spec R if and only if R is reduced. Ringed spaces form a category` in a natural way. If f : X Y is a continuous map of topological spaces, and is a presheaf on X, we→ define a F 1 presheaf f on Y as follows: f (U)= (f − (U). It is easy to see that, if ∗F ∗F F F 120 CHAPTER 3. LOCALIZATION is a sheaf, then so is f . Suppose that x X. Then, if U is an open subset of ∗F ∈ 1 Y containing f(x), there is the homomorphism f (U) (f − (U)) x. ∗F →F →F These fit together to define a homomorphism of stalks (f )f(x) x. ∗F →F Definition 3.7.6. Let (X, ) and (Y, ) be ringed spaces. A morphism OX OY (X, ) (Y, ) consists of a pair (f, f ∗), where f : X Y is a continuous OX → OY → function and f ∗ : Y f X is a homomorphism of sheaves of rings. If X ∗ and have someO extra→ structure,O for example if they are k-algebras, thenO OY we require that f ∗ is a k-algebra homomorphism as well. Isomorphisms of ringed spaces are then defined in the usual way. The homomorphism f ∗ determines a way to pull back sections of from OY Y to X. Typically, we omit both the sheaves and the homomorphism f ∗ from the notation and simply refer to f : X Y as a morphism of ringed spaces. Note that, given x X, there is an induced→ homomorphism on the stalks, ∈ via f ∗ Y,f(x) (f X )f(x) X,x. O −→ ∗O → O If both X and Y are locally ringed spaces, then a morphism f is required to satisfy: for all x X, the homomorphism f ∗ induces a local homo- ∈ morphism Y,f(x) X,x. Typically, X and Y are k-algebras and the homomorphismsO k→ O /m and k O /mO are isomorphisms for all → OX,x x → OY,y y x X,y Y . In this case, f ∗ : /m /m is an isomorphism ∈ ∈ OY,f(x) f(x) → OX,x x for all x X, and the “values” of s and f ∗s (their images in Y,f(x)/mf(x) and ∈/m respectively) agree. Thus, if sections of areO specified by OX,x x OX their values, then the pullback f ∗ of a section s Y (U) is just the pullback of the function s: U k in the usual sense. ∈ O → Proposition 3.7.7. If f : R S is a ring homomorphism, then f induces → a morphism of locally ringed spaces Spec S Spec R. Moreover, every homomorphism of locally ringed spaces Spec S → Spec R is of this form. → Proof. We have seen in Lemma 1.6.13 that a ring homomorphism ϕ induces a continuous function f : Spec S Spec R. Given an open set of Spec R → of the form Xs = Spec Rs, ϕ induces a homomorphism Rs Sϕ(s), i.e. 1 → Spec R(Xs) Spec S(f − (Xs)) and these fit together to give Spec R O → O 1 O → f Spec S. If q Spec S and p = f − (q), then the induced homomorphism on ∗O ∈ stalks is the homomorphism Rp Sq, which is local. Thus f is a morphism of locally ringed spaces. → 3.7. SCHEMES AND RINGED SPACES 121

Conversely, suppose that f : Spec S Spec R is a morphism of locally → ringed spaces. Then R = ( ) and S = (Spec S), by The- OSpec R OSpec R OSpec S orem 3.5.6. Thus f ∗ yields a homomorphism ϕ: R S. Our goal is to show that the homomorphism of ringed spaces induced→ by ϕ is f. If q Spec S ∈ and p = f(q), then there is also a homomorphism of local rings f ∗ : R S , p → q which is compatible with the homomorphism f ∗ = ϕ: R S. Thus f ∗ = ϕ → p is the homomorphism Rp Sq induced from ϕ, and there is a commutative diagram → ϕ R p S p −−−→ q

iR iS x ϕ x R S,  −−−→  where iR : R Rp is the canonical homomorphism, and similarly for iS. We →1 claim that ϕ− (q)= p. To see this, we use the fact that f is local to conclude 1 that ϕp− (qSq)= pRp. Thus,

1 1 1 1 1 1 ϕ− (q)= ϕ− (iS− (qSq)) = iR− (ϕp− (qSq)) = iR− (pRp)= p.

It follows that ϕ∗ : Spec S Spec R is the same function as f. We must → show that the pullback f ∗ is the same as that induced by ϕ. It suﬃces to check this on open sets of the form Spec Rs, where it follows from the fact that f ∗ : Rs Sϕ(s) is compatible with f ∗ = ϕ: R S and thus is the homomorphism→ induced by ϕ. →

Ringed spaces provide an efficient way to describe spaces which are glued together out of local models in the same way that a manifold is glued out of coordinate charts. For example, let U and V be open subsets of Rn, and suppose Φ: U V is an isomorphism of locally ringed spaces (U, U∞) (V, V∞) (respecting→ the R-algebra structures). We claim (the analogueC of→ Proposi-C tion 3.7.7) that such a Φ is the same thing as a diffeomorphism Ψ: U V . → Clearly, a diffeomorphism Ψ induces an isomorphism of locally ringed spaces. Conversely, given an isomorphism Φ of locally ringed spaces, we claim that the underlying function Ψ: U V is a diffeomorphism. It suffices to show 1 → that Ψ and Ψ− are C∞ functions, which follows from the fact that the pull- backs Ψ∗x of the coordinate functions x ∞(V ) are again C∞. By the i i ∈ CV remarks in Definition 3.7.6, the pullback Φ∗ must then be Ψ∗. Now let (X, ) be a locally ringed space of R-algebras. We say that OX X is a C∞ manifold of dimension n if X is locally isomorphic to an open 122 CHAPTER 3. LOCALIZATION subset of Rn, in other words, for every x X, there exists an open set U ∈ containing x, an open subset V of Rn, and an isomorphism of ringed spaces

(U, X U) (V, V∞). To see that this is the same as the usual definition, noteO that| it→ impliesC that there is an open cover U of X, open subsets V in { α} α Rn for every α, and isomorphisms of locally ringed spaces Φ : (U , U ) α α OX | α → (Vα, V∞α ) for every α, such that the induced homeomorphisms gαβ = Φα 1 C ◦ Φβ− are diffeomorphisms from the appropriate open subset Vαβ of Vβ to the corresponding subset Vβα of Vα. Note that gαβ gβγ = gαγ. Conversely, such a collection of homeomorphisms defines the structure◦ of a ringed space on X. In fact, very generally we have the following procedure for constructing a ringed space from local models:

Theorem 3.7.8. Let X be a topological space and = U : α A U { α ∈ } an open cover of X. Suppose that, for each α A, we are given a sheaf of rings on U , and, for all α, β A, isomorphisms∈ of ringed spaces OUα α ∈ ϕ : U U U U satisfying ϕ ϕ = ϕ . Then there is a αβ OUβ | α ∩ β → OUα | α ∩ β αβ ◦ βγ αγ sheaf of rings X on X and isomorphisms ϕα of ringed spaces (Uα, X Uα) (U , ) suchO that the induced isomorphisms over U U areO the| ϕ →. α OUα α ∩ β αβ Moreover, and the ϕ are unique up to isomorphism in the obvious sense. OX α Proof. This tedious but essentially straightforward argument is left to the reader. (Use for example Theorem 3.7.1.)

In our situation, we make the following deﬁnition:

Definition 3.7.9. A scheme is a locally ringed space (X, X ) which is locally isomorphic to an affine scheme, i.e. for every x XO, there exists an ∈ open set U containing x, an affine scheme V , and an isomorphism of ringed spaces (U, X U) (V, V ). There is a similar definition for spaces locally isomorphicO to| affine→ algebraicO sets.

Thus (roughly speaking) a scheme is a ringed space obtained by gluing together affine schemes. One slight difference is that, in case X is a C∞ manifold and Uα, Uβ are two open subsets of X isomorphic as locally ringed n spaces (i.e. diffeomorphic) to open subsets of R , then it is clear that Uα Uβ is again isomorphic to an open subset of Rn. However, if X is a scheme∩ and Uα, Uβ are two open subsets of X isomorphic as locally ringed spaces to affine schemes, then it is no longer clear that Uα Uβ is isomorphic to an affine scheme, and in fact it is not always true. However,∩ it is true under a mild 3.8. PROJ AS A SCHEME 123 hypothesis on X (X is separated), which is the analogue for schemes of the Hausdorff property for manifolds. We conclude with a method for constructing schemes starting with affine schemes and gluing data, which is just a restatement of Theorem 3.7.8 in this context: Theorem 3.7.10. Let X be a topological space and = U : α A an U { α ∈ } open cover of X. Suppose that, for each α A, we are given a homeomorphism ϕ : U V , where V is an affine∈ scheme, and, for all α, β A, α α → α α ∈ isomorphisms Gαβ of ringed spaces from the appropriate open subset Vαβ of Vβ to the corresponding subset Vβα of Vα such that the homeomorphisms 1 gαβ induced by Gαβ satisfy: gαβ = ϕα ϕβ− . Assume further that, for all α,β,γ, G G = G . Then there is◦ a unique sheaf of rings on X αβ ◦ βγ αγ OX and isomorphisms Φα of ringed spaces (Uα, X Uα) (Vα, Vα ) such that 1 O | → O G = Φ Φ− for all α, β A. In particular, (X, ) is a scheme. αβ α ◦ β ∈ OX Proof. Define a sheaf on U by (U)= (ϕ (U)), with the obvious OUα α OUα OVα α restriction maps. The Gαβ then define isomorphisms of ringed spaces over U U satisfying the hypotheses of Theorem 3.7.8. α ∩ β Note that the extra condition Gαβ Gβγ = Gαγ , which is automatically satisfied by the underlying homeomorphisms,◦ is necessary because an isomorphism of ringed spaces is not necessarily specified by the underlying homeomorphism of topological spaces.

3.8 Proj as a scheme

Recall that, if R = n 0 Rn is a graded ring, then Proj R is the set of ≥ homogeneous prime ideals of R which do not contain R+. We will show that there is a natural schemeL structure on Proj R, and in fact that Proj R is obtained by gluing together aﬃne schemes in a way which generalizes the n construction of projective space Pk . The closed sets for the Zariski topology on Proj R are of the form V+(I), where I is a homogeneous ideal. In particular, a basis for the Zariski topology on Proj R is given by the open sets D (f) = Proj R V (f), where f is a + − + homogeneous element of R. Clearly, D+(f) is the intersection of Spec Rf with the subset Proj R of Spec R. In particular (or directly) D+(fg) = D+(f) D+(g). It is easy to see that the sets D+(f), f R+ form an open cover of∩ Proj R. In fact: ∈ 124 CHAPTER 3. LOCALIZATION

Proposition 3.8.1. If f , α A generate R , then D (f ), α A form an α ∈ + + α ∈ open cover of Proj R.

Proof. Note that Proj R α A D+(fα) = α A V+(fα). Hence it suffices − ∈ ∈ to show that, if the fα, α A generate R+, then α A V+(fα) = . But if ∈S T ∈ ∅ p α A V+(fα), then fα p for all α, and hence R+ p. ∈ ∈ ∈ T ⊆ WeT now write down constructions for Proj R which are the analogues of the discussion in 1.7. More explicit connections are developed in the exercises. To show that§ Proj R has a scheme structure, we begin by showing that D+(f) is homeomorphic to an affine scheme. First we claim that, if f R is homogeneous of degree d, then the ring Rf is graded (by Z). It ∈ k suffices to show that, if r Rn, then the degree n kd of the element r/f ∈ k ℓ − is well-defined. But if r/f = s/f for some s Rm, then there exists a > 0 such that f a(f ℓr) = f a(f ks). So ad + aℓd +∈ n = ad + ak + m, and n ak = m aℓ as desired. Let (R ) be the subring of homogeneous degree − − f 0 zero elements of Rf .

Proposition 3.8.2. Let f R . The open subset D (f) is homeomorphic ∈ d + to Spec(Rf )0.

Proof. We begin by constructing a bijection Φf from D+(f) to Spec(Rf )0. inh If I D+(f), let I = IRf (Rf )0. If I = p is prime, then pRf is a prime ideal∈ in R , by (iii) of Proposition∩ 3.2.1. Thus Φ (p) = pinh = pR (R ) f f f ∩ f 0 is also prime. Conversely, suppose that q is a prime ideal in (R ) . For all n 0, deﬁne f 0 ≥ p = r R : rd/f n q , n { ∈ n ∈ } where d = deg f, and let

q = p = pn. n 0 M≥ We claim that p is a homogeneouse prime ideal not containing f (hence p does not contain R+), and that these two constructions are inverse. 2n 2d First, pn is a subgroup of Rn for every n: if r, s pn, then (r + s) /f is a sum of multiples of expressions of the form r∈asb/f 2d, where a + b = 2n and hence either a or b is at least n. If say a n, then rasb/f 2d = a n b d n d 2n 2d ≥n d 2 (r − s /f )(r /f ) q. Thus (r + s) /f = [(r + s) /f ] q. Since q is prime, (r + s)n/f d ∈ q and thus r + s p by definition. Next,∈ it is clear ∈ ∈ n 3.8. PROJ AS A SCHEME 125 that R p p for every n and k, so that Rp p and p is an ideal of R. k n ⊆ n+k ⊆ Clearly, f / p. Thus p does not contain R . Since p is homogeneous, to see ∈ + that p is prime, it suffices to show that, if r Rn, s Rm, and rs pn+m, then either r p or s p . This follows immediately∈ ∈ from the fact∈ that q ∈ n ∈ m is prime. If q is a prime ideal in (Rf )0 and q = p is as above, then an element n d dn of pRf (Rf )0 is of the form r/f , with r pnd. By definition, r /f = n d∩ n ∈ (r/f ) q. As q is prime, r/f q. Thuse pRf (Rf )0 q. Conversely, if r/f n q∈, then r p and so r/f∈n pR (R ∩) . Thus⊆pR (R ) = q. ∈ ∈ nd ∈ f ∩ f 0 f ∩ f 0 To go the other way, if p is a homogeneous prime ideal of R not containing d n f and q = pRf (Rf )0, suppose that r Rn is such that r /f q. Writing rd/f n = s/f m ∩where s p, we see that∈ f N rd p for some N∈ 0. Since ∈ ∈ ≥ f / p, it follows that rd p and hence that r p. This shows that q p, and∈ the other containment∈ is clear. ∈ ⊆ This shows that Φf : D+(f) Spec(Rf )0 is a bijection. We muste show → that it is a homeomorphism for the Zariski topologies. The open sets of D+(f) of the form D+(f) D+(g)= D+(fg), where g is homogeneous of degree e, are a basis for the topology∩ of D (f). Given p D (f), p D (fg) if and + ∈ + ∈ + only if g / p. But since f / p, g / p if and only if gd/f e / pR (R ) . This ∈ ∈ ∈ ∈ f ∩ f 0 says that Φf (D+(fg)) is the basic open set Spec[(Rf )0]gd/f e of Spec(Rf )0. Conversely, given a basic open set of Spec(Rf )0, of the form Spec(Rf )0 n n − V (r/f ) for some r Rnd, then given q Spec(Rf )0, q / V (r/f ) if and only ∈ ∈ 1 ∈ n if r / p and hence if and only if r / p. Thus Φ− (Spec(R ) V (r/f )) = ∈ nd ∈ f f 0 − D+(r). It follows that Φf is a homeomorphism.

We now use the above to give Proj R the structure of a locally ringed space.

Theorem 3.8.3. Let X = Proj R. There is a natural sheaf of rings X on X such that the restrictions (D (f), D (f)) are isomorphic viaOΦ as + OX | + f locally ringed spaces to Spec(Rf )0.

Proof. We show that the conditions of Theorem 3.7.10 are met. Suppose that f R and g R . Then there are canonical isomorphisms [(R ) ] d e = ∈ d ∈ e f 0 g /f ∼ [(Rg)0]f e/gd ∼= (Rfg)0 (exercise). These isomorphisms give identiﬁcations of the corresponding sheaves on open sets of the form D (f) D (g). Com- + ∩ + patibility on triple intersections is checked in a similarly way, using three homogeneous elements f,g,h of R. 126 CHAPTER 3. LOCALIZATION 3.9 Zariski local properties

We have discussed local properties of rings and modules in 3.4. However, § in many ways this name is inaccurate, for the local rings Rp and Rm do not in general correspond to open subsets of Spec R. Instead, one could look at properties which are satisfied on an open cover of Spec R and ask if they then hold globally. More precisely, consider a property P for rings R or R-modules M. We will only consider properties P which are stable under localization, in other words, if P holds for R (or for an R-module M), then it holds for all 1 1 rings of the form S− R (or all modules S− M). For example, the properties that M be free, or projective, or finitely generated, or finitely presented, are all stable under localization. The topological analogue would be a property which, if it holds for a space X, or for a sheaf on X, holds for all open subsets U of X, or for the restriction of to everyF open subset. A natural F question is when a sort of local converse holds: if a property of spaces holds for every sufficiently small open subset of X, or equivalently for some open cover of X, does it hold for X? In the case X = Spec R, the natural open covers to use are those whose elements are of the form Xf for an appropriate collection of f R. By compactness, we need only consider the case of a finite cover. Thus∈ we are led to the following definition:

Deﬁnition 3.9.1. Let P be a property of rings or of R-modules which is stable under localization. Then P is a Zariski local property provided that P holds for a ring R (or an R-module M) if and only if there exist f ,...,f R 1 n ∈ such that (f1,...,fn) = R, and P holds for Rfi for every i (or P holds for

Mfi for every i).

Proposition 3.9.2. For an R-module M, the following properties are Zariski local:

(i) M =0;

(ii) M is ﬁnitely generated;

(iii) M is ﬁnitely presented.

Proof. (i) Let f1,...,fn R be such that (f1,...,fn) = R, and suppose that M = 0 for all i. Then,∈ for m M, there exists a > 0 such that fi ∈ i 3.9. ZARISKI LOCAL PROPERTIES 127

ai a1 an fi m = 0 for all i. As f1 ,...,fn also generate the unit ideal, so that there exist g R such that 1 = g f ai m, and hence m = 0 for all m M. i ∈ i i i ∈ (ii) With f1,...,fn R as above, suppose that Mfi is ﬁnitely generated for every i and choose m∈ ,...,mP M such that the images m /1 generate 1 k ∈ j Mfi for every i, which is possible since we can always enlarge a set of generators. Denote by Rk M the corresponding R-module homomorphism, and let Q be the cokernel.→ Thus

Rk M Q 0 → → → is exact. Since Rk M Q 0 fi → fi → fi → is also exact, Qfi = 0 for all i. Hence by (i) Q = 0, and M is generated by m1,...,mk.

(iii) Since Mfi is ﬁnitely presented for all i, Mfi is ﬁnitely generated for all i, and so by (ii) there is an exact sequence

0 K Rk M 0 → → → → for some R-module K. Localizing at fi, we obtain an exact sequence

0 K Rk M 0. → fi → fi → fi →

By Exercise 2.9(iii), Kfi is finitely generated for every i. Thus K is finitely generated, by (ii) again. It follows that M is finitely presented.

Exercises

Exercise 3.1. Give a proof of (ii) of Proposition 3.1.2. Conclude that, if 1 1 ϕ: S− R S− R is an isomorphism of R-algebras, then ϕ = Id. → Exercise 3.2. Let R = k[x, y]/(xy), where k is an algebraically closed ﬁeld. Show that R is reduced. By the Nullstellensatz the maximal ideals of R correspond to points on the union of the x- and y-axes.

(i) Show that R is isomorphic to the following subring of k[x] k[y]: ⊕ R = (f,g) k[x] k[y] : f(0) = g(0) . ∼ { ∈ ⊕ } 128 CHAPTER 3. LOCALIZATION

(ii) Suppose that m is a maximal ideal of R corresponding to the point (a, 0), a = 0. What is Rm? What if m corresponds to the point (0, a), a = 0? To6 the point (0, 0)? 6 Exercise 3.3. Let R = R[x], and let f = x2 +1 R. What is Spec R? ∈ Spec Rf ? Is f a unit in R?

Exercise 3.4. Argue carefully that Rr ∼= R[x]/(rx 1). (Use Proposi- tion 3.1.2.) − In particular, if R = A(X) is the affine coordinate ring of the affine n algebraic set X Ak and f A(X), show that Xf is an affine algebraic subset of An+1, defined⊆ by I(X∈) and fx 1 in the obvious way, and its k n+1 − affine coordinate ring is A(X)f . Exercise 3.5. (i) Let S be a multiplicatively closed subset of a ring R, 1 and let M be a finitely generated R-module. Show that S− M = 0 if and only if there exists an s S such that sM = 0. ∈ (ii) Let I be an ideal in R, and let S =1+ I = r R : r 1 mod I . { ∈ 1 ≡ } Show that S is a multiplicative subset of R and that S− I is contained 1 in the Jacobson radical of S− R. (iii) Using (i) and (ii) and Nakayama’s lemma, give a direct proof of the fact that, if M is a finitely generated R-module and M = IM, then there exists an r R such that r 1 mod I and rM = 0. (This also follows quite easily∈ from the Cayley-Hamilton≡ theorem and was used to give one of the proofs of Nakayama’s lemma.) Exercise 3.6. Let S be a multiplicative subset of R. Show that the image 1 of Spec S− R in Spec R is f S(Spec R)f . Is this always an open subset of Spec R? ∈ T Exercise 3.7. (i) Let R be an integral domain, with quotient field K, and 1 let S be a multiplicative subset of R not containing 0. Show that S− R K in an obvious way. ⊆

(ii) Using (i), for each maximal ideal m of R, we can view Rm as a subring of K. Show that R = Rm. m maxSpec R ∈ \ (Clearly R is included in the intersection. Now localize at every maximal ideal n.) 3.9. ZARISKI LOCAL PROPERTIES 129

Exercise 3.8. Consider the subring k[x, y][x/y] of the field of rational functions k(x, y). Show that k[x, y][x/y] is not a localization of k[x, y]. In fact, show that if 1/f k[x, y][x/y] for some f k[x, y], then f k∗. ∈ ∈ ∈ Note that k[x, y][x/y] is contained in the subring k[x, y][1/y] of k(x, y), which is a localization of k[x, y], and that both rings are finitely generated k- algebras. Discuss the geometry corresponding to the sequence of morphisms Spec k[x, y][1/y] Spec k[x, y][x/y] Spec k[x, y]. → → Exercise 3.9. Let R be a PID with quotient field K. Suppose that S is a subring of K containing R and that S is a finitely generated R-algebra. Show that S is a localization R for some f R. (Suppose that S = R[t ,...,t ] f ∈ 1 n with ti = ri/si. We may assume that ri and si are relatively prime. Show that 1/s S.) i ∈ Exercise 3.10. Let f : R S be a ring homomorphism, let p Spec R, → ∈ and let S be the localization of S at the multiplicative set f(R p). Show p − that k(p) R S = Sp/pSp, and hence that a prime ideal in k(p) R S may be identified⊗ with a prime ideal in S containing pS and disjoint from⊗ f(R p). 1 − Conclude that (f ∗)− (p) = Spec(k(p) R S). In particular, the fiber over p is ⊗ 1 itself an affine scheme. Show that (f ∗)− (p)= if and only if k(p) S = 0. ∅ ⊗R In the above notation, show that the following are equivalent:

1 (i) (f ∗)− (p) = , i.e. there exists a prime ideal q in S such that q R = p; 6 ∅ ∩ (ii) p =(pS) R; ∩ (iii) k(p) S = 0. ⊗R 6 ((i) = (ii): p (pS) R q R = p. (ii) = (iii): if p = (pS) R, ⇒ ⊆ ∩ ⊆ ∩ ⇒ ∩ then (pS) (R p) = , so that pSp is a proper ideal of Sp. (iii) = (i): by the ﬁrst∩ part− of this∅ exercise.) ⇒

Exercise 3.11. Let f : R S be a ring homomorphism. Show that S is faithfully flat over R, in the→ terminology of Exercise 2.34, if and only if S is flat over R and f ∗ : Spec S Spec R is surjective. (If S is faithfully flat → over R, then for all p Spec R, k(p) S = 0, and we are done by the ∈ ⊗R 6 previous exercise. Conversely, if f ∗ : Spec S Spec R is surjective, show → that mS = S for every maximal ideal m of R.) 6 130 CHAPTER 3. LOCALIZATION

Exercise 3.12. Is the property of being an integral domain a local property? Show that the ring R is reduced if and only if, for all prime ideals p, the ring Rp is reduced, if and only if, for all maximal ideals m, the ring Rm is reduced. Thus being reduced is a local property. Exercise 3.13. Let R be an integral domain and M an R-module. An element m M is a torsion element if there exists an r R with r = 0 such ∈ ∈ 6 that rm = 0. Define the torsion submodule τ(M) to be the set of all torsion elements of M. We say that M is torsion free if τ(M) = 0. Show that τ(M) is a submodule of M and that M/τ(M) is torsion free. Show further that if 0 M ′ M M ′′ is exact, then 0 τ(M ′) τ(M) τ(M ′′) is exact. → → → → → → Is τ an exact functor? Show that the dual M ∨ of an R-module M is always torsion free. Exercise 3.14. Continuing the above exercise, show that τ(M) is the kernel of the map M K R M, where K is the fraction field of R. If M is a finitely generated→ R-module,⊗ show that M is torsion free if and only if M is isomorphic to a submodule of Rn for some n. Again assuming that M is a finitely generated R-module, show that τ(M) = Ker(M M ∨∨). Is this → still true if M is not finitely generated? (What is HomZ(Q, Z)?) Exercise 3.15. Show that, if R is an integral domain, then for every mul- 1 tiplicative subset S of R and every R-module M, we have τ(S− M) = 1 S− τ(M). Thus M is torsion free if and only if Mp is torsion free for all prime ideals p of R, if and only if Mm is torsion free for all maximal ideals m of R. Exercise 3.16. Let R be a Noetherian local domain with maximal ideal m and residue field k. Denote by K the field of quotients of R. Let M be a finitely generated R-module. Show that

dim (M/mM) = dim (M k) dim (M K), k k ⊗R ≥ K ⊗R and moreover that M is free if and only if

dim (M K) = dim (M/mM) = dim (M k). K ⊗R k k ⊗R (Let d = dim (M/mM). Use Nakayama’s lemma to ﬁnd a surjection Rd k → M and hence a surjection Kd M K. If Q is the kernel and d = → ⊗R dimK (M RK), use the fact that K is a ﬂat R-module to show that Q R K = 0, and then⊗ since Rd is torsion free conclude that Q = 0.) ⊗ 3.9. ZARISKI LOCAL PROPERTIES 131

Exercise 3.17. Let f : R R′ be a ring homomorphism, where R and R′, → and let M and N be R-modules. Show that the two R′-modules (M ⊗R N) R R′ and (M R R′) R′ (N R R′) are isomorphic. In the special case ⊗ 1 ⊗ ⊗ ⊗ where R′ = S− R for a multiplicative subset S of R, conclude that

1 1 1 S− (M N) = S− M −1 S− N. ⊗R ∼ ⊗S R Exercise 3.18. (i) If N is an R-module such that, for each s S, the 1 ∈ homomorphism s: N N has an inverse s− , then N is naturally an 1 × → S− R-module. 1 1 (ii) Show that S− M has the following universal property: if N is an S− R- module and f : M N is an R-module homomorphism, then there is a 1 → 1 unique S− R-module homomorphism fˆ: S− M N such that fˆ(m/s) = → f(m)/s, and in fact the homomorphism f fˆ sets up an isomorphism 1 7→ HomR(M, N) ∼= HomS−1R(S− M, N). (This follows from Exercise 2.21 in Chapter 2. Compare also the next exercise.) Exercise 3.19. Let M and N be R-modules and let S be a multiplicative subset of R. Then there is a natural homomorphism from

1 1 1 S− Hom (M, N) Hom −1 (S− M,S− N). R → S R If M is finitely generated, then this homomorphism is injective. If moreover M is finitely presented, then this homomorphism is an isomorphism. (First show that, if M = Rn, then the map is always an isomorphism. Then show, for example if M is finitely presented, by using exactness properties of localization and Hom, that there is a commutative diagram −1 −1 n −1 m 0 S HomR(M,N) S HomR(R ,N) S HomR(R ,N) −−−−→ −−−−→ −−−−→

−1 −1 −1 n −1 −1 m −1 0 HomS−1R(S M,S N) HomS−1R((S R) ,S N) HomS−1R((S  R) ,S N), −−−−→ y −−−−→ y −−−−→ y and use the result for free R-modules, together with a diagram chase.) 1 Show by an example that this homomorphism from S− HomR(M, N) to 1 1 HomS−1R(S− M,S− N) need not be injective if M is not finitely generated. More generally, let T be an R-algebra. Then there is a natural homomorphism Hom (M, N) T Hom (M T, N T ). R ⊗R → T ⊗R ⊗R If T is flat over R, then this homomorphism is injective if M is finitely generated and is an isomorphism if M is finitely presented. 132 CHAPTER 3. LOCALIZATION

Exercise 3.20. constantrank2 Suppose that Rk Rn M 0 is an exact → → → sequence of R-modules, where the homomorphism Rk Rn corresponds to the k n matrix A. Suppose that, for some d n, the→ (d + 1) (d + 1) minors× of A are all zero, and that, for every maximal≤ ideal m of×R, there exists a d d minor of A which does not lie in m. Show that M is projective. (Combine× Exercise 2.23 with Proposition 3.4.4.)

Exercise 3.21. Let R be a ring such that, for all r R, r2 = r, i.e. every element of R is an idempotent. For example, it is easy∈ to see that F [x]/(x2 2 − x) is such a ring, where F2 is the ﬁeld with two elements. Show the following:

(i) For all r R, 2r = 0. ∈ (ii) Every ﬁnitely generated ideal in R is principal. (It is enough to show that (r, s) is principal, generated by an appropriate combination of r and s.)

(iii) If I is a ﬁnitely generated ideal, and hence I =(r) is principal by (ii), then there exists an ideal J such that, as R-modules, R ∼= I J. In particular, every ﬁnitely generated ideal is projective. ⊕

(iv) For every ideal I of R, the quotient R/I is a projective R-module if and only if I is principal and hence if and only if I is ﬁnitely generated.

(v) Every R-module is ﬂat.

(vi) If p is a prime ideal, then R/p ∼= F2. In particular, p is maximal. (vii) If m is a maximal ideal, then the natural homomorphism R R is → m surjective and its kernel is m. In particular, Rm ∼= R/m. (viii) Let I be an ideal of R. Show that, for every maximal ideal m of R, I = R if I m, and that I =0 if I m. In particular, for every m ∼ m 6⊆ m ⊆ maximal ideal m of R, Im and (R/I)m ∼= Rm/Im are ﬁnitely generated and projective Rm-modules.

Exercise 3.22. Continuing the previous exercise, let R = F2[x1, x2,... ] be the polynomial ring in inﬁnitely many variables, subject to the relations x2 = x for all i and x x = 0 for i = j. i i i j 6 3.9. ZARISKI LOCAL PROPERTIES 133

(i) Let I be the maximal ideal (x1, x2,... ). Show that I is not ﬁnitely

generated but that it is projective. (In fact, I = i∞=1(xi), so (iii) of the last exercise applies.) Further show that R/I is ﬁnitely generated but not projective. L

(ii) For every prime (hence maximal) ideal m of R, show that Im and (R/I)m are finitely generated, in fact finitely presented, and projective Rm- modules. (iii) By considering I (R/I), show that none of the following is a local property of an R⊕-module M: M is finitely generated, M is finitely presented, M is projective, M is finitely generated and projective, M is finitely presented and projective. (iv) Show that the maximal ideals of R are exactly the ideal I described above and the principal ideals (1 + xn), i> 0. Exercise 3.23. Is the Noetherian property a local property of rings? Exercise 3.24. Suppose that R is a UFD and that S is a multiplicative 1 subset of R. Show that S− R is a UFD, and that every irreducible element 1 of S− R, up to associates, is of the form r/1 where r R and no multiple of r lies in S. (However, being a UFD is not a local property∈ of a ring R, as we shall see in Chapter 5.) Exercise 3.25. With X = R with the usual topology, which of the following presheaves on X are sheaves? (a) (U) = Z, if U = R, and (U) = 0 otherwise; all restriction homo- Fmorphisms are zero. F (b) (U)=0, if U = R, and (U)= Z otherwise (U = ); all restriction F F 6 ∅ homomorphisms ρUV are the identity, or zero if U = R. (c) (U) is the group of continuous and bounded functions from U to R. F (d) (U) = Z, if 0 U, and is zero otherwise; all restriction homomor- F ∈ phisms ρUV are the identity, if both (U) and (V ) are Z, and are zero otherwise. F F (e) (U)= Z, if 0, 1 U, and is zero otherwise; all restriction homomor- F ∈ phisms ρUV are the identity, if both (U) and (V ) are Z, and are zero otherwise. F F 134 CHAPTER 3. LOCALIZATION

Exercise 3.26. Let M be the sheaf on Spec R associated to the module M. Show that its stalk at a point x X corresponding to p is M . ∈ p f Exercise 3.27. Let R be a graded ring, and let f,g R be homogeneous of ∈ degrees d, e respectively. Let (Rf )0 be the ring of homogeneous elements of degree zero in Rf . Argue carefully that [(Rf )0]gd/f e ∼= (Rfg)0.

Exercise 3.28. Let R be the graded ring k[x0,...,xn] with its usual grading. We use the notation of the proof of Proposition 3.8.2.

(a) Identify the ring (Rx0 )0 with the polynomial ring k[y1,...,yn], where yi = xi/x0. If I is a homogeneous ideal of R, describe the corresponding inh ideal I of k[y1,...,yn]. (b) If q is a prime ideal in k[y ,...,y ] (R ) , what is the corresponding 1 n ∼= x0 0 graded ideal p = q in R?

Compare with the proof of Proposition 1.7.7 in Chapter 1. e Exercise 3.29. Let f : R S be a graded homomorphism of graded rings (i.e. f(R ) S for all n).→ Show that n ⊆ n 1 U = q Proj S : f − (q) does not contain R { ∈ +} is an open subset of Proj S, and that there is an induced morphism of ringed spaces f ∗ : U Proj R, where U is given the ringed space structure U. → OProj S| Chapter 4

Integral Ring Homomorphisms

4.1 Deﬁnition of an integral homomorphism

One attempt to study finite algebraic extensions Q K is to try and reduce ⊆ mod p for various primes p. To do so, we would use Z instead of Q, and then we need to find an appropriate subring o of K playing the role of the subring Z of Q, as well as a prime ideal p in o generalizing the ideal (p) in Z. One motivation for the definition of integral extensions is to try to generalize this idea, to a relative situation f : R S, where f need not be an inclusion. As → we will see, the correct generalization is important in the geometric setting as well. Definition 4.1.1. Let f : R S be a homomorphism, and let R = f(R) S. Then s S is integral over→ R if there exists a monic polynomial p(x) ⊆ R[x] such that∈ p(s) = 0 in S. In other words, there exist r ,...,r R such∈ 1 n ∈ that n n 1 s + f(r )s − + + f(r )=0. 1 ··· n Proposition 4.1.2. Let f : R S be a homomorphism of commutative rings, with R = f(R), and let s →S. Then the following are equivalent: ∈ (i) s is integral over R. (ii) The subring R[s] of S is finitely generated as an R-module, i.e. it is isomorphic to a quotient of Rn for some n. (iii) There exists a subring T of S containing R[s] which is an R-algebra and which is finitely generated as an R-module.

135 136 CHAPTER 4. INTEGRAL RING HOMOMORPHISMS

(iv) There exists an R[s]-module M, which is finitely generated as an R- module, and which is faithful as an R[s]-module, i.e. if x R[s] and xm =0 for all m M, then x =0. ∈ ∈ Proof. (i) = (ii): If p(s) = 0 for some monic polynomial p(x) R[x] of ⇒ n 1 ∈ degree n, then R[s] is generated over R by 1,s,...,s − . (ii) = (iii): Take T = R[s]. ⇒ (iii) = (iv): Take M = T , which by assumption is a finitely generated R-module⇒ and contains 1. Then x 1=0 = x = 0, so that T is faithful. · ⇒ (iv) = (i): By the Cayley-Hamilton theorem (Theorem 2.7.6) applied ⇒ to the R-module M, the ideal I = R, and the homomorphism M M given by multiplication by s, there exists a monic polynomial q(x) →R[x] such that q(s) M = 0. Since M is a faithful R[s]-module, q(s) = 0,∈ where · q(x)= f(q(x)). Thus s is integral over R. Corollary 4.1.3. Let f : R S be a homomorphism of commutative rings → and let s1,...,sn S be integral over R. Then the subring f(R)[s1,...,sn] of S is a finitely generated∈ R-module. Proof. By induction on n, the case n = 1 being dealt with in (i) = (ii) ⇒ of Proposition 4.1.2 above. Write f(R)[s1,...,sn] = f(R)[s1,...,sn 1][sn]. − By induction, f(R)[s1,...,sn 1] is a finitely generated R-module, say by − m1,...,mk. Since sn is integral over R, it is integral over f(R)[s1,...,sn 1], − so that f(R)[s1,...,sn] is generated over f(R)[s1,...,sn 1] by n1,...,nℓ. It − is then routine to see that f(R)[s1,...,sn] is generated over R by minj. Corollary 4.1.4. Let f : R S be a homomorphism of commutative rings. Then s S : s is integral over→ R is a subring of S. { ∈ } Proof. If s, t S are integral over R, then so are all elements of R[s, t] and ∈ in particular s t, st are integral over R. So the set of elements of S which are integral over± R is closed under the ring operations. Definition 4.1.5. If f : R S is a homomorphism of commutative rings, the subring s S : s is integral→ over R is called the integral closure of R { ∈ } in S. The ring S is integral over R if every element of S is integral over R, in other words if the integral closure of R in S is S. If R is a subring of S and f is the inclusion, we will call S an integral extension of R. The ring R is integrally closed in S if the integral closure of R is f(R). If R is an integral domain, then we say that R is integrally closed if it is integrally closed in its quotient field. 4.1. DEFINITION OF AN INTEGRAL HOMOMORPHISM 137

Corollary 4.1.6. Let f : R S and g : S T be homomorphisms of com- → → mutative rings, and suppose that S is integral over R and that T is integral over S. Then, via g f, T is integral over R. ◦ n n 1 Proof. Let t T . There exist s1,...,sn S such that t + g(s1)t − + + g(s ) =∈ 0. Now the subring f(R)[s∈,...,s ] of S is a finitely gen- ··· n 1 n erated R-module, and since t is integral over f(R)[s1,...,sn], the subring g f(R)[s ,...,s , t] of T is a finitely generated f(R)[s ,...,s ]-module ◦ 1 n 1 n (here of course we identify the si with their images in T ). It follows as in the proof of Corollary 4.1.3 that g f(R)[s1,...,sn, t] is a finitely generated R-module. ◦ Corollary 4.1.7. Let f : R S be a homomorphism of commutative rings → and let R be the set of elements of S which are integral over R. Then R is integrally closed in S. e e Next we give some examples. 1. The ring R is integral over itself via the identity homomorphism. 2. Let f : R S be a homomorphism of commutative rings, and suppose → that S is integral over R. If I is an ideal of of R and J is an ideal of S such that f(I) J, then S/J is integral over R/I. In particular, applying this to R⊆= S, f = Id and I = 0, the quotient homomorphism R R/J is an integral homomorphism. → 3. If R is a UFD and K is its quotient field, then R is integrally closed in K. The proof is the same as the argument that the integers are integrally closed in Q: suppose that p/q K is integral over R, where p, q R are relatively prime. Clearing denominators∈ in a relation of the form∈ n n 1 p p − + r + + r =0, q 1 q ··· n where the r R, we find that q pn and thus q is a unit. Hence p/q R. i ∈ | ∈ 4. The subring Z[i] of Q(i) is integral over Z, since i is integral over Z (it is a root of the monic polynomial x2 + 1). In fact, as we shall see, the integral closure of Z in Q(i) is exactly Z[i]. On the other hand, the ring Z[2i] is integral over Z, but it is not integrally closed in Q(i). For a related example, the integral closure of Z in Q(√ 3) contains Z[√ 3], but is strictly larger, since 1 (1+ √ 3) is a root of− x2 x + 1. − 2 − − 138 CHAPTER 4. INTEGRAL RING HOMOMORPHISMS

5. Let k be a field and let x, y be algebraically independent over k. Then k[x, y]/(y2 x3)= R is easily checked to be an integral domain (y2 x3 is irreducible).− It is not integrally closed in its quotient field however:− if t = y/x, then t / R (why?) but t2 = y2/x2 = x3/x2 = x, so that t ∈ satisfies t2 x = 0. In particular, the quotient of an integral domain which is integrally− closed by a prime ideal need not be integrally closed.

We have the following straightforward lemma which relates integral elements in localizations:

Lemma 4.1.8. Let f : R S be a homomorphism, and let M be a multiplicative subset of R. →

1 (i) For all s S and m M, the element s/f(m) (f(M))− S is integral 1∈ ∈ ∈ over M − R if and only if there exists an n M such that f(n)s is integral over R. ∈

1 1 (ii) If S is integral over R, then (f(M))− S is integral over M − R.

1 (iii) If R is the integral closure of R in S, then the integral closure of M − R 1 1 in (f(M))− S is (f(M))− R. e Proof. To see (i), identify an element of R with its image in S and an element 1 1e 1 of M − R with its image in M − S. Suppose that s/m M − S is integral 1 ∈ over M − R. There exists a monic equation

s k r s k 1 r + 1 − + + k =0. m m1 m ··· mk Multiplying by mk(m m )k gives a monic equation for (m m )s with 1 ··· k 1 ··· k coeﬃcients in R. Thus (m1 mk)s is integral over R. Conversely, if ns is integral over R, then there is··· some monic equation

k k 1 (ns) + r (ns) − + + r =0. 1 ··· k k 1 Dividing by (nm) in M − S gives a monic equation for s/m with coeﬃcients 1 1 in M − R. Thus s/m is integral over M − R. Given (i), (ii) and (iii) are immediate.

Corollary 4.1.9. Let R be an integral domain. Then the following are equivalent: 4.1. DEFINITION OF AN INTEGRAL HOMOMORPHISM 139

(i) R is integrally closed.

(ii) For all prime ideals p of R, Rp is integrally closed.

(iii) For all maximal ideals m of R, Rm is integrally closed.

Proof. Let K be the quotient field of R and let R be the integral closure of R in K. Then R is integrally closed if and only if the natural map R R is → surjective, if and only if for every prime ideal p ofe R, the map R (R) is p → p surjective. Now the previous lemma implies that (R)p is the integral closuree R of R in K, and the map R R is surjective if and only if eR is p p p → p p integrally closed. Thus R is integrally closed if ande only if, for all prime idealsf p of R, Rp is integrally closed,f and a similar argument handles the case of maximal ideals. One of the main reasons to be interested in integral extensions is to study how integers behave in finite field extensions. Lemma 4.1.10. Let R be an integral domain with quotient field k, and let K be an algebraic extension of k. For all α K, there exists an r R,r =0 such that rα is integral over R. ∈ ∈ 6 Proof. For α K, irr(α,k,x) is a monic irreducible polynomial with coefficients in k, and∈ so after multiplying by a suitable element of R, there exists n an irreducible polynomial f(x) R[x] with f(α) = 0, say a0α + +an = 0, ∈ n 1 ··· where we may assume that a = 0. Multiplying by a − , we see that a α is 0 6 0 0 the root of a monic polynomial with coefficients in R. Thus a0α is integral over R. Lemma 4.1.11. Let R be an integral domain, integrally closed in its quotient field k. Let K be an algebraic extension of k and let α K. Then α is integral ∈ over R if and only if irr(α,k,x) R[x]. ∈ Proof. If irr(α,k,x) R[x], then α is integral over R since irr(α,k,x) is monic. Conversely, let∈ α be integral over R. After enlarging K, we can assume that K is normal over k. Let σ1,...,σn be the distinct embeddings of k(α) in K. If f(x) R[x] is a monic polynomial such that f(α) = 0, ∈ f(σi(α)) = 0 as well, and so σi(α) is integral over R. Then

N irr(α,k,x)= (x σ (α))p − i i Y 140 CHAPTER 4. INTEGRAL RING HOMOMORPHISMS for some N (if k has characteristic p, otherwise we omit the exponent). Thus the coefficients of irr(α,k,x), which by definition lie in k, are integral over R, since they lie in k[σ1(α),...,σn(α)]. Since R is integrally closed in k, the coefficients of irr(α,k,x) lie in R. Corollary 4.1.12. Suppose that R is an integral domain, integrally closed in its quotient field k, and that K is a finite separable extension of k. Let R be the integral closure of R in K. Then there exists a basis α1,...,αn of K over k such that R Rα . In other words, R is an R-submodule of a ⊆ i i finitelye generated R-submodule of K. In particular, if R is Noetherian, then P R is a finitely generatede R-module. e Proof. Choose an arbitrary basis β ,...,β for K over k. After multipliying e 1 n by suitable elements of R, we can assume that βi R for every i. Let Tr: K k be the trace map. Since α, β = Tr(αβ∈) is a nondegenerate → h i pairing, there exists a dual basis α1,...,αn, in other wordse Tr(αiβj)= δij for all i, j. Given γ R, write γ = c α for c k. Then γβ = c α β ∈ j j j j ∈ i j j j i and so Tr(γβ ) = c . On the other hand, γβ R since R is a subring, and i i P i ∈ P thus as in the precedinge proof ci = Tr(γβi) is integral over R and lies in k. It follows that c R for all i and thus that R e Rα .e i ∈ ⊆ i i For example, in case R is a PID, say Z,e thenPR is a finitely generated, free R-module, necessarily of rank equal to [K : k]. e 4.2 The going up theorem and dimension

Theorem 4.2.1 (Going up theorem). Let R be a subring of S and suppose that S is integral over R. Suppose that p is a prime ideal of R. Then there exists a prime ideal q of S such that q R = p. Moreover, if q′ is another ∩ prime ideal of S with q′ q and q′ R = q R = p, then q′ = q. ⊆ ∩ ∩ Corollary 4.2.2. With R, S as above, suppose that p p p is a 1 ⊆ 2 ⊆···⊆ n sequence of prime ideals in R, and that, for m < n, q q q is a 1 ⊆ 2 ⊆···⊆ m sequence of prime ideals in S, with qi R = pi for i m. Then there exist prime ideals q q , with q ∩ q , such that≤ q R = p for all m+1 ⊆···⊆ n m ⊆ m+1 i ∩ i i n. ≤ Proof. It suﬃces to consider the case m = 1, n = 2. Then the natural map R/p S/q is an inclusion, and S/q is integral over R/p . Since R R/p 1 → 1 1 1 → 1 4.2. THE GOING UP THEOREM AND DIMENSION 141

is surjective, the image p2 of p2 in R/p1 is prime. Applying Theorem 4.2.1 to the homomorphism R/p1 S/q1, there exists a prime ideal q2 S/q1 such that q (R/p )= p . Taking→ q to be the inverse image of q in⊆S gives the 2 ∩ 1 2 2 2 required prime ideal q2.

To prove the going up theorem, we begin with a series of lemmas:

Lemma 4.2.3. Let R and S be integral domains with R S and suppose ⊆ that S is integral over R. Then R is a ﬁeld if and only if S is a ﬁeld.

Proof. If R is a field and S is integral over R, let s S,s = 0. Then R[s] S ∈ 6 ⊆ is a finitely generated R-module and thus is a finite dimensional vector space over the field R. But an integral domain which is finite dimensional over a field is again a field. Thus R[s] is a field. It follows that s has an inverse in R[s] and so in S, so that S is a field. Conversely, suppose that S is a field. 1 Given r R,r = 0, r has an inverse r− S which is integral over R. Thus ∈ 6 ∈ there exist a R such that i ∈ 1 n 1 n 1 (r− ) + a (r− ) − + + a =0. 1 ··· n n 1 Multiplying the above by r − , we find that

1 n 1 r− = (a + + a r − ) R. − 1 ··· n ∈ Thus R is a ﬁeld as well.

Lemma 4.2.4. Let S be a ring and let R be a subring of S such that S is integral over R. If q is a prime ideal in S, then q is maximal if and only if p = q R is maximal in R. ∩ Proof. Apply the previous lemma to R/p S/q. ⊆ Lemma 4.2.5. Let S be an integral domain which is integral over the subring R. Suppose that I is an ideal of S such that I R = (0). Then I =0. ∩ Proof. Let α I and suppose that α = 0. Since α is integral over R, there ∈ 6 n n 1 exist n> 0 and a0,...,an 1 R such that α + an 1α − + + a0 = 0. We − − can suppose that n is chosen∈ as small as possible. Now ···

n n 1 a0 = (α + an 1α − + + a1α) R I, − − ··· ∈ ∩ 142 CHAPTER 4. INTEGRAL RING HOMOMORPHISMS

since I is an ideal, so that a0 = 0. Thus

n n 1 n 1 n 2 α + an 1α − + + a1α =0= α(α − + an 1α − + + a1). − ··· − ··· n 1 n 2 Since R is an integral domain and α = 0, α − + an 1α − + + a1 = 0. − But this contradicts the choice of n. Thus6 I = (0). ··· Corollary 4.2.6. Let S be a ring and let R be a subring of S such that S is integral over R. Suppose that q q are two prime ideals in S such that 1 ⊆ 2 q R = q R = p, say. Then q = q . 1 ∩ 2 ∩ 1 2 Proof. Apply the previous lemma to the integral extension S/q1 of R/p and the ideal in S/q1 which is the image of q2. Proof of the going up theorem. We have already proved the last statement of Theorem 4.2.1. If R is a subring of S, with S integral over R, and p is a prime ideal of R, consider the inclusion Rp Sp. Then Sp is integral over Rp, by Lemma 4.1.8. We summarize the rings⊆ involved in the following diagram (here j and jp are the natural inclusions):

j R p S p −−−→ p

iR iS x j x R S.  −−−→  1 Let m be a maximal ideal of Sp. Then, by Lemma 4.2.4, m Rp = jp− (m) ∩ 1 is a maximal ideal in Rp, necessarily equal to pRp. Set q = iS− (m). Then 1 1 1 1 1 1 q R = j− (q)= j− (i− (m)) = i− (j− (m)) = i− (pR )) = p. ∩ S R p R p We now give some applications of the going up theorem. The ﬁrst is a consequence for the induced map on spectra: Proposition 4.2.7. Let f : R S be an integral ring homomorphism. Then → the induced function f ∗ : Spec S Spec R is closed, i.e. for every closed → subset Z of Spec S, f ∗(Z) is a closed subset of Spec R. Proof. First assume that f is injective, so that R is identiﬁed with a subring of S, and that Z = Spec S. In this case, the going up theorem implies that f ∗ is surjective, and hence that f ∗(Spec S) = Spec R and in particular is closed. In the general case, suppose that Z = V (J) is a closed subset of Spec S, where J is an ideal in S. Then Z is the image of Spec S/J via the natural 4.2. THE GOING UP THEOREM AND DIMENSION 143

homomorphism S S/J, and f ∗ Z is the map Spec S/J Spec R induced → | → f π 1 by the composition π f : R S S/J. Let I = f − (J). Then π f = g π′, ◦ −→ −→ ◦ ◦ where π′ : R R/I is the quotient homomorphism and g : R/I S/J is the induced injection.→ Moreover, g is integral by (2) after Corollary→ 4.1.7. Then, by the ﬁrst part of the proof, the map Spec S/J Spec R/I is surjective. It → follows that f ∗(Z) is the image of Spec R/I in Spec R, and hence is a closed subset of Spec R.

Warning: If S is an integral extension of R, then the map Spec S Spec R is closed and surjective. It need not, however, be open (Exercise 4.20).→

Remark 4.2.8. We are most interested in Proposition 4.2.7 in the geometric setting: suppose that ϕ: X Y is a morphism of affine algebraic sets, induced from a homomorphism→ of k-algebras f : A(Y ) A(X). We shall → say that ϕ is integral or finite if f is an integral homomorphism. In this case, ϕ is a closed map with finite fibers, by Exercise 4.10.

A second geometric consequence of the going up theorem is as follows:

Definition 4.2.9. Let R be a commutative ring. A chain of prime ideals in R is a strictly increasing sequence p p p of prime ideals p in 0 ⊂ 1 ⊂···⊂ n i R. With a chain as above, the length of the chain is n. The Krull dimension dim R of R is the largest nonnegative integer n such that there exists a chain of length n (or dim R = if there is no bound on the length of a chain). ∞ Example 4.2.10. If R is an integral domain, then dim R = 0 if and only if 0 is a maximal ideal, if and only if R is a field. If R is a PID which is not a field, then dim R = 1, by Lemma 1.4.16. In particular, dim Z = 1 and dim k[x] = 1, where k is a field. Clearly, dim k[x ,...,x ] n, by considering 1 n ≥ the chain pi =(x1,...,xi). More generally, dim R[x] dim R + 1. However, for a general ring R, the best one has in this direction≥ is

dim R +1 dim R[x] 2 dim R +1. ≤ ≤ If R is Noetherian, then in fact dim R[x] = dim R + 1. We shall see directly that dim k[x1,...,xn]= n.

Corollary 4.2.11. If R is a subring of S such that S is integral over R, then dim S = dim R. 144 CHAPTER 4. INTEGRAL RING HOMOMORPHISMS

Proof. If p p p is a chain of prime ideals in R, then by the going 0 ⊂ 1 ⊂···⊂ n up theorem there exists a chain q q q of prime ideals in S with 0 ⊂ 1 ⊂···⊂ n qi R = pi. Thus dim S dim R. On the other hand, if q0 q1 qn is∩ a chain of prime ideals≥ in S, then by the last statement⊂ of⊂···⊂ the going up theorem, if we set pi = qi R, then pi+1 properly contains pi. Thus p p p is a chain of∩ prime ideals in R, and dim R dim S. It 0 ⊂ 1 ⊂···⊂ n ≥ follows that dim R = dim S.

4.3 Integrally closed domains and the going down theorem

The going down theorem is in some sense dual to the going up theorem. However, it requires stronger hypotheses: R needs to be an integral domain, integrally closed in its quotient ﬁeld K. First, we state a preliminary result, concerning the Galois theory of integral extensions.

Proposition 4.3.1. Let R be an integral domain, integrally closed in its quotient ﬁeld K. Let L be a ﬁnite normal extension of K, and let S be a subring of L which is integral over R. Further assume that, for all σ Aut(K/k), σ(S) = S. For example, by Lemma 4.1.11, if S is the integral∈ closure of R in K, then σ(S) = S for all σ Aut(K/k). If p is a prime ∈ ideal of R, then Aut(K/k) operates transitively on the set of ideals q S such that q R = p. ∈ ∩ Proof. First we claim that we can reduce to the case where p and q are maximal, as follows. Localize R and S at p. In this case, pRp is a maximal ideal of Rp. The prime ideals Q of Sp such that Q Rp = pRp are in one-to- one correspondence with the prime ideals q of S such∩ that q (R p) = and such that (qS ) R = pR . In this case (qS ) R R∩= p,− but also∅ p ∩ p p p ∩ p ∩ (qS ) S = q, so that p ∩ q R =(qS ) S R =(qS ) R ∩ p ∩ ∩ p ∩ =(qS ) R R = p. p ∩ p ∩ Conversely if q R = p and we set q = qS , then it is easy to check that ∩ p q Rp = pRp. In other words there is a one-to-one correspondence between prime∩ ideals Q of S such that Q R = pR and prime ideals q of S such that p ∩ p p 4.3. THE GOING DOWN THEOREM 145 q R = p, and this correspondence clearly preserves the action of Aut(L/K). ∩ So we can assume throughout that p is maximal. Suppose then that p is a maximal ideal of R and that q, q′ are two prime ideals of S such that q R = q′ R = p. By the going up theorem, both ∩ ∩ q and q′ are maximal in S. Supppose that q′ is not of the form σ(q) for some σ Aut(L/K). By the Chinese remainder theorem (Exercise 1.7), ∈ since the σ(q) and q′ are comaximal, there exists an s S such that s 1 1 ∈ ≡ mod σ− (q) for all σ Aut(L/K), but s 0 mod q′. Now by the properties of the norm ∈ ≡ pN r = N (s)= σ(s) K. L/K ∈ σ ! Y As r is integral over R and in K, r R. By construction, s 0 mod q′, ∈ ≡ and so, using σ = Id, r 0 mod q′, so that r q′ R = p. But also ≡ ∈ ∩ σ(s) 1 mod q for all σ, so that r 1 mod q, and thus r / p. This is a ≡ ≡ ∈ contradiction. So q′ = σ(q) for some σ Aut(L/K). ∈ Theorem 4.3.2 (Going down theorem). Let R be an integrally closed domain and let S be an integral domain containing R as a subring which is integral over R. Let p p be two prime ideals in R, and let q be a prime ideal in 1 ⊇ 2 1 S such that q1 R = p1. Then there exists a prime ideal q2 in S such that q q and q ∩ R = p . 1 ⊇ 2 2 ∩ 2 Corollary 4.3.3. With R and S as above, suppose that p p p 1 ⊇ 2 ⊇···⊇ n is a chain of prime ideals in R, and that q q q is a chain of 1 ⊇ 2 ⊇···⊇ m prime ideals in S, with m < n, such that qi R = pi for i m. Then there exist prime ideals q q with q ∩ q , such that≤ q R = p for m+1 ⊇···⊇ n m+1 ⊆ m i ∩ i all i.

Proof of Theorem 4.3.2. Let k be the quotient field of R and let K be the quotient field of S. Note that K is an algebraic extension of k. We suppose first that K is a finite normal extension of k and that S is the integral closure of R in K. By the going up theorem, there exist prime ideals q′ q′ in S such 2 ⊆ 1 that q′ R = p . Now q R = q′ R = p . Thus there is a σ Aut(K/k) i ∩ i 1 ∩ 1 ∩ 1 ∈ such that σ(q1′ )= q1. Set q2 = σ(q2′ ), so that p2 = σ(p2)= σ(q2′ R)= q2 R. Thus q q are as desired. ∩ ∩ 1 ⊇ 2 Next assume that K is a finite extension of k, not necessarily normal. Let F be a splitting field for K over k, and let S′ be the integral closure of S in F . Thus S S′. Choose an ideal q′ in S′ such that q′ S = q , which ⊆ 1 1 ∩ 1 146 CHAPTER 4. INTEGRAL RING HOMOMORPHISMS

is again possible by the going up theorem since S′ is integral over S. Using the above, we can ﬁnd q′ q′ in S′ such that q′ R = q , and then we can 2 ⊆ 1 2 ∩ 2 take q = q′ S. 2 2 ∩ In the general case, we may assume by an argument similar to that above that S is integrally closed in K. Consider the set of all pairs (T, I), where T is a subring of S containing R, integrally closed in its quotient ﬁeld, and I is a prime ideal of T such that I q1 T and I R = p2. We order two elements (T , I ) and (T , I ) in the obvious⊆ ∩ way: (T ∩, I ) (T , I ) if T T 1 1 2 2 1 1 ≤ 2 2 1 ⊆ 2 and I I . If (T , I ) is a chain, then it is routine to check that 1 ⊆ 2 { α α }

Tα, Iα α α ! [ [ is an upper bound. Thus there is a maximal element (T0, I0). If K0 is the quotient ﬁeld of T , then K is an algebraic extension of T . If K = K , we 0 0 6 0 can make a ﬁnite extension of K0 and argue as in the previous case that (T0, I0) is not a maximal element. Thus K = K0 and so T0 = S, and we are done.

Remark 4.3.4. If R S satisfy the hypotheses of the going down theorem, and moreover R is Noetherian⊆ and S is a ﬁnitely generated R-algebra, then one can show that Spec S Spec R is an open map. →

4.4 More on dimension

If dim R = n, then there exists a maximal chain of prime ideals in R of length n, and every maximal such chain has length at most n. However, there may exist maximal chains of length less than n (Exercises 4.14 and 4.15). Thus, we may sometimes need to focus attention on chains containing a speciﬁc prime ideal p.

Definition 4.4.1. Let p be a prime ideal in R. Then the height of p, written ht p, is the maximum length k of a chain of strictly increasing prime ideals p p = p whose final member is p. Clearly, ht p = dim R . 0 ⊂···⊂ k p The coheight of p, written coht p, is the maximum length k of a chain of strictly increasing prime ideals p0 = p pk whose first member is p. Clearly, coht p = dim R/p. ⊂···⊂ 4.4. MORE ON DIMENSION 147

For an arbitrary ideal I of R, we let the height of I be the smallest height of a prime ideal containing I: ht I = min ht p : I p , and deﬁne the coheight of I to be the largest coheight of a prime{ ideal⊆ containing} I, so that coht I = dim R/I.

Clearly, ht p+coht p is the maximal length of a chain of prime ideals in R containing p. More generally, ht I + coht I is the maximal length of a chain of prime ideals in R, one of which contains I. Thus:

Lemma 4.4.2. For every ideal I of R, ht I + coht I dim R. In particular, dim R/I dim R. ≤ ≤ To relate these deﬁnitions to the going up and going down theorems, we have:

Lemma 4.4.3. Let S be a ring and R a subring of S such that S is integral over R. Let q be a prime ideal of S and let p = q R. Then ht q ht p and ∩ ≤ coht q = coht p. If in addition R is an integrally closed integral domain, then ht q = ht p.

Proof. Let q q = q be a chain of prime ideals in S of length 0 ⊂ ··· ⊂ k k = ht q. By the going up theorem, if pi = qi R, then the pi are distinct. It follows that p p = p is a chain of prime∩ ideals terminating in p, 0 ⊂···⊂ k so that ht p k = ht q. Moreover, R/p S/q is an integral extension, so that coht p =≥ dim R/p = dim S/q = coht q→. If in addition R and S satisfy the hypotheses of the going down theorem, then every chain p0 pk = p of prime ideals terminating in p lifts to a chain q q⊂···⊂= q. Thus ht q ht p, and so ht q = ht p. 0 ⊂···⊂ k ≤ An example of an integral ring extension as above where ht q = ht p is given in Exercise 4.19. 6 Let us turn now to the question of when all chains have the same length:

Lemma 4.4.4. For a ring R of dimension n, the following are equivalent:

(i) Every maximal chain of prime ideals in R has length n.

(ii) For every prime ideal p of R, ht p + coht p = n.

Proof. Clearly, the length of a maximal chain of prime ideals containing p is ht p + coht p. From this, the proof of the lemma is clear. 148 CHAPTER 4. INTEGRAL RING HOMOMORPHISMS

The property that every maximal chain of prime ideals has the same length is not stable under localization (as follows from Exercise 4.14, for example). Here is a related property, which is stable under localization:

Deﬁnition 4.4.5. The ring R is catenary if, for every pair of prime ideals p, q in R with p q, every maximal chain ⊆ p = p p = q 0 ⊂···⊂ k has the same length.

Lemma 4.4.6. If every maximal chain of prime ideals in R has length n, then R is catenary.

Proof. If p0 = p pk = q is any maximal chain, then clearly ht p + k + coht q is the length⊂···⊂ of a maximal chain in R. Thus k = n ht p coht q is − − the same for every chain.

At the end of the chapter, we shall show that, if R is an integral domain which is a finitely generated k-algebra for some field k, then every maximal chain of prime ideals in R has the same length, and so, for every prime ideal p in R, ht p + coht p = dim R. In particular, every localization of a quotient of k[x1,...,xn] is catenary. There is also the following geometric interpretation of height and coheight. Suppose that R = A(X) is the affine coordinate ring of the affine algebraic variety X, and that Z is a closed subvariety of X; thus both A(X) and A(Z) are integral domains. We let dim X = dim A(X). n As we shall see later, dim Ak = n, lending at least some plausibility to this definition. If Z = V (p), where p is a prime ideal in A(X), then

coht p = dim A(X)/p = dim A(Z) = dim Z.

Thus, ht p = dim A(X) coht p = dim X dim Z, which by deﬁnition is the − − codimension of Z in X.

4.5 Noether normalization, version 1

There are many versions of the Noether normalization lemma. We prove one of them here and use it to deduce the Nullstellensatz. A related version will be proved in the next section. 4.5. NOETHER NORMALIZATION, VERSION 1 149

We begin by recalling some standard terminology. Let K be an extension field of the field k. The elements α ,...,α K are algebraically inde- 1 r ∈ pendent over k if the homomorphism evα1,...,αn : k[x1,...,xr] K defined by x α is injective. In this case there is an extension of→ ev to i 7→ i α1,...,αn a homomorphism k(x1,...,xr) K. The elements α1,...,αr K are a transcendence basis for K over →k if they are algebraically independent∈ over k and if K is an algebraic extension of k(α1,...,αr). In this case, every two transcendence bases for K over k have the same number of elements, and this number is called the transcendence degree tradeg(K/k) of K over k. If R is a k-algebra which is an integral domain and α ,...,α R, then we 1 r ∈ will say that α1,...,αr are algebraically independent over k or that they are a a transcendence basis for R over k if the corresponding statements hold in the quotient field K of R, and we will let tradeg(R/k) = tradeg(K/k).

Theorem 4.5.1 (Noether normalization lemma, version 1). Let R = k[y1,...,yN ]/p be an integral domain, finitely generated over a field k. Then there exist x ,...,x R, algebraically independent over k, such that R is integral over 1 d ∈ the subring k[x1,...,xd]. We will just prove the theorem under the assumption that k is infinite and sketch an argument for the case of a finite field at the end. In case k is infinite, the proof shows that we can take x1,...,xd to be images of k-linear combinations of y1,...,yN ; in fact, almost all such choices will satisfy the conclusion of the theorem. Geometrically, this has the following corollary: Corollary 4.5.2. Let k be an algebraically closed field and let X be an affine n algebraic variety in Ak . Then there exists a nonnegative integer d and a n d linear projection π : Ak Ak such that π∗ induces an integral extension k[x ,...,x ] A(X). In→ particular, d = dim X and π X : X Ad is a 1 d → | → k closed morphism with finite fibers, by Remark 4.2.8. There is also a projective analogue of the above, which we shall discuss later. Proof of the normalization lemma. The proof is by induction on N, the number of generators for R as a k-algebra. The case N = 0 is clear. So assume inductively that the theorem has been proved for all k-algebras R which can be generated by less than N elements. Suppose that R = k[α1,...,αN ]. If the αi are algebraically independent over k, we can simply take xi = αi. Thus we may assume that the αi are not algebraically independent over k. 150 CHAPTER 4. INTEGRAL RING HOMOMORPHISMS

We shall show that, provided k is infinite, there exist α1′ ,...,αN′ 1 which − are k-linear combinations of α1,...,αN , such that R = k[α1′ ,...,αN′ 1, αN ] − and αN is integral over k[α1′ ,...,αN′ 1]. Then by induction there exist − x1,...,xd k[α1′ ,...,αN′ 1], algebraically independent over k, such that ∈ − k[α1′ ,...,αN′ 1] is integral over k[x1,...,xd]. By Corollary 4.1.6, since R = − k[α1′ ,...,αN′ 1, αN ] is integral over k[α1′ ,...,αN′ 1] and k[α1′ ,...,αN′ 1] is in- − − − tegral over k[x1,...,xd], R is integral over k[x1,...,xd]. Thus, assume that the αi are not algebraically independent over k. Then there is a polynomial P (t1,...,tN ) k[t1,...,tN ], not identically zero, such ∈ D that P (α1,...,αN ) = 0. Write P = ν=0 Pν, where each Pν is homogeneous of degree ν and PD = 0. Since k is infinite, there exist λ1,...,λN k such 6 P ∈ that PD(λ1,...,λN ) = 0. In particular, if p(t) = PD(λ1,...,λN 1, t), then − p(λ ) = 0, so that 6 p(t) is not the zero polynomial, and hence p(t)=0 N 6 for at most finitely many t. Since k is infinite, there exists a λ′ = 0 such N 6 that p(λ′ ) = P (λ ,...,λ′ ) = 0. Since P is homogeneous, we can, after N D 1 N 6 D replacing λi by λi/λN′ , assume that PD(λ1,...,λN 1, 1) = 0. Now let αi′ = − 6 αi λiαN for i < N, so that αi = αi′ +λiαN . Clearly R = k[α1′ ,...,αN′ 1, αN ]. Now− − D

0= P (α1,...,αN )= Pν(α1,...,αN ). ν=0 X The leading term is

PD(α1,...,αN )= PD(α1′ + λ1αN ,...,αN′ 1 + λN 1αN , αN ). − − D Expanding this out in powers of αN gives PD(λ1,...,λN 1, 1)αN + lower or- − der terms in αN (with coefficients in k[α1′ ,...,αN′ 1]). Likewise, for ν < D, − Pν(α1′ + λ1αN ,...,αN′ 1 + λN 1αN , αN ) is a polynomial in αN with coef- − − ficients in k[α1′ ,...,αN′ 1] of degree at most ν < D. Thus expanding out − 0 = P (α1,...,αN ) in terms of αN and the αi′, i < N, gives a polynomial of degree D in αN with coefficients in k[α1′ ,...,αN′ 1], whose leading coefficient − PD(λ1,...,λN 1, 1) lies in k∗. Dividing by this leading coefficient gives a − monic polynomial for αN with coefficients in k[α1′ ,...,αN′ 1]. It follows that − αN is integral over k[α1′ ,...,αN′ 1], as claimed. − In case k is finite, the overall structure of the proof is the same, but di instead of choosing α′ = α λ α , one takes α′ = α + α for large degrees i i − i N i i N di. This is worked out in the course of the proof of the second version of the normalization lemma below. 4.5. NOETHER NORMALIZATION, VERSION 1 151

The most important consequence for us of the normalization lemma is a proof of the Nullstellensatz. We begin with the following lemma:

Lemma 4.5.3. Let k be a field and let K be an extension field which is finitely generated as a k-algebra. Then K is a finite extension of k. In particular, if k is algebraically closed, R is a finitely generated k-algebra and m is a maximal ideal of R, then the natural inclusion k R/m is an isomorphism. → Proof. By the normalization lemma, there exist t ,...,t K, algebraically 1 r ∈ independent over k, such that, as a ring, K is integral over k[t1,...,tr]. On the other hand, since K is a field, it follows that k[t1,...,tr] is a field as well, by Lemma 4.2.3, and so r = 0. But then K is an algebraic extension of k. Since K is also finitely generated as a k-algebra, it is a finite extension of k. If k is algebraically closed, then K must equal k. Thus, for R a finitely generated k-algebra and m a maximal ideal of R, the map k R/m is an → isomorphism.

Theorem 4.5.4 (Hilbert Nullstellensatz). Let k be an algebraically closed field and let m be a maximal ideal in k[x1,...,xn]. Then m =(x1 a1,...,xn a ) for some a ,...,a k. − − n 1 n ∈ Proof. First note that the ideal (x a ,...,x a ) is the kernel of the 1 − 1 n − n evaluation map k[x ,...,x ] k defined by f(x ,...,x ) f(a ,...,a ). 1 n → 1 n 7→ 1 n For example, this is clear if (a1,...,an) =(0,..., 0), and more generally an easy version of Taylor’s theorem implies that every f k[x1,...,xn] can be written in the form f(a ,...,a )+g(x ,...,x ), where∈g (x a ,...,x 1 n 1 n ∈ 1 − 1 n − an). Now let m be an arbitrary maximal ideal of k[x1,...,xn]/m. By the previous lemma, the map k k[x1,...,xn]/m is an isomorphism. Thus, for all i, there exists an a k→such that x a mod m, and so x a m. i ∈ i ≡ i i − i ∈ It follows that m contains the maximal ideal (x1 a1,...,xn an) and so is equal to (x a ,...,x a ). − − 1 − 1 n − n For a field which is not algebraically closed, there is the following result:

Corollary 4.5.5. Let k be a ﬁeld and let I be an ideal in k[x1,...,xn]. Then I is not the unit ideal if and only if there exist a ,...,a k¯ such that, for 1 n ∈ all f I, f(a ,...,a )=0. ∈ 1 n 152 CHAPTER 4. INTEGRAL RING HOMOMORPHISMS

Proof. Clearly, if there exist a ,...,a k¯ such that f(a ,...,a ) = 0 for all 1 n ∈ 1 n f I, then I is not the unit ideal. Conversely, suppose that I = R. Then I is∈ contained in a maximal ideal m, and we may as well assume that6 I = m. It will suffice to show that there is a finite extension K of k and a ,...,a K 1 n ∈ such that f(a ,...,a ) = 0forall f m. By Lemma 4.5.3, k[x ,...,x ]/m = 1 n ∈ 1 n K is a finite extension of k, which we may view as a subfield of k¯. Let ai be the image of xi in K. Then the homomorphism eva1,...,an : k[x1,...,xn] K has as kernel m by definition. Hence f(a ,...,a )=0 for all f m. → 1 n ∈ We turn now to the usual statement of the Nullstellensatz.

Corollary 4.5.6. Let k be an algebraically closed ﬁeld. Let I be an ideal in k[x1,...,xn] and deﬁne, as in Chapter 1,

V (I)= z kn : f(z)=0 for all f I . { ∈ ∈ }

Let g k[x1,...,xn] and suppose that g(z)=0 for all z V (I). Then there exists∈ a positive integer m such that gm I. ∈ ∈ Proof. Suppose that gm / I for every positive integer m. We will ﬁnd a point ∈ z0 V (I) such that g(z0) = 0. As usual let k[x1,...,xn]g be the localization of k∈[x ,...,x ] at the multiplicative6 set S = 1,g,g2,... . Since I S = , 1 n { } ∩ ∅ Ik[x1,...,xn]g is a proper ideal of k[x1,...,xn]g, and hence is contained in a m maximal ideal M. Let m = M k[x1,...,xn]. Note that I m but that g / m for every positive m. We claim∩ that the prime ideal m is⊆ maximal. In fact,∈ we have a sequence of injections k k[x ,...,x ]/m k[x ,...,x ] /M. ⊆ 1 n ⊆ 1 n g But k[x1,...,xn]g = k[x1,...,xn, 1/g] is a ﬁnitely generated extension of k, and so, by Lemma 4.5.3, the inclusion of k in k[x1,...,xn]g/M is an isomorphism. It follows that k[x1,...,xn]/m ∼= k and hence that m is a maximal ideal of k[x1,...,xn]. Thus, by the Nullstellensatz, m corresponds to a point z kn. Since I m, z V (I). But g / m, so that g(z ) = 0, as 0 ∈ ⊆ 0 ∈ ∈ 0 6 claimed.

The last application of the normalization lemma is to prove the following theorem of Noether, which implies in particular that the integral closure of the affine coordinate ring of an algebraic variety is also an affine coordinate ring, i.e. is finitely generated over k. If k has characteristic zero, both this statement and the more general statement below follow from Corollary 4.1.12. 4.5. NOETHER NORMALIZATION, VERSION 1 153

Theorem 4.5.7 (Noether). Let R be an integral domain, finitely generated over a field k, and let K be the quotient field of R. Suppose that L is a finite extension of K, and that R is the integral closure of R in L. Then R is finitely generated as an R-module. e e Proof. We shall use repeatedly the fact that, if R is Noetherian and M is a finitely generated R-module, then every submodule of M is again finitely generated. In particular, if L is contained in a larger field L′, and R′ is the integral closure of R in L′, then if R′ is a finitely generated R-module, then so is R. In particular, we may assume that L is a normal extension of K. If L is normal, and G = Aut(L/K), then we have K LG L, L is a separable ⊆ ⊆ extensione of LG, and LG is a purely inseparable extension of K. Let R be the integral closure of R in LG. Clearly R is the integral closure of R in L. We know by Corollary 4.1.12 that R is a finitely generated R-module. Ifb we can show that R is a finitely generated Re-module, then it follows asb in the proof of Corollary 4.1.3 that R is ae finitely generated R-module.b Thus we may assume inb what follows that L is a purely inseparable extension of K. By the normalization lemma,e there exists a subring k[x ,...,x ] R, 1 d ⊆ where the xi are algebraically independent, such that R is integral over k[x1,...,xd]. Since R is integral over k[x1,...,xd], it is clear that R is the integral closure of k[x1,...,xd] in L. If we can show that R is a finitely generated k[x1,...,xed]-module, then it is a fortiori a finitely generatede R- module. Thus we may assume throughout that R = k[x1,...,xe d] and hence that K = k(x1,...,xd) is the field of rational functions in x1,...,xd. The assumption that L is purely inseparable over K implies that, if p = char k, then there exists some m > 0 such that L = K(α1,...,αn), where pm m q αi K. Let q = p , and let αi = fi/gi, where fi,gi R. There is ∈ th ∈ some finite extension k′ of k which contains q roots of all of the coefficients appearing in the fi and gi. Let yi be elements in an algebraic closure of K q satisfying yi = xi. Since the xi are algebraically independent over k, the yi are algebraically independent over k′ (if Q(y1,...,yn) = 0 for some nonzero polynomial Q, raise Q to a large power of p to get a nonzero polynomial in x1,...,xn with coefficients in k). Let L′ = k′(y1,...,yn). It is easy to see th that fi and gi become q powers in L′, and hence that L′ contains a subfield isomorphic to L. Thus we will view L as a subfield of L′. If R′ = k′[y1,...,yn], then clearly L′ is the quotient field of R′. Moreover, R R′ L′. Since ⊆ ⊆ R′ is a polynomial algebra, it is a UFD and hence is integrally closed in L′. Thus R R′. But R′ is generated over R by the y , which satisfy the ⊆ i e 154 CHAPTER 4. INTEGRAL RING HOMOMORPHISMS

q monic polynomial t x = 0, and a finite number of elements of k′ which − i are algebraic over k and hence integral over R. Thus R′ is generated as an R-algebra by a finite number of elements which are integral over R, and so R′ is a finitely generated R-module. It follows that the submodule R is also a finitely generated R-module. e Noether’s theorem and Corollary 4.1.12 imply that, for any integral domain R which is finitely generated over Z or over a field, the integral closure R is a finitely generated R-module. However, Akizuki has constructed coun- terexamples to this statement in general, even for Noetherian rings R. In fact,e the integral closure of a Noetherian domain may fail to be Noetherian.

4.6 Noether normalization, version 2

Throughout this section, k is a ﬁeld, not necessarily algebraically closed. Theorem 4.6.1 (Noether normalization lemma, version 2). Let R be an integral domain which is a ﬁnitely generated k-algebra and let I I 1 ⊆ 2 ⊆ Ik be an increasing sequence of ideals of R with Ik = R. Then there exist··· ⊆x ,...,x R, algebraically independent over k, such6 that: 1 d ∈

(i) R is integral over k[x1,...,xd];

(ii) For all i, 1 i k, there exists a nonnegative integer hi such that I k[x ,...,x≤ ]=(≤ x ,...,x )k[x ,...,x ]. i ∩ 1 d 1 hi 1 d The case k = 1 and I1 = (0) reduces to the first version of the normalization lemma, but without the statement that we can choose the xi to be linear combinations of a given set of generators for R over k. Before we prove this second version of the normalization lemma, let us use it to prove various statements about affine algebraic sets. Corollary 4.6.2. dim k[x ,...,x ]= d. Moreover, for all i with 0 i d, 1 d ≤ ≤ ht(x ,...,x )= i and coht(x ,...,x )= d i. 1 i 1 i − Proof. As we have essentially already seen, dim k[x1,...,xd] d (consider the chain 0 (x ) (x , x ) (x ,...,x )). To see≥ the opposite ⊂ 1 ⊂ 1 2 ⊂ ··· ⊂ 1 d inequality, apply the normalization lemma to the ring R = k[x1,...,xd] and to a chain 0 p1 pk of prime ideals in R. There thus exist t ,...,t R, algebraically⊂ ⊂ ··· independent ⊂ over k, such that R is integral 1 n ∈ 4.6. NOETHER NORMALIZATION, VERSION 2 155 over k[t ,...,t ], and p k[t ,...,t ]=(t ,...,t ) for some nonnegative 1 n i ∩ 1 n 1 hi integer h . By the going up theorem, p k[t ,...,t ] is strictly contained i i ∩ 1 n in pi+1 k[t1,...,tn]. Thus hi < hi+1. But then hi is a strictly increasing sequence∩ of nonnegative integers with h = 0 and{ }h d. It follows that 0 k ≤ k d, and hence every chain of prime ideals in k[x1,...,xd] has length at most≤ d. It follows that dim k[x ,...,x ] d, and hence dim k[x ,...,x ]= d. 1 d ≤ 1 d To see the second statement, note that coht(x ,...,x ) = dim k[x ,...,x ]/(x ,...,x ) = dim k[x ,...,x ]= d i. 1 i 1 d 1 i d+1 d − Clearly ht(x ,...,x ) i, by considering the chain 0 (x ) (x , x ) 1 i ≥ ⊂ 1 ⊂ 1 2 ⊂ (x1,...,xi). On the other hand, by Lemma 4.4.2, ht(x1,...,xi) + ···coht( ⊂x ,...,x ) d, and thus ht(x ,...,x ) d (d i)= i. It follows that 1 i ≤ 1 i ≤ − − ht(x1,...,xi)= i. Here is another corollary, which describes the dimension of the affine coordinate ring A(X) in case X is an affine variety. Corollary 4.6.3. Let R be an integral domain which is a finitely generated k-algebra. Then dim R = tradeg(R/k). Proof. By either version of the normalization lemma, there exists a subring of R of the form k[x1,...,xd], where the xi are algebraically independent over k, such that R is integral over k[x1,...,xd]. By Corollary 4.2.11, dim R = dim k[x1,...,xd] = d, by the previous corollary. On the other hand, since the quotient field of R is algebraic over k(x1,...,xd), tradeg(R/k) = d as well. Thus dim R = tradeg(R/k). Corollary 4.6.4. Let R be an integral domain which is a finitely generated k-algebra, and let p be a prime ideal in R. Then ht p + coht p = dim R. In particular, R is catenary.

Proof. There exists a subring of R of the form k[x1,...,xd], where the xi are algebraically independent over k, such that R is integral over k[x1,...,xd], and such that p k[x ,...,x ] = (x ,...,x ). Thus dim R = d. Ap- ∩ 1 d 1 h plying Lemmas 4.4.3 and 4.6.2, ht p = ht(x1,...,xh) = h, and coht p = coht(x ,...,x )= d h. Thus ht p + coht p = h + d h = d = dim R. 1 h − − Proof of the normalization lemma, version 2. Let R be an integral domain, ﬁnitely generated over k, and let I1 Ik be an increasing sequence of ideals in R with I = R. ⊆···⊆ k 6 156 CHAPTER 4. INTEGRAL RING HOMOMORPHISMS

Step I: It suﬃces to prove the theorem for R = k[y1,...,yn] a polynomial algebra. In fact, if the theorem has been established for such, write a general R as k[y1,...,yn]/J0 for some ideal J0, and apply the theorem to k[y1,...,yn] and to the increasing sequence J J J , where J is the inverse image 0 ⊆ 1 ⊆···⊆ k i of I in k[y ,...,y ]. Thus there exist x′ ,...,x′ k[y ,...,y ], algebraically i 1 n 1 m ∈ 1 n independent over k, such that k[y1,...,yn] is integral over k[x1′ ,...,xm′ ] and k[x1′ ,...,xm′ ] Ji = (x1′ ,...,xh′ i ). Let xi be the image of xi′ for i > h0. Clearly, ∩ k[x′ ,...,x′ ]/J k[x′ ,...,x′ ]= k[x′ ,...,x′ ]/(x′ ,...,x′ ) = k[x′ ,...,x′ ]. 1 m 0∩ 1 m 1 m 1 h0 ∼ h0+1 m

Thus the ring k[xh′ 0+1,...,xm′ ] injects into R, so that xh0+1,...,xm are algebraically independent over k. Since k[y1,...,yn] is integral over k[x1′ ,...,xm′ ],

R is integral over k[xh0+1,...,xm]. Finally, the image of k[x1′ ,...,xm′ ] Ji in R is just k[x ,...,x ] I , and this ideal is clearly the same ideal∩ as h0+1 m ∩ i (x1′ ,...,xh′ i )/(x1′ ,...,xh′ 0 )=(xh0+1,...,xhi ). Thus (after reindexing), the theorem holds for R.

Step II: We thus assume that R = k[y1,...,yn]. Next assume that k = 1, i.e. that we have a single ideal I, which we further assume to be principal; I = (x ) for some x k[y ,...,y ]. If x = 0, then we can just take y = x and 1 1 ∈ 1 n 1 i i h = h1 = 0. So we will assume that x1 = 0. We claim that there exist positive 6 d integers d , i 2, such that, if we set x = y y i , i 2, then k[y ,...,y ] i ≥ i i − 1 ≥ 1 n is integral over k[x1,...,xn] and k[x1,...,xn] I = x1k[x1,...,xn]. Since di ∩ yi = xi + y1 , if we can show that y1 is integral over k[x1,...,xn] then all of the yi are integral over k[x1,...,xn]. Write x1 = P (y1,...,yn) for some nonzero polynomial P . Then

0= x x = P (y ,...,y ) x = P (y , x + y d ,...,x + ydn ) x . 1 − 1 1 n − 1 1 2 1 2 n n − 1

ν1 νn Write P (y1,...,yn) = ν aνy1 yn , where ν = (ν1,...,νn) is a multi- index, so that the above equation··· becomes P F (y )= a yν1 (x + yd2 )ν2 (x + ydn )νn x =0. 1 ν 1 2 1 ··· n 1 − 1 ν X Set f(ν) = ν1 + d2ν2 + + dnνn. Then the terms in the above expression ··· f(ν) which involve y1 alone (with no powers of xi) are of the form aν y1 , for a = 0, and the terms which involve factors which are powers of some of ν 6 4.6. NOETHER NORMALIZATION, VERSION 2 157

f(ν) a the xi are all of the form p(x1,...,xn)y1 − for a strictly positive integer a. Thus, if we can choose the di so that the positive integers f(ν), for aν = 0, are f(µ6) all distinct, then, as a polynomial in y1, F (y1) has leading term aµy1 , where µ is the unique multi-index such that a = 0 which maximizes the function µ 6 f(ν). Thus, after dividing by aµ, F (y1) becomes a monic polynomial in y1 with coeﬃcients in k[x1,...,xn], and so y1 is integral over k[x1,...,xn]. i To see that such a choice is possible, choose for example di = N , where N is a natural number larger than all of the νi such that aν = 0. Then 2 n 6 f(ν) = ν1 + N ν2 + + N νn, where the νi are the base N digits of f(ν) and hence are uniquely··· determined. Since all of the yi are integral over k[x1,...,xn], k[y1,...,yn] is integral over k[x1,...,xn]. But then x1,...,xn must be algebraically independent over k, since tradeg(k[x1,...,xn]/k) = tradeg(k[y1,...,yn]/k) = n. Finally, we must check that k[x1,...,xn] I = x1k[x1,...,xn]. Clearly x k[x ,...,x ] k[x ,...,x ] I. Conversely,∩ if x g = h k[x ,...,x ] I 1 1 n ⊆ 1 n ∩ 1 ∈ 1 n ∩ with g k[y ,...,y ], then g = h/x k(x ,...,x ) and g is integral over ∈ 1 n 1 ∈ 1 n k[x1,...,xn]. Thus g k[x1,...,xn] since k[x1,...,xn] is integrally closed. Hence x g x k[x ,...,x∈ ]. 1 ∈ 1 1 n Step III: We continue to assume that R = k[y1,...,yn] and that k = 1, but now I is an arbitrary ideal. The proof is by induction on n, the case n = 0 being clear, and n = 1 follows from Step II since then I is necessarily principal. As in Step II, we may assume that I = 0. Choose x = 0, x I. The proof of Step II shows that there exist6 t ,...,t 1 6 1 ∈ 2 n ∈ k[y1,...,yn] such that x1, t2,...,tn are algebraically independent over k, k[y ,...,y ] is integral over k[x , t ,...,t ], and such that k[x , t ,...,t ] 1 n 1 2 n 1 2 n ∩ x1k[y1,...,yn] = x1k[x1, t2,...,tn]. Applying the inductive hypothesis to the ring k[t ,...,t ] and the ideal I k[t ,...,t ], there exist x ,...,x 2 n ∩ 2 n 2 n ∈ k[t2,...,tn], algebraically independent over k, such that k[t2,...,tn] is integral over k[x2,...,xn] and

I k[t ,...,t ] k[x ,...,x ]= I k[x ,...,x ]=(x ,...,x )k[x ,...,x ], ∩ 2 n ∩ 2 n ∩ 2 n 2 h 2 n say. Since k[y1,...,yn] is integral over k[x1, t2,...,tn] and k[x1, t2,...,tn] is integral over k[x1, x2,...,xn], k[y1,...,yn] is integral over k[x1,...,xn] and hence x1, x2,...,xn are algebraically independent over k. Finally, note that x I for 1 i h, and thus (x ,...,x )k[x ,...,x ] I k[x ,...,x ]. i ∈ ≤ ≤ 1 h 2 n ⊆ ∩ 1 n Write p I k[x1,...,xn] as x1q +r for some r k[x2,...,xn]. Since x1 I, x q I ∈and∩ hence r I k[x ,...,x ]=(x ,...,x∈ )k[x ,...,x ]. It follows∈ 1 ∈ ∈ ∩ 2 n 2 h 2 n 158 CHAPTER 4. INTEGRAL RING HOMOMORPHISMS that I k[x ,...,x ] (x ,...,x )k[x ,...,x ] and hence I k[x ,...,x ]= ∩ 1 n ⊆ 1 h 1 n ∩ 1 n (x1,...,xh)k[x1,...,xn].

Step IV: It now suffices to consider the case where R = k[y1,...,yn] and k is arbitrary. The proof will be by induction on k; the case k = 1 is Step III. Assume that Step IV has been established for the integer k 1. Given − a sequence I I I for k[y ,...,y ], use the inductive hypothesis 1 ⊆ 2 ⊆···⊆ k 1 n to find t1,...,tn, algebraically independent over k, such that k[y1,...,yn] is integral over k[t ,...,t ] and I k[t ,...,t ]=(t ,...,t )k[t ,...,t ] for 1 n i ∩ 1 n 1 hi 1 n i k 1. Let r = hk 1 and let I = Ik k[tr+1,...,tn]. By Step III, we − can≤ find− x ,...,x k[t ,...,t ], algebraically∩ independent over k, such r+1 n ∈ r+1 n that k[tr+1,...,tn] is integral over k[xr+1,...,xn] and I k[xr+1,...,xn] = (x ,...,x )k[x ,...,x ] for some integer a. Set x∩ = t for i r. r+1 a r+1 n i i ≤ Since k[y1,...,yn] is integral over k[t1,...,tn] and k[t1,...,tn] is integral over k[t1,...,tr, xr+1,...,xn] = k[x1,...,xn], k[y1,...,yn] is integral over k[x ,...,x ] and x ,...,x are algebraically independent over k. For i 1 n 1 n ≤ k 1, let p I k[x ,...,x ]. Then we can write p = q + q , where − ∈ i ∩ 1 n 1 2 q1 (x1,...,xhi )k[x1,...,xn] I and q2 k[xhi+1,...xn] k[thi+1,...tn]. Thus∈ q I k[t ,...t ]⊆ = 0. It follows∈ that I ⊆k[x ,...,x ] 2 ∈ i ∩ hi+1 n i ∩ 1 n ⊆ (x1,...,xhi )k[x1,...,xn], and, since the opposite containment is clear, Ii k[x ,...,x ]=(x ,...,x )k[x ,...,x ]. As for I k[x ,...,x ], it contains∩ 1 n 1 hi 1 n k ∩ 1 n (x1,...,xr)k[x1,...,xn] since Ik contains Ik 1, and Ik contains xr+1,...,xa − as well. Conversely, if p Ik k[x1,...,xn], write p = q1 + q2, where q (x ,...,x )k[x ,...,x∈] I∩ and q k[x ,...x ] k[t ,...,t ]. 1 ∈ 1 a 1 n ⊆ 2 ∈ a+1 n ⊆ r+1 n Thus q2 Ik k[tr+1,...,tn] k[xa+1,...xn] = 0. Setting hk = a, it follows that I ∈k[x ∩,...,x ]=(x ,...,x∩ )k[x ,...,x ], and we are done. k ∩ 1 n 1 hk 1 n

Exercises

Exercise 4.1. Let R be an integral domain. Show that, if R is integrally closed, then so is R[x], where x is an indeterminate. More generally, for an arbitrary integral domain R, the integral closure of R[x] in its ﬁeld of fractions is R[x], where R is the integral closure of R.

Exercise 4.2.e Verify thate the integral closure of Z in Q(i) is Z[i]. More generally, if d is a square free integer, what is the integral closure of Z in Q(√d)? 4.6. NOETHER NORMALIZATION, VERSION 2 159

Exercise 4.3. Let k be a field and let x, y be algebraically independent over k. Show that y2 x3 is irreducible in k[x, y], so that k[x, y]/(y2 x3) = R is an integral domain.− Show that R is not integrally closed in its− quotient field: if t = y/x, then show that t / R but that t2 = y2/x2 = x3/x2 = x R. ∈ ∈ Finally argue that the integral closure of R in its quotient field is k[t].

Exercise 4.4. Suppose that R is a subring of S and that S is integral over R. If r R and r is a unit in S, show that r is a unit in R. Show also that the Jacobson∈ radical of R is the intersection of R with the Jacobson radical of S.

Exercise 4.5. Let R be a ring and let S1, S2 be two R-algebras which are integral over R. Show that S S and S S are integral over R. 1 × 2 1 ⊗R 2

Exercise 4.6. Let R′ be a commutative ring, let G be a ﬁnite group of G automorphisms of R′ and let R =(R′) . Show that R′ is integral over R. Suppose further that S′ is a multiplicative subset of R′ such that σ(S′)= G S′ for all σ G, and set S = (S′) . Finally let N ′ be an R′-module such ∈ that G acts on N ′ via group automorphisms and satisﬁes σ(rn)= σ(r)σ(n) for all r R′, n N ′, and σ G. Show there is a natural isomorphism ∈ ∈ ∈ 1 G 1 G (S′)− N ′ ∼= S− (N ′) .

In particular, if R′ is an integral domain with fraction ﬁeld K, taking S′ = G R′ 0 and S = R 0 , show that K is the quotient ﬁeld k of R. −{ } −{ } Exercise 4.7. (a) Let f : R S be a surjective homomorphism of rings. Show (directly) that dim S →dim R. ≤ (b) Let f : R S be an integral homomorphism of rings. Show that dim S dim R. → ≤ Exercise 4.8. (i) Let R be a ring contained in a ring S. Let

f = s S : sS R . { ∈ ⊆ } Show that f is an ideal of S contained in R, and that, if I is a subset of R such that I is an ideal of S, then I f. Thus f is the largest subset of S ⊆ which is an ideal both in S and in R. Finally show that R = S if and only if f = S if and only if f = R. (ii) What is f in case R = Z and S = Z[i]? What if R = k[t2, t3] S = k[t]? ⊆ 160 CHAPTER 4. INTEGRAL RING HOMOMORPHISMS

(iii) Now suppose that R is an integral domain and that S = R is the integral closure of R in its quotient field. In this case f = fR is called the conductor of 1 R. Let M be a multiplicative subset of R. Show that fM −1Re M − fR, and 1 ⊇ hence that, if M fR = , then M − R is integrally closed. Assume further ∩ 6 ∅ 1 that R is a finitely generated R-module. Show that fM −1R = M − fR and that 1 M − R is integrally closed if and only if M f = . ∩ R 6 ∅ e Exercise 4.9. (i) Let R1 and R2 be rings. Describe all prime ideals in the product R R . What does this say about dim(R R ) in terms of dim R 1 × 2 1 × 2 1 and dim R2? (ii) Let R be a ring and let I be an ideal contained in the radical √0 of R. Show that dim R = dim(R/I). Exercise 4.10. Let S be an R-algebra which is a finitely generated R- module, so that in particular S is integral over R. Suppose that S is generated by r elements as an R-module. Let p be a prime ideal in R. Then there are at most r prime ideals q of S such that q R = p. (Localizing at p, we may ∩ assume that p and the q are maximal. Thus if q1,..., qn are distinct prime ideals with q R = p, the q are comaximal and by the Chinese remainder i ∩ i theorem the map S S/q → i i Y is surjective, with pS contained in the kernel. Now pS is not necessarily prime, but S/pS is a vector space of dimension at most r over R/p = kp. On the other hand, if kqi = S/qi, show that

dim (S/pS) [k : k ] n. kp ≥ qi p ≥ i X Thus n r.) ≤ Exercise 4.11. Let f : R S be a homomorphism of rings. We say that f → has the going up property if, whenever p1 and p2 are prime ideals of R with 1 p1 p2 and q1 is a prime ideal of S such that f − (q1)= p1, then there exists ⊆ 1 a prime ideal q of S with q q and such that f − (q )= p . 2 1 ⊆ 2 2 2 (i) Show that the composition of two homomorphisms with the going up property has the going up property. (ii) Show that, if S is integral over R, then f has the going up property. (The going up theorem is the special case where f is an inclusion.) 4.6. NOETHER NORMALIZATION, VERSION 2 161

(iii) Show that the following are equivalent:

(a) f : R S has the going up property. → 1 (b) For every prime ideal q of S, if p = f − (q), then the induced function f ∗ : Spec(S/q) Spec(R/p) is surjective. → (c) If q is a prime ideal of S and p f ∗( q ) = f ∗(V (q)), by Exer- ∈ { } cise 1.19, then V (p)= p f ∗( q )= f ∗(V (q)). { } ⊆ { }

(iv) Suppose that f ∗ : Spec S Spec R is closed (i.e. if X is a closed subset → of Spec S, then f ∗(X) is a closed subset of Spec R). Show that f : R → S has the going up property. Conversely, suppose that Spec S is a Noetherian topological space and that f : R S has the going up → property. Show that f ∗ is closed. (Use the fact, which we shall prove in the next chapter, that every closed subset of Spec S is a union of ﬁnitely many irreducible closed subsets of Spec S, which are thus necessarily of the form V (q), by Exercise 1.24, and (c) above.) Exercise 4.12. Let f : R S be a homomorphism of rings. We say that → f has the going down property if, whenever p1 and p2 are prime ideals of R 1 with p1 p2 and q1 is a prime ideal of S such that f − (q1)= p1, then there ⊇ 1 exists a prime ideal q of S with q q and such that f − (q )= p . Thus, 2 1 ⊇ 2 2 2 if f is an injection, R is an integrally closed domain, and S is an integral domain which is integral over R, then f has the going down property. (i) Show that the following are equivalent:

(a) f has the going down property; 1 (b) For every prime ideal q of S, if p = f − (q), then the induced function f ∗ : Spec S Spec R is surjective. q → p (c) For every prime ideal p of R, if q is a prime ideal of S containing 1 pS and minimal with respect to this property, then f − (q)= p.

((a) (b) is straightforward. To see that (c) = (a), use the fact ⇐⇒ ⇒ that, if p2 is a prime ideal of R, then there exists a minimal prime ideal q2 of S contained in q1 and containing p2S, by applying Exercise 1.6 to S/p2S.) (ii) Show that a ﬂat homomorphism f : R S has the going down property. (By (i), it suﬃces to show that, given→ a prime ideal q of S, with 162 CHAPTER 4. INTEGRAL RING HOMOMORPHISMS

1 p = f − (q), then f ∗ : Spec S Spec R is surjective. Use the fact q → p that R S is flat, and hence faithfully flat by Exercise 2.35, and p → q thus that f ∗ : Spec S Spec R is surjective by Exercise 3.11.) q → p Exercise 4.13. Let R = k[t2, t3] R = k[t]. Show that R is not a flat ⊆ 2 3 R-module. (If R were flat, argue that, with m = (t , t ), the module Rm would be a free rank one Rm-module, ande hence we would havee e e dimk(Rm/mRm) = dimk(R/mR)=1.

But show that dimk(R/mRe) = 2.)e e e

Exercise 4.14. (i) Lete p ande q be two prime ideals of R, neither one of which is contained in the other. Show that S = R (p q) is a multiplicative subset 1 − ∪ of R and that the prime ideals of S− R are in one-to-one correspondence with the prime ideals of R which are either contained in p or in q. Use this to construct an example of a ring with two maximal chains of prime ideals of diﬀerent lengths. (ii) With R, p, and q as above, show that the prime ideals of R/p q are in one-to-one correspondence with the prime ideals of R which contain· either p or q. What does this say about dim R/p q? ·

Exercise 4.15. Let k be a field and let R = k[x](x) be the localization of k[x] at the maximal ideal (x). Show that (x)R is the unique nonzero prime ideal of R and that R[1/x] is a field, the quotient field K of R. Now let R[t] be the polynomial ring in one variable with coefficients in R. Show that every maximal ideal of R[t] is either of the form (x, f(t)), where f(t) R[t] is such that its image in (R/(x))[t]= k[t] is irreducible, or of the form∈xg(t) 1, where g(t) R[t] and xg(t) 1 is an irreducible polynomial − ∈ − in K[t]. In particular, the ideal (xt 1) is a maximal ideal in R[t]; show that the chain (0) (xt 1) is a maximal− chain of prime ideals in R[t]. On the ⊆ − other hand, argue that dim R[t] 2. (In fact, dim R[t] = 2.) ≥

Exercise 4.16. Let k be an inﬁnite ﬁeld, and let f(x1,...,xn) k[x1,...,xn], where f is nonconstant. Show that, after a linear change of coordina∈ tes, we may assume that f / k[x1,...,xn 1] and that the ring k[x1,...,xn]/(f) is in- ∈ − tegral over the image of k[x1,...,xn 1]. In particular, dim k[x1,...,xn]/(f)= − n 1. − 4.6. NOETHER NORMALIZATION, VERSION 2 163

Exercise 4.17. Let R be an integral domain and let S be a finitely generated R-algebra containing R as a subring which is also an integral domain. Show that there exist f R, f = 0, and x ,...,x S, algebraically independent ∈ 6 1 n ∈ over R, such that Sf is integral over R[x1,...,xn]f . (To see this, let α1,...,αk generate S over R and let K be the quotient field of R and let M = R 0 , 1 1 −{ } so that K = M − R. Then M − S is a finitely generated K-algebra, so by 1 the normalization lemma there exist y1,...,yn M − S, algebraically inde- 1 ∈ pendent over K, such that M − S is integral over K[y, ...,yn]. In particular, the α are integral over K[y ...,y ]. Now find an f R 0 such that j , n ∈ −{ } y = x /f for some x S and such that fα is integral over R[x ,...,x ].) i i i ∈ j 1 n Exercise 4.18. (i) Let R be a ring, and suppose that a finite group G acts on R as a group of ring homomorphisms. Let RG be the fixed subring of R and let S RG be the subring of R generated by the elementary symmetric ⊆ functions in the elements g(r),g G, as r ranges over all elements of R. Show that R is integral over S and∈ hence over RG. (Compare Exercise 4.6.) (ii) In the above situation, if R is an integrally closed integral domain, show that RG is also integrally closed. (iii) In the situation of (i), if p is a prime ideal of RG, by following one of the lemmas used in the proof of the going down theorem, show that G acts transitively on the set of prime ideals P of R such that P RG = p. Thus Spec RG = (Spec R)/G and maxSpec RG = (maxSpec R)/G.∩ (iv) (Noether) If in the situation of (i) k is a field, R is a k-algebra, finitely generated over k, and G acts via k-algebra homomorphisms, then RG is also finitely generated over k. Thus (using (iii)), the quotient of an affine algebraic set by a finite group is again naturally an affine algebraic set. (If G R = k[α ,...,α ], let S′ R be generated by the elementary symmetric 1 n ⊆ functions in the αi. Thus S′ is a finitely generated k-algebra, and in particular it is Noetherian. Now R = S′[α1,...,αn], where the αi are integral over S′, G and hence R is a finitely generated S′-module. Since R is an S′-submodule G of R, it is also a finitely generated S′-module, and a set of generators for R as an S′-module will also generate it as an S′-algebra. From this conclude that RG is a finitely generated k-algebra.) (v) Suppose that k is a field of characteristic different from 2 and that G = Z/2Z acts on R = k[x1,...,xn] as follows: the nontrivial element of G sends n xi to xi. In other words, G is induced by the action of Id on Ak . Show that the− invariant subring RG of R is generated by x x , i ± j. In particular, i j ≤ 164 CHAPTER 4. INTEGRAL RING HOMOMORPHISMS

G 2 2 G for n = 2, R = k[x1, x1x2, x2]. Show that, in this case, R is isomorphic to k[x, y, z]/(xz y2), and is not isomorphic to a polynomial ring. − Exercise 4.19. (An example related to Noether normalization and going down.) Let k be a ﬁeld and let R = k[x1, x2, x3]/(x1x3, x2x3). Note that (x x , x x )=(x , x ) (x ). Draw a picture of V (x x , x x ) A3. Let 1 3 2 3 1 2 ∩ 3 1 3 2 3 ⊆ k t1 = x1 + x3 and t2 = x2. Show that R is integral over the subring k[t1, t2] and that t1 and t2 are algebraically independent over k. (For example, you can use the isomorphism R/x3R = k[x1, x2] for the last part.) What is 2∼ the morphism V (x1x3, x2x3) Ak corresponding to the inclusion of rings k[t , t ] R? → 1 2 ⊆ Let p = (x1, x2). Show that p is a prime ideal of R of coheight 1, but that the height of p is zero. (Use the fact that any prime ideal in k[x1, x2, x3] containing (x , x ) (x ) must contain either (x , x ) or (x ).) Thus 1 2 ∩ 3 1 2 3 ht p + coht p =1 = dim R =2. 6 Moreover, p k[t1, t2]=(t2), so that ht(p k[t1, t2]) = ht p. In particular, the going down∩ theorem does not hold for the∩ integral6 extension k[t , t ] R; 1 2 ⊆ here k[t1, t2] is integrally closed but R is not an integral domain. Exercise 4.20. (Another example where going down fails.) Let S be the integral domain k[x1, x2] and let R be the subring (and hence integral domain) R = f(x , x ) S : f(x , 0) = f(x , 1) . { 1 2 ∈ 1 1 } (i) Show that R is a ﬁnitely generated k-algebra and that S is integral over R, in fact is its integral closure. Identify S as the coordinate ring of X = A2/ , where is the equivalence relation (x , 0) (x , 1). (It k ∼ ∼ 1 ∼ 1 may be easier to do (i) for S0 = k[t] and R0 the subring consisting of all polynomials f(t) such that f(0) = f(1). In this case, R0 is generated by u = t(t 1) and v = t2(t 1), with v2 = u3 + uv and t = v/u.) − − (ii) Let q2 =(x1, x2) S and let p2 = q2 R. Let p1 =(x1x2 + x2 1) R. Show that p , p ,⊆ and q are all prime∩ ideals and that p p −but∩ that 1 2 2 1 ⊆ 2 there is no prime ideal q1 q2 with p1 = q1 R. Thus the going down theorem does not hold for⊆ the integral extension∩ R S; here both R and S are integral domains, and in fact S is integrally⊆ closed, but R is not integrally closed. Draw a picture of V (x1x2 + x2 1) and V (x1, x2) 2 − 2 as subsets of Ak, as well as the corresponding subsets of X = Ak/ , and explain geometrically why the going down theorem fails. ∼ 4.6. NOETHER NORMALIZATION, VERSION 2 165

1 Exercise 4.21. Show that, if R is catenary, then so are S− R for every multiplicative subset S of R, and R/I, for every ideal I in R. Chapter 5

Ideals in Noetherian Rings

The main focus in this chapter is to try to describe arbitrary ideals in a Noetherian ring in terms of prime ideals. The prototype for this result is the fact that every ideal in the ring of integers in an algebraic number field factors uniquely into a product of prime ideals. Indeed, this result, which is a substitute for unique factorization in the ring itself, is one of the origins of ring theory. For a general Noetherian ring, there are somewhat weaker results. For example, a radical ideal is an intersection of finitely many prime ideals, and the prime ideals which appear are uniquely determined. This n result is the algebraic analogue of the fact that an algebraic set in Ak is a union of finitely many irreducible algebraic subsets.

5.1 Irreducible sets and radical ideals

We begin with the geometric picture:

n Proposition 5.1.1. Every algebraic set X Ak can be written as X = X X , where the X are irreducible.⊆ Moreover the above expression 1 ∪···∪ n i is unique up to order if we assume that, for all i = j, X X . 6 i 6⊆ j

Deﬁnition 5.1.2. The Xi in Proposition 5.1.1 are the irreducible components of X. They are unique up to order.

More generally, we have the following result, a very special case of primary decomposition, which holds for a general Noetherian ring and implies Proposition 5.1.1:

166 5.1. IRREDUCIBLE SETS AND RADICAL IDEALS 167

Proposition 5.1.3. Let R be a Noetherian ring. Every proper radical ideal J in R can be written as J = p1 pn, where the pi are prime ideals. Moreover the above expression is unique∩···∩ up to order if we assume that, for all i = j, p p . 6 i 6⊆ j Proof. Consider the collection of all proper radical ideals in R which are not an intersection of prime ideals. If this collection is nonempty, there must be a maximal element J0 in this set. Clearly a prime ideal is an intersection of prime ideals, so that J0 cannot be prime. Thus there exist f,g R, f,g / J , such that fg J . Consider the ideal (J , f). If it were the∈ unit ∈ 0 ∈ 0 0 ideal, then 1 = r + sf with r J0 and s R, and so g = gr + sfg J0, a contradiction. Likewise (J ,g)∈ is a proper ideal.∈ Clearly J (J , f) ∈(J ,g). 0 0 ⊆ 0 ∩ 0 We claim that in fact J0 =(J0, f) (J0,g). For if r (J0, f) (J0,g), write r = r + h f = r + h g, with r ,r∩ J . Then ∈ ∩ 1 1 2 2 1 2 ∈ 0 r2 =(r + h f)(r + h g)= r r + r h g + r h f + h h fg J . 1 1 2 2 1 2 1 2 2 1 1 2 ∈ 0 Since J is a radical ideal, r J . Now 0 ∈ 0 J = J = (J , f) (J ,g)= (J , f) (J ,g). 0 0 0 ∩ 0 0 ∩ 0 But (J0, f) andp (J0,gp) are radical idealsp properly containingp J0 and thus they are both intersections of prime ideals. So J is also an intersection of p p 0 prime ideals, a contradiction. It follows that every proper radical ideal is an intersection of prime ideals. To see the uniqueness, recall that, by Exercise 1.5 in Chapter 1, for any ring R, if p is a prime ideal in R and I1,...,Ik are ideals such that p I I , then there exists an i such that I p. Suppose that ⊇ 1 ∩···∩ k i ⊆ p p = q q . 1 ∩···∩ k 1 ∩···∩ ℓ Applying Exercise 1.5 to this situation, it follows that, for all i, there exists a j such that p q and likewise q p for some a. Thus p p . So i ⊆ j j ⊆ a i ⊆ a if there are no containment relations among the pi, we must have pi = pa and thus pi = qj = pa. Thus every pi is a qj. The converse then holds by symmetry.

For example, suppose that X = V (f) for f k[x1,...,xn]. As k[x1,...,xn] a ∈ is a UFD, we can factor f = p 1 par and then X = V (p ) V (p ), 1 ··· r 1 ∩···∩ r where the V (pi) are the irreducible components of X. We can partially generalize the above to the context of aﬃne schemes. First, a preliminary deﬁnition: 168 CHAPTER 5. IDEALS IN NOETHERIAN RINGS

Definition 5.1.4. Let X be a topological space. Then X is a Noetherian topological space if it satisfies the descending chain condition on closed subsets: If X X is a decreasing sequence of closed subsets of X, then 1 ⊇ 2 ⊇··· the sequence is eventually constant: there exists an N such that Xk = Xk+1 for all k N. ≥ Lemma 5.1.5. If R is a Noetherian ring, or more generally if R satisfies the ascending chain condition on radical ideals, then Spec R is a Noetherian topological space.

Proof. By Exercise 1.23 in Chapter 1, the closed sets Xk are of the form Xk = V (Ik), where the Ik = √Ik are radical ideals with Ik Ik+1. By hypothesis, there exists an N such that I = I for all k ⊆ N. Thus k k+1 ≥ X = X for all k N as well. k k+1 ≥ A very straightforward argument, left to the reader, shows the following:

Proposition 5.1.6. Let X be a Noetherian topological space and let Y be a n closed subset of X. Then Y = i=1 Yi, where the Yi are irreducible closed subsets of X. If in addition we require that, for i = j, Yi is not contained in S 6 Yj, then the Yi are unique up to order and are called the irreducible components of Y .

If X is an affine algebraic set, then the irreducible components of X are identified with the irreducible components of Spec A(X). The following, either in the case of affine algebraic sets or Noetherian topological spaces, is also straightforward and its proof is left to the reader:

Proposition 5.1.7. Let X be a Noetherian topological space and suppose that the irreducible components of X are X1,...,Xn. If Z is any irreducible closed subset of X, then Z is contained in Xi for at least one i. In particular, the Xi are exactly the maximal irreducible closed subsets of X.

One way to describe the irreducible components of Spec R, where R is Noetherian, or of X, where X is an aﬃne algebraic set, is as follows:

Proposition 5.1.8. Let R be a Noetherian ring. Then there are only ﬁnitely many minimal prime ideals in R. If p1,..., pk are the minimal primes in R, then p p = √0. Moreover, the irreducible components of Spec R are 1 ∩···∩ k exactly the closed sets V (pi). 5.2. ASSOCIATED PRIMES 169

More generally, if I is an ideal in R, then there are only ﬁnitely many prime ideals in R containing I and minimal with respect to this condition. If p ,..., p are the minimal such primes in R, then p p = √I, and 1 k 1 ∩···∩ k the irreducible components of V (I) are exactly the closed sets V (pi).

Proof. First assume I = 0. Applying Proposition 5.1.3 to the radical ideal √0, we can write √0 = p1 pk, where the prime ideals pi satisfy no containment relations. If p is∩···∩ any prime ideal of R, then √0 p, and hence p p p. By Exercise 1.5, p p for some i. In particular,⊆ if p is 1 ∩···∩ k ⊆ i ⊆ minimal, then p = p for some i. Thus, the set p ,..., p contains all of the i { 1 k} minimal primes. Since there are no containments among the pi, all of the pi are minimal, so that p1,..., pk is exactly the set of minimal primes. The ﬁnal statement is then{ clear by} the correspondence between prime ideals of R and irreducible closed subsets of Spec R (Exercise 1.24 in Chapter 1). The claim for an arbitrary ideal I follows from the case I = 0 by passing to the ring R/I.

We will give another proof of a more general version of Proposition 5.1.8 in the next section. Our goal will now be to generalize the statement of Proposition 5.1.3 to ideals which are not necessarily radical ideals. As we shall see, there is a somewhat technical generalization, although we will lose the uniqueness. As is often the case, the arguments will be greatly simpliﬁed by considering a generalization, to showing that a submodule of a ﬁnitely generated R-module is equal to an intersection of submodules of M of a very special type.

5.2 Associated primes

Deﬁnition 5.2.1. Let R be a ring and M an R-module. An element r R is a zero divisor on M if there exists m M, m = 0, such that rm = 0.∈ The element r is nilpotent on M if there exists∈ an n>6 0 such that rnM = 0.

Deﬁnition 5.2.2. Let R be a ring and M an R-module. The support of M, Supp M, is the set p Spec R : M =0 . { ∈ p 6 } In terms of the sheaf M on Spec R deﬁned by M, p Supp M if and only if the stalk of M at the point p Spec R is nonzero. ∈ ∈ f f 170 CHAPTER 5. IDEALS IN NOETHERIAN RINGS

Example 5.2.3. If I is an ideal of R, then Supp(R/I) = V (I), since (R/I)p = 0 if and only if Ip = Rp if and only if I (R p) = if and only if I does not contain p if and only if p / V (I). ∩ − 6 ∅ In particular, Supp R = V (0) = Spec R. ∈ The following lemma is clear from the exactness property of localization:

Lemma 5.2.4. If 0 M ′ M M ′′ 0 is an exact sequence of R- → → → → modules, then Supp M = Supp M ′ Supp M ′′. ∪ Lemma 5.2.5. Suppose that M is a ﬁnitely generated R-module. Then Supp M = V (Ann M). In particular, Supp M is a closed subset of Spec R. Proof. We begin with the following lemma:

Lemma 5.2.6. Let 0 M ′ M M ′′ 0 be an exact sequence of → → → → R-modules. Then V (Ann M)= V (Ann M ′) V (Ann M ′′). ∪ Proof. An easy argument shows that Ann M Ann M ′ Ann M ′′, and hence ⊆ ∩

V (Ann M ′ Ann M ′′)= V (Ann M ′) V (Ann M ′′) V (Ann M). ∩ ∪ ⊆

Moreover, clearly Ann M ′ Ann M ′′ Ann M. Thus · ⊆

V (Ann M) V (Ann M ′ Ann M ′′)= V (Ann M ′) V (Ann M ′′). ⊆ · ∪

It follows that V (Ann M)= V (Ann M ′) V (Ann M ′′). ∪ We return to the proof of Lemma 5.2.5. The proof is now by induction on the number of generators of M. If M is generated by one element, then M ∼= R/I, and so Supp M = V (I)= V (Ann M). In general, if M is generated by n elements, there is an exact sequence 0 M ′ M M ′′ 0, where M ′ = → → → → ∼ R/I and M ′′ is generated by n 1 elements. Thus by induction Supp M = − Supp M ′ Supp M ′′ = V (Ann M ′) V (Ann M ′′)= V (Ann M). ∪ ∪ Deﬁnition 5.2.7. Let R be a ring and M an R-module. A prime p is an associated prime of M if there exists an m M with Ann m = p, if and only if there exists an injective homomorphism of∈ R-modules R/p M. The set of all associated primes of M is denoted (somewhat unfortunately)→ Ass M. If M = R/I, one sometimes (incorrectly) says that an associated prime of R/I is an associated prime of I and writes Ass I for the set of all associated primes of R/I. 5.2. ASSOCIATED PRIMES 171

Example 5.2.8. 1) If R is an integral domain and M is torsion free, then Ass M = (0) . { } 2) If M = R/p, where p is a prime ideal, then Ass R/p = p . To see this, note that clearly p Ass R/p. Conversely, if x R/p {and} x = 0, then { } ∈ ∈ 6 Ann x = p. Thus Ass R/p = p . { } Lemma 5.2.9. If p is a maximal element in the set of all ideals Ann m : m M, m = 0 , then p is a prime ideal. In particular, if R is Noetherian{ and∈M =0,6 then} Ass M = . 6 6 ∅ Proof. Suppose that p = Ann m, m = 0, and that p is maximal among all 6 ideals of R of this form. Suppose that rs p and that r / p. Then by deﬁnition rm = 0. Clearly p Ann rm, and∈ so by maximality∈ p = Ann rm. 6 ⊆ As s Ann rm, s p. Thus p is prime. ∈ ∈ Corollary 5.2.10. The set of r R which are zero divisors on M is the union of the associated primes of M∈ .

Proof. If p is an associated prime of M, then p = Ann m for some m, necessarily nonzero, and hence p is contained in the set of all zero divisors on M. Conversely, suppose that r is a zero divisor on M. Thus rm = 0 for some m R, m =0. If I = Ann m, then r I and there exists a maximal element ∈ 6 ∈ p in the set of all ideals of the form Ann m : m M, m = 0 and which contain I. Such an ideal p satisﬁes the{ hypotheses∈ of the above6 } lemma, and hence is an associated prime of M. Thus r is contained in the union of the associated primes of M.

Next we discuss the behavior of Ass M under localization:

Lemma 5.2.11. Suppose that R is Noetherian. Let S be a multiplicative subset of R and let M be an R-module. Then

1 1 Ass S− M = S− p : p Ass M, p S = . { ∈ ∩ ∅} 1 Proof. Suppose that p Ass M, p S = . Then S− p is a prime ideal 1 ∈ ∩ ∅ 1 in S− R. Moreover, an injection R/p M gives an injection S− (R/p) = 1 1 1 1 → 1 ∼ S− R/S− p S− M. Thus S− p Ass S− M. → ∈ 1 To see the opposite inclusion, let q be a prime ideal of S− R, necessarily 1 1 of the form S− p, and suppose that q Ass S− M. Then there is an injection ∈ 172 CHAPTER 5. IDEALS IN NOETHERIAN RINGS

1 1 1 1 S− (R/p) = S− R/S− p S− M. Since R is Noetherian, the ﬁnitely gener- ∼ → ated R-module R/p is ﬁnitely presented. Thus, by Exercise 3.19 in Chapter 3, 1 1 1 HomS−1R(S− (R/p),S− M) ∼= S− HomR(R/p, M), where the above isomorphism is induced by the natural homomorphism 1 1 HomR(R/p, M) HomS−1R(S− (R/p),S− M). Hence there exists a ho- → 1 1 momorphism ϕ: R/p M and an s S such that ϕ/s: S− (R/p) S− M → ∈ 1 → which is an injective homomorphism of S− R-modules. It is then easy to check that ϕ(x) = 0 if and only if there exists a t S such that tx = 0 in R/p. As p S = , t is not a zero divisor on R/p,∈ and hence x = 0. Thus ∩ ∅ ϕ: R/p M is injective, so that p Ass M. → ∈ Corollary 5.2.12. If R is Noetherian and M is an R-module, then Ass M Supp M, and the minimal elements of Supp M are the same as the minimal⊆ elements of Ass M.

Proof. First suppose that p Ass M. With S = R p, and hence S p = , the above lemma implies that∈ pR Ass M , and in− particular M =∩ 0. Thus∅ p ∈ p p 6 p Supp M. ∈ Conversely, suppose that p Supp M, so that Mp = 0. Since Rp is Noetherian, Ass M = . By the∈ previous lemma, there exists6 a prime ideal p 6 ∅ p0 such that p0 Ass M and (R p) p0 = , i.e. p0 p. It follows that every prime ideal∈ in Supp M contains− ∩ an element∅ of Ass⊆M, and so the two sets have the same collections of minimal elements.

We can prove a more precise result in case M is ﬁnitely generated. First, the following gives a partial description of how associated primes behave over exact sequences:

Lemma 5.2.13. If 0 M ′ M M ′′ 0 is an exact sequence of R- → → → → modules, then Ass M ′ Ass M Ass M ′ Ass M ′′. If M = M ′ M ′′, then ⊆ ⊆ ∪ ⊕ Ass M = Ass M ′ Ass M ′′. ∪

Proof. The inclusion Ass M ′ Ass M is obvious. Suppose that p Ass M. ⊆ ∈ Then there exists a submodule N of M isomorphic to R/p. If N M ′ = 0, ∩ 6 let x N M ′. As we have seen in 2) of Example 5.2.8, Ann x = p, ∈ ∩ and hence p Ass M ′. Otherwise, N M ′ = 0 and the projection of N ∈ ∩ into M ′′ is injective. Thus M ′′ contains a submodule isomorphic to R/p, and so p Ass M ′′. The last statement is clear since, on the one hand, ∈ 5.2. ASSOCIATED PRIMES 173

Ass M Ass M ′ Ass M ′′, and since both M ′ and M ′′ are isomorphic to ⊆ ∪ submodules of M, we have Ass M ′ Ass M and Ass M ′′ Ass M. Thus ⊆ ⊆ Ass M = Ass M ′ Ass M ′′. ∪ Proposition 5.2.14. Let R be Noetherian and let M =0 be a ﬁnitely gener- 6 ated R-module. Then there exists a strictly increasing sequence of submodules

0 M M = M, ⊂ 1 ⊂···⊂ n such that, for all i> 0, Mi/Mi 1 = R/pi, where pi is a prime ideal. − ∼ Proof. Since M = 0, there exists an associated prime p of M and a sub- 6 1 module M1 ∼= R/p1. If M = M1, we are done. Otherwise, apply the above to M/M1 to ﬁnd a prime p2 of R and a submodule M2 ∼= R/p2 of M/M1, and let M2 be the inverse image of M 2 in M. Continuing in this way, we ﬁnd a strictly increasing equence of submodules 0 M1 M2 . Since M is Noetherian R-module, this sequence must terminate.⊂ ⊂ ⊂ ···

The prime ideals pi in the preceding proposition are far from unique in general (see Exercise 5.8). Nonetheless, we can say the following:

Lemma 5.2.15. Suppose that R is a ring and that M is an R-module for which there exists a strictly increasing sequence of submodules

0 M M = M, ⊂ 1 ⊂···⊂ n such that, for all i > 0, Mi/Mi 1 = R/pi, where pi is a prime ideal in R. − ∼ Then:

(i) Supp M = V (p ). In particular, p Supp M for every i. i i i ∈ (ii) Ass M pS,..., p . ⊆{ 1 n} It follows that the minimal elements of Supp M are the minimal elements of p ,..., p , and, if R is Noetherian, then these are just the minimal { 1 n} elements of Ass M.

Proof. To see (i), by repeatedly applying Lemma 5.2.4, we see that Supp M =

i Supp R/pi = i V (pi), by Example 5.2.3. Part (ii) follows similarly from Lemma 5.2.13 and Example 5.2.8. The ﬁnal statement follows from Corol- laryS 5.2.12. S 174 CHAPTER 5. IDEALS IN NOETHERIAN RINGS

Corollary 5.2.16. If R is Noetherian and M is a finitely generated R- module, then Ass M is finite. Corollary 5.2.17. Suppose that R is Noetherian. Then there are only a finite number p ,..., p of minimal prime ideals of R, and √0= p { 1 n} 1 ∩···∩ pn. More generally, for any ideal I of R, there are only a finite number of prime ideals of R containing R and minimal with respect to this property, and √I is the intersection of these finitely many prime ideals. Proof. Taking M = R, we see that there are only a finite number of minimal prime ideals of R. Clearly, the intersection of all of the prime ideals of R, namely √0, is the same as the intersection of the minimal prime ideals. This proves the first claim, and the second follows by looking at R/I.

5.3 Primary decomposition

We now look at the special case where the R-module M has just one associated prime. Throughout this section, we suppose that R is Noethe- rian and that M is a ﬁnitely generated R-module.

Proposition 5.3.1. Ass M = p has just one element if and only if M =0 and, for all r R, either multiplication{ } by r is injective or r is nilpotent6 ∈ on M. In this case, p = V (Ann M) is the set of all r R such that r is nilpotent on M. ∈

Proof. First suppose that Ass M = p . In particular, M = 0. Let m M, m = 0. Then Ass(R m) = {and} Ass(R m) Ass6 M = p , so∈ that Ass(6 R m) = p as· well.6 Thus∅ p is the unique· ⊆ minimal element{ } of · { } Supp(R m) = V (Ann m). But √Ann m is the intersection of all of the prime ideals· in V (Ann m), and thus √Ann m = p. Hence, if r p, then n ∈ r m = 0 for some n> 0. Applying the above to a ﬁnite set m1,...,mk of generators of M, we see that if r p, then there exists an N{ > 0 such that} rN m = 0 for all i and hence rN m∈= 0 for all m M. If on the other hand i ∈ r / p, then, for all nonzero m M, r / Ann m. Thus multiplication by r is injective.∈ Moreover, the proof∈ shows that,∈ in this case, p = V (Ann M). Conversely suppose that M = 0 and, for all r R, either multiplication by r is injective or r is nilpotent6 on M. Deﬁne ∈

p = r R : multiplication by r is nilpotent . { ∈ } 5.3. PRIMARY DECOMPOSITION 175

If r,s / p, then neither r nor s is a zero divisor on M, and hence multiplica- ∈ tion by rs is injective. Thus p is a prime ideal of R. Let p1,..., pn be the associated primes of M. If r / p, then multiplication by{r is injective,} and so r / p for any i. Thus p ∈ p for every i. On the other hand, if r p, ∈ i i ⊆ ∈ then rnM = 0, and so rn p for every i. Thus r p as well. Hence p p ∈ i ∈ i ⊆ i for every i, and so p = pi. Thus there is just one associated prime, namely p.

Deﬁnition 5.3.2. If Ass M = p has just one element, then we say that M { } is p-coprimary. The submodule N of M is p-primary if the quotient M/N is p-coprimary. A submodule N of M is a primary submodule if N is p-primary, where p is necessarily V (Ann(M/N)) and hence is uniquely determined by N and M.

In particular, an ideal q of R is p-primary if, for every r R, either ∈ multiplication by r on R/q is injective or r is nilpotent on R/q, and √q = p. We say that q is a primary ideal if q = R and, for every r R, either 6 ∈ multiplication by r on R/q is injective or r is nilpotent on R/q. Equivalently, the ideal q of R is a primary ideal if q = R and, for all r, s R, if rs q, then either s q or rn q for some n>6 0, i.e. every zero divisor∈ in the∈ ring ∈ ∈ R/q is nilpotent.

Lemma 5.3.3. Let q be a primary ideal. Then √q = p is a prime ideal, and it is the smallest prime ideal containing q.

Proof. Clearly √q is contained in every prime ideal containing q, so it suﬃces to prove that √q = p is a prime ideal. If rs √q = p, then for some n> 0, rnsn q and thus either sn q or rnm q∈for some m > 0. Thus either r p∈or s p, so that p is a prime∈ ideal. ∈ ∈ ∈ Example 5.3.4. Suppose that R is an integral domain and that p = (t) is a prime ideal. Then one can show that the p-primary ideals are exactly the ideals of the form (tn) for n> 0 (Exercise 5.5).

In general, a p-primary ideal q need not be of the form pm for some m> 0, and conversely an ideal of the form pm, where p is prime, need not be 2 2 primary. For example, with R = k[x1, x2] and I =(x1, x2), R/I = k[x2]/(x2), and every zero divisor is nilpotent. Thus I is primary, but √I =(x1, x2) and n I is not of the form (x1, x2) for any integer n > 0. As another example, let R = k[x , x , x ]/(x x x2) and let p = (x , x )R. Then R/p = k[x ] 1 2 3 1 2 − 3 1 3 2 176 CHAPTER 5. IDEALS IN NOETHERIAN RINGS

2 2 2 and so p is prime. Note that x3 = x1x2 p . However x1 / p (why?) and 2 2 ∈ ∈ x2 / √p = p, so that p is not primary. ∈However, we can say the following:

Lemma 5.3.5. Let q be an ideal of R such that √q = m is a maximal ideal. Then q is m-primary. Proof. Let p be a prime ideal of R/q. Then the inverse image of p is a prime ideal in R containing q and thus containing m. Thus the inverse image of p in R is equal to m, and so there is a unique prime ideal of R/q, the image of m. It follows that every element of R/q is either a unit or nilpotent. Thus q is m-primary. Returning to the case of modules, we have:

Lemma 5.3.6. If N1 and N2 are two p-primary submodules of M, then N1 N2 is a p-primary submodule of M. Hence the intersection of ﬁnitely many∩ p-primary submodules of M is a p-primary submodule of M. Proof. Since N N N , N N = M, and hence Ass(M/N N ) = . 1 ∩ 2 ⊂ i 1 ∩ 2 6 1 ∩ 2 6 ∅ On the other hand, there is an injective homomorphism from M/N1 N2 to (M/N ) (M/N ). Applying Lemma 5.2.13, we have ∩ 1 ⊕ 2 Ass(M/N N ) Ass(M/N M/N ) = Ass(M/N ) Ass(M/N )= p . 1 ∩ 2 ⊆ 1 ⊕ 2 1 ∪ 2 { }

Thus Ass(M/N1 N2)= p , so that N1 N2 is a p-primary submodule of M. ∩ { } ∩ Lemma 5.3.7. Let N be a submodule of M with N = M, and let S be a 6 1 multiplicative subset of R. If S p = and N is p-primary, then S− N = 1 ∩ 6 ∅ S− M. If S p = , then N is a p-primary submodule of M if and only if 1 ∩ 1 ∅ 1 S− N is an S− p-primary submodule of S− M. In this case, the preimage of 1 1 S− N in M under the natural homomorphism M S− M is exactly N. → 1 1 Proof. By Lemma 5.2.11, Ass S− (M/N)= S− p : p Ass(M/N), p S = { ∈ 1 ∩ . In particular, if Ass(M/N)= p and S p = , then Ass S− (M/N)= , ∅} 1 1 { 1} ∩ 6 ∅ 1 1 ∅ and hence S− (M/N)= S− M/S− N = 0. Thus S− N = S− M. 1 1 If S p = and Ass(M/N) = p , then Ass S− (M/N) = S− p . ∩1 ∅ 1 { } 1 { } Hence S− N is an S− p-primary submodule of S− M. Conversely, suppose 1 1 that Ass S− (M/N) = S− p . Then Ass(M/N) is contained in the set of { } 1 1 all prime ideals P of R such that S− P = S− p and S P = . It then ∩ ∅ 5.3. PRIMARY DECOMPOSITION 177 follows from Proposition 3.2.1 that P = p. Thus Ass(M/N) p , and ⊆ { } hence Ass(M/N) = p since M/N = 0. Finally, suppose that m M and { } 6 1 ∈ that the image of m is contained in S− N. Then there exists s S such that sm m. Thus sm = 0 in M/N. If m / N, then s is a zero divisor∈ on ∈ ∈ M/N, and hence s is nilpotent on M. By Proposition 5.3.1, s p. But this contradicts S p = . ∈ ∩ ∅ Theorem 5.3.8 (Existence of primary decompositions). If N is a submodule of M with M = N, then there exist ﬁnitely many primary submodules Q ,...,Q such that6 N = Q Q . 1 r 1 ∩···∩ r Proof. After replacing M by M/N, we may assume that N = 0. Next we use the following lemma: Lemma 5.3.9. Suppose that p Ass M. Then there exists a submodule Q ∈ of M, possibly 0, such that Ass(M/Q)= p and p / Ass Q. { } ∈ Assuming Lemma 5.3.9, let us the proof of Theorem 5.3.8. Suppose that Ass(M/N)= p ,..., p . Applying Lemma 5.3.9 to M and p , there exists { 1 n} i a submodule Q of M such that Ass(M/Q ) = p and p / Ass Q . In i i { i} i ∈ i particular, Qi is pi-primary. Let Q = Q1 Qn. Since Q Qi, Ass Q Ass Q Ass M for every i, and thus Ass Q∩···∩n (Ass M p⊆ )= . Thus⊆ i ⊆ ⊆ i=1 −{ i} ∅ Q = 0. T Proof of Lemma 5.3.9. Let be the set of all submodules P of M such that I p / Ass P . Then = , since for example 0 (because Ass 0 = ). Thus,∈ has a maximalI 6 element∅ Q. By assumption,∈p I/ Ass Q. In particular,∅ Q = MI , so that Ass(M/Q) = . We claim that Ass(∈ M/Q) = p . Let 6 6 ∅ { } P Ass(M/Q), let Q′ be a submodule of M/Q isomorphic to R/P, and let ∈ Q be the inverse image of of Q′ in M. From the exact sequence 0 Q 0 → → Q Q′ 0, we see that Ass Q Ass Q Ass Q′ = Ass Q P . We claim 0 → → 0 ⊆ ∪ ∪{ } that P Ass M. For, if not, then since Ass Q0 Ass M, P / Ass Q0. In this case, Ass∈ Q Ass Q, and so p / Ass Q . But⊆ then Q ∈ , contradicting 0 ⊆ ∈ 0 0 ∈ I the maximality of Q. Likewise, if P = p, then Ass Q Ass Q P again 6 0 ⊆ ∪{ } implies that p / Ass Q0, so that again Q0 , contradicting the maximality of Q. Thus P =∈ p, and so Ass(M/Q)= p∈. I { } We turn now to the uniqueness question for primary decompositions. There are some obvious ways to eliminate potential sources of non-uniqueness. We begin with a preliminary deﬁnition. 178 CHAPTER 5. IDEALS IN NOETHERIAN RINGS

Deﬁnition 5.3.10. Let N be a submodule of M and let N = Q Q 1 ∩···∩ r be a primary decomposition of N. We say that the primary decomposition is minimal if

(i) For all i, Qi does not contain j=i Qj; 6 (ii) For all i = j, Ass(M/Q ) = Ass(T M/Q ). 6 i 6 j Lemma 5.3.11. For every submodule N of M, there exists a minimal primary decomposition of N.

Proof. By Theorem 5.3.8, there exists some primary decomposition N =

Q1 Qr of N. If j=i Qj Qi, then Q1 Qr = j=i Qj, and we ∩···∩ 6 ⊆ ∩···∩ 6 can replace Q1,...Qr by a proper subset. Continuing until this is no longer possible, we may assumeT that (i) of Deﬁnition 5.3.10 holds.T Now suppose that Ass(M/Q ) = Ass(M/Q ) for some i = j. By Lemma 5.3.6, Q Q i j 6 i ∩ j is again primary. We may thus amalgamate all of the Qj with a common associated prime to achieve (ii) of Deﬁnition 5.3.10. We then have the following partial uniqueness result for primary decompositions:

Theorem 5.3.12 (Uniqueness of primary decomposition). Suppose that N is a submodule of M and that N = Q1 Qr and N = Q1′ Qs′ are two minimal primary decompositions of∩···∩N. Let Ass(M/Q ) =∩···∩p and let i { i} Ass(M/Q′ )= p′ . Then j { j} (i) p ,..., p = p′ ,..., p′ = Ass(M/N). In particular, r = s. { 1 r} { 1 s}

(ii) If Qi is such that pi is a minimal prime in Ass(M/N), then Qi is the inverse image in M of the kernel of the natural homomorphism M/N (M/N) . Hence, every such Q is equal to Q′ for some k, → pi i k and conversely. In other words, the submodules Qi which correspond to minimal primes in Ass(M/N) are independent of the choice of minimal primary decomposition.

Proof. As before, we may assume that N = 0. To prove (i), it suﬃces to show that Ass M = p ,..., p . Since 0 = Q Q , the homomorphism { 1 r} 1 ∩···∩ r M (M/Q ) is injective. It follows that Ass M Ass(M/Q ) = → i i ⊆ i i p1,..., pr . Conversely, we must show that pi is contained in Ass M for { L } S every i. Let Pi = j=i Qj. Then Pi Qi = 0, but Pi = 0 since the 6 ∩ 6 T 5.4. ARTINIAN RINGS 179 primary decomposition is minimal. Thus the homomorphism P M/Q i → i is injective, and identiﬁes Pi with a nonzero submodule of M/Qi. Thus Ass Pi Ass(M/Qi) = pi . It follows that Ass Pi = pi . Since Pi is also a submodule⊆ of M, p { Ass} M for every i. Thus Ass{M}= p ,..., p as i ∈ { 1 r} claimed. Now suppose that p is a minimal element of Ass M. If i = j, then i 6 pj (R pi) = , since otherwise we would have pj pi, contradicting ∩ − 6 ∅ ⊆ 1 the minimality of pi. Thus, by Lemma 5.3.7, with S = R pi, S− Qj = 1 − 1 S− M = Mpi for j = i, and Qi is the preimage of the submodule S− Qi of 1 6 S− M under the natural homomorphism. But 0 = Q1 Qr, so that 1 1 1 ∩···∩ 0 = S− Q1 S− Qr = S− Qi. Thus, Qi is exactly the kernel of the homomorphism∩···∩M M . → pi Let us state the above results explicitly for the case M = R/I, the main case of interest:

Corollary 5.3.13. Every ideal I in R is an intersection I = q1 qr of primary ideals. If the primary decomposition is minimal in the∩···∩ sense of Definition 5.3.10, then the prime ideals pi = √qi are exactly the associated primes of I, and hence are independent of the choice of primary decomposition. Finally, if pi is a minimal prime in Ass(R/I), then qi is independent of the particular choice of primary decomposition. The minimal primes (sometimes called isolated) in Ass(R/I) correspond to the components of V (I). We can think of the difference between pi and qi as reflecting a sort of “generic multiplicity” along the component V (pi). For example, if pi = (p) is principal, then a typical pi-primary ideal is of the form (pk) for some positive integer k, which we view as the multiplicity of the component V (pi) in V (I). In general, for a minimal prime pi of

Ass(R/I), the Rpi -module (R/I)pi has ﬁnite length (Exercise 5.10), and the length ℓ((R/I)pi ) is again some measure of the multiplicity of the component V (pi). The primes in Ass(R/I) which are not minimal are called embedded primes. They measure nilpotence in R/I along proper subvarieties of the components of Spec(R/I).

5.4 Artinian rings

In this section we shall describe the structure of Artinian rings. In particular, we show that R is Artinian if and only if it is Noetherian and of dimension 180 CHAPTER 5. IDEALS IN NOETHERIAN RINGS zero.

Lemma 5.4.1. Let R be an Artinian ring. Then there are only ﬁnitely many maximal ideals in R.

Proof. Let be the set of all ﬁnite intersections m1 mn of maximal ideals in R.S If n is a minimal element of this set, then∩···∩n = m m for 1 ∩···∩ k some collection m1,..., mk of maximal ideals. Let m be an arbitrary maximal ideal. We claim that m = mi for some i. Indeed, since

n m = m m m n ∩ 1 ∩···∩ k ∩ ⊆ and n m , we must have n m = n. Thus m m1 mk. Since m is maximal∩ ∈ and S therefore prime,∩ we must have m ⊇m for∩···∩ some i. But since ⊇ i mi is maximal, in fact m = mi.

Lemma 5.4.2. Let R be an Artinian ring. Then every prime ideal p in R is maximal. In particular an Artinian integral domain is a ﬁeld.

Proof. If p is a prime ideal, then R/p is an Artinian integral domain. Choose r R/p,r = 0. Then we have the decreasing sequence of ideals (r) (r2) ∈ 6 ⊇ ⊇ (r3) . It follows that there exists a k > 0 with (rk)=(rk+1). Thus rk =⊇crk ···+1 for some c R/p. Since R/p is an integral domain, cr = 1 in R/p and thus r is a unit. Hence∈ R/p is a ﬁeld, and so p is maximal.

Theorem 5.4.3. A ring R is Artinian if and only if it is Noetherian and every prime ideal is maximal. In other words, R is Artinian if and only if it is Noetherian and of dimension zero.

Proof. We have seen that, if R is Artinian, then every prime ideal is maximal. Next we claim: Lemma 5.4.4. If R is Artinian, then the radical N = √0 of R is a nilpotent ideal, in other words N m = (0) for some m> 0.

Proof. Note that the statement is clear if R is Noetherian. In general, consider the decreasing sequence N N 2 N 3 . Thus there exists a ⊇ ⊇ ⊇ ··· k 1 such that N k = N k+1 = = I, say. Note that I2 = N 2k = I. If I =≥ 0, deﬁne a collection of ideals··· of R by: 6 I = J R : JI =0 . I { ⊂ 6 } 5.4. ARTINIAN RINGS 181

Choose a minimal element J of . By definition, there exists r J such 0 I ∈ 0 that rI = 0, and thus (r) . As(r) J0, (r) = J0. Moreover (rI)I = 2 6 ∈ I ⊆ r(I )= rI = 0, and so rI . But also rI (r)= J0, and so by minimality rI = (r). Thus6 there exists∈ Is I such that⊆ r = rs, and so r = rsn for all ∈ n> 0. But s I N = √0, and so for some n> 0, sn = 0. It follows that r = 0 as well,∈ a contradiction.⊆ Thus I = N k = (0). Returning to the proof of Theorem 5.4.3, we know that every prime ideal of R is maximal and that there are only finitely many such, say m1,..., mk. k Then N = i=1 mi is the radical of R. By the previous lemma, there exists an n> 0 such that N n = 0. Thus T n mn m N n = (0). i ⊆ i ⊆ i i ! Y \ n So i mi = 0 for some n > 0. Now choose nonnegative integers ai with aj ai n, and ai < n for some i. Let I = j mj . Then if ai < n we can≤Q consider m I, and I/m I is a vector space over R/m . Since I is i · i · i a submodule of R, it is an Artinian R-module,Q and thus so is the quotient I/m I. But I/m I is also a module over the field R/m , and it is clearly i · i · i Artinian as an R/mi-module. It is elementary to see that a vector space over a field is Artinian if and only if it is finite-dimensional, if and only if it is Noetherian, if and only if it has a composition series. Thus I/m I has such. i · Now we can choose a finite sequence

I = R I = m I I = m I I = mn =0. 0 ⊇ 1 i1 0 ⊇ 2 i2 1 ···⊇ ℓ i i Y

Since each Ij/Ij+1 is an Artinian module over the ﬁeld R/mij+1 and thus over R, each Ij/Ij+1 has a composition series over R. Then R itself has a composition series over R, so that R is Noetherian. Conversely, suppose that R is Noetherian and that every prime ideal in R is maximal. Choose a primary decomposition of (0):

(0) = q q . 1 ∩···∩ k

By hypothesis pi = √qi is maximal. Since R is Noetherian, there exists an n> 0 such that pn q for every i, and thus i ⊆ i pn pn q q = (0). 1 ··· k ⊆ 1 ∩···∩ k 182 CHAPTER 5. IDEALS IN NOETHERIAN RINGS

So pn pn = (0). By the above arguments, there exists a ﬁnite sequence 1 ··· k I = R I = p I I = p I I = pn =0. 0 ⊇ 1 i1 0 ⊇ 2 i2 1 ···⊇ ℓ i i Y

Moreover each Ij/Ij+1 is a Noetherian module over the ﬁeld R/pij+1 and thus over R, and so each Ij/Ij+1 has a composition series over R. Then R also has a composition series over itself, so that R is Artinian. Proposition 5.4.5. Let R be a local Noetherian ring with maximal ideal m. Then either, for all n 0, mn = mn+1, or there exists an n 0 such that mn = mn+1, in which case≥ mn =6 (0) and R is Artinian. ≥ Proof. Note that, if for all n 0, mn = mn+1, then R is not Artinian. Suppose that mn = mn+1 for some≥ n 6 0. By Nakayama’s lemma, since ≥ mn is ﬁnitely generated, mn = (0). If p is a prime ideal of R, then since p contains mn, p m. As m is maximal, m = p. In particular every prime ⊇ ideal of R is maximal. By Theorem 5.4.3, R is Artinian. Note that we can use Proposition 5.4.5 to construct examples of Artinian rings as follows: Lemma 5.4.6. Let R be a Noetherian ring and let m be a maximal ideal in R. Then, for every positive integer N, R/mN is an Artinian local ring. Proof. For every positive integer N, the ring R/mN has a unique maximal ideal, the image of m, and some power of it is zero. Thus, Proposition 5.4.5, R/mN is an Artinian local ring. We further remark that in an Artinian local ring, there is a unique prime ideal and so every zero divisor is nilpotent. Let us give a structure theorem for more general Artinian rings: Proposition 5.4.7. Let R be an Artinian ring. Then R is isomorphic to a n product i=1 Ri, where each Ri is a local Artinian ring, and the factors Ri are unique up to order and isomorphism. Q Proof. Let m1,..., mn be the distinct maximal ideals of R. As we have seen, mk = (0) for some k. Now the mk are comaximal, since for i = j i i i 6 Q k k k k √ mi + mj = mi + mj = mi + mj = R = R. r q q q p 5.4. ARTINIAN RINGS 183

Thus by the Chinese remainder theorem

R (R/mk) → i i Y k is surjective with kernel i mi = (0). Hence R is isomorphic to a product of local Artinian rings. To see uniqueness, letQ (0) = q q be a minimal primary decompo- 1 ∩···∩ n sition of (0). Then √qi is necessarily minimal, since every prime ideal of R is maximal, and so the primary decomposition is unique by the uniqueness n theorem. On the other hand, given an isomorphism R ∼= i=1 Ri, where each Ri is a (nonzero) local Artinian ring, the kernel ki of the induced map R R is primary since every zero divisor in R is nilpotent.Q Since → i i

ki = R1 Ri 1 0 Ri+1 Rn, ∼ ×···× − ×{ } × ×···× for i = j the radicals of ki and kj are distinct, ki is not contained in i=j kj, 6 6 and (0) = k1 kn. It follows that (0) = k1 kn is a minimal ∩···∩ ∩···∩ T primary decomposition of (0) and thus that after relabeling the ki = qi, proving uniqueness. For use in the section on Dedekind domains, we shall prove the following lemma:

Lemma 5.4.8. Let R be an Artinian local ring with maximal ideal m and residue ﬁeld k = R/m. Then the following are equivalent:

(i) Every ideal in R is a principal ideal;

(ii) The maximal ideal m is principal;

(iii) dim m/m2 1. k ≤ In case one of the above equivalent conditions holds, every nonzero ideal of R is equal to mn for some n 0. ≥ Proof. Clearly (i) = (ii) and (ii) = (iii) (if m =(r), then the surjection R m defined by r⇒induces a surjection⇒ R/m m/m2.) So we must show → → that (iii) = (i). ⇒ 2 2 First, if dimk m/m = 0, then m = m , so by Nakayama’s lemma (recall m is finitely generated since R is Noetherian) m = 0 and R is a field. Thus 184 CHAPTER 5. IDEALS IN NOETHERIAN RINGS

(i) is trivially satisﬁed. Otherwise suppose that m = (r). Let I be an ideal of R, where we suppose that I = (0). Then (since mN = (0) for some N) there exists an n such that I 6mn but I is not contained in mn+1. Choose s I mn+1. Since I mn,⊆ we can write s = arn for some a R. Since ∈ − ⊆ ∈ s / mn+1, a / m. Thus a is a unit, since R is local, so that rn I. It follows that∈ mn I∈ mn and so I = mn. Clearly I = (rn), so that ∈I is principal, ⊆ ⊆ and we have also seen that I is a power of m.

For example, the ring k[x]/(xN ) satisﬁes the conclusions of Lemma 5.4.8, N but the ring k[x1,...,xd]/m does not, where m =(x1,...,xd) and d,N > 1. In general, we can make the following deﬁnition:

Definition 5.4.9. Let R be a local ring with maximal ideal m and residue 2 field k = R/m. Then the embedding dimension of R is dimk m/m (which can be 0 or . If m is generated by d elements (for example if R is Noetherian, then m is∞ at least finitely generated), then the embedding dimension of R is at most d, as the surjection Rd m induces a surjection (R/m)d m/m2. The name comes from the implicit→ function theorem. In general, the→ ring R can have embedding dimension much larger than its actual dimension. (For a local Noetherian ring R, we shall see that the embedding dimension is always at least dim R.)

5.5 Fractional ideals and invertible modules

This section is preliminary to our discussion of Dedekind rings in the next section. Throughout this section, we do not assume that R is Noetherian.

Definition 5.5.1. Suppose that R is an integral domain and that K is its field of quotients. A fractional ideal of R is an R-submodule M of K such that there exists an r R,r = 0 with rM R. ∈ 6 ⊆ For example, a finitely generated R-submodule M of K is a fractional ideal. If M is a fractional ideal contained in R, then M is an integral ideal (of course this is the same as saying that M is an ideal of R). Given λ K∗, ∈ we denote the ideal λR = (λ) and call it a principal fractional ideal. Given M and M two fractional ideals, it is easy to check that M + M , M M , 1 2 1 2 1 ∩ 2 and M1M2 are fractional ideals, where M1M2 is defined in the obvious way, as 5.5. FRACTIONAL IDEALS AND INVERTIBLE MODULES 185

generated by all products of elements of M1 with elements of M2. Moreover, if we deﬁne the ideal quotient (M1 : M2) by (M : M )= λ K : λM M , 1 2 { ∈ 2 ⊆ 1} then it is easy to check that (M1 : M2) is an R-submodule of K. We claim that, if M = 0, then (M : M ) is in fact a fractional ideal. We must 2 6 1 2 show that there exists an r R,r = 0 such that r(M1 : M2) R. Choose α M , α = 0. Then λα∈ M 6 for all λ (M : M ).⊆ Now choose ∈ 2 6 ∈ 1 ∈ 1 2 s R,s = 0 such that sM1 R. It follows that (αs)λ sM1 R for all λ ∈ (M :6 M ), so that (αs)(M⊆ : M ) R. ∈ ⊆ ∈ 1 2 1 2 ⊆ Deﬁnition 5.5.2. With R and K as above, an R-submodule M of K is invertible if there exists an R-submodule M ′ of K such that MM ′ = R.

1 For example, (λ) is invertible and its inverse is (λ− ). Lemma 5.5.3. An invertible R-submodule M of K is ﬁnitely generated, and hence is a fractional ideal. n Proof. Suppose that MM ′ = R for some M ′, and write 1 = i=1 λiµi with λi M and µi M ′. We claim that M is generated by λ1,...,λn. Given ∈ ∈ P m M, m = m 1 = (mµ )λ . As mµ MM ′ = R, m is in the ∈ · i i i i ∈ R-submodule of M generated by λ1,...,λn and thus M is generated by P λ1,...,λn.

Lemma 5.5.4. If M and M ′ are two fractional ideals such that MM ′ = R, then M ′ =(R : M). Thus, if M is invertible, then its inverse is unique, and 1 necessarily equal to (R : M), which we also denote by M − . In general M is invertible if and only if M(R : M) = R. More generally, if M1 and M2 are 1 fractional ideals such that M2 is invertible, then (M1 : M2)= M1M2− .

Proof. If MM ′ = R, then by deﬁnition M ′ (R : M). On the other hand, by deﬁnition M(R : M) R, so that ⊆ ⊆ (R : M)= R(R : M)= MM ′(R : M)= M ′M(R : M) M ′R = M ′. ⊆ 1 So M ′ = (R : M). A similar argument shows that, if M2M2− = R, then 1 (M1 : M2)= M1M2− .

Lemma 5.5.5. If M1,...,Mn are fractional ideals such that M1 Mn is invertible, then each M is invertible. In particular, if M M is a··· principal i 1 ··· n fractional ideal, then each Mi is invertible. 186 CHAPTER 5. IDEALS IN NOETHERIAN RINGS

Proof. Let M = M M . Then 1 ··· n

1 1 M − Mj Mi = M − M = R, j=i ! Y6 so that Mi is invertible. Deﬁnition 5.5.6. Let be the set of invertible fractional ideals of R. The is a group under multiplication,I with identity element R. There is a mul- I tiplicative homomorphism K∗ deﬁned by λ (λ), with kernel R∗, the → I 7→ group of units of R. The quotient of be the subgroup of principal ideals is called the ideal class group of R. I

More generally, we discuss invertible R-modules M, for an arbitrary ring R:

Definition 5.5.7. Let M be a finitely generated R-module. We say that M is invertible if there exists an R-module M ′ such that M M ′ = R. (We ⊗R ∼ will see shortly that in this case M ′ is also finitely generated, and that, if R is an integral domain, then M is isomorphic, not uniquely, to an invertible fractional ideal in K.)

Theorem 5.5.8. A ﬁnitely generated R-module M is invertible if and only if M is projective and, for all prime ideals p of R, Mp ∼= Rp.

Proof. Suppose first that M is invertible and let M ′ be an R-module such that M M ′ = R. We shall show that M is projective and, for all prime ⊗R ∼ ideals p of R, Mp ∼= Rp. Assume first that R is local, with maximal ideal m and residue field k. Thus M k = M/mM. We shall show that M is a free ⊗R rank one R-module and hence that Mp ∼= Rp for all primes p. To see this, it follows from general properties of the tensor product that

(M/mM) (M ′/mM ′) = (M M ′) k = R k = k. ⊗k ∼ ⊗R ⊗R ⊗R

Thus dimk(M/mM) = dimk(M ′/mM ′) = 1. Choose x M such that x maps onto a generator of the one-dimensional k-vector space∈ M/mM. By Nakayama’s lemma, since M is ﬁnitely generated by hypothesis, x generates M as an R-module, so that M = R/I, where I = r R : rx =0 . If r I, ∼ { ∈ } ∈ then rM = 0 so that r(M M ′)= rR = 0, and hence r = 0. Thus I = (0) ⊗R and M ∼= R. 5.5. FRACTIONAL IDEALS AND INVERTIBLE MODULES 187

In the general case, let p be a prime ideal of R. Then Rp is a local ring and M M ′ = (M M ′) R = R . p ⊗Rp p ∼ ⊗R ⊗R p ∼ p Thus Mp is an invertible module over the local ring Rp and so by the above argument Mp is a free rank one Rp-module for every p. Next, we have: Lemma 5.5.9. An invertible R-module M is flat. In fact M is faithfully flat, in other words M is flat and M N =0 if and only if N =0. ⊗R

Proof. Since Mp is a free Rp-module for every p, and hence ﬂat, it follows that M is a ﬂat R-module. If M N = 0, then ⊗R

0= M ′ M N = R N = N, ⊗R ⊗R ⊗R so that N = 0.

Lemma 5.5.10. Suppose that M is an invertible R-module and suppose that M ′ is an R-module such that M M ′ = R. Then M ′ is ﬁnitely generated. ⊗R ∼ Hence M ′ is also invertible.

n Proof. Write 1 = m m′ with m M and m′ M ′. We claim that i=1 i ⊗ i i ∈ i ∈ M ′ is generated by the m′ . Consider the exact sequence P i n R M ′ Q 0 → → → deﬁned by the mi′ . We must show that Q = 0. By the ﬂatness of M, the sequence n M M M ′ M Q 0 → ⊗R → ⊗R → is still exact. By construction M R M ′ = R and the generator 1 lies in the n ⊗ ∼ image of the map M M M ′. Thus M Q = 0. By Lemma 5.5.9, → ⊗R ⊗R Q = 0. Thus M ′ is generated by m1′ ,...,mn′ .

Lemma 5.5.11. Suppose that M is an invertible R-module. Then M is ﬁnitely presented.

Proof. Since M is ﬁnitely generated, there is an exact sequence

0 N Ra M 0. → → → → 188 CHAPTER 5. IDEALS IN NOETHERIAN RINGS

If M ′ is such that M M ′ = R, then by the previous lemma M ′ is invertible, ⊗R ∼ and so it is a ﬂat R-module. Thus the sequence

a 0 M ′ N (M ′) M M ′ 0 → ⊗R → → ⊗R → is also exact. Since M R M ′ is free, this exact sequence splits and so ⊗ a a M ′ N is a quotient of (M ′) . Since M ′ is finitely generated, (M ′) is also ⊗R finitely generated, and so M ′ N is finitely generated as well. Now M ⊗R is finitely generated, and so M M ′ N is also finitely generated. But ⊗R ⊗R M M ′ N = R N = N, and so N is finitely generated. It follows ⊗R ⊗R ∼ ⊗R that M is finitely presented. Returning to the proof of Theorem 5.5.8, by Proposition 3.4.4, since M is finitely presented and Mp is free for all prime ideals p, M is projective. Thus we have shown that, if M is invertible, then M is projective and, for all prime ideals p, Mp is a free rank one Rp-module. Conversely, suppose that M is projective and that, for all prime ideals p, Mp is a free rank one Rp-module. We must show that M is invertible. Set M ′ = Hom (M, R). The evaluation map M Hom (M, R) R defined by R × R → (m, ϕ) ϕ(m) is R-bilinear and so induces a homomorphism f : M R M ′ R. We7→ will show that f is an isomorphism. It suffices to show that,⊗ for all→ prime ideals p, the induced map (M R M ′)p Rp is an isomorphism. Now (M M ) M M . Since M⊗is projective→ and finitely generated, it R ′ p ∼= p Rp p′ is finitely⊗ presented⊗ by Proposition 3.4.4, and by Exercise 3.19 the map

HomR(M, R) HomRp (Mp, Rp) p → is an isomorphism since M is finitely presented. Thus the induced map (M M ′) R can be identified with the evaluation map ⊗R p → p M Hom (M , R ) R . p ⊗Rp Rp p p → p Since Mp ∼= Rp, it is easy to see (cf. Exercise 2.18 in Chapter 2) that this evaluation map is an isomorphism for all p. Thus M R M ′ R is an isomorphism, and we are done. ⊗ → Corollary 5.5.12. A finitely generated R module M is invertible if and only if M is finitely presented and, for all prime ideals p of R, Mp ∼= Rp. Proof. This is immediate from Theorem 5.5.8 and Proposition 3.4.4. 5.5. FRACTIONAL IDEALS AND INVERTIBLE MODULES 189

Remark 5.5.13. Clearly, in the statements of Theorem 5.5.8 and Corol- lary 5.5.12, we can replace the condition that, for all prime ideals p of R, Mp ∼= Rp, by the weaker condition that, for all maximal ideals m of R, Mm ∼= Rm. Deﬁnition 5.5.14. Let R be a commutative ring. Deﬁne Pic R to be the set of all isomorphism classes of invertible R-modules M.

Proposition 5.5.15. For a commutative ring R, Pic R is an abelian group under tensor product. Moreover the inverse of an element M of Pic R is given by the dual module HomR(M, R).

Proof. If M and M are two invertible R-modules, and M M ′ = R, then 1 2 i ⊗R i ∼ (M M ) (M ′ M ′ ) = R as well. Clearly R is an identity. If M is 1 ⊗R 2 ⊗R 1 ⊗R 2 ∼ invertible and M M ′ = R, then M ′ is invertible as well by Lemma 5.5.9 ⊗R ∼ and so every element M of Pic R has an inverse, necessarily unique and given by the dual module, as we have seen in the proof of Theorem 5.5.8.

The link with the ideal class group is given by the following:

Theorem 5.5.16. If R is an integral domain with quotient ﬁeld K, then Pic R is isomorphic to the ideal class group of R.

Proof. First we deﬁne a homomorphism from Pic R to the ideal class group of R. If M is an invertible R-module, then it is projective and therefore torsion free (since it is a submodule of a free module). Thus the homomorphism M M K is an inclusion. For every prime ideal p of R, → ⊗R M K =(M R ) K = R K = K. ⊗R ⊗R p ⊗Rp p ⊗Rp Thus there exists some inclusion M K, and in particular M is isomorphic to a fractional ideal of K. (Not every→ fractional ideal of K is an invertible R-module, of course, and so a fractional ideal of K isomorphic to M is not a priori an invertible fractional ideal.) We next show that a fractional ideal isomorphic to M is unique up to multiplication by a principal fractional ideal:

Lemma 5.5.17. If R is an integral domain and M1, M2 are two fractional ideals in K, then M1 and M2 are isomorphic as R-modules if and only if there exists λ K∗ such that M = λM . Moreover in this case every isomorphism ∈ 2 1 from M to M is given by multiplication by some λ K∗. 1 2 ∈ 190 CHAPTER 5. IDEALS IN NOETHERIAN RINGS

Proof. Clearly, if M = λM for some λ K∗, then multiplication by λ 2 1 ∈ induces an R-module isomorphism from M1 to M2. Conversely, suppose that ϕ: M1 M2 is an isomorphism of R-modules. For α M1, α = 0, and for all r →R,r = 0, ϕ(rα)= rϕ(α), and hence ∈ 6 ∈ 6 ϕ(rα)/rα = ϕ(α)/α.

Choosing β M1, β = 0, it follows that α/β = r/s for some nonzero r, s R, and hence that∈ rα =6 sβ. Thus ∈ ϕ(α)/α = ϕ(rα)/rα = ϕ(sβ)/sβ = ϕ(β)β, for all nonzero α, β M1. Setting λ = ϕ(α)/α for α = 0, it follows that λ is independent of the∈ choice of α, and that ϕ(α) = λα6 for every nonzero α M1. Of course, the same is true for α = 0. Thus M2 = λM1 and ϕ is multiplication∈ by λ.

Lemma 5.5.18. If R is an integral domain and M1, M2 are two fractional ideals which are invertible R-modules, then M M = M M . 1 2 ∼ 1 ⊗R 2 Proof. Multiplication is bilinear, and thus there is an induced homomorphism f : M M M M . Clearly this homomorphism is surjective. To see 1 ⊗R 2 → 1 2 that it is injective, it suﬃces to check that, for every prime ideal p, the induced surjective homomorphism (M1 R M2)p (M1M2)p is injective. Note that M M = 0. Since M M is a⊗ submodule→ of K, it is torsion free, 1 2 6 1 2 and thus (M1M2)p = 0 for every prime ideal p, because M1M2 = 0. Since (M M ) K, it is6 a torsion free R -module. Moreover 6 1 2 p ⊆ p (M M ) = (M ) (M ) = R R = R . 1 ⊗R 2 p ∼ 1 p ⊗Rp 2 p p ⊗Rp p ∼ p The map (M M ) = R (M M ) is surjective and hence nonzero. 1 ⊗R 2 p ∼ p → 1 2 p Since (M M ) is a torsion free R -module, the map (M M ) = R 1 2 p p 1 ⊗R 2 p ∼ p → (M1M2)p is injective. Thus f : M1 R M2 M1M2 is injective, and so it is an isomorphism. ⊗ → Lemma 5.5.19. If R is an integral domain and M is a fractional ideal which is an invertible R-module, then M is an invertible fractional ideal.

Proof. There exists an R-module M ′ such that M M ′ = R, and we may ⊗R ∼ assume that M ′ is chosen to be a fractional ideal of R by the ﬁrst part of the proof. Thus the fractional ideal MM ′ is isomorphic to M R M ′ = R, and ⊗ 1 so, by Lemma 5.5.17, MM ′ = (λ) for some λ K∗. Thus M(λ− M ′) = R, so that M is invertible. ∈ 5.5. FRACTIONAL IDEALS AND INVERTIBLE MODULES 191

Putting this together, we see that, given an invertible R-module M, M is isomorphic to a fractional ideal I(M) by the first paragraph of the proof, and I(M) is invertible by Lemma 5.5.19. Moreover, the image of I(M) in /K∗ is independent of the choice of fractional ideal, so that there is a well-definedI map F : Pic R /K∗. By Lemma 5.5.18, F is a homomorphism. Finally, F is injective since→ I a principal fractional ideal is isomorphic to R. It remains to show that F is surjective, or in other words that every invertible fractional ideal is also an invertible R-module. We have seen that every invertible fractional ideal M is finitely generated. Next we claim that M is projective: Lemma 5.5.20. If R is an integral domain and M is an invertible fractional ideal, then M is projective.

n Proof. If M ′ is such that MM ′ = R, write 1 = i=1 mimi′ for some mi M, n ∈ m′ M ′. We have a homomorphism f : R M deﬁned by f(r1,...,rn)= i ∈ → P i rimi, and f is surjective (if m M, then mmi′ R and m = i(mmi′ )mi.) n ∈ ∈ Deﬁne g : M R by g(m)=(mm′ ,...,mm′ ). Then P → 1 n P

f g(m)= (mm′ )m = m, ◦ i i i X and so M is a direct summand of Rn. Thus M is projective.

Since M is ﬁnitely generated and projective, for all prime ideals p of R, M is free, say of rank n. Thus M K Kn. But M K K K, p p Rp ∼= p Rp Rp which is the localization of K at the⊗ set of nonzero elements⊗ of ⊆R, and⊗ thus is isomorphic to K. It follows that n = 1 and Mp is free of rank one for all p. Thus, by Theorem 5.5.8, M is an invertible R-module. This completes the proof of Theorem 5.5.16.

The group Pic R is closely connected to unique factorization in R. In particular, if R is a UFD then Pic R = 0. The converse does not quite hold. For example, the ring R = k[x, y, z]/(xy z2) (where k is a ﬁeld) is not a UFD, − as one can check by showing that the images x, y, z are irreducible elements no two of which are associates. Hence the relation xy = z2 in R shows that unique factorization fails. However, one can show that Pic R = 0. On the other hand, if R is a Noetherian integral domain, all of whose localizations at maximal ideals are UFDs, then R is a UFD if and only Pic R = 0. Thus Pic R is the global obstruction for unique factorization to hold. We do not 192 CHAPTER 5. IDEALS IN NOETHERIAN RINGS yet have a large supply of rings R such that Pic R is nontrivial, but we will explain shortly how to ﬁnd such examples.

5.6 Dedekind domains

Dedekind domains encompass rings of integers in an algebraic number field as well as those in an algebraic function field. We begin with a characterization in the local setting: Lemma 5.6.1. Let R be a local Noetherian integral domain of dimension one. Let m be the maximal ideal of R, K the quotient field of R and k = R/m the residue field. Then the following are equivalent: (a) R is integrally closed.

(b) m is principal. (c) The embedding dimension of R is 1.

(d) Every nonzero ideal of R is a power of m.

(e) There exists a t R such that every nonzero ideal of R is equal to (tn) for some n. ∈ Proof. First note that if I is a nonzero ideal of R then √I = m, since √I is an intersection of prime ideals and m is the unique nonzero prime ideal of R. Thus, since R is Noetherian, there exists an n > 0 such that mn I. ⊆ In particular I is m-primary. Also, for all n > 0, mn = mn+1 since R is not Artinian. 6 (a) = (b): Choose r m such that r = 0. There exists an n> 0 such n ⇒ n 1 ∈ 6 n 1 that m (r) and m − is not contained in (r). Choose s m − (r). ⊆ 1 1 ∈ − Let t = r/s K∗. Since s = rt− , t− / R for otherwise s (r). Moreover 1 s ∈ 1 1 n 1 ∈ 1 ∈ t− m = r m = r (sm) r m r (r) = R. Thus t− m is an ideal of R. 1 ⊆ ⊆ 1 Moreover t− m is not contained in m, for otherwise t− would be integral 1 over R and hence lie in R. Thus t− m = R and so m = (t) is a principal ideal. (b) = (c): Since m is principal and nonzero, dim m/m2 = 1. ⇒ k (c) = (d): If I is a nonzero ideal of R, then mn I for some n > 0, and so I/⇒mn is an ideal in the Artinian ring R/mn. Since⊆ the embedding dimension of R is 1, it follows that the embedding dimension of R/mn is at 5.6. DEDEKIND DOMAINS 193 most 1 and hence by from Lemma 5.4.8 that every ideal in R/mn is a power of the maximal ideal. Thus I/mn =(mN +mn)/mn and so I = mN +mn = ma where a = min n, N . { } (d) = (e): Choose t m m2, which is possible since m = 0. Then (t) = mn⇒for some n, but∈ clearly− we must have n = 1. Since6 m = (t), mn =(tn), implying (e). (e) = (a): A PID is a UFD and in particular it is integrally closed. ⇒ In case R is not necessarily local, we make the following deﬁnition:

Deﬁnition 5.6.2. Let R be an integral domain Then R is a Dedekind domain if every nonzero ideal I in R is equal to a product of (not necessarily distinct) prime ideals p1 pn, and R is not a ﬁeld. (By convention the unit ideal is the empty product.)···

For example, a PID R is a Dedekind domain, since every nonzero I is of the form (r), and if we factor r into a product p1 pn of irreducible elements, then I =(r)=(p ) (p ) is a product of prime··· ideals. In particular, Z and 1 ··· n k[t] are Dedekind domains. For another example, if R is a Dedekind domain and S is a multiplicative 1 1 subset of R such that S− R is not a ﬁeld, then S− R is a Dedekind domain. 1 1 Indeed, every nonzero ideal of S− R is of the form S− I for some ideal I 1 1 1 of R, so that if I = p1 pn, then S− I = S− p1 S− pn. Now either 1 1 1 ··· 1 ··· 1 1 S− pi = S− R or S− pi is a prime ideal of S− R. If all of the S− pi = S− R, 1 then S− I is the unit ideal, and otherwise it is a nonempty product of prime ideals. The following set of characterizations of a Dedekind domain is the main result of this section:

Theorem 5.6.3. Let R be an integral domain which is not a ﬁeld. Then the following are equivalent:

(i) R is a Dedekind domain.

(ii) The nonzero fractional ideals of R are a group under multiplication.

(iii) R is Noetherian, integrally closed, and dim R =1.

(iv) R is Noetherian and, for every nonzero prime ideal p of R, Rp is a PID. 194 CHAPTER 5. IDEALS IN NOETHERIAN RINGS

(v) R is Noetherian and, for every nonzero prime ideal p of R, there exists a t Rp, not zero or a unit, such that every ideal in Rp is of the form (tk)∈.

Proof. (i) = (ii): We begin with the following lemma on unique factorization into invertible⇒ prime ideals:

Lemma 5.6.4. Let R be an integral domain and let p1,..., pa, q1,..., qb be not necessarily distinct invertible prime ideals in R such that p1 pa = q q . Then a = b and after reordering p = q for all i. ··· 1 ··· b i i Proof. After reordering, we can assume that p1 is a minimal element of the set p ,..., p , i.e. does not contain p if p = p . Since q = p { 1 a} i i 6 1 j j i i ⊆ p1, there must exist a j such that qj p1, for otherwise we could choose r q p for every j, in which case r⊆ r q p ,Q a contradiction.Q j ∈ j − 1 1 ··· b ∈ j j ⊆ 1 After reordering we can assume that j = 1, and that q1 is minimal among all the q p . In particular it is minimal. By symmetryQ there exists a k such j ⊆ 1 that pk q1. But then pk p1 and by minimality pk = p1 = q1. Multiply 1 ⊆ ⊆ by p1− and conclude by induction (on a, say). Theorem 5.6.5. Let R be a Dedekind domain. Then every nonzero prime ideal p of R is maximal, and p is an invertible fractional ideal. Proof. First we claim that, if p is an invertible prime ideal of R, then p is maximal. It suﬃces to show that, if a R p, then (p, a) is the unit ∈ − 2 ideal. We can factor (p, a)= i Pi, and we can also write (p, a ) = j Qj, for prime ideals Pi, Qj. Now by assumption (p, a)/p = a(R/p) is a nonzero principal ideal. In particular itQ is an invertible ideal of R/p. Moreover pQ P ⊆ i for all i, since it is contained in the product, and Pi/p is a prime ideal of R/p which is nonzero since a / p. Thus we have a factorization ∈

a(R/p)= (Pi/p), i Y so that each Pi/p is an invertible prime ideal in R/p, and similarly

2 a (R/p)= (Qj/p). j Y By the uniqueness result of Lemma 5.6.4 applied to invertible prime ideals 2 in R/p, we see that we have factored a (R/p) both as j(Qj/p) and as Q 5.6. DEDEKIND DOMAINS 195

2 i(Pi/p) , and so after reordering the Qj’s we can assume that, for all i, Q2i 1/p = Q2i/p = Pi/p. Since all of the ideals in question contain p, it Q − 2 2 2 follows that Q2i 1 = Q2i = Pi. But then (p, a )= Qj = Pi =(p, a) . − j i In particular p (p, a2) (p, a)2 (p2, a), and so every r p can be ⊆ ⊆ ⊆ Q Q ∈ written as r = s + ta with s p2 and t R. Thus ta p, and since a / p by hypothesis t p. It follows∈ that p ∈ p2 + p (a), and∈ since clearly p2∈+ p (a) p, we have∈ p = p2 + p (a)= p ⊆(p +(a)).· As p is invertible, we · ⊆ · · can cancel it to obtain R = p +(a). Thus (p, a) is the unit ideal and so p is maximal if p is invertible. Now let p be an arbitrary nonzero prime ideal. To show that p is invertible and maximal, it suﬃces by the above to show that p is invertible. Choose r p, r = 0 and factor (r)= p . Note that each p is invertible since (r) ∈ 6 i i i is, and so each pi is maximal. Since (r) p we have pi p, and by an Q ⊆ i ⊆ argument like that given above it follows that pi p for some i. As pi is ⊆ Q maximal, pi = p, and hence p is both invertible and maximal

Corollary 5.6.6. Let R be a Dedekind domain. Then every nonzero ideal I in R can be factored uniquely into a product of prime ideals, and is invertible.

Proof. The uniqueness follows from Lemma 5.6.4 and the fact that every nonzero prime ideal is invertible, which also implies that I is invertible.

Corollary 5.6.7. Let R be a Dedekind domain. Then every nonzero frac- ni tional ideal M in R can be uniquely written as a product i pi , where the pi are distinct prime ideals and ni Z is nonzero. Moreover every nonzero fractional ideal M is invertible. ∈ Q

Proof. Given M, ﬁnd r R with r = 0 such that rM = I is an ideal of R, and factor I and (r)∈ into prime ideals.6 This gives the existence of the factorization and the invertibility of M, and the uniqueness is standard.

To complete the proof of the implication (i) = (ii) of Theorem 5.6.3, note that every fractional ideal is invertible, and they⇒ form a group by the remarks in Definition 5.5.6. (ii) = (i): Let R be an integral domain such that every nonzero ideal is invertible.⇒ Then R is Noetherian, since by Lemma 5.5.3 an invertible ideal is finitely generated. Suppose that R is an integral domain such that every nonzero ideal is invertible. We claim that every nonzero ideal of R is a product of maximal ideals, and in particular of prime ideals. If not, 196 CHAPTER 5. IDEALS IN NOETHERIAN RINGS since R is Noetherian there exists a maximal element in the set of ideals which are not products of maximal ideals, say I. In particular I is not itself maximal, so I is contained in a maximal ideal m. Now m is invertible, with 1 1 1 m− = (R : m) R, and I m− I R. Also, I = m− I or we would have ⊇ ⊆ ⊆ 6 mI = I, in which case Im = 0 which would imply I = 0 since R is an integral 1 domain. Thus m− I is an ideal of R strictly containing I, and so is a product of maximal ideals. But then so is I, a contradiction. Thus every nonzero ideal in R is a product of maximal ideals and so R is a Dedekind domain, so that (ii) = (i). Note that the proof shows that a Noetherian domain such ⇒ that every maximal ideal is invertible is a Dedekind domain. Before we do the implication (i) = (iii), we introduce some notation. Suppose that R is a Dedekind domain.⇒ For a nonzero fractional ideal M, if p is a prime ideal we can define an integer vp(M) by the formula

M = pvp(M), p Y where the product is taken over all prime ideals p. Thus vp(M) = 0 for almost all p, and vp(M)=0 for all p if and only if M = R.

Proposition 5.6.8. Let R be a Dedekind domain and let M1 and M2 be two nonzero fractional ideals of R. Then for all prime ideals p of R,

1 (1) vp(M1M2) = vp(M1)+ vp(M2). Thus vp(M − ) = vp(M) for all M and p. −

(2) M M if and only if, for all p, v (M ) v (M ). 1 ⊆ 2 p 1 ≥ p 2 1 (3) v ((M : M )) = v (M M − )= v (M ) v (M ). p 1 2 p 1 2 p 1 − p 2 (4) v (M + M ) = min v (M ), v (M ) . p 1 2 { p 1 p 2 } (5) v (M M ) = max v (M ), v (M ) . p 1 ∩ 2 { p 1 p 2 }

Proof. (1) is clear. For (2), clearly M R if and only if vp(M) 0 for all p. 1 ⊆ ≥ Thus M M if and only if M M − R if and only if v (M ) v (M ) 0 1 ⊆ 2 1 2 ⊆ p 1 − p 2 ≥ for all p if and only if, for all p, vp(M1) vp(M2). This also implies (3), since 1 ≥ (M1 : M2)= M1M2− . As for (4) and (5), M1 + M2 is the smallest fractional ideal containing both M and M and M M is the largest fractional ideal 1 2 1 ∩ 2 5.6. DEDEKIND DOMAINS 197

contained in M1 and M2. On the other hand, by (2) the smallest fractional ideal containing both M1 and M2 is necessarily

min vp(M1),vp(M2) p { }, p Y proving (4), and (5) is similar.

We can restrict v to principal ideals (α), α K∗; we shall just denote p ∈ vp((α)) by vp(α), and shall also view vp as a function from K∗ to Z. (Occa- sionally, we shall adopt the convention that v (0) = .) p ∞

Corollary 5.6.9. The function vp : K∗ Z is a surjective homomorphism satisfying v (α + α ) min v (α ), v (α→) provided that α + α = 0 (the p 1 2 ≥ { p 1 p 2 } 1 2 6 ultrametric inequality).

Proof. The function vp is a homomorphism, by (1) of Proposition 5.6.8. It is surjective, since there exists an α p p2 (otherwise p = p2, which implies since p is invertible that p = R is not∈ a− prime ideal). To see the ultrametric inequality, suppose that α + α = 0. Then, since α + α (α )+(α ), 1 2 6 1 2 ∈ 1 2 vp(α1 + α2) vp((α1)+(α2)) = min vp(α1), vp(α2) , where we have used (2) and (4) of Proposition≥ 5.6.8. { }

Example 5.6.10. Suppose that R = A(X) is the affine coordinate ring of the variety X defined over the algebraically closed field k, and is in addition a Dedekind domain. In this situation, we say that X is a smooth (affine) curve. Then the nonzero prime ideals of R are exactly the maximal ideals of R, and hence correspond to points of X. Given f R, let (f)= mni , ∈ i xi where the mx are maximal ideals corresponding to distinct points xi. Write i Q vxi for vmxi . Then vxi (f)= ni, and in fact Proposition 5.6.8(ii) implies that ni ni+1 f mxi mxi , in other words that ni is the order of vanishing of f at ∈ − ni xi. Geometrically, instead of a fractional ideal i mxi , where the ni are not necessarily positive integers, we think of a divisor, which by definition is a Q finite formal sum i nixi of points of X with integer coefficients. Let K(X) be the quotient field of A(X). Every rational function f K(X) has an P ∈ ni associated divisor of zeroes and poles. To ask if a fractional ideal i mxi is principal is then to ask if the divisor i nixi is the divisor associated to a rational function. Q P 198 CHAPTER 5. IDEALS IN NOETHERIAN RINGS

We return to the proof of Theorem 5.6.3. (i) = (iii): We know that, if R is a Dedekind domain, then R is Noetherian⇒ and every nonzero prime ideal in R is maximal. Thus, since R is not a ﬁeld, dim R = 1. We must show that R is integrally closed. Suppose that α K∗ is integral over R. Then there exists a monic equation for α: ∈ n n 1 α + r α − + + r =0, 1 ··· n where r R are not all 0. Thus, using Corollary 5.6.9, i ∈ n i nvp(α) min vp(riα − ) = min(vp(ri)+(n i)vp(α)). ri=0 ri=0 ≥ 6 6 − Fixing an i for which the minimum is attained, we have nv (α) v (r )+(n i)v (α). p ≥ p i − p Thus iv (α) v (r ) 0 and so v (α) 0 for all p. So α R. p ≥ p i ≥ p ≥ ∈ (iii) = (iv): Since R is a Noetherian integral domain of dimension one, ⇒ integrally closed, the same is true for Rp, for every nonzero prime ideal p. By Lemma 5.6.1 (e), Rp is a PID for every nonzero prime ideal p. To complete the proof of Theorem 5.6.3, Lemma 5.6.1 implies that (iv) and (v) are equivalent. We will be done if we show that (iv) = (i). Let ⇒ I be a nonzero ideal of R. Let I = q q be a minimal primary 1 ∩···∩ n decomposition of I and let pi = √qi. For every p, Rp is a PID and thus pRp is a minimal nonzero prime ideal of Rp. It follows that the only prime ideal contained in p is (0), so that every nonzero prime ideal p is maximal. Thus qi and qj are comaximal if i = j and so q1 qn = q1 qn, i.e. I is a product of primary ideals. (This6 is true for every∩···∩ Noetherian··· domain of dimension one.) Localize qi at pi: (qi)pi is an ideal in Rpi , and so by hypothesis and N N the equivalence of (iv) and (v) it is of the form (piRpi ) = pi Rpi for some N 0. Since q and pN are both p -primary, it follows that ≥ i i i (q ) R = q = pN R R = pN . i pi ∩ i i pi ∩ i N So qi = pi and I is a product of prime ideals. Thus a Dedekind domain R is a Noetherian integral domain which is locally a PID. If p is a nonzero prime ideal of R and t R is such that pR = ∈ p (t), or equivalently such that vp = 1, then t is called a local uniformizing element at p. We have the following easy fact about unique factorization in a Dedekind domain: 5.7. HIGHER DIMENSIONS 199

Proposition 5.6.11. A Dedekind domain R is a UFD if and only if it is a PID if and only if the ideal class group of R is trivial.

Proof. We know that a PID is a UFD. Conversely, if R is a UFD, a nonzero element π R is irreducible if and only if (π) is prime. Let p be a nonzero ∈ prime ideal of R and choose r p with r = 0. Factor r into a product π π where each π is irreducible.∈ Since p 6 is prime, there exists an i such 1 ··· n i that π p. Thus (π ) p. But as π is irreducible, (π ) is prime, and as i ∈ i ⊆ i i R has dimension one (πi)= p. So every prime ideal in R is principal and so every ideal is principal. To see the last equivalence, if the class group is trivial then every ideal 1 is principal. Conversely, every fractional ideal is of the form r− I for some ideal I, so if every ideal is principal then so is every fractional ideal.

5.7 Higher dimensions

The arguments in the proof of (iv) = (i) of Theorem 5.6.3 can be generalized to a certain extent to higher dimensions.⇒

Deﬁnition 5.7.1. Let R be a Noetherian domain. Then R is locally factorial if, for every maximal ideal m of R, the local ring Rm is a UFD.

Note that, if R is locally factorial, then Rm is integrally closed for every maximal ideal m, since it is a UFD, and hence R is integrally closed.

Theorem 5.7.2. Let R be a locally factorial Noetherian domain. Then

(i) Every height one prime ideal p is invertible.

(ii) If I is an ideal in R such that every associated prime of I has height one, and in particular I has no embedded primes, then I is a product of height one prime ideals, and in particular I is invertible.

(iii) Conversely, if I is an invertible ideal, then every associated prime of I has height one.

Proof. (i) For every maximal ideal m, either p is not contained in m, in which case p = R , or p m, in which case p = pR is a height one prime ideal m m ⊆ m m in Rm. In this second case, pRm is a minimal nonzero prime ideal in the UFD 200 CHAPTER 5. IDEALS IN NOETHERIAN RINGS

Rm, and hence is principal. Thus, in both cases, pm ∼= Rm is free of rank one, and hence p is invertible by Corollary 5.5.12 and the following remark. (ii) Suppose that I = q q is the minimal primary decomposition 1 ∩···∩ k of I. Let pi = √qi, so that by hypothesis pi is a height one prime ideal in R. The ring Rpi is a local Noetherian domain of dimension one which is integrally closed, and hence Rpi is a PID in which every nonzero ideal is a Ni power of piRpi . Thus qiRpi = pi Rpi for some positive integer Ni. Hence pNi (pNi R ) R =(q R ) R = q . i ⊆ i pi ∩ i pi ∩ i Thus, pN1 pNk pN1 pNk q q = I. 1 ··· k ⊆ 1 ∩···∩ k ⊆ 1 ∩···∩ k We claim that in fact I = pN1 pNk . It suﬃces to check that equality holds 1 ··· k after localizing at every maximal ideal m. If pi is not contained in m, then q R = pNi R = R . If p m, then p R is a minimal nonzero prime ideal i m i m m i ⊆ i m in the UFD Rm, so that piRm =(ti) for some irreducible ti Rm. Moreover, mi ∈ qiRm is (ti)-primary, and so by Exercise 5.5 qiRm = (ti ) for some positive Ni mi integer mi. By localizing at piRm, we see that pi Rpi = pi Rpi , and hence m = N . Finally, if I = i : p m , we must show that i i { i ⊆ } Ni Ni (ti )= (ti ). i I i I \∈ Y∈ This is an easy consequence of the fact that, if a and b are two relatively prime elements in a UFD, then (a) (b)=(ab). ∩ (iii) Suppose that I is an invertible ideal and that p is an associated prime of I. Let m be a maximal ideal containing p. Clearly, p is a height one prime ideal in R if and only if pRm is a height one prime ideal in Rm. Thus, after replacing R by Rm, we may assume that R is a UFD and that I = (t) is a principal ideal in R. In this case, by Exercise 5.5, all of the associated primes of (t) have height one.

5.8 Extensions of Dedekind domains

The main applications of Dedekind domains arise by taking rings of integers in number fields or finite extensions of k[x]. Proposition 5.8.1. Let R be a Dedekind domain with quotient field K and let L be a finite separable extension of K. If R is the integral closure of R

e 5.8. EXTENSIONS OF DEDEKIND DOMAINS 201 in L, then R is a finitely generated R-module, and hence it is a Dedekind domain. e Proof. We have seen that there exist α , , α L such that R Rα + 1 ··· n ∈ ⊆ 1 +α . Thus R is a submodule of a finitely generated R-module and so, since ··· n R is Noetherian, R is a finitely generated R-module and in particulare it is also Noetherian. Sincee R is an integral extension of R, dim R = dim R = 1, and R is integrally closede by Corollary 4.1.7. Thus R is a Dedekind domain. e e Of course, by Noether’s theorem, if R is an affine coordinate ring, then e e the integral closure of R in any finite, not necessarily separable extension of K is a finitely generated R-module and thus is Noetherian (and hence a Dedekind domain). In fact, for the final statement of Proposition 5.8.1, we need only assume that L is a finite, not necessarily separable extension of K. Theorem 5.8.2 (Krull-Akizuki). Let R be a Dedekind domain with quotient field K and let L be a finite extension of K. If R is the integral closure of R in L, then R is a Dedekind domain. e Proof. If L is the maximal separable extension of K contained in L, then sepe L is a purely inseparable extension of Lsep. By Proposition 5.8.1, the integral closure of R in Lsep is a finitely generated R-module and a Dedekind domain. Replacing R by its integral closure in Lsep, it suffices to prove the theorem under the assumption that L is a purely inseparable extension of K. Since L is a purely inseparable extension of K, there exists a power q of the characteristic of K such that, for all α L, αq K. Thus we can assume ∈1/q ∈ that L is contained in the subfield L′ = K of the algebraic closure K of K, where by definition

1/q q L′ = K = α K : α k . { ∈ ∈ } Since q is a power of the characteristic, α αq is a ﬁeld isomorphism from 7→ q L′ to K. In particular, the subring R′ = α L′ : α R is isomorphic to R, and hence is a Dedekind domain. { ∈ ∈ } Let R be the integral closure of R in L. Since every α R has an q′ ∈ irreducible polynomial over K of the form x a, a R, where q′ q, R is a − ∈ | subring ofe R′, and in fact R = R′ L. To prove that R is a Dedekinde domain, ∩ it suﬃces to prove that every nonzero ideal I in R is invertible. Let J =eIR′. Since R′ is a Dedekind domain,e J is invertible and hencee J(R′ : J)= R′, by Lemma 5.5.4. Thus there exist r J and s (Re′ : J) such that r s = 1. i ∈ i ∈ i i i P 202 CHAPTER 5. IDEALS IN NOETHERIAN RINGS

We can write each r as a u with a R′ and u I. After absorbing i j ij ij ij ∈ ij ∈ the terms aij into the si, using the fact that (R′ : J) is a fractional ideal, P we may assume that we have found ri I and si (R′ : J) such that q q ∈ ∈ i risi = 1. Hence i ri si = 1 as well. Write this as i riti = 1, where q 1 q q q q 1 q ti = r − s L, since ri L and s (L′) = K L. Since ti = r − s , P i i ∈ P ∈ i ∈ ⊆ P i i q q q t I I s (Is ) R′, i ⊆ i ⊆ i ⊆ since s (R′ : J) and I J. Thus t I L R′ = R and so t (R : I). i ∈ ⊆ i ⊆ ∩ i ∈ Since we can write 1 = r t with r I and t (R : I), it follows that i i i i ∈ i ∈ R I(R : I). As, trivially, I(R : I) R, it follows thate I(R : I)= R.e Thus, P again⊆ by Lemma 5.5.4, I is invertible.⊆ e e e e e e e For the rest of this section, we fix the following notation: R is a Dedekind domain with quotient field K, L is a finite separable extension of K with [L : K]= n, and R is the integral closure of R in L. We want to study how ideals in R factor in R, and it is enough to study prime ideals. Let p be a prime ideal of R ande write e r p Pei e R = i , i=1 Y e where the Pi are distinct prime ideals in R and ei > 0. Note that Pi is a factor of pR if and only if P pR, in which case we say that P divides p. i| i e Lemma 5.8.3.e With notation ase above, the following are equivalent:

(i) Pi divides p.

(ii) pR P . ⊆ i (iii) p P R. e⊆ i ∩ (iv) p = P R. i ∩

Proof. (ii) = (i): It follows from Lemma 5.6.8(i) that vPi (pR) vPi (Pi)= 1. (iii) = ⇒(ii): Obvious. (iv) = (iii): Obvious. (i) =≥ (iv): If r ⇒ ei ⇒ ⇒ pR = i=1 Pi Pi, then p (pR) R Pi R. But bothe p and Pi R are nonzero prime⊆ ideals of R⊆, necessarily∩ ⊆ maximal,∩ so that if p P ∩ R ⊆ i ∩ thene pQ= P R. e i ∩ 5.8. EXTENSIONS OF DEDEKIND DOMAINS 203

Thus Pi divides p if and only if Pi lies over p. We assume that P divides p, and consider the extension R/p R/P . i ⊆ i Since R is a finitely generated R-module, the field R/Pi is a finite extension of R/p. We let fi = [R/Pi : R/p]. In classical terminology, ei is callede the ramificatione index of Pi over p and fi is called thee residue field degree. If ei = 1 for all i we say thate p is unramified in R, whereas p ramifies if ei > 1 for some i. If pR = Pe for a prime ideal P of R and f = [R/P : R/p] = 1, then we say p is totally ramified in R. If ei =efi = 1 for all i we say that p splits completelye in R, whereas if p = P for a primee ideal Pe of R, then p is undecomposed. e In the geometrice case, where R = A(X) is the affine coordinatee ring of a variety of dimension one defined over an algebraically closed field k, R is again an affine coordinate ring, and R/Pi ∼= R/p ∼= k for every prime ideal Pi lying above p. Thus the residue field degree is always one. In general,e if t is a local uniformizing parametere for p and w is a local uniformizing ei parameter for Pi, then we can write t = uw , where u / Pi. Thus in the case where R = A(X), R = A(Y ), and k = C, locally∈ around the points x X and y Y corresponding to the maximal ideals p and P , the ∈ ∈ i morphism f : X Y correspondinge to the inclusion R R is a branched → ⊆ cover with ei sheets. This is the origin of the term “ramified.” e Lemma 5.8.4. With notation as above,

eifi = n. i=1 X Proof. The ring R is a torsion free finitely generated R-module. If we localize at p, then Rp is a torsion free finitely generated Rp-module and thus it is free of rank d, say, sincee R is a PID. Moreover R K Kd, and also R p p Rp ∼= p Rp e ⊗ 1 ⊗ K ∼= R R K, which is the localization (R 0 )− R of R at R 0 . Now by ⊗ −{ } −{ } 1 the universal property of localization, theree is a map ϕ: (R 0 )− Re L, −{ } → definede by (α,r) α/r. Since R is an integral domain,e e ϕ is injective. By Lemma 4.1.10, for7→ every β L, there exists an r =0,r R such thate rβ is ∈ 6 ∈ integral over R, so that β = α/r efor some α R and r R 0 . Hence ϕ ∈ ∈ d −{ } is surjective as well and thus is an isomorphism. Thus K ∼= L as K-vector spaces, and so d = [L : K]. It follows that Rp ise a free module of rank n over Rp. e 204 CHAPTER 5. IDEALS IN NOETHERIAN RINGS

˜ Now Rp/(pRp)Rp ∼= (R/pR)p since localization commutes with taking quotients, and (R/pR) = R/˜ pR since the elements of R p are already p ∼ − invertiblee on R/pRe. Thus R/peR is a vector space of dimension n over the e e e e1 er residue ﬁeld k = R/p. On the other hand, R/pR = R/P1 Pr , and since ei ··· the Pi are distincte e maximale ideals,e the ideals Pi are comaximal. By the Chinese remainder theorem, e e e

R/Pe1 Per = R/Pei . 1 ··· r ∼ i i M e e ei ei Since p Pi for every i, R/Pi is also a vector space over k = R/p. It thus ⊆ ei ei suﬃces to prove that dimk R/Pi = eifi. There is a ﬁltration of R/Pi by the a images of the ideals Pi ,a

e √d, if d 2, 3 mod 4, α = ≡ 1+√d , if d 1 mod 4. ( 2 ≡ Here x2 d, if d 2, 3 mod 4, irr(α, Q, x)= 2 − 1 d ≡ x x + − , if d 1 mod 4, ( − 4 ≡ with discriminant 4d if d 2, 3 mod 4 and d if d 1 mod 4. For another example, suppose that k ≡is a field of characteristic≡ = 2 and that f(x) 6 ∈ 5.8. EXTENSIONS OF DEDEKIND DOMAINS 205 k[x] = R is a square free polynomial. Then R = k[x, y]/(y2 f(x)) is − integrally closed and is the integral closure of R in L = k(x)[y]. We have the following result of Kummer: e Theorem 5.8.5 (Kummer). Suppose that R is a Dedekind domain and that R = R[α] is the integral closure of R in a finite separable extension L = K[α]. Let f(x) = irr(α,x,K), which is a monic irreducible polynomial of degree n withe coefficients in R. Let p be a nonzero prime ideal of R and let f¯(x) be the image of f(x) in (R/p[x]. Suppose that, in the PID (R/p)[x], f¯(x) has the following factorization into distinct prime powers:

r ei f¯(x)= h¯i(x) , i=1 Y where h¯ ,..., h¯ are distinct irreducible polynomials in (R/p)[x] and h 1 r i ∈ R[x] is some lift of h¯i. Then there are exactly r prime ideals Pi in R lying above p and they are of the form Pi =(p, hi(α)), which is independent of the choice of lift hi. Moreover e r ei pR = Pi . i=1 Y e Proof. Note that, with our assumptions, R ∼= R[x]/(f(x)) under the homomorphism evα which sends x to α. First we show that the prime ideals in R lying above p are exactly the ideals Pie= (p, hi(α)). Note that there is a surjective homomorphism R[x] (R/p)[x]/(h¯ (x)). Its kernel is clearly → i (ep, hi(x)), and the kernel contains f(x), since h¯i(x) divides f¯(x). Thus there is an induced surjective homomorphism R (R/p)[x]/(h¯ (x)); let its kernel → i be the prime ideal Pi. Then Pi is the image of (p, hi(x)) in R, and hence Pi =(p, hi(α)). e Conversely, suppose that P is a prime ideal of R such thateP R = p. ∩ We must show that P = Pi for some i. Let α be the image of α in R/P. Since R = R[α] and P R = p, it follows that R/p ise a subring of R/P and ∩ that R/P =(R/p)[α]. We have a commutative diagram e e e R[x] evα R e −−−→ e  evα  (R/p)[x] (R/p)[α], y −−−→ y 206 CHAPTER 5. IDEALS IN NOETHERIAN RINGS where the left hand vertical arrow is reduction mod p and the right hand ¯ arrow is the map R R/P = (R/p)[α]. If hi(x) = irr(α,R/p, x), then Ker ev = (h¯ (x)), and→ h¯ (x) f¯(x) since f¯(α) = f(α) = 0. Thus h¯ is an α i i | i irreducible factor ofe f¯, ande P contains Pi = (p, hi(α)). As we have seen above, P is a prime ideal in R, and hence P = P . Moreover, if P = P , i i i 6 j then (h¯i(x)) = (h¯j(x)) and hence h¯i(x) = h¯j(x) since the h¯i(x) are monic. Thus i = j, and the Pi are exactlye the distinct prime ideals lying over p. ei ei Let Qi = (p, hi(x) ). Then √Qi = Pi and Pi Qi Pi. Thus ′ ei ⊆ ⊆ Q = P for some positive integer e′ e . Since √Q = P and the P are i i i ≤ i i i i comaximal, the Qi are comaximal as well, and hence Q1 Qr = Q1 Qr. Clearly ∩· · ·∩ ···

r pR Q Q = Q Q (pR, h¯ (α)ei )= pR, ⊆ 1 ∩···∩ r 1 ··· r ⊆ i i=1 Y r e e e since h¯ (α)ei f(α) = 0 mod p. Thus i=1 i ≡ Q ′ pR = Q Q = Pei . 1 ··· r i i Y e So we must show that ei′ = ei. It is enough to show that i ei′ fi = n = ′ ei e f , since e′ e . On the one hand, since pR = Q = P , it i i i i ≤ i i iP i i follows from Lemma 5.8.4 that e′ fi = n. On the other hand, since f(x) is P i i Q Q monic, n = deg f(x) = deg f¯(x)= e deg h¯ , ande deg h¯ = [R/P : R/p]= P i i i i i fi. Thus n = eifi as well, so that e′ = ei. This concludes the proof of i P i Kummer’s theorem. e P Recall that, for every polynomial f(x) K[x], there is the discriminant ∈ 2 df K, generalizing the usual discriminant b 4ac of the quadratic poly- ∈ 2 − nomial ax + bx + c, and with the same property, that df = 0 if and only if f(x) has a multiple root. We then have the following corollary to Kummer’s theorem:

Corollary 5.8.6. With notation and hypotheses as in Theorem 5.8.5, the prime p ramiﬁes in R if and only if p divides the principal ideal (df ). In particular, there are only ﬁnitely many prime ideals in R which ramify in R. e e 5.8. EXTENSIONS OF DEDEKIND DOMAINS 207

Remark 5.8.7. In the general case, where R is not necessarily of the form R[α], one can still deﬁne the discriminant of the extension as an ideal in R and show that the primes of R which ramifye in R are exactly the primes dividing the discriminant. In particular, there are still only ﬁnitely many prime ideals in R which ramify in R. e

Example 5.8.8. Let R = Z and Re = Z[√ 5]. Then we are in the situation of Kummer’s theorem with f(x) = x2 +5.− Let p be a prime number, and consider the prime ideals in Ze[√ 5] dividing (p). Since Q[√ 5] is a − − Galois extension of Q of degree 2, we have efr = 2. Thus there are three possibilities: e = 2, f = r = 1 (the ramiﬁed case); e = f = 1,r = 2 ((p) splits completely); e = r = 1, f = 2 ((p) is undecomposed). The discriminant of f(x) is 20, so the primes which ramify are p = 2, 5. Since f(x) (x + 1)2 mod− 2, the unique prime ideal P lying above (2) is equal ≡ 2 to (2, 1+ √ 5). Likewise, since f(x) x2 mod 5, we have 5Z[√ 5] = P2 − ≡ − 5 with P5 = (5, √ 5) = (√ 5). Thus P5 is principal. On the other hand, − − 2 2 2 P2 is not principal: if P2 = (λ), then P2 = (λ ) = (2), so that 2 = βλ for some β Z[√ 5]. However, given α = a + b√ 5 Z[√ 5], we have the norm function∈ −N(α)= N(a + b√ 5) = αα = a2−+5∈b2, and−N is multiplicative. Thus we would have 4 = N(−β)N(λ)2. Clearly N(λ) = 1 is impossible, because then λλ = 1 and (λ) is the unit ideal. So N(λ) 2, and hence N(λ) = 2. But there are no integral solutions to the equation≥ a2 +5b2 = 2. Thus the ideal class group of Z[√ 5] is nontrivial (which follows from the − fact that Z[√ 5] is not a UFD in any case), and P2 deﬁnes an element of order 2 in the− ideal class group. In fact, it turns out that the ideal class group of Z[√ 5] is isomorphic to Z/2Z, and hence that the image of P2 is a generator. − It is easy to give examples of the other possible cases. Since

f(x)= x2 +5 x2 1=(x + 1)(x 1) mod 3, ≡ − − there are two ideals P = (3, 1+√ 5) and P′ = (3, 1 √ 5) lying over (3), 3 − 3 − − and P3P3′ = 3Z[√ 5]. One can check directly that P3 is not principal. In fact, since 1 + √ 5− P and 1+ √ 5 P as well, it is easy to see directly − ∈ 3 − ∈ 2 that P2P3 = (1+ √ 5), and hence that P3 deﬁnes the same element of the − 1 2 ideal class group as P2− = P2. Finally, since x + 5 is irreducible mod 11, (11) is undecomposed in Z[√ 5], and in fact the unique prime ideal lying − above (11) is the ideal 11Z[√ 5]. − 208 CHAPTER 5. IDEALS IN NOETHERIAN RINGS

Example 5.8.9. Let k be an algebraically closed field with char k = 2. Let 6 R = k[x, y]/(y2 f(x)), where f(x) k[x] has distinct roots. Then R is an integrally closed− Noetherian domain∈ of dimension one, and hence is a Dedekind domain. In this case, one can show that the ideal class group of R is trivial in case f(x) has at most two roots, and is infinite otherwise. In fact, if f(x) has at most two roots, one can check that R is a localization of k[t] and hence is a PID. If f(x) has degree three, it is a classical result that the 2 2 ideal class group of R is identified with the points of X = V (y f(x)) Ak, together with one additional point. − ⊆

Exercises

Exercise 5.1. (i) Prove Proposition 5.1.6. (ii) Give another proof of Proposition 5.1.3 by combining Proposition 5.1.6 with Exercises 1.24 and 1.23.

Exercise 5.2. Let R be Noetherian, let M be a ﬁnitely generated R-module, and let I be an ideal of R. Then Supp M V (I) if and only if there exists an n> 0 such that InM = 0. ⊆

Exercise 5.3. Let M and N be two ﬁnitely generated R-modules. Show that Supp(M N) = Supp M Supp N. ⊗R ∩ Exercise 5.4. Let R be Noetherian, let M be a ﬁnitely generated R-module, and let I be an ideal of R. Then either there exists an r I which is not a zero divisor on M, or I is contained in the annihilator of∈ an element of M. (If every element of I is a zero divisor on M, then I is contained in the union of the associated primes of M, and hence is contained in an associated prime ideal.)

Exercise 5.5. Let R be a Noetherian integral domain. Suppose that (t)= p is a prime ideal. (i) Show that (tn) is a p-primary ideal for all n> 0. (ii) Show that every p-primary ideal is of the form (tn) for some n> 0. (First show that, if t is not a unit, then every r R can be written as htn with ∈ n h / (t), by considering the ascending chain of ideals In = g R : gt (r) . Now∈ note that, if q is p-primary and r = htn q, with h{ / ∈p, then tn∈ q.)} ∈ ∈ ∈ 5.8. EXTENSIONS OF DEDEKIND DOMAINS 209

k ni (iii) If r = i=1 pi , where each ideal (pi) is a prime ideal (for example if R is a UFD), show that (pi) is a minimal associated prime of (r), that the (pi) Q n1 nk are all of the associated primes of (r), and that (r)=(p1 ) (pk ) is the unique primary decomposition of (r). ∩···∩ 2 Exercise 5.6. Let R = k[x1, x2] and let I =(x1, x1x2) R. Draw a picture of V (I). What is √I? What is the support of √I/I = ⊆V (Ann √I/I)? Show that I = q q , where q = p = (x ) and q = p2, where 1 ∩ 2 1 1 1 2 2 p2 = (x1, x2). Argue that the ideals q1 and q2 are primary, and hence that I = q q is a minimal primary decomposition of I. What are the associated 1 ∩ 2 primes of I? Which are minimal, and which are embedded? 2 Show that, for every λ k, the ideal (x1, λx1 + x2) is a primary ideal and 2 ∈ that I = (x1) (x1, λx1 + x2). Finally show that, if λ = λ′, then the ideals 2 ∩ 2 6 (x1, λx1 + x2) = (x1,λ′x1 + x2). This gives an inﬁnite number of diﬀerent primary decompositions6 of the ideal I.

Exercise 5.7. Let R = k[x1, x2, x3], let p1 = (x1, x2), p2 = (x1, x3), and m =(x1, x2, x3). Finally, let I = p1p2.

(i) Describe V (p1), V (p2), V (m), and V (I) geometrically. Is I = p1 p2? Is I = √I? Is p p equal to its radical? ∩ 1 ∩ 2 2 (ii) Show that I = m p1 p2 is a minimal primary decomposition of I. Describe the associated∩ ∩ primes of I, as well as the minimal associated primes of I. Explain why not all associated primes of I can be minimal. Exercise 5.8. Let R be a ring, p a prime ideal of R, and M an R-module. ֒ Suppose that there exists a submodule N of M and an inclusion R/p → M/N, i.e. p Ass(M/N). Show that p Supp M. Conversely, if p Supp M, show∈ that there exists a submodule ∈N of M such that p Ass(M/N∈). ∈ (Choose m M whose image in Mp is nonzero and let N = p m M. Show that the homomorphism∈ R/p M/N defined by r r m is· injective.)⊆ → 7→ · Argue that, if R is Noetherian, M is a finitely generated R-module, and p Supp M, then there exists a filtration of M by submodules M , ∈ i 0= M M M M = M, 0 ⊆ 1 ⊆ 2 ⊆···⊆ n such that Mi+1/Mi ∼= R/pi, where the pi are prime ideals and pi = p for some i. Conclude that the pi are not uniquely determined by M, and also that there exists an exact sequence 0 M ′ M M ′′ 0 of finitely → → → → generated R-modules such that Ass M = Ass(M ′) Ass(M ′′). 6 ∪ 210 CHAPTER 5. IDEALS IN NOETHERIAN RINGS

Exercise 5.9. Let R be a Noetherian ring and let M be a ﬁnitely generated R-module. Show that the following are equivalent:

(i) M has ﬁnite length.

(ii) Every prime in Ass M is maximal.

(iii) Every prime in Supp M is maximal.

((i) = (ii): Use (ii) of Lemma 5.2.15. (ii) = (iii): Use Corollary 5.2.12. ⇒ ⇒ (iii) = (i): Use (i) of Lemma 5.2.15.) ⇒ Similarly, show that, if R is arbitrary and M is an R-module such that there exists a strictly increasing sequence of submodules

0 M M = M, ⊂ 1 ⊂···⊂ n such that, for all i > 0, Mi/Mi 1 = R/pi, where pi is a prime ideal in R, − ∼ then M has ﬁnite length if and only if every prime in Supp M is maximal.

Exercise 5.10. Let R be a Noetherian ring and let M be a ﬁnitely generated R-module. Suppose that p Ass M. Show that M has ﬁnite length as an ∈ p Rp-module if and only if p is a minimal prime of Ass M.

Exercise 5.11. Give another proof of the existence theorem for primary decompositions as follows. Suppose that R is Noetherian and that M is a finitely generated R-module. (i) Define a submodule N of M to be irreducible if it is not an intersection of two submodules N1, N2 of M with neither N1 nor N2 equal to N. Show (by standard arguments) that every submodule N = M of a finitely generated R-module M is the intersection of finitely many6 irreducible submodules. (ii) Show that an irreducible submodule Q of M is primary, as follows: we may assume that Q = 0. Suppose that multiplication by r on M is not injective, and choose m such that rm = 0 but m = 0. Then we must show that r is nilpotent on M. Let M = m M6 : rnm = 0 . Since the n { ∈ } Mn are increasing, there exists an n such that Mn = Mn+1. For such an n, consider R m rnM. If x R m rnM, then x = am = rnu, say, with r(am) = ·rn+1∩u = 0. Thus∈u ·M ∩ = M , so that rnu = 0. Hence ∈ n+1 n R m rnM = 0. Since m R m, it follows that rnM = 0, and hence that r is· nilpotent∩ on M. ∈ · 5.8. EXTENSIONS OF DEDEKIND DOMAINS 211

Exercise 5.12. Let R be a reduced Noetherian ring. Let S be the multiplicative subset of R consisting of all elements which are not zero divisors. 1 Show that S− R is a product of ﬁelds as follows: 1 (i) Show that S− R is a reduced Noetherian ring in which every element is either a divisor of zero or a unit. (ii) Let R be a reduced Noetherian ring in which every element is either a divisor of zero or a unit. Show that R is Artinian. (If (0) = P1 Pn is a minimal primary decomposition of (0), argue that every prime∩···∩ ideal P is equal to Pi for some i.) (iii) If an Artinian local ring is reduced, show that it is a ﬁeld. Exercise 5.13. Let R be a Dedekind domain and let I be a nonzero ideal in R. Show that every ideal in R/I is generated by one element. Conclude that every ideal of R can be generated by at most two elements. Exercise 5.14. (Chinese remainder theorem) Let R be a Dedekind domain and let I1,...,In be proper ideals in R. Given r1,...,rn R, show that there exists an r R such that r r mod I for every ∈i if and only if ∈ ≡ i i r r mod I + I for all i = j. (This is the statement that i ≡ j i j 6 R R/I R/(I + I ) → i → i j i i

ni M ∼= R/pi . i M Exercise 5.16. Let R be a Dedekind domain and let M be a finitely generated R-module. Show that M is a flat R-module if and only if M is a projective R-module if and only if M is torsion free. (Localize at each prime p and use the fact that Rp is a PID, together with the classification of modules over a PID.) Thus every finitely generated R-module M is isomorphic to a direct sum M M , where M is torsion and M is torsion free. 1 ⊕ 2 1 2 212 CHAPTER 5. IDEALS IN NOETHERIAN RINGS

Exercise 5.17. Let R be a Dedekind domain, let K be the quotient ﬁeld of R and let M1 and M2 be two nonzero fractional ideals in K.

(i) Show that there exist r ,r K∗ such that r M and r M are rela- 1 2 ∈ 1 1 2 2 tively prime integral ideals, in other words r1M1 + r2M2 = R. (ii) Show that M M = R (M M ). (We can assume that the M 1 ⊕ 2 ∼ ⊕ 1 2 i are integral and relatively prime, by the above. Consider the map M M R deﬁned by addition.) 1 ⊕ 2 → (iii) Let M be a ﬁnitely generated torsion free R-module of rank n. Show that there is an inclusion M Rn, and a free submodule of rank n contained in M. →

(iv) Show that if M be a finitely generated torsion free R-module of rank n, then M = I I for some nonzero ideals I in R. (By (iii), there ∼ 1 ⊕···⊕ n i exists a free rank n R-module Rn with M Rn and such that Rn/M ⊆ is a torsion R-module. By Exercise 5.15, there exist prime ideals pi n mi such that R /M is isomorphic to i R/pi . Show that we can then n m write R /M ∼= i=1 R/Ji for some ideals J1 J2 Jm where m is the minimal number of generators of RnL/M. Thus| m|···|n. Find a surjective n mL n ≤ n n map R i=1 R/Ji = R /M and lift it to a map R R . Argue that this→ map is surjective, by localizing and using Nakayam→ a’s lemma, L n m and is thus an isomorphism. Thus M = J J R − .) ∼ 1 ⊕···⊕ m ⊕ (v) Conclude that every finitely generated torsion free R-module M of rank n 1 n is isomorphic to R − I for some ideal I in R. Show also that the ⊕ image of I in the ideal class group of R is a well-defined invariant of M. Exercise 5.18. Let R be a Dedekind domain and let M be a finitely generated torsion R-module. Show that M has finite length. Exercise 5.19. Let R be a Dedekind domain with only finitely many prime ideals. Then R is a PID. (If p1,..., pn are the prime ideals of R, we must show that pi is principal for all i. Find an element r R such that r pi, r / p2, and r / p for i = j, and factor (r) into a product∈ of primes.) ∈ ∈ i ∈ j 6 Exercise 5.20. (Strong approximation for Dedekind domains) Let R be a Dedekind domain with quotient field K, and let p1,..., pk be distinct nonzero prime ideals in R. Show that, for all α ,...,α K and n ,...,n Z, there 1 k ∈ 1 k ∈ 5.8. EXTENSIONS OF DEDEKIND DOMAINS 213 exists β K such that v (β α ) n for all i and v (β) 0 for all nonzero ∈ pi − i ≥ i q ≥ prime ideals q such that q = pi for all i. (First assume that α 6 R for all i and that n 0, and apply the i ∈ i ≥ Chinese Remainder Theorem to find the appropriate β R. In the general case, there exists t R 0 such that tα R; write down∈ the inequalities ∈ −{ } i ∈ to be satisfied by tβ, possibly after enlarging the set p1,..., pk to include those primes p such that v (t) = 0.) { } p 6 Conclude that, with p ,..., p as above and for a ,...,a Z, there 1 k 1 k ∈ exists an α K such that vpi (α) = ai for all i and such that vq(α) 0 for all nonzero prime∈ ideals q such that q = p for all i. ≥ 6 i Exercise 5.21. Let R be a Dedekind domain with quotient field K, and let p1,..., pk be distinct nonzero prime ideals in R. Let

S = α K : v (β) 0 if q = p . { ∈ q ≥ 6 i} Show that S is a localization of R if and only if there exist positive integers n1 nk ni such that the ideal p1 pk is principal. ··· 1 (First suppose that S = M − R for some multiplicative subset M of R; in particular, every element of S is of the form r/s with s M. Using the ∈ previous exercise, there exists α S with vpi (α) < 0 for all i. If α = r/s for some r, s R such that 1/s S∈, show that v (s) > 0 if and only if q = p ∈ ∈ q i for some i. Conversely, if s R and v (s) > 0 if and only if q = p for some ∈ q i i, show that S = Rs.)

Exercise 5.22. Let R, S, p1,..., pk and

S = α K : v (β) 0 if q = p { ∈ q ≥ 6 i} be as in the previous exercise. Let α S be such that v (α) < 0 for all i. ∈ pi (i) Show that S = R[α]. (Localize at q for every q Spec R.) ∈

(ii) If ι: R S is the inclusion, show that ι∗ is a homeomorphism from Spec S to→ Spec R p ,..., p . −{ 1 k} (iii) Show that S is a Dedekind domain.

Exercise 5.23. Let k be a ﬁeld of characteristic unequal to 2, and let R = k[x]. We have seen that the ring R = k[x, y]/(y2 f(x)) is an integral domain − e 214 CHAPTER 5. IDEALS IN NOETHERIAN RINGS if and only if f(x) is not a square in k[x], and is integrally closed if and only if f(x) has no square factors (i.e. if g(x) is an irreducible polynomial in k[x], then g2 does not divide f). In case k is algebraically closed and f(x) has no square factors, describe all of the prime ideals of R and the prime ideals in R which lie over them. Which of them are ramiﬁed and which split completely? e Exercise 5.24. Let d be a square free integer, and let

√d, if d 2, 3 mod 4, α = ≡ 1+√d , if d 1 mod 4, ( 2 ≡ so that Z[α] is the integral closure of Z in Q[√d]. Let

x2 d, if d 2, 3 mod 4, f(x) = irr(α, Q, x)= 2 − 1 d ≡ x x + − , if d 1 mod 4. ( − 4 ≡ As usual, we define a + b√d = a b√d, so that f(x)=(x α)(x α). − − − (i) Let P be a prime ideal of Z[α] lying over (p). Show that P is also a prime ideal of Z[α], and that P = P if and only if (p) ramifies or is undecomposed in Z[α]. Conclude that PP =(p2) if (p) is undecomposed, and that PP =(p) if (p) splits or ramifies. (ii) Let n Z. Show that n α is not contained in any ideal of the form ∈ − pZ[α], and thus that the principal ideal (n α) has at most one prime factor (possibly occurring to a power) lying above− any given prime ideal (p) Z, and no factor corresponding to an undecomposed prime ideal. Suppose⊆ that ni ni (n α) = i Ppi for distinct primes pi. Conclude that f(n) = i pi . Conversely,− a factorization of f(n) into prime powers gives relations in the ideal class group.Q Illustrate with d = 5 and 0 n 4. Q − ≤ ≤ Chapter 6

Quasiprojective varieties

Throughout this chapter, k denotes an algebraically closed ﬁeld n n n n unless otherwise noted. We abbreviate Ak and Pk by A and P respectively.

6.1 Projective and quasiprojective varieties

Let X be an affine algebraic variety, or more generally an affine algebraic set, corresponding to the reduced k-algebra A(X). More generally, we could consider an affine scheme of the form X = Spec R, where R is an arbitrary finitely generated k-algebra, and hence is of the form k[x1,...,xn]/I for some ideal I (where I is a prime ideal if and only if X is a variety and I is a radical ideal if and only if X is reduced). We call such a scheme X an affine scheme of finite type over k and call R its affine coordinate ring A(X). Let x X ∈ be a (closed) point, and let mx A(X) be the corresponding maximal ideal. The local ring of X at x is the⊆ local ring = A(X) . OX,x mx As we have seen in Chapter 3, affine schemes form a category where the morphisms are morphisms of locally ringed spaces. In fact, every morphism ϕ: X Y is induced by a homomorphism A(Y ) A(X). If A(X) and A(Y ) → → are k-algebras, it is natural to require the homomorphism A(Y ) A(X) to be a k-algebra homomorphism. In terms of spaces, this exactly say→s that the diagram X Y → ցSpec k ւ

215 216 CHAPTER 6. QUASIPROJECTIVE VARIETIES commutes. We say that ϕ is a morphism of k-schemes. For X and Y reduced aﬃne schemes, we can describe morphisms more directly as follows.

Proposition 6.1.1. Let X and Y be reduced aﬃne schemes of ﬁnite type over k, and let ϕ: X Y be a function which is continuous in the Zariski → topology. Then the following are equivalent:

(i) ϕ is a morphism of k-schemes.

(ii) For all f A(Y ), f ϕ A(X), and f f ϕ = ϕ∗(f) deﬁnes a ∈ ◦ ∈ 7→ ◦ k-algebra homomorphism ϕ∗ : A(Y ) A(X). →

(iii) For all x X and f Y,ϕ(x), f ϕ X,x, and f f ϕ deﬁnes a k-algebra∈ homomorphism∈ O ◦ ∈ O. 7→ ◦ OY,ϕ(x) → OX,x

Proof. (i) = (ii): Clearly, ϕ∗ is a homomorphism of k-algebras. (Com- ⇒ pare also Chapter 1, Proposition 1.6.2.) (ii) = (iii): By deﬁnition, ⇒ ϕ∗(m ) m for every x X. Thus ϕ∗ induces a homomorphism ϕ(x) ⊆ x ∈ A(Y )mϕ(x) = Y,ϕ(x) A(X)mx = X,x, which is just composition by ϕ at the level ofO germs→ of functions. (iii)O = (i): Choosing an embedding Y An, it is enough to show that the composition⇒ X An is a mor- ⊆ → phism, and by considering the projections it is enough to consider the case Y = A1. In this case, if t is the coordinate on A1, corresponding to the 1 1 identity Id: A A , since ϕ∗t for every x X, it follows that, for → ∈ OX,x ∈ all x X, there exists a Zariski open set U containing x such that ϕ U is equal∈ to an element ψ (U). In particular, ϕ is a morphism. | U ∈ OX Similar results apply for projective varieties or schemes. Finally, we make the following basic deﬁnition:

Definition 6.1.2. A quasiprojective variety is an open subset of a projective variety. It is naturally a ringed space. Quasiprojective schemes are defined similarly, as are quasiaffine varieties and schemes.

6.2 Local rings and tangent spaces

Let X be a quasiprojective scheme and x X. We wish to discuss various properties of X at x that can be deﬁned via∈ the local rings . OX,x 6.2. LOCAL RINGS AND TANGENT SPACES 217

Deﬁnition 6.2.1. The scheme X is reduced at x X if the local ring ∈ OX,x is reduced.

The following elementary proposition holds in the more general context of Noetherian schemes, and its proof is left as an exercise (Exercise 6.1).

Proposition 6.2.2. Let X be a quasiprojective scheme. Then the set

x X : X is reduced at x { ∈ } is an open subset of X.

There are many other subsets of X which can be deﬁned in a similar way. For example:

Deﬁnition 6.2.3. Let X be a quasiprojective variety and let x X. Then X is normal at x if the local ring is integrally closed. ∈ OX,x Proposition 6.2.4. Let X be a quasiprojective variety. Then the set

x X : X is normal at x { ∈ } is an open subset of X.

Proof. We may assume that X is affine, with affine coordinate ring R = A(X). Let R be the integral closure of R in the field of quotients K of R.

Then X is normal at x if and only if Rmx is integrally closed, where mx is the maximale ideal corresponding to x, if and only if Rmx = Rmx , where Rmx is the integral closure of Rmx in K. But, by Lemma 4.1.8, Rmx = (R)mx , where R is the integral closure of R in K. Moreover, by Noether’sg theoremg (Theorem 4.5.7), R is a ﬁnite R-module, as is the quotient R/Rg . Wee also have ane exact sequence e e 0 R (R) (R/R) 0. → mx → mx → mx →

Thus, Rmx is integrally closed ife and onlye if (R/R)mx = 0 if and only if x / Supp(R/R). Since Supp(R/R) is a closed subset of Spec R, and hence of∈ maxSpec R, x X : X is normal at x is ane open subset of X. { ∈ } e e 218 CHAPTER 6. QUASIPROJECTIVE VARIETIES

The preceding proposition does not hold in the generality of Noetherian schemes, since for a general Noetherian ring R, R need not be a ﬁnite R- module. We turn now to another fundamental local propertye of X, that X is smooth at x (also called regular or nonsingular). We need some preliminary deﬁnitions and results.

Definition 6.2.5. The Zariski cotangent space Tx(X)∨ is the finite-dimen- 2 sional k-vector space mx/mx. (Note that we get the same answer viewing mx as a maximal ideal in A(X) or as the corresponding maximal ideal mx X,x.) The differential df of an element f A(X) is the image of f f(x) mO/m2 . x ∈ − ∈ x x The Zariski tangent space Tx(X) is the k-vector space dual of the Zariski 2 cotangent space: Tx(X)=(mx/mx)∨.

To understand the geometric meaning of the tangent space and to compute it, we begin by relating it to another object:

Deﬁnition 6.2.6. Let R be an S-algebra. An S-derivation D : R R is a function D satisfying: For all r, r ,r R and s S, → 1 2 ∈ ∈

(i) D(r1 + r2)= D(r1)+ D(r2);

(ii) D(r1r2)= r1D(r2)+ r2D(r1);

(iii) D(sr)= sD(r).

The set of all S-derivations of R is DerS(R); it is an R-module. More generally, if M is an R-module, we can define an S-derivation of R with coefficients in M, as a function D : R M satisfying (i)–(iii) above; it is again an R- module. The R-module of→ all S-derivations of R with coefficients in M is then denoted DerS(R, M).

Note that, if D Der (R, M), then D(1) = D(12)=1D(1)+ 1D(1) = ∈ S D(1) + D(1), and so D(1) = 0. Thus D(s 1) = 0 for all s S. · ∈ The following is a simple example of DerS(R, M):

Lemma 6.2.7. For every S[x1,...,xn]-module M, DerS(S[x1,...,xn], M) ∼= M n. More precisely, for every m ,...,m M, there is a unique S-derivation 1 n ∈ D : S[x ,...,x ] M such that D(x )= m for all i. 1 n → i i 6.2. LOCAL RINGS AND TANGENT SPACES 219

∂ Proof. The partial derivatives : S[x1,...,xn] S[x1,...,xn] are clearly ∂xi → k-derivations. More generally, given an S[x1,...,xn]-module M and an ele- ∂ ∂ ment m M, the function m : S[x1,...,xn] M defined by m (f)= ∈ ∂xi → ∂xi ∂f n ∂ m is a derivation from S[x ,...,x ] to M. Hence, D = m is ∂x · 1 n i ∂x i i=1 i X a derivation from S[x1,...,xn] to M such that D(xi) = mi for all i. This proves existence, and uniqueness is clear since the xi generate S[x1,...,xn] as a S-algebra. Proposition 6.2.8. Let X be an affine scheme of finite type over k, and let x X correspond to the maximal ideal m . Then the Zariski tangent space ∈ x Tx(X) is isomorphic to the module Derk( X,x,k), where k is viewed as the -module /m . O OX,x OX,x x 2 Proof. Concretely, the claim is that a k-linear function on mx/mx is the same thing as a k-linear function D : X,x k satisfying D(fg) = f(x)D(g)+ g(x)D(f). The correspondence isO as follows:→ 2 Given a linear function ψ : mx/mx k, let D(f)= ψ(dfx)= ψ(f f(x)). Clearly, D is k-linear and D(1) = ψ(0)→ = 0. Finally, − D(fg)= ψ(fg f(x)g(x)) = ψ(fg fg(x)+ fg(x) f(x)g(x)). − − − Since fg fg(x)= f(g g(x)), where g g(x) m , and f f(x) mod m , − − − ∈ x ≡ x fg fg(x) f(x)(g g(x)) mod m2 . Thus − ≡ − x D(fg)= D(f(x)(g g(x))) + D((f f(x))g(x)) − − = f(x)D(g g(x)) + g(x)D(f f(x)) = f(x)D(g)+ g(x)D(f). − − In the other direction, suppose that D : X,x k is a k-derivation. If O 2 → f,g mx, then D(fg) = 0. Thus, D is zero on mx, and by restriction induces ∈ 2 a k-linear function ψ : mx/mx k. It is then straightforward to check that these two constructions are inverse.→

Here is a direct way to describe Tx(X):

Lemma 6.2.9. Suppose that A(X)= k[x1,...,xn]/I, where I =(f1,...,fk). n Thus, if X is reduced we can identify X with V (f1,...,fk) A . Let x X An. Then ⊆ ∈ ⊆ n ∂f T (X) = (λ ,...,λ ) kn : j (x)λ =0 for all j . x ∼ { 1 n ∈ ∂x i } i=1 i X 220 CHAPTER 6. QUASIPROJECTIVE VARIETIES

Proof. Using the previous lemma, it suﬃces to prove the analogous statement for derivations. Let D Der ( ,k). By composing D with the ∈ k OX,x surjection k[x ,...,x ] = An (k[x ,...,x ]/I) = , there is 1 n mx O ,x → 1 n mx ∼ OX,x an induced homomorphism Der ( ,k) Der ( An ,k), which is injec- k OX,x → k O ,x tive since the homomorphism An,x X,x is surjective. By Lemma 6.2.7, n O → O n ∂ ∂ Der ( An ,k) = k . An element D = λ is in the image of k O ,x ∼ · ∂x i ∂x i=1 i i=1 i M X n ∂f Der ( ,k) if and only if D annihilates I, if and only if j (x)λ = 0 k OX,x ∂x i i=1 i for all j. This establishes the formula of the lemma. X ∂f Corollary 6.2.10. With notation as above, dim T (X)= n rank j (x) . k x − ∂x i

Remark 6.2.11. We will see later that, for every x X, dimk Tx(X) dim X. ∈ ≥

Deﬁnition 6.2.12. Let X be a quasiprojective scheme, and let x X. ∈ We deﬁne dimx X = dim X,x. If X is a variety. or more generally if all components of X have theO same dimension, and x X, then dim X = ∈ x dim X. If x X is such that dimk Tx(X) = dimx X, then X is smooth at x. X is smooth∈if it is smooth at every x X. We let ∈ Sing X = x X : X is not smooth at x . { ∈ } By Remark 6.2.11,

Sing X = x X : dim T (X) > dim X . { ∈ x } 2 Example 6.2.13. If X is a variety of dimension one, then dimk mx/mx =1 for all x X if and only if A(X) is a Dedekind ring, if and only if A(X) is normal,∈ if and only if is normal for every x X. More generally, if OX,x ∈ dim X = 1, then X is smooth at x if and only if X,x is normal (or equivalently a PID). However, if dim X > 1, then it turnsO out that smoothness at x is a stronger condition than the normality of . OX,x

Example 6.2.14. Given f(x1,...,xn) k[x1,...,xn], deg f > 0, let X = V (f) An. Then, as we shall see, dim∈X = n 1 (in fact a proof is also ⊆ − 6.2. LOCAL RINGS AND TANGENT SPACES 221 given in Exercise 4.16 using the normalization lemma). Let x X, so that ∈ f(x) = 0. Then

∂f dim Tx(X)= n 1 if, for some i, (x) =0, − ∂xi 6  ∂f dim Tx(X)= n if (x) = 0 for every i.  ∂xi   ∂f ∂f More succinctly, Sing X = V (f, ,..., ). From this it is easy to see ∂x1 ∂xn that “most” f satisfy Sing X = . For example, if f is linear, V (f) is always ∅ empty. On the other hand, if X = V (x1x2), Sing X = V (x1x2, x2, x1) = 2 3 2 3 2 (0, 0) . For X = V (x1 x2), Sing X = V (x1 x2, 2x1, 3x2) = 0, 0 (if {char k }=2, 3). On the other− hand, if X is the aﬃne− scheme− corresponding{ } to 6 2 2 k[x1, x2]/(x1), Sing X = V (x1, 2x1)= V (x1)= X as a set. Thus it is possible for Sing X = X.

We can generalize the case of hypersurfaces as follows:

Proposition 6.2.15. Let X be an aﬃne variety. Then Sing X is a proper closed subset of X.

Proof. If X = V (f ,...,f ) An and dim X = d, then Sing X is the subset 1 k ⊆ of X defined by f1,...,fk and the (n d) (n d) minors of the matrix ∂f − × − j (x) , and hence it is a closed subset of X. We must show that it is a ∂xi proper closed subset. Let K(X), the function field of X, be the field of quotients of the integral domain A(X). By Corollary 4.6.3, tradeg(K(X)/k) = dim X = d. We need the following algebraic fact:

Claim 6.2.16. dimK(X)(Derk(K(X),K(X))) = tradeg(K(X)/k) = dim X = d.

Proof of the claim. We use the fact that K(X) is separably generated over k, i.e. that there exists a pure transcendental subextension k(t ,...,t ) 1 d ⊆ K(X) such that K(X) is a separable algebraic extension of k(t1,...,td). d (Appendix?) By Lemma 6.2.7, Derk(k(t1,...,td),K(X))) ∼= K(X) , and hence dim (Der (k(t ,...,t ),K(X))) = d. Now suppose that α K(X) K(X) k 1 d ∈ 222 CHAPTER 6. QUASIPROJECTIVE VARIETIES and that f(t) = irr(α,k(t ,...,t ), t). If D Der (K(X),K(X))), then it is 1 d ∈ k easy to see that

0= D(0) = D(f(α))=(Df)(α)+ f ′(α)Dα, where (Df)(t) is the polynomial obtained by applying D to the coefficients of f (which lie in k(t ,...,t )). Since α is separable over k(t ,...,t ), f ′(α) = 0. 1 d 1 d 6 Thus Dα = (Df)(α)/f ′(α). − In particular, D is uniquely specified by its restriction D0 to k(t1,...,td), i.e. every element D0 in Derk(k(t1,...,td),K(X)) has at most one extension to an element of Der (K(X),K(X)). Conversely, if D Der (k(t ,...,t ),K(X)), k 0 ∈ k 1 d define D : k(t ,...,t )[t] K(X) by the formula 0 1 d → D (g(t))=(D g)(t)+ g′(t)( (Df)(α)/f ′(α)). e 0 0 − It is easy to see that D is a k-derivation of k(t ,...,t )[t] which is zero on e 0 1 d (f(t)), and hence descends to a k-derivation on K(X). Repeating this construction for a sequencee of extensions k(t ,...,t ) k(t ,...,t )[α ] 1 d ⊆ 1 d 1 ⊆···⊆ k(t1,...,td)[α1,...,αℓ] = K(X), we see that Derk(k(t1,...,td),K(X)) ∼= Derk(K(X),K(X)) as K(X)-vector spaces, and in particular both sides have dimK(X) = d. Returning to the proof of Proposition 6.2.15, note next that, if D ∈ Derk(K(X),K(X)) and f/g K(X), where f,g A(X), then by the quotient rule D(f/g)=(gDf ∈fDG)/g2, and thus∈ D is determined by its − restriction to A(X). Conversely, every element of Derk(A(X),K(X)) ex- tends uniquely to an element of Derk(K(X),K(X)) via the quotient rule, so that Derk(K(X),K(X)) ∼= Derk(A(X),K(X)) by restriction. Viewing A(X) as k[x1,...,xn]/(f1,...,fk) and arguing as in the proof of Lemma 6.2.9, since Derk(A(X),K(X)) is the same as the space of k-derivations from k[x1,...,xn] to K(X) which are zero on I, we see that n ∂f Der (A(X),K(X)) = (λ ,...,λ ) K(X)n : λ j = 0 for all i , k ∼ { 1 n ∈ i ∂x } i=1 i X where we view the polynomials ∂fj as elements of the quotient A(X) K(X). ∂xi In particular, ⊆ ∂f d = dim Der (K(X),K(X)) = n rank j , K(X) k − ∂x i 6.2. LOCAL RINGS AND TANGENT SPACES 223

∂f where again we view the matrix j as having entries in the field K(X). ∂x i ∂f In particular, there exists an (n d) (n d) minor µ of j which is − × − ∂xi nonzero as an element of K(X), and thus as an element of A(X). It follows that X V (µ) is a proper closed subset of X, and clearly Sing X X V (µ). Thus Sing∩ X is a proper closed subset of X. ⊆ ∩ Remark 6.2.17. The proof shows in general that, if X is a quasiprojective scheme, then Sing X is a closed subset of X (not necessarily proper), and that Sing X is a proper closed subset if X is a variety. One can define the analogue of Sing X for a general scheme using the concept of a regular local ring, but it is not necessarily a closed subset of X, even for a Noetherian affine scheme. Remark 6.2.18. If X is just assumed to be an scheme of finite type over k, one can show (see Chapter 7) that, if X is smooth at x, then X is reduced at x and there is exactly one component Xi of X containing x. Moreover, in this case, x is a smooth point of Xi. Conversely, if X is reduced at x, x is contained in a unique component Xi of X and Xi is smooth at x, then x is a smooth point of X. In particular, if X is a reduced quasiprojective scheme, then Sing X is a proper closed subset of X. Let us discuss the computation of Sing X in the case of a projective variety defined by a homogeneous ideal. First, the case of a hypersurface:

Lemma 6.2.19. Let f k[x0,...,xn] be homogeneous of degree d. Then ∈ ∂f for every i with 0 i n, is homogeneous of degree d 1, and, as an ≤ ≤ ∂xi − algebraic set, ∂f ∂f Sing X = V+(f, ,..., ). ∂x0 ∂xn If k has characteristic zero, or more generally if the characteristic of k does not divide d, then in fact ∂f ∂f Sing X = V+( ,..., ). ∂x0 ∂xn ∂f Proof. Clearly is homogeneous of degree d 1. Let x Pn. For simplicity, ∂xi − ∈ we shall just consider the case where x U , the standard aﬃne open subset ∈ 0 224 CHAPTER 6. QUASIPROJECTIVE VARIETIES where x = 0. Thus x = (1,c ,...,c ), and V (f) U = V (g), where 0 6 1 n + ∩ 0 g(y1,...,yn)= f(1,y1,...,yn). By Example 6.2.14, ∂g ∂g Sing X U0 = V (g, ,..., ). ∩ ∂y1 ∂yn ∂g ∂f Clearly = (1,y1,...,yn). It now suﬃces to show that ∂yi ∂xi ∂f ∂g ∂g (1,y1,...,yn) V (g, ,..., ). ∂x0 ∈ ∂y1 ∂yn

But, plugging in (x0,...,xn)=(1,y1,...,yn) to Euler’s identity n ∂f x = d f i ∂x · i=0 i X gives

∂f n ∂g ∂g ∂g (1,y ,...,y )= y dg V (g, ,..., ). ∂x 1 n − i ∂y − ∈ ∂y ∂y 0 i=1 i 1 n X ∂f ∂f Finally, if d is invertible in k, then f V+( ,..., ), and hence ∈ ∂x0 ∂xn ∂f ∂f ∂f ∂f V+( ,..., )= V+(f, ,..., ). ∂x0 ∂xn ∂x0 ∂xn

In general, we have the following: Proposition 6.2.20. Let X Pn be a projective variety of dimension d, n ⊆ where P has coordinates x0,...,xn, and suppose that I(X)=(f1,...,fk). Then Sing X is deﬁned by f1,...,fk together with the (n d) (n d) minors ∂f − × − of i . Equivalently, x X is a smooth point if and only if, for some ∂x ∈ j n+1 ∂fi lift of x to a point c =(c0,...,cn) k , the rank of the matrix (c) ∈ ∂xj is n d. − Proof. The proof involves the same ingredients as the previous arguments and is left to the reader. 6.3. DERIVATIONS, DIFFERENTIALS, AND TANGENT SPACES 225 6.3 Derivations, diﬀerentials, and tangent spaces

In this section (which we shall not use elsewhere) we try to clarify the relationship between various objects introduced above and their generaliza- tions. Given a variety X and a point x X, we have deﬁned the Zariski 2 ∈ cotangent space mx/mx and its dual, the Zariski tangent space Tx(X). We showed that Tx(X) ∼= Derk(A(X),k), where k is viewed as an A(X)-module via the isomorphism A(X)/mx ∼= k. However, it is also natural to look at TX = Derk(A(X), A(X)), which as we shall see is a ﬁnite A(X)-module. More generally, if R is an S-algebra, then DerS(R, R) = DerS(R) is an R- module, and in fact a Lie algebra via the bracket

[D ,D ](r)= D (D (r)) D (D (r)). 1 2 1 2 − 2 1

In the case of an aﬃne variety X, we call the A(X)-module TX (or rather the associated sheaf) the tangent sheaf of X. If X is smooth, TX is projective of rank equal to dim X, and plays the role of the (space of sections of) the tangent bundle of X, whereas the ﬁnite-dimensional k-vector space Tx(X) is the tangent space to X at x. For an S-algebra R, we can view the association M DerS(R, M) as a functor from R-modules to R-modules. The content of7→ the next theorem is that this functor is representable by a universal object together with a universal derivation:

1 Theorem 6.3.1. If R is an S-algebra, there exists an R-module ΩR/S and a derivation d: R Ω1 , such that, for every R-module M and derivation → R/S D Der (R, M), there exists a unique R-module homomorphism ϕ: Ω1 ∈ S R/S → M such that D = ϕ d. In particular, Der (R, M) = Hom (Ω1 , M). ◦ S ∼ R R/S 1 Definition 6.3.2. The R-module ΩR/S is the module of Kähler differentials of R over S.

We shall prove Theorem 6.3.1 shortly. Before doing so, let us record some obvious consequences of the deﬁning property:

Lemma 6.3.3. For an S-algebra R,

1 1 (i) HomR(ΩR/S , R)=(ΩR/S )∨ = DerS(R). 226 CHAPTER 6. QUASIPROJECTIVE VARIETIES

(ii) In case A(X) is an aﬃne variety and x X, Hom (Ω1 , A(X)) = ∈ A(X) A(X)/k m 1 TX and (with k ∼= A(X)/ x) HomA(X)(ΩA(X)/k,k) = Tx(X). In par- 1 1 ticular, Tx(X) is the k-vector space dual of ΩA(X)/k/mxΩA(X)/k.

(iii) If R = S[x ,...,x ] is a polynomial ring over S, then Ω1 = Rdx 1 n R/S ∼ 1 ⊕ n ∂f Rdx . Given f S[x ,...,x ], df = dx . ···⊕ n ∈ 1 n ∂x i i=1 i X Proof. (i) and (ii) are immediate from the deﬁnitions. As for (iii), it follows 1 from Lemma 6.2.7 and the universal properties of ΩR/S .

Proof of Theorem 6.3.1. One approach, modeled after our construction of 1 the tensor product, is to define ΩR/S as the quotient of the free module on R, where we take as basis the set of symbols df : f R , by the submodule { ∈ } generated by the relations d(sr) sdr, for r R,s S, d(r1 +r2) dr1 dr2, and d(r r ) r dr r dr , for r −,r R. Instead∈ we∈ shall proceed− differently.− 1 2 − 2 1− 1 2 1 2 ∈ The basic idea, coming from differential geometry, is that the tangent bundle of a manifold M is the normal bundle to the diagonal in the product M M. Dualizing, the cotangent bundle of M is the conormal bundle (the dual× of the normal bundle) of the diagonal in M M. We now give an algebraic analogue of this construction. × The S-bilinear function R R R defined by (f,g) fg induces an S- × → 7→ linear homomorphism R S R R, sending f g to fg. Viewing R S R as an R-algebra by the homomorphism⊗ → r r 1, the⊗ homomorphism R⊗ R R 7→ ⊗ ⊗S → is in fact R-linear (and likewise for the R-algebra structure on R R defined ⊗S by r 1 r). Let I = Ker(R S R R). In particular, I is an ideal of R 7→R and,⊗ for all f,g R, f⊗ g →g f I. Geometrically, R R ⊗S ∈ ⊗ − ⊗ ∈ ⊗S corresponds to the fiber product Spec R Spec S Spec R and I is the ideal of the diagonal embedding of Spec R in the× fiber product. 2 Since R S R R is R-linear, I is an R-submodule of R S R, as is I . 1 ⊗ 2 → 1 ⊗ Let ΩR/S = I/I . As we shall see, ΩR/S can thus be viewed as the conormal bundle of the diagonal in Spec R Spec R. Let df be the image of ×Spec S f 1 1 f I in I/I2. ⊗ − ⊗ ∈ Claim 6.3.4. d: R Ω1 is a derivation. → R/S 6.3. DERIVATIONS, DIFFERENTIALS, AND TANGENT SPACES 227

Proof. Since d(fg)=(fg) 1 1 (fg), we have ⊗ − ⊗ fdg + gdf d(fg)=(fg) 1 f g +(fg) 1 g f (fg) 1+1 (fg) − ⊗ − ⊗ ⊗ − ⊗ − ⊗ ⊗ =(fg) 1 f g g f +1 (fg) ⊗ − ⊗ − ⊗ ⊗ =(f 1 1 f)(g 1 1 g)=(df)(dg). (6.1) ⊗ − ⊗ ⊗ − ⊗ Thus fdg + gdf d(fg) I2. It follows that fdg + gdf = d(fg) in I/I2. − ∈ Claim 6.3.5. I/I2 is generated as an R-module by dR. Proof. Note that f g =(fg) 1 (f 1)(g 1 1 g) ⊗ ⊗ − ⊗ ⊗ − ⊗ =(fg) 1 fdg. ⊗ − Thus f g =( f g ) 1 f dg . i ⊗ i i i ⊗ − i i i i i X X X In particular, if f g I, then f g = 0, and hence f g = i i ⊗ i ∈ i i i i i ⊗ i fidgi RdR. − i ∈ P P P PNow suppose that M is an R-module and that D : R M is a derivation. If there exists an R-linear homomorphism ϕ: Ω1 →M such that Dr = R/S → ϕ(dr), then ϕ is determined on dR, and hence by Claim 6.3.4 is unique. We must therefore prove existence. Let D : R M be a derivation. Deﬁne a homomorphism ϕ: R R → ⊗S → M induced from the bilinear homomorphism (f,g) gDf, and hence in particular ϕ(f g)= gDf. Note that ϕ(df)= ϕ(f 7→1 1 f)= Df. If we ⊗ ⊗ − ⊗ e can show that ϕ(I2) = 0, then we can just take ϕ to be the homomorphism I/I2 = Ω1 M induced by the restriction of ϕ to I. By Claim 6.3.4, I is R/Se → e e generated over eR by elements of the form df, and hence I2 is generated by elements of the form df dg. But, using the identitye · df dg =(f 1 1 f)(g 1 1 g) · ⊗ − ⊗ ⊗ − ⊗ =(fg) 1 f g g f +1 (fg), ⊗ − ⊗ − ⊗ ⊗ proved in the course of proving (6.1), we have that ϕ(df dg)= D(fg) gDf fDG =0. · − − Thus we have constructed the desired homomorphism ϕ. e 228 CHAPTER 6. QUASIPROJECTIVE VARIETIES

Let us record the basic properties of the module of K¨ahler diﬀerentials:

Proposition 6.3.6. (i) (Base change.) If S′ is an S-algebra and R′ = R S′, then there is a canonical isomorphism ⊗S 1 1 Ω ′ ′ = Ω R′. R /S ∼ R/S ⊗R (ii) (Localization.) If M is a multiplicative subset of R, then there is a canonical isomorphism

1 1 1 ΩM −1R/S ∼= M − ΩR/S .

(iii) (First exact sequence.) Given ring homomorphisms S R T , there is an exact sequence of T -modules → →

Ω1 T Ω1 Ω1 0. R/S ⊗R → T/S → T/R → (iv) (Second exact sequence = conormal sequence.) If J is an ideal of R, then there is an exact sequence of R/J-modules

J/J 2 d Ω1 (R/J) Ω1 0. −→ R/S ⊗R → (R/J)/S → Here d: R Ω1 induces the homomorphism J Ω1 , and the → R/S → R/S composition J Ω1 Ω1 (R/J) is zero on J 2, hence induces → R/S → R/S ⊗R J/J 2 Ω1 (R/J). → R/S ⊗R Proof. These are all straightforward applications of the universal property of 1 ΩR/S , and we shall be brief.

(i): Let M ′ be an R′-module. It suffices to show that there is a natural 1 isomorphism Hom ′ (Ω R′, M ′) = Der ′ (R′, M). But (Exercise 2.21) R R/S ⊗R ∼ S 1 1 Hom ′ (Ω R′, M ′) = Hom (Ω , M ′) = Der (R, M ′), R R/S ⊗R ∼ R R/S ∼ S so it suffices to show that there is a natural isomorphism DerS(R, M ′) ∼= Der ′ (R′, M). Given D Der (R, M ′), the function (r, s) R S′ sDr S ∈ S ∈ × 7→ is S-bilinear, and thus induces an S-linear homomorphism D′ : R S′ = ⊗S R′ M ′, which is easily checked to be an S′-derivation. Conversely, given → an S′-derivation D′ : R′ M ′, define D(r) = D′(r 1). Then D : R M ′ is an S-derivation, and these→ constructions are inverse.⊗ → 6.3. DERIVATIONS, DIFFERENTIALS, AND TANGENT SPACES 229

1 1 (ii): If N is an M − R-module, then we must show that DerS(M − R, N) ∼= − 1 1 HomM 1R(M − ΩR/S , N). By Exercise 3.18, 1 1 1 HomM −1R(M − ΩR/S , N) ∼= HomR(ΩR/S , N) ∼= DerS(R, N), 1 so we must show that DerS(R, N) ∼= DerS(M − R, N). Let D : R N be a derivation. We define D(r/s) by the quotient rule: D(r/s)=(→sDr − rDs)/s2. Clearly, this is the unique possible definition for a derivation on 1 M − R extending D. We must check that this is well-defined and does indeed 1 give an S-derivation on M − R. To see that it is well-defined, suppose that r/s = r′/s′, i.e. that there exists t M such that trs′ = tr′s. Applying D, we see that ∈

s′rDt + trDs′ + ts′Dr = sr′Dt + tsDr′ + tr′Ds. 2 2 We must show that (sDr rDs)/s = (s′Dr′ r′Ds′)/(s′) . It suffices to check that − − 2 2 2 2 t (s′) (sDr rDs)= t s (s′Dr′ r′Ds′). − − This is a straightforward computation. The fact that this extension is again a derivation follows from the fact that r r rr r r r r D ′ = D ′ = D ′ + ′ D , s · s′ ss′ s s′ s′ s which is again a straightforward exercise in the definitions. 1 (iii): Let M be a T -module. Using the isomorphism HomT (ΩR/S R T, M) ∼= 1 ⊗ HomR(ΩR/S , M) and the universal property, it suffices to show that there is an exact sequence 0 Der (T, M) Der (T, M) Der (R, M). → R → S → S Here, DerR(T, M) is naturally a submodule of DerS(T, M) and is clearly the kernel of the restriction homomorphism DerS(T, M) DerS(R, M). 1 → (iv): Again using the universal property of ΩR/S , the content of this statement is that, for an R/J-module M, there is an exact sequence 0 Der (R/J, M) Der (R, M) Hom (J/J 2, M). → S → S → R 2 The map DerS(R, M) HomR(J/J , M) is defined by restriction: given D Der (R, M), then→ by restriction D defines a homomorphism J M, ∈ S → which is R-linear and vanishes on J 2 since JM = 0. The kernel of restriction consists of those derivations which vanish on J, and hence is the image of DerS(R/J, M). 230 CHAPTER 6. QUASIPROJECTIVE VARIETIES

Corollary 6.3.7. If R is a localization of a finitely generated S-algebra, 1 then ΩR/S is a finitely generated R-module. If R is a localization of a finitely 1 presented S-algebra, then ΩR/S is a finitely presented R-module. Remark 6.3.8. Later, we will show that, if X is a smooth subvariety of An (or more generally a smooth variety Y ) and J is the ideal I(X) defining X, 2 2 1 then J/J is projective, the homomorphism J/J ΩAn/k k[x1,...,xn] A(X) is 1 → ⊗ injective, and the quotient ΩA(X)/k is projective. Dualizing, there is an exact sequence of projective A(X)-modules

2 0 T TAn A(X) (J/J )∨ 0. → X → ⊗k[x1,...,xn] → → 2 n Thus it is appropriate to call (J/J )∨ the normal bundle of X in A and J/J 2 the conormal bundle. To relate the module of K¨ahler diﬀerentials to the discussion to smooth points, we will prove the following:

Theorem 6.3.9. Let k be an algebraically closed ﬁeld, let R = k[x1,...,xn]/I be an integral domain of dimension d, where I = (f1,...,fk), and let m be a maximal ideal in R, corresponding to the point p V (I) = X. Then the following are equivalent: ∈ (i) The point p is a smooth point of X. ∂f (ii) The matrix i (p) has rank n d. ∂x − j 1 (iii) The Rm-module ΩRm/k is free. Finally, we always have dim T (X) dim X. k p ≥ Proof. The equivalence of (i) and (ii) is clear from Corollary 6.2.10. To see the equivalence of (i) and (iii), note that, by (ii) of Lemma 6.3.3,

dim T (X) = dim Ω1 k = dim Ω1 k, k p k R/k ⊗R k Rm/k ⊗Rm 1 where we have used the localization property to identify ΩRm/k with the localization Ω1 . Next we use: R/km 1 Claim 6.3.10. Let K be the ﬁeld of quotients of R. Then ΩK/k is a K-vector space of dimension d. 6.4. ELEMENTARY PROJECTIVE GEOMETRY 231

Proof of the claim. The argument is very similar to that for the claim in Proposition 6.2.15 and we shall just sketch it. As there, the main point is that K is separably generated over k, and hence that we can write K as a ﬁnite separable extension of a pure transcendental extension k(y1,...,yd) of k. Now Ω1 = d k[y ,...,y ]dy , and hence Ω1 = k[y1,...,yd]/k i=1 1 d i k(y1,...,yd)/k d k y ,...,y dy K k y ,...,y α K k y ,...,y t / f t i=1 ( 1 d) i. If =L ( 1 d)[ ], then = ( 1 d)[ ] ( ( )), where f ′(α) = 0. The conormal sequence then gives L 6 d K f d Kdy Kdt Ω1 0. · −→ i ⊕ → K/k → i=1 M Moreover, the image of f in the direct sum has a last component equal to f ′(α)dt, and in particular is nonzero. Thus the quotient has dimension d.

Returning to the proof of the theorem, and applying Exercise 3.16, it 1 follows that the ﬁnitely generated Rm-module ΩRm/k is free if and only if

dim (Ω1 k) = dim (Ω1 K). k Rm/k ⊗Rm K Rm/k ⊗Rm

As we have seen, the left hand side is dimk Tp(X), and the right hand side is 1 1 dimK(ΩK/k) = dim X. Thus ΩRm/k is free if and only if p is a smooth point of X. The ﬁnal statement follows easily from the ﬁrst part of Exercise 3.16.

Remark 6.3.11. Assuming only that X = V (I) is an aﬃne k-scheme, not necessarily a variety, one can show the following: if k has characteristic zero, m 1 then corresponds to a smooth point of X if and only if ΩRm/k is a free Rm-module. If k has positive characteristic, then m corresponds to a smooth 1 point of X if and only if ΩRm/k is a free Rm-module of rank equal to dim Rm, 1 and, if X is reduced, then it is enough to assume that ΩRm/k is a free Rm- module. However, if I = (xp) k[x], where k has characteristic p, then 1 ⊆ ΩRm/k is a free Rm-module of rank one, whereas X is a (non-reduced) point.

6.4 Elementary projective geometry

In this section, we discuss the geometry of linear spaces and quadrics in Pn. Roughly speaking, every construction in linear algebra has a counterpart in projective geometry. 232 CHAPTER 6. QUASIPROJECTIVE VARIETIES

Subspaces: Let V = kn+1 and identify Pn with P(V ). Every vector subspace W of V of dimension d + 1 gives rise to a closed subvariety P(W ) of P(V ), which by deﬁnition is a linear subspace of P(V ) of dimension d. If dim W = 1, so that W is a line in V , dim P(W ) = 0 and P(W ) is a point of P(V ), namely the point corresponding to the line W . If dim W = 2, then dim P(W )=1 1 and P(W ) ∼= P ; in this case we call P(W )a line in P(V ). Planes are deﬁned similarly. If dim W = n, say W is the kernel of the linear form ℓ: V k, then P(W )= V (ℓ) has dimension n 1, and is called a hyperplane in P→(V ). − If W and W are two vector subspaces of V , then clearly P(W ) P(W )= 1 2 1 ∩ 2 P(W1 W2). Thus the intersection of two linear spaces is linear (with the convention∩ that = P( 0 ) is a linear subspace of P(V )). If W and W are ∅ { } 1 2 two vector subspaces of V such that W1 W2 = 0 , then P(W1) P(W2)= . If in addition the natural homomorphism∩ W {W} V is an∩ isomorphism∅ 1 ⊕ 2 → (i.e. V is an internal direct sum of W1 and W2), then the linear spaces P(W1) and P(W2) are said to be complementary. Note that in this case dim W1 + dim W = n+1, so that dim P(W )+dim P(W )= n 1. Conversely, if P(W ) 2 1 2 − 1 and P(W2) are two linear subspaces of P(V ) such that P(W1) P(W2)= and dim P(W ) + dim P(W )= n 1, then P(W ) and P(W ) are∩ complementary.∅ 1 2 − 1 2 On the other hand, if dim P(W ) + dim P(W ) n, then P(W ) P(W ) is 1 2 ≥ 1 ∩ 2 never empty, and in fact P(W1) P(W2) is a linear space of dimension at least d + d n, where d = dim∩P(W ). In terms of the codimensions, this 1 2 − i i says that

codim P(W ) P(W ) codim P(W ) + codim P(W ), 1 ∩ 2 ≤ 1 1 with equality holding for “most” choices of W1 and W2. Linear independence and span: Given a subset X of P(V ), we denote by X the smallest linear subspace of P(V ), and call it the span or the h i linear span of X. For X = p ,...,p , we also write p ,...,p for the { 1 k} h 1 ki span. Clearly, if X V 0 is any subset whose image in P(V ) is X, then ⊆ −{ } X = span X. We say that p ,...,p span P(V ) if p ,...,p = P(V ). The h i 1 k h 1 ki points p1,...,pk aree linearly independent or in general position if, for every choice of liftse of the p to v V , the v are linearly independent, if and only i i ∈ i if dim p ,...,p = k 1. h 1 ki − Automorphisms: The group GL(n +1,k) acts on kn+1 via linear changes of coordinates, and hence on Pn as a group of automorphisms. The subgroup of homotheties k∗ Id = λ Id : λ k∗ (the center of GL(n +1,k)) acts · { · ∈ } 6.4. ELEMENTARY PROJECTIVE GEOMETRY 233

trivially, and the quotient P GL(n+1,k)= GL(n+1,k)/k∗ Id acts faithfully · on Pn. We shall show later: Proposition 6.4.1. The automorphism group of Pn is exactly P GL(n+1,k) as described above. By contrast, the automorphism group of An, for k > 1, is much larger than the “obvious” group of affine transformations of An (i.e. x A x + b, 7→ · where A GL(n, k) and b kn), and in fact is infinite-dimensional in any reasonable∈ sense. ∈ In general, one goal of projective geometry is to classify projective varieties up to projective equivalence, i.e. up to the action of P GL(n +1,k), and more generally to study those properties of projective varieties which are defined up to projectively equivalence. n One can also study the “abstract” projective geometry of PK, where K is n n+1 an arbitrary field. In other words, P is a set ((K 0 )/K∗), with certain K −{ } distinguished subsets (points, lines, planes, and linear spaces more generally) as well as certain relations among them (e.g. incidence) and various properties satisfied by these subsets and relations (e.g. there is exactly one line through two distinct points). An automorphism in this sense is one which preserves the distinguished sets and which also preserves incidence. The fundamental n theorem of projective geometry states that the automorphism group of PK in this sense is the semidirect product of P GL(n +1,K) with the group of all field automorphisms of K. Linear maps: A major concern of linear algebra is with linear maps W V between two vector spaces. There are related constructions in projective→ geometry, but the situation is a little more complicated. Every linear map W V factors canonically into a surjection followed by an injection, and it is best→ to consider these two cases separately. The case of an injection W V is → essentially just the case of an isomorphism from P(W ) onto a linear subvariety of P(V ). So we shall just consider the case of a surjection p: W V . In → this case, if v U = Ker p, we cannot define the image of p(v) as a point of P(V ). Instead,∈ there is a well-defined functionp ˆ: P(W ) P(U) P(V ). It is easy to see thatp ˆ really just depends on U, not on the− specific→ choice of surjection with kernel U. We have: Proposition 6.4.2. Let p: W V be a linear surjection of vector spaces with kernel U, and let pˆ: P(W )→ P(U) P(V ) be the induced function. − → Suppose that dim W = n and dim V = d, so that dim U = n d. Let U ′ be a − 234 CHAPTER 6. QUASIPROJECTIVE VARIETIES complement to U, so that W is an internal direct sum W = U U ′ and p U ′ ∼ ⊕ | is an isomorphism from U ′ to V . Then:

(i)p ˆ is a morphism.

(ii) The ﬁbers of pˆ are aﬃne spaces modeled on U. More precisely, given 1 v V , then pˆ− (v) = U, u P(U), where u is the unique element of ∈ h i − U ′ such that p(u)= v.

(iii) If V has coordinates x0,...,xd and U0,...,Ud is the standard aﬃne 1 { n d } open cover of P(V ), then pˆ− (Ui) = Ui A − , in such a way that the 1 ∼ × n d restriction of pˆ to pˆ− (U ) is identiﬁed with the projection U A − i i × → Ui.

(iv) The restriction of pˆ to P(U ′) deﬁnes an isomorphism P(U ′) P(V ). → Proof. We leave these straightforward arguments to the reader.

We can use the restriction of projection to a subvariety X to study X:

Proposition 6.4.3. Let X be a subvariety of P(V ) of dimension r. Then there exist linear coordinates x0,...,xn on V with the following property: If p: An+1 Ar+1 is the linear projection p(x ,...,x )=(x ,...,x ) and → 0 n 0 r U = Ker p, then P(U) X = and pˆ induces a morphism X Pr with ﬁnite ﬁbers. ∩ ∅ →

Proof. Let C(X) be the affine cone over X, defined by the homogeneous ideal I = I(X). Then C(X) is an affine variety of dimension r + 1, with affine coordinate ring A(C(X)) = R(X). By the Noether normalization lemma, there exist linear coordinates x0,...,xn on V such that the induced homomorphism k[x0,...,xr] A(C(X)) is injective, and A(C(X)) is integral over → r+1 k[x0,...,xr]. In particular, if π : C(X) A is the corresponding morphism, then π is induced by the linear projection→ p and π has finite fibers. But C(X) is invariant under multiplication by k∗, and π is equivariant under 1 the action of k∗. Let v C(X) U = 0 = π− (0). Then, for all λ k∗, ∈ ∩ { 1} ∈ λ v C(X) U = 0 as well. Since π− (0) is finite, v = 0. It follows that P(·U)∈ X = ∩. The{ final} statement is then clear. ∩ ∅ For example, suppose that X = V (f) is a hypersurface of degree d in Pn, where f is irreducible. Choose y Pn X. We can choose linear coordinates ∈ − 6.4. ELEMENTARY PROJECTIVE GEOMETRY 235

so that y = V (x0,...,xn 1), and the statement that y / X is equivalent to − ∈ the statement that f has degree d in xn, i.e. that

d d 1 f(x0,...,xn)= c0x + c1(x0,...,xn 1)xd− + + cd(x0,...,xn 1), − ··· − where c0 k∗ and ci(x0,...,xn 1) k[x0,...,xn 1] is homogeneous of degree ∈ n n 1 − ∈ − i. Letp ˆ: P y P − be the projection. We see that, given a = − { n}1 → (a0,...,an 1) P − , there exists an x X such thatp ˆ(x) = a, and that − ∈ 1 ∈ the cardinality ofp ˆ− (a) is at most d. In fact, as we shall see, for generic a, n 1 i.e. for a in some nonempty Zariski open subset of P − , the cardinality of 1 pˆ− (a) is exactly d. In the above situation, where X = V (f) is a hypersurface, it is also n n 1 interesting to look at the projectionp ˆ: P y P − , where y X. The 1 −{ }→ ∈ fiberp ˆ− (a) = a, y y consists of the (projective) line joining y and a, h i−{ } n 1 minus the point y. In this case,p ˆ: X y P − has finite fibers if and −{ } → only if X contains no line through y. Since every projective line through y either meets X in at most d points, one of which is y, or is contained in y, if X contains no line through y, then the cardinality of every fiber ofp ˆ X y | −{ } is at most d 1. − Projective cones: Let X be a subvariety (or algebraic subset) of Pn, and let C(X) An+1 be the affine cone over X (Definition 1.7.9), defined by ⊆ n the homogeneous ideal I = I(X). Let x0,...,xn be linear coordinates on P n+1 n+1 and on A . We can view A as a the standard affine open subset Un+1 of Pn+1. Thus U = Pn+1 V (x ), where the hyperplane V (x ) is n+1 − + n+1 + n+1 naturally identified with Pn. The closure C(X) of C(X) in Pn+1 is then a new projective algebraic set, the projective cone over X.

Proposition 6.4.4. (i) C(X) V+(xn+1)= X under the natural identiﬁ- n ∩ cation of V+(xn+1) with P .

n+1 (ii) Let v = (0,..., 0, 1) Un+1 P . Then C(X) is the union x X v, x ∈ ⊆ ∈ h i of all lines joining v to a point of x Pn. ∈ S

(iii) Let I k[x0,...,xn] be the homogeneous ideal of X. Then C(X) = V (I)=⊆ V (I k[x ,...,x ]), where I k[x ,...,x ] is the homoge- + + · 0 n+1 · 0 n+1 neous ideal in k[x0,...,xn+1] generated by I.

(iv) If X is a variety of dimension d, then C(X) is a variety of dimension d +1. 236 CHAPTER 6. QUASIPROJECTIVE VARIETIES

Proof. (i), (ii): These follow from the definition. (iii), (iv): First assume that X is a variety, i.e. X is irreducible of dimension d. Then I is prime, and since A(C(X)) = k[x0,...,xn]/I, C(X) is an affine variety of dimension d+1. Thus C(X) is also a variety of dimension d+1, proving (iv). Moreover, clearly C(X) V (I k[x ,...,x ]). On the other hand, ⊆ + · 0 n+1 k[x ,...,x ]/I k[x ,...,x ] = (k[x ,...,x ]/I)[x ], 0 n+1 · 0 n+1 ∼ 0 n n+1 (exercise) and (k[x0,...,xn]/I)[xn+1] is an integral domain of dimension d+2. Thus V (I k[x ,...,x ]) is a projective variety of dimension d + 1, and + · 0 n+1 since it contains C(X) they must be equal. This proves (iii) in case X is irreducible. The general case follows from the fact that, if X = X X , 1 ∪ 2 then C(X)= C(X1) C(X2), and by induction on the number of components of X. ∪ The point v = (0,..., 0, 1) is called the vertex of the cone C(X). Except in somewhat degenerate cases, a projective cone has exactly one vertex. Note that, ifp ˆ: Pn+1 v Pn is the projection, thenp ˆ(C(X)) = X. In particular, the dimension− { } → of the image ofp ˆ restricted to C(X) is dim C(X) 1. − Conversely, let Y Pn+1 be a projective variety and let X = pˆ(Y ) be the Zariski closure⊆ ofp ˆ(Y ). We shall show in the next chapter that X is irreducible. Suppose that X has dimension dim Y 1. Then Y is contained ≤ − in the irreducible algebraic subset C(X), and we shall see that C(X) has dimension at most equal to dim Y . Thus Y = C(X), and so Y is a projective cone. More generally, if X is a closed subset of Pr, r n, with I = I(X) k[x ,...,x ], then I generates the homogeneous ideal≤I k[x ,...,x ], and⊆ 0 r · 0 n+1 we can consider V+(I) = V+(I k[x0,...,xn+1]) as before. In this case V (I) is an iterated cone over X,· and contains the linear space x = = + 0 ··· xr = 0, which we call the vertex of V+(I) as before. Of course, this whole process can be reversed: if Y Pn is of the form V (f ,...,f ), where ⊆ + 1 k f1,...,fk k[x0,...,xr], r < n, then Y is a (generalized) projective cone ∈ r n over V+(f1,...,fk) P . Thus to say that Y is a projective cone in P is to say that can be⊆ defined by equations in fewer than n + 1 homogeneous variables. Duality: The space P(V ) is, as a set, the set of lines in V . Thus it is a moduli space: a geometric object, whose points can be thought of as the space of all geometric objects of a certain kind. (There are more technical conditions we should impose, having to do with families of lines, as well.) The 6.4. ELEMENTARY PROJECTIVE GEOMETRY 237

vector space V ∨ is the set of linear forms on V . A point of P(V ∨) is then an equivalence class of linear forms on V ; two linear forms are equivalent if and only if they differ by a nonzero scalar, if and only if they define the same hyperplane in V , if and only if they define the same hyperplane in V ∨. Thus, a point of P(V ∨) is the same as a hyperplane of V , so that P(V ∨) is also a moduli space. Note that, dually, since V ∼= V ∨∨, a point p of P(V ) defines a hyperplane in P(V ∨). In fact, the set of all hyperplanes H containing p is a hyperplane in P(V ∨), and every hyperplane in P(V ∨) is of this form. Quadratic forms: For simplicity, we assume that the characteristic of k is not equal to two. A symmetric k-bilinear form B(x, y) on kn kn is then × the same thing as a homogeneous quadratic polynomial q(x)= i j cijxixj. Often, however, it is more convenient to begin with a symmetric n≤ n ma- P × trix A = (aij) and consider the corresponding quadratic polynomial q(x) = a x x . Of course, since char k = 2, the c determine the a and vice- i,j ij i j 6 ij ij versa. We shall refer to Q(x) or to the projective hypersurface Q = V+(q) asP a quadric. After a linear change of coordinates, we can assume by diago- 2 2 nalizing B(x, y) that q(x)= x0 + + xr. Here, r + 1 is the rank of q or of Q. ··· If Q = V (q) where q(x) = x2 + + x2, then Sing Q = V (x ,...,x ). + 0 ··· r + 0 r Thus Sing Q is a linear space of dimension n r. In particular, Q is smooth if and only if q has rank n +1. If Q is singular,− then it is a projective cone over a smooth quadric in Pr, with vertex equal to the linear space Sing Q. It is easy to describe the low rank cases: q has rank one if and only if, for some 2 choice of coordinates, q = x0, and so Q = V+(q) is a hyperplane (counted with multiplicity two, in the appropriate sense). q has rank two if and only if, for some choice of coordinates, q = x2 + x2 =(x + √ 1x )(x √ 1x ), 0 1 0 − 1 0 − − 1 and so Q = V+(q) is a union of two hyperplanes. In all other cases q and Q are irreducible. A quadric in P2 is called a conic. A smooth conic in P2 is projectively 2 2 2 2 equivalent to V+(x0 + x1 + x2), and also to C = V+(x1 x0x2). It is easy 1 − to verify directly that C is the image of P via the morphism (s0,s1) 2 2 7→ (s0,s0s1,s1). Thus C is a special case of a rational normal curve, to be described in full generality below. For quadrics in P3, beyond the cases of rank one and rank two described above, there are quadrics of rank three and four. A quadric of rank three has a unique singular point; it is the cone over a smooth conic, and the singular point is the vertex of the cone. A quadric of rank four is smooth; it 238 CHAPTER 6. QUASIPROJECTIVE VARIETIES

2 2 2 2 is projectively equivalent to V+(x0 +x1 +x2 +x3), and also to Q = V+(x1x2 1 −1 x0x3). As we shall see, Q is the image of the Segre embedding of P P as a subvariety of P3. The surface Q has two different families of lines,× each parametrized by a copy of P1: for a point a = (a , a ) P1, there is the 0 1 ∈ family of lines La = V+(a1x0 a0x1, a0x3 a1x2), as well as the family of lines M = V (b x b x , b x− b x ). One− can check that, for all a, b P1, b + 1 0 − 0 2 0 3 − 1 1 ∈ L M consists of exactly one point, whereas if a = a′ and b = b′, then a ∩ b 6 6 L L ′ = M M ′ = . We leave these facts as exercises. a ∩ a b ∩ b ∅ Beyond the integer valued invariant of the rank, a quadratic polynomial has no further invariants. If we consider the set of nonzero quadratic polyno- (n+2) 1 mials mod k∗ as the projective space P 2 − of all symmetric n n matrices, and thus as a moduli space in its own right, the moduli space of×all quadrics in Pn, then there is a Zariski open subset consisting of smooth quadrics, all of which are projectively equivalent. The set of singular quadrics is the hypersurface det A = 0, where a quadric q is given by the homogenous coordinates q(x)= i,j aijxixj, and A =(aij) is the corresponding symmetric matrix. All linear spaces of a given dimension in Pn, and hence all hyperplanes, are projectivelyP equivalent. Likewise, all smooth quadrics in Pn are projectively equivalent. One can also check that any two sets of three distinct points in P1 are projectively equivalent. Already for homogeneous polynomi- 1 als f(x0, x1) of degree four, defining four points in P , one can see that not all of the sets V+(f) are projectively equivalent. In fact, the classical cross ratio of four points in P1 is invariant under P GL(2,k) and distinguishes projectively inequivalent configurations of four points in P1. Likewise, for cubic 2 polynomials f(x0, x1, x2), defining cubic curves in P , there is a classical invariant, the j-invariant, which is invariant under P GL(2,k) and distinguishes projectively inequivalent plane cubic curves. A fundamental object of study in algebraic geometry is the set of all projective varieties of a certain type, modulo projective equivalence.

6.5 Products

Let X and Y be projective (or more generally quasiprojective) varieties. We would like to realize X Y as a projective or quasiprojective variety, in × much the same way that we analyzed the product of aﬃne varieties in 2.9. In the projective case, if X Pn corresponds to the homogeneous ideal§ I(X) k[x ,...,x ], and Y ⊆ Pm corresponds to the homogeneous ideal ⊆ 0 n ⊆ 6.5. PRODUCTS 239

I(Y ) k[y ,...,y ], then we could consider the homogeneous ideal I(X)+ ⊆ 0 m I(Y ) k[x ,...,x ,y ,...,y ], with quotient R(X) R(Y ). However, this ⊆ 0 n 0 m ⊗k is not the right construction; in fact, V+(I(X)+I(Y )) has dimension d+e+1, where d = dim X and e = dim Y . (See the exercises for a discussion of the meaning of V+(I(X)+ I(Y )).) Instead, we will consider two diﬀerent but related approaches. First construction: To endow the set Pn Pm with a topology, we deﬁne × an element f k[x0,...,xn,y0,...,ym] to be bihomogeneous of degree (a, b) ∈ a b if, for all λ,µ k∗, f(λ x, µ y) = λ µ f(x, y). If f is bihomogeneous of some degree, then∈ · ·

V (f)= (x, y) Pn Pm : f(x, y)=0 ++ { ∈ × } is a well-defined subset of Pn Pm. If f ,...,f are bihomogeneous, the × 1 k subset V++(f1,...,fk) is defined in a similar way. It is standard to see that the sets V++(f1,...,fk) satisfy the axioms for the closed sets of a topology on Pn Pm, which we define to be the Zariski topology. × Lemma 6.5.1. With the Zariski topology on Pn Pm as defined above, × (i) If X is a closed subset of Pn and Y is a closed subset of Pm, then X Y is a closed subset of Pn Pm. In particular, the projections π :×Pn Pm Pn and π : Pn ×Pm Pm are continuous. 1 × → 2 × → (ii) For all x Pn and y Pm, the inclusions of the slices x Pm and Pn y into∈ Pn Pm∈define homeomorphisms i and j{ from} × Pn and ×{ } × x y Pm respectively onto their images.

(iii) Pn Pm is an irreducible topological space. More generally, if X is a × subvariety of Pn and Y is a subvariety of Pm, then, with the induced topology, X Y is an irreducible topological space. × (iv) If An Pn and Am Pm are inclusions of standard aﬃne open subsets, then the⊆ inclusion A⊆n Am Pn Pm is a homeomorphism onto an × → × open subset of Pn Pm. × (v) The diagonal is a closed subset of Pn Pn. × Proof. (i): If f k[x ,...,x ] is homogeneous of degree d, then, viewing f ∈ 0 n as an element of k[x0,...,xn,y0,...,ym], f is bihomogeneous of degree (d, 0). 240 CHAPTER 6. QUASIPROJECTIVE VARIETIES

In this way, if X = V+(f1,...,fk) and Y = V+(g1,...,gℓ), it is easy to check that X Y = V (f ,...,f ,g ,...,g ). × ++ 1 k 1 ℓ (ii): If f k[x0,...,xn,y0,...,ym], f is bihomogeneous of degree (d, e) and x Pn, then∈ evaluating f at x is well-defined up to a scalar and gives a ∈ homogeneous polynomial of degree e in k[y0,...,ym]. Thus, if Z is a closed subset of Pn Pm, then Z ( x Pm) is a closed subset of x Pm when it is identified× as a topological∩ { } × space with Pm. Conversely,{ if}Y ×is a closed subset of Pm, then Pn Y is a closed subset of Pn Pm, such that n m × n n m × (P Y ) ( x P )= x Y . Hence ix : P P P is a homeomorphism onto× its∩ image,{ }× and similarly{ }× for j : Pm Pn→ Pm×. y → × (iii): Examining the proof of Lemma 2.9.6 shows more generally: Lemma 6.5.2. Let X and Y be irreducible topological spaces. Suppose that we are given a topology on the product X Y such that, for all x X and × ∈ y Y , the inclusions ix : Y X Y defined by t (x, t), and jy : X X∈ Y defined by s (s,y) →are homeomorphisms× onto7→ their images. Then→ X × Y is irreducible.7→ × (iii) then follows from (ii). (iv): This is very similar to the proof of Proposition 1.7.7, and we leave the details to the reader. (iv): It is clear that the diagonal in Pn Pn is the set V (x y x y , 0 × ++ i j − j i ≤ i, j n), where the polynomials xiyj xjyi k[x0,...,xn,y0,...,yn] are bihomogeneous≤ of degree (1, 1). − ∈

We can also give Pn Pm the structure of a locally ringed space, and in × fact a scheme. This can be done directly along the lines of giving Pn (or more generally Proj R, where R is a graded ring) a scheme structure. However, it is clear that there is exactly one such structure on Pn Pm which is compatible n m n+m × with the scheme structure on A A ∼= A , for inclusions of standard affine open subsets An Pn and A×m Pm. A similar discussion holds for a general product X Y ⊆, where X and⊆Y are quasiprojective schemes. × Second construction: In this approach, we define Pn Pm directly as a projective variety, and then check that this construction×agrees with the previous one. To do so, consider (n + 1)(m + 1) algebraically independent elements z , 0 i n and 0 j m. Let Pnm+n+m be the corresponding ij ≤ ≤ ≤ ≤ projective space. Define a homomorphism ϕ: k[zij, 0 i n, 0 j m] k[x ,...,x ,y ,...,y ] via z x y . ≤ ≤ ≤ ≤ → 0 n 0 m ij 7→ i j 6.5. PRODUCTS 241

Proposition 6.5.3. With ϕ as above,

(i) Ker ϕ = √I, where I is the ideal generated by the homogeneous quadratic polynomials z z z z , for all 0 i, k n and 0 j, ℓ m. ij kℓ − iℓ kj ≤ ≤ ≤ ≤ (ii) If σ : Pn Pm Pnm+n+m is deﬁned by × → ((a ,...,a ), b ,...,b )) (a b , 0 i n, 0 j m), 0 n 0 m 7→ i j ≤ ≤ ≤ ≤

then σ is injective and its image is V+(Ker ϕ) = V+(I). In particular, σ(Pn Pm) is a Zariski closed subset of Pnm+n+m. × (iii) The topology induced on σ(Pn Pm) by the Zariski topology on Pnm+n+m is identiﬁed via σ with the Zariski× topology on Pn Pm deﬁned via bihomogeneous polynomial functions. ×

Proof. (i), (ii): Clearly, z z z z Ker ϕ. Thus I Ker ϕ, and hence ij kℓ − iℓ kj ∈ ⊆ V (Ker ϕ) V (I). Moreover, ϕ = σ∗. In particular, f Ker ϕ if and + ⊂ + ∈ only if σ∗f = 0, if and only if V (f) contains Im σ. Thus Im σ V+(Ker ϕ). If (c ) V (I), choose (k,ℓ) such that c = 0. After multiplying⊆ by an ij ∈ + kℓ 6 element of k∗, we may assume that ckℓ = 1. Then for all k,ℓ, cij = ciℓckj. Set a = (c0ℓ,...,cnℓ) (with the understanding that ak = ckℓ = 1) and b = (ck0,ck1,...,ckm) (again, with bℓ = 1). Then (cij) = σ(a, b). Moreover, if (a, b) is any point of Pn Pm such that σ(a, b)=(c ), then we must have × ij ak, bℓ = 0. Normalizing a and b so that ak = bℓ = 1, we see that ciℓ = aibℓ = ai and that6 c = a b = b . Thus σ is injective. Moreover, V (I) Im σ. Since kj k j j + ⊆ Im σ V (Ker ϕ) V (I) as well, we must have equality everywhere. In ⊆ + ⊆ + particular, since V+(I)= V+(Ker ϕ) and Ker ϕ is a prime ideal, Ker ϕ = √I.

(iii): If f k[zij, 0 i n, 0 j m], then σ∗f k[x0,...,xn,y0,...,ym] is bihomogeneous∈ of≤ degree≤ (≤d,d).≤ Moreover, every∈ bihomogeneous polynomial of degree (d,d) in k[x0,...,xn,y0,...,ym] arises in this way: if say M = xa0 xan yb0 ybm is a bihomogeneous monomial of degree (d,d), so 0 ··· n 0 ··· m that i ai = j bj = d, the it is clear that M is the pullback under σ of a monomial of degree d in the zij, in many different ways in general. It thus P P suffices to show that, if Z = V++(f1,...,fk) where the fi are bihomogeneous, then Z = V++(g1,...,gN ) where the gi are bihomogeneous of equal degree. Since V++(f1,...,fk)= V++(f1) V++(f2) V++(fk), it suffices to consider the case k = 1. Suppose∩ for example∩···∩ that f(x, y) is bihomogeneous d e d e of degree (d, e) with d > e. Then clearly V++(f) = V++(y0− f,...,ym− f), 242 CHAPTER 6. QUASIPROJECTIVE VARIETIES

d e where each polynomial yi − f is bihomogeneous of degree (d,d). This proves that the Zariski topology on Pn Pm is the pullback of the induced topology on σ(Pn Pm), as claimed. × × Remark 6.5.4. (1) In fact, one can show that I = Ker ϕ in the above notation, and hence I is a prime ideal. (2) Giving σ(Pn Pm) the scheme structure it inherits as a projective variety, it is easy to check× that σ induces an isomorphism of schemes from Pn Pm to σ(Pn Pm), where the scheme structure on Pn Pm was described at× the × × end of the discussion of the ﬁrst construction.

Definition 6.5.5. The morphism σ : Pn Pm Pnm+n+m is the Segre em- × → bedding of Pn Pm. × Example 6.5.6. In case n = m = 1, the Segre embedding is a morphism from P1 P1 to P3, where P3 has coordinates t = z , t = z , t = z , t = × 0 00 1 01 2 10 3 z11. The only nonzero degree two expression of the form zijzkℓ ziℓzkj is the term z z z z = t t t t . Since V (t t t t ) is a smooth− quadric 00 11 − 01 10 0 3 − 1 2 + 0 3 − 1 2 in P3, it follows that every smooth quadric in P3 is projectively isomorphic to the image of P1 P1 via the Segre embedding. × Given quasiprojective varieties X Pn and Y Pm, the above defines X Y as a quasiprojective variety⊆ in Pnm+n+m.⊆ It is easy to see that X Y is× the categorical product of X and Y in the category of quasipro- × jective varieties; in fact, this reduces to the corresponding statement for products of affine varieties. We leave this as an exercise. As a consequence of Lemma 6.5.1, we have:

Corollary 6.5.7. If X is a quasiprojective variety, the diagonal is a closed subset of X X. × In the terminology of abstract schemes, this says that the scheme X is separated. Here the property of being separated is a weak analogue of the Hausdorﬀ condition for topological spaces, as can be seen from the following:

Corollary 6.5.8. Suppose that X and Y are two quasiprojective varieties and that f1, f2 : X Y are two morphisms such that there exists a nonempty open subset U X→with f U = f U. then f = f . ⊆ 1| 2| 1 2 Proof. We begin with a topological lemma: 6.5. PRODUCTS 243

Lemma 6.5.9. Let X be an irreducible topological space and U a nonempty open subset of X. Then U is dense, i.e. U = X. Proof. X is the union of the two closed sets U and X U. Since U = , − 6 ∅ X U = U, and since X is irreducible, X = U. − 6 Returning to the proof of the corollary, consider the product morphism (f1, f2): X Y Y . If ∆ Y Y is the diagonal, then ∆ is closed, → × ⊆ 1 × by Corollary 6.5.7. Then (f1, f2)− (∆) is closed and contains the nonempty 1 1 open subset U. Thus (f1, f2)− (∆) contains U = X and so (f1, f2)− (∆) = X. This says that f (x)= f (x) for all x X. 1 2 ∈ The Veronese embedding. We describe a related construction, the Vero- n+d n 1 nese embedding v : P P( n )− (also called the d-uple embedding). It is → given as follows: for each (n + 1)-tuple of nonnegative integers (a0,...,an) such that a = d, let M be the corresponding monomial xa0 xan . i i a0,...,an 0 ··· n Suppose that za0,...,an are algebraically independent over k, and consider the P n+d ( n ) 1 projective space P − corresponding to k[za0,...,an ]. There is a morphism n+d n 1 a0 a v : P P( n )− defined by v(c ,...,c )=(c c n ). Dually there is → 0 n 0 ··· n the homomorphism ϕ: k[za0,...,an ] k[x0,...,xn] defined by ϕ(za0,...,an ) = a → x 0 xan . 0 ··· n We omit the proof of the following: n+d n ( ) 1 Proposition 6.5.10. The morphism v is an embedding of P P n − . → In particular, Im v is a projective variety. More precisely, Im v = V+(Ker ϕ). Finally, the ideal Ker ϕ can be described by an explicit set of generators, which are of degree two. The Veronese embedding is related to iterated products of Pn as follows: we can embed Pn in the product (Pn)d as the diagonal, and then embed n d n n (P ) in PN via repeated use of the Segre embedding. The image of P will n + d be contained in a linear space of dimension 1, and the induced n − n+d n 1 morphism P P( n )− is the Veronese embedding. For example, following → the diagonal embedding P1 P1 P1 with the Segre embedding P1 P1 P3 realizes P1 as a conic in a hyperplane→ × in P3. × → One very important special case of the Veronese embedding is the case n = 1, so that v : P1 Pd is defined by: → d d 1 d 1 d (x , x ) (x , x − x ,...,x x − , x ). 0 1 7→ 0 0 1 0 1 1 244 CHAPTER 6. QUASIPROJECTIVE VARIETIES

d If the image P has homogeneous coordinates (u0,...,ud), then the image of v is contained in V (u u u u : i + j = k + ℓ), and it is easy to + i j − k ℓ check that in fact Im v = V+(uiuj ukuℓ : i + j = k + ℓ). For example, 1 2 − 1 if d = 2, v(P ) = V+(u0u2 u1) is a smooth conic. If d = 3, v(P ) = 2 − 2 V+(u0u2 u1,u0u3 u1u2,u1u3 u2); we called this curve the twisted cubic curve in Exercise− 1.31.− In general,− the image of v is called the rational normal curve of degree d.

6.6 Grassmannians

Grassmannians are an important example of a moduli space. As a set, G(k, n) is the set of vector subspaces of kn of dimension k. If we are interested in the projective case, we denote the set of all linear subspaces of Pn of dimension k by G(k, n). Given the correspondence between linear subspaces of Pn of dimension k and vector subspaces of kn+1 of dimension k + 1, it is clear that G(k, n) = G(k +1, n + 1). Our goal will be to outline how G(k, n) can be given the structure of a projective variety. n Let V be a vector subspace of k of dimension k. Choose a basis v1,...,vk of V , and consider the vector v v k kn. If w ,...,w is another 1 ∧···∧ k ∈ 1 k basis of V , then there exists an invertible k k matrix A = (aij) such that ×V wi = j aijvj, and hence P w w = (det A)v v . 1 ∧···∧ k 1 ∧···∧ k

Thus, the line through the origin containing v1 vk, or equivalently n k n (∧···∧) 1 the point in the projective space P( k ) = P k − corresponding to v 1 ∧ vk, is independent of the choice of basis and depends only on V . The ···∧ k n V function ρ: G(k, n) P( k ) defined by V v1 vk is called the Plücker embedding →of G(k, n). The image of ρ7→is the∧···∧ set of decomposable elements of P( k kn), i.e.V those elements of P( k kn) which are the image of k n n a vector in k of the form v1 vk, where v1,...,vk k are linearly independent.V One can show that∧···∧ this set isV a closed algebraic∈ subvariety of P( k kn),V described by a set of homogeneous quadratic polynomials (the “Plücker relations”). WeV will give another description of a scheme structure on G(k, n) via an affine open cover. Fix once and for all the standard basis e1,...,en of kn. Now suppose that V is a vector subspace of kn of dimension k, and that v1,...,vk is a basis of V . Let vi = j rijej. Then there are k linearly P 6.6. GRASSMANNIANS 245 independent rows of the k n matrix R = (r ). Suppose that the first k × ij columns are in fact linearly independent. Then there is a unique choice of basis of V for which the matrix R has the form (I B). Here B is a k (n k) | × − matrix, uniquely determined by V . In particular, the subset U1,...,k of G(k, n) consisting of V for which the the first k columns of the matrix R are linearly k(n k) independent is naturally affine space A − . In general, if the columns indexed by the integers i1 < i2 < < ik are linearly independent, we ··· k(n k) find a subset Ui1,...,ik of G(k, n) which is again identified with A − . The intersections Ui1,...,ik Uj1,...,jk can be identified with the Zariski open subset k(n k) ∩ of A − where det X = 0, where X is a k k matrix whose entries are either 6 k(n×k) certain of the coordinate functions on A − or else 0 or 1. Moreover, with 1 this notation, the transition functions are of the form X− , and by Cramer’s 1 k(n k) rule for an inverse matrix, X− is a morphism from V (det X) A − to itself. Thus G(k, n) has the structure of a smooth scheme, covered⊆ by open k(n k) subsets isomorphic to A − . It is easy to check that, with this scheme structure, the Plücker embedding ρ is an injective morphism. In fact, it is an isomorphism from G(k, n) to a closed subvariety of P( k kn). V Exercises

Exercise 6.1. Let R be a Noetherian ring. Show that

p Spec R : R is reduced ∈ p } is an open subset of Spec R. (Identify this set with the complement of Supp √0.)

Exercise 6.2. Let R = k[x, y]/(xy). Compute Derk(R, R) and Derk(R,k), and show that the natural homomorphism Derk(R, R) Derk(R,k) is neither injective or surjective. →

Exercise 6.3. (i) Let d be a positive integer. Suppose that k has characteristic zero, or that the characteristic of k does not divide d. Show that the “Fermat hypersurface” V (xd + + xd ) is a smooth hypersurface in Pn. + 0 ··· n (ii) Let k be a field of characteristic 2. For every n 1, write down an equation for a smooth quadric in Pn. ≥ (iii) Let k be a field of characteristic not equal to 2. Let f be a homogeneous polynomial of degree three in two variables such that f(1, 0) = 0. Show that 6 246 CHAPTER 6. QUASIPROJECTIVE VARIETIES the cubic curve C in P2 (with coordinated (x, y, z) defined by C = V (y2z f(x, z)) + − is singular if and only if the cubic polynomial f(t, 1) has a multiple root, if 1 and only if V+(f) consists of at most two points in P . Exercise 6.4. Let X and Y be quasiprojective varieties (or more generally k- schemes). For (x, y) X Y , show that T (X Y ) = T X T Y . (For ex- ∈ × (x,y) × ∼ x ⊕ y ample, it is easy to check directly that Der( ,k) = Der( ,k) OX,x ⊗k OY,y ∼ OX,x ⊕ Der( Y,y,k).) In particular, X Y is smooth at (x, y) if and only if X is smoothO at x and Y is smooth at×y. Exercise 6.5. For an S-algebra R and f R, the element df of Ω1 defines ∈ R/S a homomorphism R Ω1 via r rdf. Dualizing, for every R-module M, → R/S 7→ there is a homomorphism DerS(R, M) HomR(R, M) = M. Describe this homomorphism explicitly. →

Exercise 6.6. Verify that, if I is an ideal in k[x0,...,xn], then k[x ,...,x ]/I k[x ,...,x ] = (k[x ,...,x ]/I)[x ]. 0 n+1 · 0 n+1 ∼ 0 n n+1 (Deﬁne homomorphisms in both directions.) Exercise 6.7. Let X Pn correspond to the homogeneous ideal I(X) ⊆ ⊆ k[x ,...,x ], and Y Pm correspond to the homogeneous ideal I(Y ) 0 n ⊆ ⊆ k[y0,...,ym]. Deﬁne the projective join J(X,Y ) of X and Y as follows: view Pn and Pm as complementary subspaces of Pn+m+1, and let J(X,Y ) be the union of all the lines x, y , where x X and y Y . Show that h i ∈ ∈ J(X,Y )= V+(I(X)+ I(Y )). Exercise 6.8. Show that X Y is the categorical product of X and Y in the category of quasiprojective× varieties. Exercise 6.9. Show that the homogeneous coordinate ring of the Segre em- n m bedding of P P is the graded subring S′ = k[x y , 0 i n, 0 j m] × i j ≤ ≤ ≤ ≤ of k[x0,...,xn,y0,...,ym]. Interpret this ring as follows: if S is the “bigraded ring” S = k[x ,...,x ,y ,...,y ]= k[x ,...,x ] k[y ,...,y ] 0 n 0 m 0 n ⊗k 0 m = (S ) (S ) = S , 1 d ⊗k 2 e d,e d,e 0 d,e 0 M≥ M≥ 6.6. GRASSMANNIANS 247

where S1 = k[x0,...,xn] and S2 = k[y0,...,ym] with the usual gradings, then S = d,e 0 Sd,e is a direct sum of its bihomogeneous summands. Here ≥ Sd,e = (S1)d k (S2)e is the vector space of bihomogeneous polynomials of L ⊗ degree (d, e). Then S′ = d 0 Sd,d is a graded subring of S, and it is in fact ≥ the subring generated by the xiyj. L Exercise 6.10. Let Q = V (x x x x ). + 1 2 − 0 3

(i) Show that, for a =(a0, a1) = (0, 0), the line La = V+(a1x0 a0x1, a0x3 6 1 − − a1x2) only depends on the image of a in P and lies on Q. Similarly for the family of lines M = V (b x b x , b x b x ), where b =(b , b ). b + 1 0 − 0 2 0 3 − 1 1 0 1 (ii) Show that, for all a, b P1, L M consists of exactly one point, ∈ a ∩ b whereas if a = a′ and b = b′, then L L ′ = M M ′ = . 6 6 a ∩ a b ∩ b ∅ 1 (iii) For each x Q, show that there is a unique a P such that x La, ∈ 1 ∈ 1 ∈ and a unique b P such that x Mb. If π : Q P is deﬁned by: π(x) = a, where∈a P1 is the unique∈ point such→ that x L , show ∈ ∈ a that π is a morphism, and similarly for the corresponding map using the lines Mb. Chapter 7

Graded Rings

Graded rings have a;ready made an appearance in the study of projective varieties. They also appear in the study of pairs (R, I), where R is a Noethe- rian ring and I is an ideal of R, typically a maximal ideal. We associate numerical invariants to graded rings, which are related to the degree of a variety, in the projective case, and which are connected to dimension theory in the case of a pair (R, I). One consequence is the dimension theorem for local Noetherian rings. The rest of the chapter deals with material motivated by the dimension theorem, regular local rings and completions.

7.1 Filtrations and the Artin-Rees lemma

Throughout this section, S = n 0 Sn is a graded ring, so that S0 is a ≥ subring of S and S+ = Sn is an ideal. Recall that a graded S-module n>0 L is an S-module N = n 0 such that Sn Nk Nn+k for all n and k. L≥ · ⊆ Example 7.1.1. LetLR be a ring and let I be an ideal of R. 2 n (i) Consider the R-algebra R I I = n 0 I , with the usual ⊕ ⊕ ⊕··· ≥ convention that I0 = R. If r Ik and s Iℓ, viewed as elements of the L direct sum, then the product∈r s Ik+∈ℓ. We can also view the direct sum as the subring R[It] R[t·].∈ This graded ring is denoted B (R), ⊆ I the blowup algebra of R. (The reason for this terminology is explained in the exercises and the next chapter.)

(ii) Viewing I as an ideal in R, the quotient BI (R)/IBI(R)=(R/I) 2 n n+1 ⊕ (I/I ) = n 0 I /I . We denote this graded ring by grI R. ⊕··· ≥ L 248 7.1. FILTRATIONS AND THE ARTIN-REES LEMMA 249

For example, if R = k[x1,...,xn] and m = (x1,...,xn), then grm R ∼= R with its usual grading.

Lemma 7.1.2. Let S = n 0 Sn be a graded ring. ≥ (i) The ring S is NoetherianL if and only if S0 is Noetherian and S is a ﬁnitely generated S0-algebra.

(ii) Suppose that S is Noetherian and that M = n 0 is a ﬁnitely generated ≥ graded S-module. Then, for all n 0, Mn is a ﬁnitely generated S0- module. ≥ L

Proof. (i) If S is Noetherian, then so is S0 since it is the quotient of S by S+. Moreover, the ideal S+ is finitely generated, by s1,...,sk, say, and we may assume that the si are homogeneous of degree di for all i. Hence every k element s of Sn is of the form s = risi with ri Sn di . By induction i=1 ∈ − on n, s is a sum of monomials in the si times elements of S0, so that the natural homomorphism S [s ,...s ]P S is surjective. Thus S is a finitely 0 1 k → generated S0-algebra. Conversely, if S0 is Noetherian and S is a finitely generated S0-algebra, then the Hilbert basis theorem implies that S is Noetherian. (ii) Suppose that S is generated over S0 by homogeneous elements s1, ..., sk, where deg si = di > 0, and that M is generated as an S-module by m1,...,mℓ, where we may as well take the mj to be homogeneous of degree e . For n 0, it is easy to see that M is generated as an S -module by j ≥ n 0 k sa1 sak m : a d + e = n , { 1 ··· k j i i j } i=1 X and this is a finite set. Thus Mn is a finitely generated S0-module. Corollary 7.1.3. Let R be a ring and let I be an ideal of R. If R is Noethe- rian, then BI (R) and grI R are both Noetherian. If BI (R) is Noetherian, then R is Noetherian. Proof. Suppose that R is Noetherian, and hence that I is finitely generated. If I = (r1,...,rk), then viewing I as a subset of BI (R), it follows that n BI (R)+ = n 1 I is generated over R by r1,...,rk. Thus (i) above implies ≥ that BI (R) is Noetherian. Since gr R = BI (R)/IBI (R) is a quotient of L I BI (R), grI R is Noetherian as well. Conversely, if BI (R) is Noetherian, then so is BI (R)0 = R. 250 CHAPTER 7. GRADED RINGS

Definition 7.1.4. Let R be a ring and let M be an R-module. A decreasing filtration on M is a sequence of submodules M = M0 M1 M2 . We call M a filtered R-module and denote M, together with⊇ its⊇ filtration,⊇··· by the notation (Mn). The associated graded module gr(Mn) is then by definition

n 0 Mn/Mn+1. ≥ LExample 7.1.5. If R is a ring and I is an ideal of R, then we have the n filtration Rn = I , and the associated graded ring is grI R. More generally, n for an R-module M, we can consider the filtration Mn = I M, and the n n+1 associated graded module gr(Mn) = n 0 I M/I M is a graded module over gr R. ≥ I L Definition 7.1.6. With R, I and (Mn) as above, the filtration on M is a I-filtration if IM M for all n. In this case, IkM M for all n n ⊆ n+1 n ⊆ n+k and k. The I-filtration is stable if, for all n 0, IMn = Mn+1 and hence k ≫ I Mn = Mk+n for all k 0. In other words, there exists an N 0 such that, for all k 0, IkM = M≥ . ≥ ≥ N k+N n Example 7.1.7. The filtration (Mn) defined by Mn = I M is a stable I- filtration.

Definition 7.1.8. Suppose that (Mn) and (Mn′ ) are two filtrations on the same underlying R-module M, i.e. M0 = M0′ . Then the filtrations (Mn) and (M ′ ) have bounded difference if there exist integers N , N 0 such that n 1 2 ≥ Mn+N1 Mn′ and Mn′ +N2 Mn for all n 0. Of course, we can then always assume⊆N = N by replacing⊆ N by max≥N , N . 1 2 i { 1 2} Lemma 7.1.9. If (Mn) is a stable I-filtration, then (Mn) and the filtration of M defined by InM have bounded difference.

Proof. If (Mn) is an I-filtration, and in particular if it is a stable I-filtration, n then I M Mn for all n 0. Furthermore, if for all n N, IMn = Mn+1, ⊆ ≥ n n ≥ then for all n 0, Mn+N = I MN I M. Thus the two filtrations have bounded difference.≥ ⊆ Theorem 7.1.10 (Artin-Rees lemma). Suppose that R is a Noetherian ring, I is an ideal of R and M is a finitely generated R-module. Let (Mn) be a stable I-filtration of M. Then for every submodule M ′ of M, the filtration (M ′ M ) is a stable I-filtration on M ′. ∩ n The theorem is usually stated in the following more concrete form: 7.1. FILTRATIONS AND THE ARTIN-REES LEMMA 251

Corollary 7.1.11. Suppose that R is a Noetherian ring, I is an ideal of R and M is a finitely generated R-module. Then for every submodule M ′ of n M, there exists an integer N 0 such that, for all n N, I M M ′ = n N N ≥ ≥ ∩ I − (I M M ′). ∩ Proof. This follows immediately by considering the stable I-filtration InM and using the remarks after the definition of stability (with k = n N). − To prove the Artin-Rees lemma, we begin with a preliminary definition:

Definition 7.1.12. If (Mn) is a filtered R-module, define M ∗ = n 0 Mn. ≥ n For example, with M = R and Mn = I , M ∗ = BI (R). MoreL generally, if (Mn) is an I-filtration, then M ∗ is a graded module over BI (R) (with Mn∗ = Mn) in the obvious way. Lemma 7.1.13. Let R be a Noetherian ring, with I and M as above, and let (Mn) be an I-filtration of M. Then (Mn) is a stable I-filtration if and only if M ∗ is a finitely generated graded BI (R)-module.

Proof. Since R is Noetherian, BI (R) is Noetherian as well, by Lemma 7.1.2. Since by assumption (M ) is an I-filtration of M, IkM M for all n n ⊆ k+n k, n. For n 0, consider the R-submodule M ∗ of M ∗ defined by ≥ n 2 Mn∗ = M0 M1 Mn IMn I Mn n ⊕ ⊕···⊕ ⊕ ⊕ ⊕··· n k = M I − M . k ⊕ n k=0 k n+1 M M≥ k k 1 Clearly, M ∗ is a graded B (R)-submodule of M ∗. Since I M I − M , n I n ⊆ n+1 the submodules Mn∗ are an increasing sequence of submodules of M ∗ with n 0 Mn∗ = M ∗. Moreover, Mn∗ is generated as a BI (R)-module by M0 ≥ ⊕ M1 Mn. Since M and hence the Mn are finitely generated R-modules, S ⊕···⊕ it is clear that Mn∗ is a finitely generated BI (R)-module. Thus, since BI (R) is Noetherian, if M ∗ is a finitely generated BI (R)-module, then the chain Mn∗ terminates and hence M ∗ = Mn∗ for some n. Conversely, since Mn∗ is a finitely generated BI (R)-module, we see that M ∗ is a finitely generated BI (R)-module if and only if M ∗ = Mn∗ for some n. But M ∗ = Mn∗ for some k n if and only if, for all k 0, Mn+k = I Mn, if and only if (Mn) is a stable I-filtration. ≥ 252 CHAPTER 7. GRADED RINGS

Proof of the Artin-Rees lemma. Since (Mn) is a stable I-filtration, Lemma 7.1.13 implies that M ∗ is a finitely generated BI (R)-module. Consider the induced filtration (M ′ M ). Since I(M ′ M ) IM ′ IM M ′ M ,(M ′ M ) ∩ n ∩ n ⊆ ∩ n ⊆ ∩ n+1 ∩ n is an I-filtration of M ′, and hence the R-submodule (M ′)∗ of M ∗ is in fact a BI (R)-submodule of M ∗. Since BI (R) is Noetherian and M ∗ is a finitely generated BI (R)-module, (M ′)∗ is a finitely generated BI (R)-module as well. But then, again by Lemma 7.1.13, (M ′ M ) is a stable I-filtration. ∩ n As a corollary, we have the following:

Corollary 7.1.14 (Krull’s theorem). Let R be a Noetherian ring, let I be an ideal of R and let M be a ﬁnitely generated R-module. Then

∞ InM = m M : there exists an r I such that (1 + r)m =0 . { ∈ ∈ } n=1 \ n In particular, n∞=1 I M =0 if either one of the following holds: (i) I is containedT in the Jacobson radical of R.

(ii) R is an integral domain, I = R, and M is a torsion free R-module. 6 n Proof. Let M ′ = n∞=1 I M. Then by Corollary 7.1.11 (with n = N + 1), N+1 N there exists an N 0 such that I M M ′ = I(I M M ′). But clearly k T≥ ∩ ∩ I M M ′ = M ′ for all n, so that M ′ = IM ′. Thus, by Nakayama’s lemma ∩ (cf. Exercise 3.5), there exists r I such that (1 + r)M ′ = 0. Conversely, if m M is such that there exists ∈r I with (1 + r)m = 0, then m = rm = 2 ∈ n ∈ − r m = n∞=1 I M. To see···∈ the last statement, if I is contained in the Jacobson radical of R, then 1 + r isT a unit for every r I. Thus, if (1 + r)m = 0, then m = 0, and n ∈ so n∞=1 I M = 0. If R is an integral domain and M is torsion free, then (1 + r)m = 0 if and only if m =0or1+ r = 0, i.e. r = 1. But 1 I if T n − − ∈ and only if I = R. Thus, if I = R, then again ∞ I M = 0. 6 n=1 T 7.2 Hilbert functions

Consider first the graded ring S = k[x0,...,xr]. Let M = n 0 Mn be a finitely generated graded S-module. The main example is M = ≥S/I, where L I is a homogeneous ideal. By Lemma 7.1.2, each Mn is a finitely generated 7.2. HILBERT FUNCTIONS 253

S0-module, i.e. a finite dimensional k-vector space. We define HM (n) = r + n dim M . For example, H (n)= . We will only be interested in the k n S n asymptotic behavior of H , i.e. the values H (n) for n 0. M M ≫ More generally, suppose that S = n 0 Sn is a Noetherian graded ring. ≥ Let χ be a function from the category of all finitely generated S0-modules to L Z which is additive on exact sequences, i.e. if 0 N ′ N N ′′ 0 is an → → → → exact sequence of finitely generated S0 modules, then χ(N)= χ(N ′)+χ(N ′′). For example, if S0 = k, then we could take χ(N) = dim N; more generally, if S0 is an Artin ring, we could take χ(N)= ℓ(N). It is an easy exercise that, if 0 M M M 0 → 1 → 2 →···→ n → is an exact sequence of finitely generated S0-modules and χ is an additive function in the above sense, then

n ( 1)iχ(M )=0. − i i=0 X

Let M = n 0 Mn be a finitely generated graded S-module. Thus Mn is a ≥ finitely generated S0-module for all n, again by Lemma 7.1.2. We define the Hilbert functionL

Hχ,M (n)= χ(Mn). The following example will be important in the applications to local rings:

Definition 7.2.1. Let R be a Noetherian ring, let m be a maximal ideal in R and let q be an m-primary ideal, or equivalently an ideal such that n n+1 √q = m. Set S = grq R = n 0 q /q . Thus S is again Noetherian. If ≥ n n+1 M is a finitely generated S-module, then grq M = n 0 q M/q M is a L ≥ finitely generated gr R-module, and we can define Hℓ,gr M , which we denote q Lq by Hq,M . Explicitly,

n n+1 Hq,M (n)= ℓ(q M/q M), where the length refers to the length of the ﬁnitely generated R/q-module qnM/qn+1M.

The function Hq,M is not quite additive over exact sequences. We will analyze the behavior of Hq,M over exact sequences shortly. 254 CHAPTER 7. GRADED RINGS

Before continuing, we pause for a brief digression on numerical polynomials. Given two functions f,g : N N, we write f g if f(n) = g(n) for all n 0. Clearly, is an equivalence→ relation which∼ allows us to identify to functions≫ on N if∼ they agree for all large values of n N. In what follows, ∈ we identify two functions f and g if f g. Note that, if p and q are two polynomials and p q, then p = q. ∼ ∼ A numerical polynomial p(t) Q[t] is a polynomial p such that p(n) N for all n N, n 0. More generally,∈ given any function f : N N, we shall∈ call f a numerical∈ ≫ polynomial if f p, where p is a numerical→ polynomial, ∼ necessarily unique. If f is a numerical polynomial and f p, then the degree deg f is by deﬁnition deg p. ∼ The following lemma is elementary and is left to the reader:

Lemma 7.2.2. (i) Given a function f : N N, define the difference func- → tion ∆f : N N by ∆f(n) = f(n + 1) f(n). Then f is a nonzero numerical polynomial→ of degree d if and only− if ∆f is a numerical polynomial of degree d 1. − (ii) The polynomial p(t) Q[t] is a numerical polynomial of degree at most ∈ d if and only if there exist integers a , 0 i d, such that i ≤ ≤ n t p(t)= a , i n i i=0 X − t 1 where by definition = t(t 1) (t i + 1) is a polynomial of i i! − ··· − degree i and leading coefficient 1/i!.

In (i) above, and in what follows, we adopt the convention that deg p(t)= 1 if and only if p is the zero polynomial. − Proposition 7.2.3. Let S be a graded Noetherian ring, and suppose that S = S0[s1,...,sr], with deg si =1 for all i. Suppose that M is a ﬁnitely generated graded S-module and that χ is an additive function from the category of ﬁnitely generated S0-modules to Z. Then the Hilbert function Hχ,M (n) is a numerical polynomial of degree at most r 1. −

Proof. The proof is by induction on r. If r = 0, then S = S0 and, since M is a finitely generated S -module, M = 0 forall n 0. Thus H (n)=0asa 0 n ≫ χ,M 7.2. HILBERT FUNCTIONS 255 numerical polynomial, and so by our conventions deg H (n)= 1=0 1 χ,M − − in this case. For r > 0, consider the graded ring S/srS, which is generated over S0 by s1,...,sr 1. Let K M be the kernel of multiplication by sr and let L M − ⊆ ⊆ be the cokernel of multiplication by sr. Thus, K and L are finitely generated graded modules over S/s S, and, for all n 0, there is an exact sequence r ≥ 0 K M M L 0. → n → n → n+1 → n+1 → It follows that ∆H (n)= χ(M ) χ(M ) χ,M n+1 − n = χ(L ) χ(K ) n+1 − n = H (n + 1) H (n). χ,L − χ,K By the inductive hypothesis, Hχ,L(n + 1) and Hχ,K(n) are both numerical polynomials of degree r 2. Hence Hχ,M (n) is a numerical polynomial of degree at most r 1. ≤ − − Corollary 7.2.4. Let R be a Noetherian ring, m a maximal ideal in R, q an m-primary ideal, and M a finitely generated R-module. Then:

(i) If q is generated by t elements, then Hq,M is a numerical polynomial of degree at most t 1. − n (ii) Deﬁne Pq,M (n) = ℓ(M/q M). Then ∆Pq,M = Hq,M . In particular, Pq,M is a numerical polynomial and deg Pq,M = deg Hq,M +1.

Proof. (i) Suppose that q is generated by s1,...,rt. Then grq R is generated over R/q by the images of s1,...,st, which are homogeneous of degree one n in grq R. (i) then follows from Proposition 7.2.3. As for (ii), M/q M is a ﬁnitely generated R/qn-module, hence is Artinian and has ﬁnite length. The exact sequence 0 qnM/qn+1M M/qn+1M M/qnM 0 → → → → then implies that ∆Pq,M = Hq,M . Lemma 7.2.5. With notation as in Corollary 7.2.4, let

0 M ′ M M ′′ 0 → → → → be an exact sequence of ﬁnitely generated R-modules. Then H H ′′ = q,M − q,M F , where F and Hq,M ′ are two polynomials with the same degree and leading coeﬃcient. 256 CHAPTER 7. GRADED RINGS

Proof. The exact sequence 0 M ′ M M ′′ 0 leads to an exact → → → → sequence n n n 0 M ′ q M q M q M ′′ 0. → ∩ → → → Passing to the associated gradeds, we have an exact sequence

n n+1 n n+1 n n+1 0 M ′ q M/M ′ q M q M/q M q M ′′/q M ′′ 0. → ∩ ∩ → → → n n+1 Let F (n)= ℓ(M ′ q M/M ′ q M). Clearly Hq,M Hq,M ′′ = F . ∩ ∩ − n By the Artin-Rees lemma (Theorem 7.1.10), (M ′ q M) is a stable q- ∩ n filtration of M ′. Hence the graded Bq(R)-module n 0 M ′ q M is finitely ≥ ∩ n generated over Bq(R). By Proposition 7.2.3, P (n) = ℓ(M ′/M ′ q M) is a n L n+1 ∩ numerical polynomial. Thus, ∆P (n)= ℓ(M ′ q M/M ′ q M)= F (n) is also a numerical polynomial. ∩ ∩ n n By Lemma 7.1.9, the two filtrations (M ′ q M) and (q M ′) of M ′ have ∩ n bounded difference. Thus, if P (n) = ℓ(M ′/M ′ q M) and Pq,M ′ (n) = n ∩ ℓ(M ′/q M ′) are defined as above, then there is a constant k0 such that P (n k ) P ′ (n) P (n + k ). This is only possible if P and P ′ − 0 ≤ q,M ≤ 0 q,M have the same degree and leading coefficient. But then ∆P = F and ∆Pq,M ′ = Hq,M ′ have the same degree and leading coefficient as well.

7.3 The degree of a projective variety

Proposition 7.3.1. Let X PN be a projective variety of dimension r, with ⊆ homogeneous coordinate ring R(X)= S/I(X), where S = k[x0,...,xN ] and I(X) is the homogeneous ideal of X. Let HX (n)= HR(X)(n) = dimk R(X)n. Then HX is a numerical polynomial of degree r with leading coefficient of the form d/r! for some positive integer d. Proof. By the Noether normalization lemma, there exist y ,...,y S, ho- 0 r ∈ mogeneous of degree one and algebraically independent over k, such that k[y0,...,yr] is a subring of R(X) and R(X) is integral over k[y0,...,yr]. In particular, R(X) is a finitely generated k[y0,...,yr]-module. Thus, by Proposition 7.2.3, with S = k[y0,...,yr] and χ = dimk, HX is a numerical r polynomial of degree at most r +1 1= r. Hence limn HX (n)/n exists − →∞ and is equal to the leading coefficient of HX , if deg HX = r, and is zero if deg H

Deﬁnition 7.3.2. If X PN is a projective variety of dimension r and the ⊆ leading coeﬃcient of HX is d/r!, then the positive integer d is called the degree of X.

The following examples may be computed directly from the deﬁntion:

Example 7.3.3. (1) As we saw in the course of the proof of Proposition 7.3.1, deg Pr = 1. d+r r ( ) 1 r (2) Let v : P P r − be the d-fold Veronese embedding of P . Then → deg v(Pr)= dr.

(3) Let s: Pr Ps Prs+r+s be the Segre embedding. Then × → r + s r + s deg s(Pr Ps)= = . × r s Here is an example in a similar vein, which we shall generalize shortly:

r Lemma 7.3.4. Suppose that X = V (f) P , where f(x) k[x0,...,xr] is irreducible and homogeneous of degree d.⊆ Then deg X = d. ∈

Proof. We have an exact sequence

f 0 k[x ,...,x ] × k[x ,...,x ] R(X) 0. → 0 r n −→ 0 r n+d → n+d → It follows that

d r 1 H (n + d)= HPr (n + d) HPr (n)= n − + lower order terms. X − (r 1)! −

Thus, the leading coeﬃcient of HX (n), which is the same as the leading coeﬃcient of H (n + d), is d/(r 1)!. X − We will use a slight generalization of the degree, as follows: 258 CHAPTER 7. GRADED RINGS

Proposition 7.3.5. Let S = k[x0,...,xN ] and let I be a homogeneous ideal in S with I = S. Let R = S/I, so that HS/I (n) = HR(n) = dimk(S/I)n. 6 t Suppose that V (I)= X = i=1 Xi, where the Xi are the distinct irreducible components of X, and dim Xi = r for i k, dim Xi k. In other words, dim X = r. Then S ≤

(i) Let N be a ﬁnitely generated graded R-module. Then, with HN (n) = dimk Nn, HN is a numerical polynomial of degree at most r, and the degree of HR is r.

(ii) For N = R, deg HR = r, and, if R is reduced, the leading coeﬃcient

of HR is d/r!, where d = i k deg Xi is the sum of the degrees of the components of X which have≤ dimension r. P Proof. The proof is by induction on the dimension r, beginning with the case r = 1, which by convention means that V (I)= and hence that √I = S . − ∅ + In this case (ii) is vacuous, and since R is Artinian, Nn = 0 for all finitely generated graded R-modules N and all n 0. Thus HN = 0 as a numerical polynomial, so that deg H = 1= r. We≫ thus assume that the proposition N − has been proved for all homogeneous ideals I in S such that dim V (I) < r. Let I be a homogeneous ideal in S with dim V (I)= r. Step I: Suppose that I is a prime ideal, so that R is the homogeneous coordinate ring of the projective variety X = V (I). By Proposition 7.3.1, deg HR = r, and the leading coefficient of HR is deg X/r!, proving (ii) in this case. Moreover, as in the proof of Proposition 7.3.1, there is an inclusion k[y ,...,y ] R, such that R is finite over the subring k[y ,...,y ]. Thus, 0 r ⊆ 0 r if N is a finitely generated R-module, then N is also a finitely generated k[y ,...,y ]-module. By Proposition 7.2.3, deg H r, proving (i) in this 0 r N ≤ case as well. Step II: Suppose that I = √I and that N = R = S/I. The proof is by induction on the number t of irreducible components of X = V (I); the case t = 1 follows from Step I. If X has t > 1 irreducible components, we can write X = X1 Y , where X1 is irreducible of dimension r, Y has t 1 irreducible components,∪ dim Y r, and Y X is a proper closed subset− of X and hence dim(Y X) < r.≤ Let I be the∩ homogeneous ideal of X and ∩ 1 1 let J be the homogeneous ideal of Y , so that √I1 = I1 and √J = J. Thus √I1 J = I1 J, implying that I1 J = I since V (I1 J) = X = V (I). Moreover,∩ V (I∩ + J) = X Y , which∩ has dimension less∩ than r. From the 1 1 ∩ 7.3. THE DEGREE OF A PROJECTIVE VARIETY 259 exact sequence

0 S/I J S/I S/J S/I + J 0, → 1 ∩ → 1 ⊕ → 1 → we see that HS/I1 + HS/J = HS/I + HS/I1+J . Moreover, deg HS/I1+J < r by induction on r. If dim Y n). Thus there is an exact sequence of graded R-modules

t 0 K R( e ) N 0, → → − i → → i=1 M so that H (n)+ H (n) = t H (n e ), at least for n 0. Since N is N K i=1 R − i ≫ a ﬁnitely generated k[x0,...,xN ]-module, it is a numerical polynomial, and the inequalities P t 0 H (n) H (n e ) ≤ N ≤ R − i i=1 X for all n 0 and the fact that t H (n e ) is a numerical polynomial ≫ i=1 R − i of degree r imply that deg HN r. ≤P Step IV: I is an arbitrary homogeneous ideal such that dim V (I) = r and N is a ﬁnitely generated R-module. Let J be the radical of (0) in R, i.e. the image of √I, and let R′ = R/J = S/√I. Then J is a nilpotent ideal since R is Noetherian, so that J a = 0 for some a> 0. It is easy to see that

a 1 − HN = HJiN/Ji+1N . i=0 X 260 CHAPTER 7. GRADED RINGS

i i+1 Since each J N/J N is a module over R′ = S/√I, Step III implies that deg H i i+1 r. Hence deg H r as well. In case N = R, one term in J N/J N ≤ N ≤ the sum is HR′ , and Step II implies that deg HR′ = r. Hence deg HR = r as well. This concludes the inductive step, and hence the proof.

There is a local analogue of the degree:

Definition 7.3.6. Let R be a local Noetherian ring with maximal ideal m. Then Hm,R is a numerical polynomial, and in fact we shall see that it is of degree r equal to dim R 1. Suppose that the leading coefficient is d/r! − for some integer d. We define the multiplicity of R at m to be the positive integer d. More generally, for an m-primary ideal q, we define the multiplicity e(q, R) to be the unique positive integer such that the leading coefficient of Hq,R is e(q, R)/r!, where r = deg Hq,R.

Some connections between local and global degrees are discussed in the exercises. Let us prove some basic facts about the intersection of two projective varieties. We begin intersections in aﬃne space.

Theorem 7.3.7. Suppose that X and Y are two affine varieties in An, of dimensions d and e respectively. Let Z be a component of X Y . Then dim Z d+e n. Equivalently, codim Z codim X +codim Y =2∩ n d e. ≥ − ≤ − − Proof. Let I(X) and I(Y ) be the ideals of X and Y in the affine coordinate n ring of A , i.e. in k[t1,...,tn]. Recall that the components of X Y = V (I(X) + I(Y )) correspond to the minimal prime ideals p associated∩ to I(X)+ I(Y ). So we must show that, if p is a minimal prime ideal asociated to I(X)+ I(Y ), then dim k[t ,...,t ]/p d + e n. 1 n ≥ − The idea of the proof is the so-called “reduction to the diagonal.” In other words, instead of considering X Y in An, we consider the intersection of the product X Y A2n with the∩ diagonal ∆ A2n. Although this × ⊆ ⊆ seems to complicate the picture, the advantage is that the diagonal is a subvariety of A2n of dimension n which is defined by exactly n equations: 2n n n if t1,...,tn,u1,...,un are the coordinates on A = A A , then ∆ = V (t u ,...,t u ). × 1 − 1 n − n To make this idea precise, note that, if k[t1,...,tn,u1,...,un] is the coordinate ring of A2n = An An, then the projections to the two factors correspond to the inclusions× k[t ,...,t ] k[t ,...,t ,u ,...,u ] and 1 n ⊆ 1 n 1 n 7.3. THE DEGREE OF A PROJECTIVE VARIETY 261 k[u ,...,u ] k[t ,...,t ,u ,...,u ]. Moreover ∆ = An, and this isomor- 1 n ⊆ 1 n 1 n ∼ phism is induced by projection onto either factor; in terms of coordinate rings, the composition

k[t ,...,t ] k[t ,...,t ,u ,...,u ] 1 n → 1 n 1 n k[t ,...,t ,u ,...,u ]/(t u ,...,t u ) → 1 n 1 n 1 − 1 n − n is an isomorphism, and similarly for the u-variables. Viewing k[t1,...,tn] as a subring of k[t1,...,tn,u1,...,un] deﬁnes an ideal

J(X)= I(X)k[t1,...,tn,u1,...,un];

J(Y ) is deﬁned similarly for I(Y ) using the other inclusion k[u ,...,u ] 1 n ⊆ k[t1,...,tn,u1,...,un]. Lemma 7.3.8. There is an isomorphism

k[t ,...,t ,u ,...,u ]/I(X Y )+ I(∆) = k[t ,...,t ]/I(X)+ I(Y ). 1 n 1 n × ∼ 1 n Proof. First, k[t1,...,tn,u1,...,un]/I(X Y )+I(∆) ∼= k[t1,...,tn]/J, where J is the image in I(X Y ) in k[t ,...,t× ] = k[t ,...,t ,u ,...,u ]/(t × 1 n 1 n 1 n 1 − u1,...,tn un). By Corollary 2.9.3 in Chapter 2, I(X Y )= J(X)+ J(Y ). Thus J =−I(X)+ I(Y ). × Lemma 7.3.9. dim X Y = dim X + dim Y . × Proof. By the normalization lemma, there exists a subring k[x1,...,xd] of A(X), where the xi are algebraically independent over k and A(X) is integral over k[x1,...,xd]. Similarly, there exists a subring k[y1,...,ye] of A(Y ), where the yi are algebraically independent over k and A(Y ) is integral over k[y ,...,y ]. Thus A(X) A(Y ) contains a subring isomorphic to 1 d ⊗k k[x1,...,xd,y1,...,ye], and is integral over this subring. It follows that

dim A(X Y ) = dim A(X) A(Y )= d + e = dim X + dim Y. × ⊗k

Combining the above two lemmas, we see that we must prove that if q is a minimal prime in A = k[t1,...,tn,u1,...,un]/I(X Y )+ I(∆), then dim A/q d + e n. This statement is equivalent to the× assertion that the minimal primes≥ in−A(X Y ) associated to the ideal (t u ,...,t u ) have × 1 − 1 n − n 262 CHAPTER 7. GRADED RINGS coheight at least d + e n, or equivalently height at most n (since A(X Y ) − × has the property that, for all prime ideals P of A(X Y ), ht P + coht P = dim A(X Y ) = d + e). More generally, if R is a× Noetherian ring and r ,...r ×R, then one form of Krull’s principal ideal theorem asserts that 1 n ∈ every minimal prime ideal associated to (r1,...,rn) has height at most n (Proposition 7.5.2 below). Applying this to R = A(X Y ) and r = t u × i i − i concludes the proof. Of course, no matter what the dimensions of X and Y in An, as long as they are proper subspaces it is always possible for X Y = . For example, ∩ ∅ n the two hypersurfaces V (xn) and V (xn 1) have empty intersection in A . But for projective space, the situation is− much better. The following result can be viewed as a big generalization of the fact that, in P2, every pair of lines meets: Theorem 7.3.10. Let X,Y Pn be two projective varieties, of dimensions d and e respectively. If d + e⊆ n, then X Y = . Moreover, in this case ≥ ∩ 6 ∅ every component of X Y has dimension at least d + e n. ∩ − Proof. Let C(X) An+1 be the affine cone over X, and similarly for C(Y ). Thus dim C(X)=⊆d + 1, dim C(Y )= e +1, and 0 C(X) C(Y ). By The- orem 7.3.7, every component of C(X) C(Y ) containing∈ zero∩ has dimension ∩ at least d+1+e+1 (n+1) = d+e n+1 1. Hence (C(X) C(Y )) 0 − − ≥ ∩ −{ } is nonempty. Since X Y = ((C(X) C(Y )) 0 )/k∗, X Y = . Since Pn is covered by open∩ sets U isomorphic∩ to A−{n, and} dim X∩ U 6 =∅ dim X, i ∩ i and similarly for Y , the final statement is clear. Definition 7.3.11. Let X and Y be closed subvarieties of An or of Pn, of dimensions d and e respectively. We say that X and Y intersect properly if every component of X Y has dimension d+e n. In particular, if d+e < n, then X and Y intersect∩ properly if and only if−X Y = , and if d + e = n, then X and Y intersect properly if and only if X ∩ Y is∅ finite. ∩ One of the main goals of intersection theory is the following: supposing that X and Y intersect properly, assign a positive integer i(X,Y ; Z) to each irreducible component Z of X P , the intersection multiplicity of X and Y along Z, with various nice properties.∩ For example, if there is a smooth point z of Z such that the tangent spaces TX,z and TY,z intersect properly (in which case we say that z is a transversal point of intersection, then we would like i(X,Y ; Z) = 1. Many of the other desired properties of i(X,Y ; Z) 7.3. THE DEGREE OF A PROJECTIVE VARIETY 263 can be summarized as follows: define a positive cycle of dimension r (in An or in Pn or in a general smooth quasiprojective variety) to be a finite formal linear combination X nX X of subvarieties of dimension r. A cycle of dimension r is defined similarly, where we drop the condition that the nX are positive. Thus, if X and YPare subvarieties of dimensions d and e respectively · which intersect properly, then the formula X Y = Z i(X,Y ; Z) defines the intersection product of X and Y as a positive cycle of dimension e + d n. The intersection of two positive cycles, or more generallyP of two cycles,− is then defined by linearity, provided that all subvarieties which appear with a nonzero coefficient meet properly. (Note that it is much neater to index by the codimension, since then the product of two cycles of codimensions a, b respectively has codimension a + b.) Then we can summarize the desiderata for the intersection multiplicities i(X,Y ; Z) by saying that · is commutative and associative where it makes sense. There are many definitions of i(X,Y ; Z): via Hilbert-Samuel functions, Tor, normal cones, . . . n Suppose that Z = X nX X is a cycle of dimension r in P . Thus the X are irreducible subvarieties of Pn. We define the degree deg Z to P be X nX deg X. One of the classical results of intersection theory is then the following: P n Theorem 7.3.12 (Bézout’s theorem). Let Z1 and Z2 be two cycles in P which meet properly. Then deg(Z1 · Z2) = (deg Z1)(deg Z2). This leads to the following interpretation of the degree of a projective variety X: Corollary 7.3.13. Let X Pn be a projective variety of dimension r. Sup- ⊆ pose that H is a linear subspace of Pn of dimension n r, that X H is finite, and that, for all p X H, i(X,H; p)=1. Then deg− X = #(X∩ H). ∈ ∩ ∩ In the last chapter, we shall show that “almost all” linear spaces H in Pn of dimension n r meet X in exactly deg X points. For future reference,− we shall prove the following very special case of Bézout’s theorem: Proposition 7.3.14. Let X Pn be a projective variety of dimension r, and 1 ⊆ let X = V (f), where f k[x ,...,x ] is an irreducible polynomial of degree 2 ∈ 0 n d and X1 is not contained in V (f). Suppose that, for all n 0, I(X1 X2)n = (I(X )+(f)) . Then deg(X X )= d(deg X ) = (deg X≫)(deg X ∩). 1 n 1 ∩ 2 1 1 2 264 CHAPTER 7. GRADED RINGS

Proof. By assumption, f is not a zero divisor in R(X1)= k[x0,...,xn]/I(X1). Thus there is an exact sequence

f 0 R(X ) × R(X ) k[x ,...,x ]/(I(X )+(f)) 0. → 1 −→ 1 → 0 n 1 → For n 0, this gives ≫

HX1 (n + d) HX1 (n)= HX1 X2 (n + d). − ∩

To see the leading coefficient of HX1 X2 (n), or equivalently of HX1 X2 (n+d), r 1 ∩ ∩ we look at the coefficient of n − in HX1 (n + d) HX1 (n), which is given by r 1 − the coefficient of n − in

d1 r d1 r dd1 r 1 (n + d) n = n − + lower order terms. r! − r! (r 1)! − Hence deg(X X )= d(deg X ) = (deg X )(deg X ) as claimed. 1 ∩ 2 1 1 2 Later, we shall give criteria for when I(X1 X2)n = (I(X1)+(f))n for all n 0. ∩ ≫ 7.4 The dimension theorem

Throughout this section, unless otherwise noted, R denotes a local Noetherian ring with maximal ideal m. In this case, there are three possible ways to deﬁne the dimension of R: 1. We have deﬁned dim R to be the Krull dimension of R. Note that dim R = 0 if and only if R is Artinian, if and only if m is nilpotent, if and only if (0) is m-primary.

2. Consider the numerical polynomial Hm,R deﬁned in this chapter. We let d(R) = deg Hm,R + 1. Recall our convention, that the degree of the zero polynomial is 1. Thus d(R) = 0 if and only if mn/mn+1 = 0 for all n 0, if and only− if mn = 0 for n 0, if and only if R is Artinian. ≫ ≫ 3. Let δ(R) be the minimal number of generators of an m-primary ideal of R. By convention or logic, δ(R) = 0 if and only if the zero ideal is m-primary, if and only if R is Artinian. Our goal is to prove: 7.4. THE DIMENSION THEOREM 265

Theorem 7.4.1 (Dimension theorem). In the above notation, dim R = d(R)= δ(R). In particular, if R is a local Noetherian ring, then dim R< . ∞ We shall prove the theorem by showing that d(R) δ(R) dim R d(R). We begin with the easiest inequality: ≤ ≤ ≤

Lemma 7.4.2. d(R) δ(R). ≤ Proof. Let q be an m-primary ideal of R which is generated by t elements. By Corollary 7.2.4, deg H t 1. Hence, if we can show that deg H +1= q,R ≤ − m,R deg Hq,R + 1, it follows that

d(R) = deg H + 1 = deg H +1 t m,R q,R ≤ for every choice of an m-primary ideal q, and hence that d(R) δ(R). But since R is local and q is m-primary, there exists a k > 0 such≤ that mk q m, and hence, for all n 0, mnk qn mn. Thus, there are surjective⊆ ⊆ ≥ ⊆ ⊆ homomorphisms R/mnk R/qn R/mn. → → In the notation of Corollary 7.2.4, for all n,

P (n) P (n) P (nk). m,R ≤ q,R ≤ m,R It follows that deg P deg P deg P , and hence that m,R ≤ q,R ≤ m,R

Hm,R + 1 = deg Pm,R = deg Pq,R = deg Hq,R +1.

This completes the proof of the lemma.

Before we turn to the inequality dim R d(R), we prove a preliminary lemma: ≤

Lemma 7.4.3. Let q be an m-primary ideal in R and let M be a ﬁnitely generated R-module. Suppose that r R is not a zero divisor in M. Then ∈ deg Hq,M/rM deg Hq,M 1. In particular, if r R is not a zero divisor in R, then d(R/rR≤ ) d(R)− 1. ∈ ≤ − Proof. Applying Lemma 7.2.5 to the exact sequence

r 0 M × M M/rM 0, → −→ → → 266 CHAPTER 7. GRADED RINGS we see that H H = F , where F is a numerical polynomial with q,M − q,M/rM the same degree and leading coeﬃcient as Hq,M , i.e. Hq,M = F + G, where deg G< deg Hq,M . Thus H = H F = G q,M/rM q,M − has degree at most deg H 1. q,M − Lemma 7.4.4. dim R d(R). In particular, if R is local Noetherian ring, then dim R< . ≤ ∞ Proof. The proof is by induction on d = d(R). If d = 0, then as we have seen at the beginning of the section mn = mn+1 for all n 0, so that R is Artinian, and hence dim R =0 d. ≫ Now assume that d = d(R)≤> 0, and that the lemma has been proved for all local Noetherian rings R′ with d(R′)

To complete the proof of the dimension theorem, it therefore suffices to prove: Lemma 7.4.5. δ(R) dim R. ≤ Proof. The proof is again by induction, this time on dim R (which we now know to be finite). If dim R = 0, then R is Artinian and hence by convention δ(R) = 0. Now suppose that the lemmas has been proved for all rings of dimension < dim R. We claim that it suffices to show: Lemma 7.4.6. Let R be a local ring, and let r R, where r is not a zero ∈ divisor or a unit. Then (i) δ(R) δ(R/rR)+1. ≤ (ii) dim R dim(R/rR)+1. ≥ Assuming Lemma 7.4.6, let us complete the proof of Lemma 7.4.5. Since dim R> 0, the maximal ideal m does not consist entirely of zero divisors (else m is contained in the union of the minimal prime ideals, hence is equal to a minimal prime ideal, in which case dim R = 0). Choose r m, not a zero divisor; hence r is also not a unit. Applying Lemma 7.4.6 and∈ induction to R/rR, and noting that dim(R/rR) < dim R so that the inductive hypothesis applies, we have

δ(R) δ(R/rR)+1 dim(R/rR)+1 dim R, ≤ ≤ ≤ completing the proof of Lemma 7.4.5.

Proof of Lemma 7.4.6. (i) Suppose that s1,...,sk generate an m¯ -primary ideal in R/rR, where m¯ is the maximal ideal in R/rR. Let r ,...,r R 1 k ∈ map to s1,...,sk. Then R/(r1,...,rk,r) ∼= S/(s1,...,sk) is Artinian, and hence (r1,...,rk,r) is an m-primary ideal in R. Thus, the minimal number of generators of an m-primary ideal in R is at most k +1. Choosing s1,...,sk so that k is minimal, we see that δ(R) δ(R/rR)+1. ≤ (ii) Choose a chain of strictly increasing prime ideals in R/rR, say

p¯ p¯ p¯ . 1 ⊂ 2 ⊂···⊂ t

Let pi be the image of p¯i in R. Then

p p p 1 ⊂ 2 ⊂···⊂ t 268 CHAPTER 7. GRADED RINGS is a strictly increasing chain of prime ideals in R. Moreover, r p and r is ∈ 1 not a zero divisor. Hence p1 is not a minimal prime ideal. Choosing some prime ideal p0 strictly contained in p1 produces a chain of prime ideals in R of length t + 1. In particular, if t = dim(R/rR), we see that dim R ≥ dim(R/rR)+1. Combining Lemmas 7.4.2, 7.4.4, and 7.4.5, we have now proved the dimension theorem.

7.5 Applications of the dimension theorem

As a corollary of the proof of the dimension theorem, we have:

Corollary 7.5.1. Let R be a local Noetherian ring, and let r R, where r ∈ is not a zero divisor or a unit. Then dim(R/rR) = dim R 1. − Proof. From (ii) of Lemma 7.4.6, we see that dim(R/rR) dim R 1. On ≤ − the other hand, using the dimension theorem and (i) of Lemma 7.4.6, we see that dim(R/rR) = δ(R/rR) δ(R) 1 = dim R 1. Thus dim(R/rR) = ≥ − − dim R 1. − We can use the above corollary to draw a conclusion about general (not necessarily local) Noetherian rings, which was used above in the proof of Theorem 7.3.7:

Proposition 7.5.2. Let R be a Noetherian ring, not necessarily local, and let r ,...,r R. Then every minimal prime ideal associated to (r ,...,r ) 1 k ∈ 1 k has height at most k.

Proof. Note that the statement is vacuous if one of the ri is a unit, or more generally if (r1,...,rk)= R. Let p be a minimal prime associated to (r1,...,rk). Then ht p = dim Rp. By (ii) of Lemma 5.2.11 in Chapter 5, pRp is a minimal prime in Rp associated to (r1,...,rk)Rp. But since pRp is also the maximal ideal in the local Noetherian ring Rp, it is the only prime ideal associated to (r1,...,rk)Rp. Thus, by Deﬁnition 5.3.2, (r1,...,rk)Rp is pRp-primary and is generated by k elements. Hence ht p = dim R = δ(R ) k. p p ≤ 7.5. APPLICATIONS OF THE DIMENSION THEOREM 269

The following corollary is known as the principal ideal theorem (= Haup- tidealsatz, in German): Corollary 7.5.3 (Krull’s Hauptidealsatz). Let R be a Noetherian ring, not necessarily local, and let r R, where r is not a zero divisor or a unit. Then ∈ the height of every minimal prime ideal associated to (r) is one. Proof. By the preceding proposition, we know that the height of every such prime ideal p is at most one. If the height of p is zero, then p is a minimal prime ideal of R and hence r p is a zero divisor. This contradicts the hypothesis on r. ∈ Using the above corollary, we give a characterization of Noetherian unique factorization domains. Corollary 7.5.4. Let R be a Noetherian domain. Then the following are equivalent: (i) R is a UFD; (ii) Every height one prime ideal of R is principal;

(iii) For every maximal ideal m of R, Rm is a UFD (i.e. R is locally factorial) and Pic R =0. Proof. (i) (ii): First suppose that R is a UFD. Then every height ⇐⇒ one prime ideal of R is a minimal nonzero prime ideal, hence is principal by Lemma 1.4.4. Conversely, suppose that every height one prime ideal of R is principal. By Theorem 1.4.18, it suﬃces to show that, for every irreducible r R, the ideal (r) is a prime ideal. Suppose that r is irreducible and let p be∈ a minimal prime ideal belonging to (r). By Corollary 7.5.3, p has height one. By hypothesis, p =(t) is a principal ideal. Thus r = tu for some u R, necessarily a unit. Hence (r)=(t) is a prime ideal, so that R is a UFD.∈ (i) = (iii): If R is a UFD, by Exercise 3.24, R is a UFD for every ⇒ m maximal ideal m. Moreover, every height one prime ideal in R is principal. By Theorem 5.7.2, every invertible ideal in R is a product of height one prime ideals, and is thus principal. Since every fractional ideal is of the form 1 s− I, where s R 0 and I is an invertible ideal, every fractional ideal is principal as well.∈ Thus−{ } the ideal class group of R is trivial, so that Pic R = 0. (iii) = (ii): By Theorem 5.7.2, every height one prime ideal in R is invertible,⇒ and hence is principal since ideal class group of R is trivial by the assumption that Pic R = 0. 270 CHAPTER 7. GRADED RINGS

Here is another corollary of the dimension theorem, which is concerned with the dimension of the fiber of a morphism. (We will further analyze this question in the next chapter.) In particular, it implies that, if X and Y are quasiprojective varieties and f : X Y is a flat morphism (i.e. f ∗ : Y,f(x) → 1 O → X,x is flat for all x X), then the dimension of the fibers f − (y) is constant forO all y Y and is∈ equal to dim X dim Y . ∈ − Theorem 7.5.5. Let R and S be local Noetherian rings, with maximal ideals m and n respectively, and let f : R S be a local homomorphism. Then → dim S dim R + dim(S/pS). ≤ Moreover, if the going down property holds for f (Exercise 4.12), for example if f is flat, then equality holds.

Proof. To see the ﬁrst statement, let d = dim R, and suppose that r ,...,r 1 d ∈ R generate an m-primary ideal q. Since R is Noetherian, there exists an n > 0 such that qn m, and hence mnS qS mS. Thus √qS = √mS, so that dim(S/qS) =⊆ dim(S/mS). Let e =⊆ dim(⊆S/mS) = dim(S/qS), and choose s1,...,se S whose images in S/qS generate a n¯-primary ideal, where n¯ is the image∈ of the maximal ideal n of S in S/qS. Then the ring S/(f(r1),...,f(rd),s1,...,se) is Artinian, so that f(r1),...,f(rd),s1,...,se generate an n-primary ideal of S. It follows that

dim S = δ(S) d + e = dim R + dim(S/pS). ≤ Now suppose that the going down property holds for f. Let e = dim(S/mS). Then there exists a chain of prime ideals in S, say

p p p = n, 0 ⊂ 1 ⊂···⊂ e 1 such that mS p . Since f is a local homomorphism, f − (n)= m, and thus ⊆ 0 1 1 1 1 m = f − (mS) f − (p ) f − (p )= f − (n)= m. ⊆ 0 ⊆···⊆ e 1 It follows that f − (pi) = m for every i. Let dim R = d. Then there exists a chain of prime ideals in R of the form

q q q = m. 0 ⊂ 1 ⊂···⊂ d 7.5. APPLICATIONS OF THE DIMENSION THEOREM 271

By the going down property, there exists a chain of prime ideals in S of the form q′ q′ q′ = p , 0 ⊂ 1 ⊂···⊂ d 0 1 with f − (qi′ )= qi. Thus the chain

q′ q′ q′ = p p p = n 0 ⊂ 1 ⊂···⊂ d 0 ⊂ 1 ⊂···⊂ e has length d + e. It follows that dim S d + e, and hence that dim S = d + e. ≥ Corollary 7.5.6. Let R and S be Noetherian rings, let f : R S be a 1 → homomorphism, and let q be a prime ideal of S. If p = f − (q), and q¯ is the image of q in S/pS, then

ht q ht p + ht q¯. ≤ Moreover, if f has the going down property, then equality holds.

Proof. We have ht p = dim Rp, ht q = dim Sq, and ht q¯ = dim Sq/pSq. The corollary then follows by applying the previous proposition to the local homomorphism R S induced by f. p → q We will use the above corollary as a key ingredient in the proof of: Proposition 7.5.7. Let R be a Noetherian ring. Then dim R[x] = dim R+1. Proof. We break the proof up into a series of lemmas: Lemma 7.5.8. Let R be a ring, not necessarily Noetherian, and suppose that p p are two distinct prime ideals in R[x] such that p R = p R = p, 1 ⊂ 2 1 ∩ 2 ∩ say. Then p1 = pR[x]= p[x]. Proof. Reducing mod p, we may assume that p = (0) and hence that R is an integral domain with ﬁeld of quotients K. Localize at M = R 0 R[x]. −{ } ⊆ Since p1 R = p2 R = 0 , p1 M = p2 M = , and hence the prime ∩1 ∩1 { } ∩ ∩ ∅ 1 ideals M − p1 M − p2 are two distinct prime ideals in M − (R[x]) = K[x]. ⊆ 1 Since dim K[x]=1(K[x] is a PID), M − p1 = (0). Since R[x] is an integral domain and 0 / M, it follows that p = (0) = (0)R[x]. ∈ 1 Corollary 7.5.9. Let R be a ring, not necessarily Noetherian. Then

dim R +1 dim R[x] 2 dim R +1. ≤ ≤ 272 CHAPTER 7. GRADED RINGS

Proof. To see the first inequality dim R +1 dim R[x], suppose that p ≤ 0 ⊂ p p is a chain of prime ideals in R. Then 1 ⊂···⊂ d p R[x] p R[x] p R[x] (p , x)R[x] 0 ⊂ 1 ⊂···⊂ d ⊂ d is a chain of length d + 1. Thus dim R[x] dim R + 1. To see the opposite inequality, suppose≥ that q q q is a 0 ⊂ 1 ⊂ ··· ⊂ e chain of prime ideals in R[x]. Let p = q R. Then the previous lemma i i ∩ implies that we cannot have three consecutive pi equal, since then two of the qi would be equal. Thus, there are at least (e +1)/2 distinct prime ideals in the sequence p0 p1 pe, yielding a chain of length at least (e 1)/2, so that dim R ⊆(e 1)⊆···⊆/2. Thus e 2 dim R + 1. Since this inequality− holds ≥ − ≤ for every chain of prime ideals in R[x], dim R[x] 2 dim R + 1. ≤ Lemma 7.5.10. Let R be a Noetherian ring and let p be a prime ideal of R. Then ht pR[x]=ht p. Proof. Apply Corollary 7.5.6 to f : R R[x] and to the prime ideal q = 1 → pR[x] R[x]. Here f − (q) = pR[x] p and q¯ is the image of pR[x] in ⊆ ∩ R/pR[x], and hence is the prime ideal (0). Thus ht q = ht pR[x] and ht q¯ = 0. Finally, equality holds in Corollary 7.5.6, since R[x] is a free R-module, hence is flat, and therefore has the going down property by Exercise 4.12. Thus ht pR[x]=ht p as claimed. Returning to the proof of Proposition 7.5.7, let dim R = d. Corollary 7.5.9 implies that dim R[x] dim R+1, so it suffices to prove the opposite inequal- ≥ ity. Let q0 q1 qe be a chain of prime ideals in R[x] and let pi = qi R. If all of the⊂p are⊂··· distinct, then e dim R = d. If some of them are equal,∩ i ≤ let j < e be the largest integer such that pj = pj+1. By Lemma 7.5.8, q = p R[x]. By Lemma 7.5.10, ht q = ht p . Clearly ht q j, and hence j j j j j ≥ ht pj j. On the other hand, the ideals pj, pj+2,..., pe are all distinct by the hypothesis≥ on j, and thus form a chain of length e j 1. Since dim R = d, − − ht p +(e j 1) d. Thus j − − ≤ e d + j +1 ht p d +1. ≤ − j ≤ It follows that dim R[x] dim R + 1. Thus dim R[x] = dim R + 1. ≤

Corollary 7.5.11. Let R be a Noetherian ring. Then dim R[x1,...,xn] = dim R + n. 7.6. REGULAR LOCAL RINGS 273 7.6 Regular local rings

Throughout this section, we continue to keep the convention that, unless otherwise noted, R denotes a local Noetherian ring with maximal ideal m and residue field k. We begin with another easy corollary of the dimension theorem: Lemma 7.6.1. dim R dim m/m2. ≤ k Proof. Let r ,...,r m project onto a k-basis of m/m2. Then, by Naka- 1 t ∈ yama’s lemma (cf. Corollary 2.7.3), m is generated by r1,...,rt. Thus dim R = δ(R) t = dim m/m2. ≤ k 2 Definition 7.6.2. A local Noetherian ring R is regular if dim R = dimk m/m . A Noetherian ring R (not necessarily local) is regular if Rm is regular for every maximal ideal m of R. (It turns out that this implies that the local ring Rp is regular for every prime ideal p.) For example, a local PID is regular. For a less trivial example, we have: Lemma 7.6.3. Let k be an algebraically closed field. For all maximal ideals m of k[x1,...,xn], the local ring k[x1,...,xn]m is regular, and hence k[x1,...,xn] is a regular ring. Proof. Since m =(x a ,...,x a ) for some a ,...,a k, m is generated 1− 1 n− n 1 n ∈ by n elements. On the other hand, since m is maximal, we have seen that dim k[x1,...,xn]m = dim k[x1,...,xn] = n. Thus k[x1,...,xn]m is regular.

Using Corollary 4.5.5, it is easy to extend the lemma to ﬁelds k which are not algebraically closed. For a further generalization, see Exercise 7.5. Proposition 7.6.4. Let R be a regular local ring of dimension n and let f1,...,fr R. Suppose that S = R/(f1,...,fr) and that dim S = d. Then ∈ 2 S is regular if and only if the span of the images of f1,...,fr in m/m has dimension n d. In particular, for an algebraically closed ﬁeld k, let m − n be the maximal ideal corresponding to the point p A , let f1,...,fr k[x ,...,x ], with p X = V (f ,...,f ), and suppose∈ that ∈ 1 n ∈ 1 r dim k[x1,...,xn]m/(f1,...,fr)= d.

∂fi Then k[x1,...,xn]m/(f1,...,fr) is regular if and only if the rank of (p) ∂xj is n d, if and only if p is a smooth point of X. − 274 CHAPTER 7. GRADED RINGS

Proof. Let n be the maximal ideal in the quotient R/(f1,...,fr). Then S is 2 regular if and only if dimk n/n = d. Clearly, n is the image of m, and

2 2 n/n ∼= m/(f1,...,fr)+ m . 2 Thus dimk n/n = d if and only if the span of the images of f1,...,fr in 2 m/m has dimension n d. Now suppose that R = k[x1,...,xn]m for an algebraically closed field−k, where m corresponds to the point p An, and ∂fi ∈ that f1,...,fr m. Viewing the r n matrix A = (p) as a linear map ∈ × ∂xj from kr to kn, then it is straightforward, using the k-basis of m/m2 given 2 n by the images of x1,...,xn, to identify m/(f1,...,fr)+ m with k / Im A. 2 Thus dimk n/n = d if and only if A has corank d, if and only if A has rank n d. − Remark 7.6.5. For fields k which are not algebraically closed, more precisely are not perfect, regularity and smoothness behave quite differently. See Exercise 7.6 for an example of this. In general, smoothness behaves much better than regularity; for example, it is an open condition under the appropriate hypotheses. Proposition 7.6.6. If R is a regular local ring, then R is an integral domain. Proof. We begin with the following lemma: Lemma 7.6.7. Let R be a regular local ring of dimension d, with maximal ideal m and residue field k. Let r ,...,r m generate m. Then the natural 1 d ∈ homomorphism ψ : k[x1,...,xd] grm R defined by sending xi to the image 2 → of ri in m/m is an isomorphism. Proof. Clearly ψ is surjective. Suppose that f Ker ψ, and that f = 0. Since ψ is a homomorphism of graded rings, we may∈ assume that f is homogeneous.6

Since ψ is surjective, there is an induced surjection k[x1,...,xd]/(f) grm R, and hence, for all n 0, → ≥ dim (gr R) dim (k[x ,...,x ]/(f)) . k m n ≤ k 1 d n

The left hand side of the above equation is Hm,R(n), where Hm,R is a numerical polynomial of degree d 1. The right hand side is H (n), − k[x1,...,xd]/(f) where Hk[x1,...,xd]/(f) is a numerical polynomial of degree d 2 by (i) of Propo- sition 7.3.5 (cf. also the proof of Lemma 7.3.4). But this is− clearly impossible. Hence Ker ψ = 0, ψ is injective, and hence ψ is an isomorphism. 7.7. COMPLETIONS 275

Returning to the proof of Proposition 7.6.6, choose r, s R 0 . Since n ∈ −{ a} a+1 n 0 m = 0 , there exist nonnegative integers a, b such that r m m ≥ { } ∈ − and s mb mb+1. Hence r deﬁnes a nonzero element of ma/ma+1 gr R, T ∈ − ⊆ m and similarly for b. Since grm R is isomorphic to a polynomial ring, and thus is an integral domain, rs ma+b ma+b+1. In particular rs = 0. Thus R is ∈ − 6 an integral domain. Remark 7.6.8. One can also show that a regular local ring is integrally closed, and in fact is a UFD.

7.7 Completions

The stalk of the sheaf of C∞ functions on a smooth manifold M depends only on the dimension on M, reflecting the fact that, locally, all manifolds of dimension n are isomorphic. A similar statement holds for holomorphic functions on a complex manifold. However, the local ring X,x of a variety at a point contains a substantial amount of global informatiO on, even at a smooth point. This is a reflection of the fact that varieties of dimension n are rarely locally isomorphic, even at smooth points. Completion is a process for simplifying the structure of X,x so that all local rings of smooth points become isomorphic. O Quite generally, let A be an abelian group and let A = A A A 0 ⊇ 1 ⊇ 2 ⊇ be a decreasing sequence of subgroups of A. We define a topology on A ··· be taking a neighborhood basis of a point a A to be the collection a+An of cosets containing a. With this definition, A ∈becomes a topological group (the group operations are continuous) and the An are open and closed subgroups of A (they are closed since A A is a union of cosets of A ). The topological − n n group A is Hausdorff if and only if n 0 An = 0 and it is discrete if and only ≥ if An = 0 for some n. The main examples are: A = RTis a ring, I is an ideal of R, and we set n n An = I . More generally, let M be an R-module and let An = I M. The topology on M is then called the I-adic topology. A special case is R = Z and I =(p) for a prime number p, in which case the topology on Z is called the p-adic topology. For another example, let R be an arbitrary ring, let A be either the polynomial ring R[x1,...,xn] or the formal power series ring R[[x1,...,xn]], and let I = m =(x1,...,xn). If A = R is a ring and In are a decreasing sequence of ideals in R, then the topology defined by the In makes R into a topological ring: ring 276 CHAPTER 7. GRADED RINGS multiplication is continuous. Moreover, if in addition M is an R-module and Mn are a decreasing sequence of submodules of M such that InM Mn for all n, then M becomes a topological module over the topological ring⊆ R, i.e. the multiplication R M M is continuous. × → Definition 7.7.1. Suppose that A is an abelian group with a topology defined by a decreasing sequence of subgroups A as above. Let a be a { n} { i} sequence in A. Then a is a Cauchy sequence if, for all n 0, there exists { i} ≥ an N such that, for all k,ℓ N, ak aℓ An. The topological group A is complete if it is Hausdorff and≥ every Cauchy− ∈ sequence a has a (necessarily { i} unique) limit a A. ∈ Example 7.7.2. If A = R[[x1,...,xn]] with the m-adic topology, then a sequence f (x ,...,x ) , where we write { i 1 n } f (x ,...,x )= r(i) xa1 xan , i 1 n a1,...,an 1 ··· n a1,...,an 0 X≥ is a Cauchy sequence if and only if, for all d 0, there exists an N such ≥ d that, for all k,ℓ N, fk(x1,...,xn) fℓ(x1,...,xn) m , in other words ≥(k) (ℓ) − ∈ n the coefficients ra1,...,an and ra1,...,an agree as long as i=1 ai d. Thus, (i) ≤ for every index (a1,...,an), the sequence ra1,...,an is eventually constant, (i) { } P and we can define limi ra1,...,an = ra1,...,an and limi fi(x1,...,xn) = →∞ →∞ r xa1 xan . Clearly mn = 0 . Thus R[[x ,...,x ]] is a1,...,an 0 a1,...,an 1 n n 0 1 n complete.≥ ··· ≥ { } P T A very similar discussion applies to the ring of p-adic integers Zp. Let n Z, n 0. Then n has a unique expansion to base p: n = k a pi, where ∈ ≥ i=0 i the ai are integers, 0 ai p 1, and ak = 0. Addition and multiplication are given by the usual≤ rules:≤ − 6 P

k k k i i i aip + bip = (ai + bi)p i=0 i=0 i=0 k X kX X i i k aip bip = ( (aibj)p . i=0 ! i=0 ! X X Xk i+Xj=k

Here the understanding is that the sums ai + bi as well as i+j=k(aibj) are also to be written out in base p and the overall expression should be simplified so that it is in standard form. These rules then extend to definPe addition and 7.7. COMPLETIONS 277 multiplication for all integers (positive or negative) in a straightforward way. The ring of p-adic integers Zp is then defined to be the set of all expressions i of the form i∞=0 aip . (It is not necessary to adjoin negatives; for example, { i } 1 = i∞=0(p 1)p .) It is easy to check that Zp is a PID, and in fact that − P− k every nonzero ideal in Zp is of the form (p ) for some nonnegative integer k. P Finally, Zp is complete with respect to the (p)-adic topology.

Let A be a topological group as above and let be the set of all Cauchy A sequences in A. The set is a group via pointwise addition of sequences. Let be the subgroup deﬁnedA by: An = a : there exists an N such that a A for all i N . An {{ i}∈A i ∈ n ≥ } Clearly is a subgroup of , and the decreasing sequence of subgroups An A makes into a topological group as well. Let {An} A

= ai : lim ai =0 . i N {{ }∈A →∞ } Then is also a subgroup of . In fact, N A = . N An n 0 \≥ Definition 7.7.3. We define the completion Aˆ to be the group / . Given A N a A, let ι′(a) be the constant sequence a , where a = a for all i, and ∈ { i} i let ι(a) be the image in of ι′(a). Clearly, ι: A Aˆ is a homomorphism. A → In addition, if A and B are topological groups with topologies defined by filtrations An and Bn and ϕ: A B is a homomorphism such that ϕ(An) B , then there is an induced homomorphism→ Aˆ Bˆ. ⊆ n → Proposition 7.7.4. In the above notation, ˆ (i) The homomorphism ι: A A has kernel n 0 An and has dense im- → ≥ age. T (ii) The group Aˆ is complete.

(iii) ι is an isomorphism if and only if A is complete. In particular, the ˆ natural homomorphism Aˆ Aˆ is an isomorphism. → 278 CHAPTER 7. GRADED RINGS

(iv) The subgroups A become topological groups via the ﬁltration A A . k k ∩ n With this topology, the natural homomorphism Aˆk Aˆ is an inclusion, with image equal to the image of in Aˆ, and also→ to the closure of Ak ι(Ak) in Aˆ.

1 1 Proof. (i): Clearly, (ι′)− ( n)= An and (ι′)− ( )= n 0 An. Thus, ι(a)= A N ≥ 0 if and only if ι′(a) , if and only if a n 0 An. Let x be the image ∈ N ∈ ≥ T in Aˆ of the Cauchy sequence a . It follows easily from the deﬁnitions that { i} T limn ι′(an)= ai in and hence in Aˆ. Thus ι(A) is dense in Aˆ. →∞ { } A (ii): Let x be a Cauchy sequence in Aˆ. For each n, choose a sequence { n} a in whose image in Aˆ is x . It is straightforward to check that a { i,n} A n { i,i} is a Cauchy sequence, and that, if x Aˆ is the point corresponding to a , ∈ { i,i} then limn xn = x. Thus Aˆ is complete. →∞ (iii): Clearly, if ι is an isomorphism, then A is complete. Conversely, if A is complete, then ι is injective by (i). To see that it is surjective, let x Aˆ be the image of the Cauchy sequence a . Since A is complete, there∈ { i}∈A exists a A such that limn an = a. It is then easy to check that ι(a)= x. ∈ →∞ (iv): Let k′ denote the set of all Cauchy sequences in Ak and let ′ = k′ be the subgroupA of all sequences in A which converge to 0. ClearlyN A ∩N= k Ak ′ + . On the other hand, Aˆ = ′ / ′, and the homomorphism Aˆ Aˆ Ak N k Ak N k → is given by the homomorphism ′ / ′ / . Since ′ = ′ , the Ak N → A N N Ak ∩N homomorphism k′ / ′ / is injective, and its image is ( k′ + )/ , which is the imageA ofN →in AAˆN. The last statement is left as anA exercise.N N Ak Part (iii) of Proposition 7.7.4 shows that the completion has a universal property, whose proof is left as an exercise:

Corollary 7.7.5. Let A and B be topological groups with topologies deﬁned by ﬁltrations An and Bn and let ϕ: A B is a homomorphism such that ϕ(A ) B . Suppose moreover that B →is complete. Then there is a unique n ⊆ n induced continuous homomorphism ϕˆ: Aˆ B. The homomorphism ϕˆ is surjective if and only if ϕ has dense image,→ and it is injective if and only if 1 n 0 ϕ− (Bn)= n 0 An. ≥ ≥ CorollaryT 7.7.6.T Let R be a ring, let m = (x ,...,x ) R[x ,...,x ] 1 n ⊆ 1 n and suppose that R[x1,...,xn] is given the m-adic topology. Then the completion of R[x1,...,xn] in the m-adic topology is naturally isomorphic to R[[x1,...,xn]]. 7.7. COMPLETIONS 279

Proof. Apply Corollary 7.7.5 to the homomorphism R[x ,...,x ] R[[x ,...,x ]]. 1 n → 1 n

Likewise, Corollary 7.7.5 implies that the ring Zp is isomorphic to the completion of Z with respect to the (p)-adic topology. Clearly the deﬁnition of the completion Aˆ only depends on knowing which sequences in A are Cauchy. Thus:

Lemma 7.7.7. Let A be an abelian group, and let An and An′ be two decreasing filtrations on A such that, for all n, there exists an m such that A A′ n ⊇ m and A′ A . Then the completion of A defined by the filtration A′ can be n ⊆ m n identified with the completion of A defined by the filtration An. The completion Aˆ can also be defined in a purely algebraic way: ˆ Proposition 7.7.8. There is an isomorphism A ∼= lim A/An, compatible ←−n with the homomorphisms of A to both groups.

Proof. First suppose that a is a Cauchy sequence, i.e. a point of . Define { i} A ψ( ai )=(bn), where bn A/An is the image of ak for k 0. The condition that{ }a is Cauchy guarantees∈ that b is well-defined, and≫ by construction { i} n thee projection of bn+1 A/An+1 to A/An is bn. Thus (bn) is a well-defined point of lim A/A , and∈ clearly the image of the constant sequence defined n n by a =←−a for all i is the element (a, a, a, . . . ) lim A/A . Moreover, i ∈ n n if a , then ψ( a ) = (0, 0, 0,... ), and hence←− there is an induced { i}∈N { i} homomorphism ψ : Aˆ lim A/A , compatible with the homomorphism → n n ι: A Aˆ and the diagonale ←− homomorphism A lim A/A . → → n n To find a homomorphism in the other direction,←− given (b ) lim A/A , n ∈ n n for each i choose a A such that the image of a in A/A is b . For←−i, j n, i ∈ i i i ≥ ai and aj have the same image bn in A/An, so that ai aj An for all i, j n. Thus a is a Cauchy sequence and hence defines− a point∈ of . If ≥ { i} A a′ is another such choice, then the difference a a′ A for all i, so that the i i − i ∈ i difference a a′ . Thus the image of a in Aˆ depends only on { i}−{ i}∈N { i} (bn); call this image ρ((bn)). It is easy to check that ρ is a homomorphism. Finally, the fact that ρ and ψ are inverses to each other follows easily from the construction.

We turn now to exactness properties of completions. Let 0 A′ → → A A′′ 0 be an exact sequence of abelian groups. Suppose that A is → → n 280 CHAPTER 7. GRADED RINGS

a filtration on A, making A into a topological group. Define A′ = A A′ n n ∩ (viewing A′ as identified with a subgroup of A), and let An′′ be the image of An in A′′. We thus obtain completions Aˆ′, A,ˆ Aˆ′′. Then we have:

Proposition 7.7.9. In the above situation, the sequence

0 Aˆ′ Aˆ Aˆ′′ 0 → → → → is exact.

Proof. This follows immediately from Proposition 2.3.11 and Proposition 7.7.8 (or by a direct argument).

Corollary 7.7.10. Let A have a topology deﬁned by the ﬁltration An. Then for all k, the homomorphism A Aˆ induces an isomorphism → ˆ ˆ A/Ak ∼= A/Ak. Proof. Consider the exact sequence

0 A A A/A 0. → k → → k →

This is an exact sequence of topological groups, where the topology on Ak is defined by the filtration Ak An, and that on the quotient A/Ak is defined by the image of A . In particular,∩ this image is 0 for n k. Thus the topology n ≥ on A/Ak is the discrete topology and the only Cauchy sequences are sequences which are eventually constant. It follows that A/Ak is complete, and hence [ A/Ak ∼= A/Ak. The previous proposition gives an exact sequence

0 Aˆ Aˆ A/A[ 0. → k → → k → ˆ ˆ [ It follows that A/Ak ∼= A/Ak ∼= A/Ak as claimed. We now apply the above to the case of Noetherian rings R and finitely generated R-modules M. If I is an ideal of R, we then have the I-adic topology on R, defined by In, and on M, defined by InM. Moreover, the completion Rˆ of R in the I-adic topology is a topological ring, and the completion Mˆ is an Rˆ-module. 7.7. COMPLETIONS 281

Proposition 7.7.11. Let R be a Noetherian ring, let I be an ideal of R, and let M be a ﬁnitely generated R-module. Then the I-adic topology on M is Hausdorﬀ if one of the following conditions holds:

(i) I is contained in the Jacobson radical of R, for example if R is a local ring and I = m is the maximal ideal.

(ii) R is an integral domain, I = R, and M is a torsion free R-module. 6 Proof. This follows immediately from Krull’s theorem (Theorem 7.1.14).

We now turn to how completion with respect to the I-adic topology behaves in exact sequences:

Proposition 7.7.12. Suppose that R is Noetherian and that

0 M ′ M M ′′ 0 → → → → is an exact sequence of R-modules, with M ﬁnitely generated. Fix an ideal I in R, and let Mˆ ′, M,ˆ Mˆ ′′ be the completions of M ′,M,M ′′ respectively with respect to the I-adic topology. Then the sequence

0 Mˆ ′ Mˆ Mˆ ′′ 0 → → → → is exact.

n Proof. By Lemma 7.7.9, the statement is true with the filtration I M M ′ ∩ on M ′ replacing the I-adic filtration. On the other hand, by the Artin-Rees lemma (in the form of Corollary 7.1.11), there exists an N such that, for n n N n N n N all n N, I M M ′ I − M ′, and of course I − M ′ I − M M ′. ≥ ∩ ⊆ n n ⊆ ∩ Thus the two filtrations I M ′ and I M M ′ define the same set of Cauchy ∩ sequences for M ′ and in particular they define isomorphic completions by Lemma 7.7.7. It follows that 0 Mˆ ′ Mˆ Mˆ ′′ 0 is exact. → → → → Proposition 7.7.13. Let R be Noetherian, let I be an ideal in R and let M be a finitely generated R-module. If Rˆ and Mˆ are the I-adic completions of R and M respectively, then the natural homomorphism of Rˆ-modules Rˆ R M Mˆ is an isomorphism. ⊗ → 282 CHAPTER 7. GRADED RINGS

Proof. It is easy to check that completion commutes with finite direct sums. In particular, Rn = (Rˆ)n. Thus in this case the homomorphism Rˆ Rn ∼ ⊗R → Rn is an isomorphism. Next suppose that M is a finitely generated, and hence finitely presented,c R-module, and choose an exact sequence c 0 K Rn M 0, → → → → with K finitely generated. Tensoring with Rˆ yields a commutative diagram with exact rows:

Rˆ K Rˆ Rn Rˆ M 0 ⊗R −−−→ ⊗R −−−→ ⊗R −−−→ ∼=

 n  0 Kˆ R Mˆ 0. −−−→ y −−−→ y −−−→ y −−−→ Here we have used Proposition 7.7.12 for the exactness of the bottom row and c the exactness properties of for the exactness of the top row. A diagram ⊗ chase shows that the homomorphism Rˆ R M Mˆ is surjective, for every ﬁnitely generated R-module M. In particular,⊗ the→ homomorphism Rˆ K ⊗R → Kˆ is also surjective, so that Rˆ R M Mˆ is also injective, by a diagram chase or the ﬁve lemma. ⊗ →

Corollary 7.7.14. If R is a Noetherian, I is an ideal in R and Rˆ is the I-adic completion of R, then Rˆ is a ﬂat R-module.

Proof. By Lemma 2.10.3, it suﬃces to check that the tensor product with Rˆ preserves exactness for sequences of ﬁnitely generated R-modules, and this is a consequence of Proposition 7.7.12 and Proposition 7.7.13.

Remark 7.7.15. By contrast, the I-adic completion functor M Mˆ is not 7→ in general exact. ˆ Next we compare the graded rings grI R and grIˆ R: Proposition 7.7.16. Let R be a Noetherian ring, let I be an ideal of R, and, for an R-module M, let Mˆ denote the I-adic completion of M.

(i) The homomorphism Rˆ R I Rˆ is an injection and its image is IRˆ = Iˆ. ⊗ →

(ii) For all n 0, (In)=(Iˆ)n. ≥ d 7.7. COMPLETIONS 283

n ˆ ˆn n n+1 ˆn ˆn+1 (iii) For all n 0, R/I ∼= R/I and I /I ∼= I /I . Thus grI R ∼= ˆ ≥ grIˆ R.

Proof. (i) Since Rˆ is flat over R, the homomorphism Rˆ R I Rˆ is an ˆ ⊗ˆ → ˆ injection. Clearly its image is IR. On the other hand, R R I ∼= I by Proposition 7.7.13, since I is finitely generated. ⊗ (ii): Using (i), (In)= InRˆ =(IRˆ)n =(Iˆ)n. n ˆ n ˆ n ˆ ˆn (iii): By Corollary 7.7.10 R/I ∼= R/(I ), and, by (ii), R/(I ) ∼= R/I . This proves the firstd statement of (iii). There is a commutative diagram d d 0 In/In+1 R/In+1 R/In 0 −−−→ −−−→ −−−→ −−−→ ∼= ∼= n n+1 ˆ ˆn+1 ˆˆn 0 I /I R/I R/I 0. −−−→ −−−→ y −−−→ y −−−→ n n+1 ˆn ˆn+1 Thus I /I ∼= I /I . Corollary 7.7.17. Let R be a Noetherian local ring, let m be the maximal ideal of R, and let Rˆ denote the m-adic completion of R. Then: (i) dim R = dim Rˆ. (ii) The ring R is regular if and only if the ring Rˆ is regular. Proof. (i): By Exercise 7.8, Rˆ is a local ring with maximal ideal mˆ . By definition and by (iii) of Proposition 7.7.16, Hm,R = Hmˆ ,Rˆ. Hence dim R = dim Rˆ by the Dimension theorem. (ii): By the following proposition (whose proof does not use Corollary 7.7.17), Rˆ is Noetherian. By (iii) of Proposition 7.7.16, mˆ /mˆ 2 = m/m2 and dim R = 2 ∼ 2 dim Rˆ. Thus dim R = dimk m/m if and only if dim Rˆ = dimk mˆ /mˆ . Proposition 7.7.18. Let R be a Noetherian ring, let I be an ideal of R, and let Rˆ denote the I-adic completion of R. Then Rˆ is Noetherian.

Proof. First note that grI R is Noetherian, by Corollary 7.1.3. Given r R n n+1 n n+1 ∈ with r I I , letr ¯ grI R denote the image of r in I /I . In case ∈ n − ∈ r n 0 I , we setr ¯ =0. If J is an ideal of R, let ∈ ≥ T J¯ = r¯ : r J . { ∈ } ˆ ˆ ¯ Similarly, for r R, we deﬁner ¯ grIˆ R ∼= grI R as well as J, for an ideal J ˆ ∈ ∈ ˆn of R. Note that in this case we automatically have n 0 I = 0. Since gr R is Noetherian, it clearly suﬃces to prove:≥ I T 284 CHAPTER 7. GRADED RINGS

Claim 7.7.19. Suppose that J is an ideal of Rˆ, and r ,...,r J are such 1 k ∈ that r¯1,..., r¯k generate J¯. Then r1,...,rk generate J.

To prove the claim, we may clearly assume that nor ¯i is 0. Thus, there N exists an N > 0 such that ri / Iˆ , and the homogeneous elementr ¯i of ˆ ∈ grIˆ R ∼= grI R has degree di N. Suppose that x J. Let degx ¯ = d0, d0 d0+1 ≤ ∈ i.e. x Iˆ Iˆ . Then, sincex ¯ J¯, andx ¯ and ther ¯i are homogeneous, ∈k (0)− ∈ (0) x¯ = i=1 s¯i r¯i for some homogeneous elementss ¯i of degree d0 di (if this (0) (0) −(0) ˆdi d0 is negative, then by deﬁnitions ¯i = 0). Lifts ¯i to an element si I − . P (0) ∈ k ˆd0+1 Then x i=1 si ri I J. By induction, for all n 0, we may (n) − ∈ ∩ k n (j) ≥ ﬁnd s Iˆdi d0+n, such that x ( s )r Iˆd0+n+1. For each i, i P − i=1 j=0 i i ∈ (j) − ∈ n (j) the series ∞ si is clearly convergent; let si = limn si . As x j=0 P P →∞ j=0 (j) (j) − k n ˆd0+n+1 k k n ( si )ri I , x siri = limn x ( si )ri = i=1 j=0P ∈ − i=1 →∞ − Pi=1 j=0 0. Hence x = k s r , so that r ,...,r generate J. P P i=1 i i P1 k P P Remark 7.7.20.P The proof shows more generally that if R is a ring and I n is an ideal of R such that grI R is Noetherian and n 0 I = 0, then R is Noetherian. ≥ T We now compare power series rings and their quotients to more general complete local rings.

Lemma 7.7.21. Let R be a ring, let I be an ideal of R, and suppose that r ,...,r I. Assume that R is complete with respect to the I-adic topology. 1 n ∈ Let k be a ﬁeld contained in R.

(i) There is a unique continuous homomorphism ϕ: k[[x ,...,x ]] R 1 n → such that ϕ(ri)= xi.

2 (ii) If r1,...,rn generate I mod I and the homomorphism k R/I is an isomorphism, then the homomorphism ϕ in (i) is surjective.→ Hence, in this case, R = k[[x ,...,x ]]/J for some ideal J k[[x ,...,x ]]. ∼ 1 n ⊆ 1 n

Proof. Let ϕ0 : k[x1,...,xn] R be the homomorphism deﬁned by sending r to x , and let m = (x→,...,x ) k[x ,...,x ]. Then ϕ (m) In, i i 1 n ⊆ 1 n 0 ⊆ and the completion of k[x1,...,xn] with respect to the m-adic topology is k[[x1,...,xn]], so that we can apply Corollary 7.7.5 to ϕ0 to obtain ϕ. 2 If r1,...,rn generate I mod I and the homomorphism k R/I is an n n n+1 → isomorphism, then ϕ0(m ) generates I mod I . It is then easy to see 7.8. HENSEL’S LEMMA AND THE IMPLICIT FUNCTION THEOREM285

that, given r R and N 0, there exists an f k[x1,...,xn] such that N∈+1 ≥ ∈ r ϕ0(f) I . Thus ϕ0 has dense image and so ϕ is surjective, again by Corollary− 7.7.5.∈ Corollary 7.7.22. Let R be a regular local ring with maximal ideal m, and suppose that R is complete with respect to the m-adic topology. Suppose that R contains a field k mapping isomorphically onto R/m. Let r1,...,rd m map 2 ∈ onto a basis for m/m . Then the induced homomorphism ϕ: k[[x1,...,xd]] R is an isomorphism. In particular, R is isomorphic to a formal power series→ ring. Proof. Applying the previous proposition yields a surjective homomorphism ϕ: k[[x1,...,xd]] R. If Ker ϕ = 0, choose f Ker ϕ with f = 0. Then ϕ induces a surjection→ k[[x ,...,x 6]]/(f) R. Hence∈ 6 1 d → dim(k[[x ,...,x ]]/(f)) dim R = d. 1 d ≥ But since f is not a zero divisor in the local ring k[[x1,...,xd]], dim(k[[x ,...,x ]]/(f)) = d 1 1 d − by Corollary 7.5.1. This is a contradiction. Thus Ker ϕ = 0 and ϕ is an isomorphism from k[[x1,...,xd]] to R. Remark 7.7.23. Cohen has proved that, if R is a (Noetherian) complete local ring with maximal ideal m and residue field k ∼= R/m, and if R contains a field, then R contains a coefficient field, in other words a field which maps isomorphically onto k. In particular, if R is also regular, then R is isomorphic to a formal power series ring.

7.8 Hensel’s lemma and the implicit function theorem

Theorem 7.8.1 (Hensel’s lemma). Let R be a local ring with maximal ideal m and residue ﬁeld k, such that R is complete with respect to the m-adic topology. Suppose that f(t) R[t] is a monic polynomial of degree d whose image f¯(t) k[t] factors as ∈f¯(t)=¯g(t)h¯(t), where g¯(t) and h¯(t) are monic, ∈ degg ¯(t)= d, deg h¯(t)= n d, and g¯(t) and h¯(t) are relatively prime in k[t]. Then there exist monic polynomials− g(t) and h(t) in R[t], whose images in k[t] are g¯(t) and h¯(t) respectively, such that f(t)= g(t)h(t). 286 CHAPTER 7. GRADED RINGS

Hensel’s lemma is often applied in the following situation:

Corollary 7.8.2. Let R be a complete local ring with maximal ideal m and residue ﬁeld k. Suppose that f(t) R[t] is a monic polynomial and that ∈ λ k is a simple root of f¯(t) k[t], i.e. f¯(λ)=0 and f¯′(λ) =0. Then there ∈ ∈ 6 exists an r R with r¯ = λ such that f(r)=0. If R is an integral domain, then r is unique.∈

Proof. The existence of r follows immediately from Hensel’s lemma applied to the factorization f¯(t)=(t λ)h¯(t) with h¯(λ) = 0. Given one such r, − 6 we can write f(t)=(t r)h(r). If r′ is a diﬀerent element of R such that − f(r′) = 0, then (r r′)h(r′) = 0 and hence h(r′) = 0. But thenr ¯′ = λ, for − 6 otherwise λ would be a multiple root of f¯.

Proof of Theorem 7.8.1. Let f(t) R[t] be monic of degree d and such that ∈ f¯(t)¯g(t)h¯(t), whereg ¯(t) and h¯(t) are monic, degg ¯(t) = d, deg h¯(t) = n d, andg ¯(t) and h¯(t) are relatively prime in k[t]. We will construct inductively− two sequences of monic polynomials g (t) and h (t) such that, for all { k } { k } k 1: ≥ 1. deg g (t)= d and deg h (t)= n d; k k −

2. The reductions of gk(t) and hk(t) areg ¯(t) and h¯(t) respectively;

3. g (t)h (t) f(t) mk[t]; k k − ∈ 4. g (t) g (t) mk[t] and g (t) g (t) mk[t]. k+1 − k ∈ k+1 − k ∈

Assuming that we have found the sequences gk(t) and hk(t) , (4) implies i { } { } that the coefficients of t in gk are a Cauchy sequence, and similarly for hk. Thus g (t) converges to a monic polynomial g(t) R[t] of degree d, and { k } ∈ hk(t) converges to a monic polynomial h(t) R[t] of degree n d. By (2) {and (4),} reductions of g(t) and h(t) areg ¯(t) and∈ h¯(t) respectively,− and, by (3), f(t) g(t)h(t) = limk (f(t) gk(t)hk(t)) = 0. So it suffices to find − →∞ − g (t) and h (t) satisfying (1)–(4). { k } { k } For k = 1, we can simply take arbitrary lifts g1(t) and h1(t) ofg ¯(t) and h¯(t) to monic polynomials of degrees d and n d respectively. Now suppose − inductively that we have found gk and hk satisfying (1)–(3). We proceed to construct gk+1 and hk+1. 7.8. HENSEL’S LEMMA AND THE IMPLICIT FUNCTION THEOREM287

Claim 7.8.3. For all i, 0 i n 1, there exist polynomials a¯ , ¯b k[t] ≤ ≤ − i i ∈ such that dega ¯ n d 1, deg ¯b d 1, and i ≤ − − i ≤ − i a¯i(t)¯g(t)+ ¯bi(t)h¯(t)= t . To prove the claim, note that, asg ¯(t) and h¯(t) are relatively prime in k[t], we can always ﬁnd some polynomials r ,s k[t] such that r (t)¯g(t)+ i i ∈ i s (t)h¯(t)= ti. Write r (t)= α (t)h¯(t)+¯a (t), with dega ¯ deg h¯ 1= n i i i i i ≤ − − d 1, and similarly let si(t)= βi(t)¯g(t)+¯bi(t), with deg ¯bi degg ¯ 1= d 1. Then− ≤ − −

i t = ri(t)¯g(t)+ si(t)h¯(t)=(αi + βi)¯g(t)h¯(t)+¯ai(t)¯g(t)+ ¯bi(t)h¯(t). ¯ ¯ Since i n 1 and degg ¯(t)h(t) = n, it follows that (αi + βi)¯g(t)h(t)=0 and hence≤ that− i t =¯ai(t)¯g(t)+ ¯bi(t)h¯(t) with dega ¯ n d 1, deg ¯b d 1 as claimed. i ≤ − − i ≤ − Returning to the construction of gk+1 and hk+1, lift the polynomialsa ¯i(t) and ¯b (t) in k[t] to polynomials a (t), b (t) R[t], keeping deg a n d 1 i i i ∈ i ≤ − − and deg b d 1. By construction and property (2) for g and h , i ≤ − k k a (t)g (t)+ b (t)h (t) ti mod m[t]. i k i k ≡ n 1 i k By property (3), gk(t)hk(t) f(t)= i=0− cit with ci m . If we set a(t)= n 1 n− 1 ∈ k − c a (t) and b(t) = − c b (t), then a(t), b(t) m [t], deg a(t) n i=0 i i i=0 i i ∈ ≤ − d 1, deg b(t) d 1, and P P− ≤ − P n 1 − a(t)g (t)+ b(t)h (t) c ti = g (t)h (t) f(t) mod mk+1[t]. k k ≡ i k k − i=0 X Set gk+1(t)= gk(t) b(t) and hk+1(t)= hk(t) b(t). Clearly, gk+1 and hk+1 are monic polynomials− of degrees d and n d respectively,− and the reductions ¯ − k of gk+1(t) and hk+1(t) areg ¯(t) and h(t). Moreover, gk+1 gk = b m [t], and likewise h h mk[t]. We have checked properties− (1), (2),− and∈ (4), k+1 − k ∈ and must verify property (3). Now g h f = g h (ag + bh )+ ab f. k+1 k+1 − k k − k k − k 2k k+1 Since a, b m [t], ab m [t] m [t]. By construction, gkhk f (agk + k∈+1 ∈ ⊆ k+1 − − bhk) m [t] as well. Thus gk+1hk+1 f m [t], showing that property (3) holds.∈ This completes the proof. − ∈ 288 CHAPTER 7. GRADED RINGS

Let us give some straightforward examples of Hensel’s lemma (or rather Corollary 7.8.2): Proposition 7.8.4. Let R be a complete local ring with maximal ideal m and residue field k. Suppose that the characteristic of k is different from 2. Let s R m, and suppose that the image s¯ of s in R/m = k equal to λ2 for ∈ − some λ k. Then s is a square in R. ∈ Proof. Apply Corollary 7.8.2 to the polynomial f(x) = x2 s and the root − λ of f¯. Since the roots of f¯ are λ and λ = λ sinces ¯ = 0 and the characteristic of k is different from± 2, λ is a simple6 − zero of f¯.6 Thus there exists an r R which is a root of f(x), i.e. s = r2. ∈ Example 7.8.5. (i) Let p be a prime number not equal to 2. If a Z and ∈ a is relatively prime to p, then a is a square in Zp if and only if the image of a in Z/pZ is a square. (ii) Let k be a field of characteristic different from 2. Let s =1 x k[[x]]. Thens ¯ = 1 is a square in k. Hence there exists r k[[x]] such that−r2 ∈=1 x. ∈ − In fact, r is given by the usual binomial series. Consider the polynomial f(x, y)= y2 (x3 x2)= y2 x2(x 1) k[x, y]. − − − − ∈ The polynomial f(x, y) is easily checked to be irreducible in k[x, y], and hence R = k[x, y]/(f) is an integral domain. But in k[[x, y]], f(x, y) factors as (y + xr)(y xr). It follows that, if Rˆ is the completion of R for the m- − adic topology, where m is the maximal ideal (x, y), then Rˆ is not an integral domain. We say that R is not analytically irreducible at m. (One should draw the “real picture” of Spec R, in other words the plane curve in R2 defined by y2 = x2(x 1) for the case k = R, to get a sense of why this should be true.) − Remark 7.8.6. Part (ii) of the above example shows that, if R is an integral domain, then Rˆ need not be an integral domain, even when R is the localization of a k-algebra of finite type for an algebraically closed field k. One can ask what properties of a ring R are preserved under completion. For example, if R is regular, then Rˆ is regular. On the other hand, there exist examples of Noetherian rings R which are reduced (resp. integrally closed), such that Rˆ is not reduced (resp. not integrally closed). On the other hand, Chevalley and Zariski have proved the following theorem: Theorem 7.8.7. Let R an integral domain which is the localization of a k- algebra of finite type, where k is a perfect field, let I be a prime ideal of R, and let Rˆ denote the I-adic completion of R. Then: 7.8. HENSEL’S LEMMA AND THE IMPLICIT FUNCTION THEOREM289

1. The ring Rˆ is reduced.

2. If R is integrally closed, then Rˆ is integrally closed.

A standard application of Hensel’s lemma is the following formal version of the implicit function theorem in two variables (proof left as exercise):

Proposition 7.8.8. Let k be a ﬁeld, let f(x, y) k[x, y], and suppose that f(0,y)= g(y) has a simple root λ. Then there exists∈ a unique formal power series y(x) k[[x]] with constant term y = λ such that f(x, y(x))=0. ∈ 0 There are also more general versions of the inverse and implicit function theorem. For example:

Theorem 7.8.9 (Inverse function theorem). Let k be a ﬁeld, let m =(x1,...,xn) be the maximal ideal in k[[x1,...,xn]], and let f1,...,fk m be such that the 2 ∈ images of f1,...,fk m/m are linearly independent. Let J = (f1,...,fk). Then ∈

(i) Possibly after reordering the xi, the natural continuous homomorphism

k[[x ,...,x ]] k[[x ,...,x ]] 1 n → 1 n deﬁned by x f , 1 i k, and x x for i>k, is an isomor- i 7→ i ≤ ≤ i 7→ i phism;

(ii) In particular, for k = n, the continuous homomorphism k[[x ,...,x ]] 1 n → k[[x1,...,xn]] deﬁned by xi fi is an isomorphism if and only if the ∂f 7→ Jacobian determinant i (0) is nonzero; ∂x j (ii) The ideal J is a prime ideal, and k[[x1,...,xn]]/J ∼= k[[xk+1,...,xn]].

Proof. (i): Choose an ordering of the xi so that f1,...,fk, xk+1,...,xn map onto a basis of m/m2. Part (i) follows from Corollary 7.7.22.

(ii): The Jacobian determinant is nonzero if and only if f1,...,fn map onto a basis of m/m2. The suﬃciency part of (ii) then follows from (i), and the necessity is clear. (iii): By Part (ii), there is an isomorphism ϕ: k[[x ,...,x ]] k[[x ,...,x ]] 1 n → 1 n which sends fi to xi and hence J to the prime ideal (x1,...,xk). Clearly, ϕ induces an isomorphism k[[x1,...,xn]]/J ∼= k[[xk+1,...,xn]]. 290 CHAPTER 7. GRADED RINGS

As in multivariable calculus, the inverse function theorem is essentially equivalent to the implicit function theorem: Theorem 7.8.10 (Inverse function theorem). Let k be a ﬁeld, let m = (x1,...,xn) be the maximal ideal in k[[x1,...,xn]], and let f1,...,fn k m be ∂f − ∈ such that the (n k) (n k) matrix i (0) is the identity. Then there − × − ∂x k+j exist ϕk+1,...,ϕn m k[[x1,...,xk]] such that fi(x1,...,xk,ϕk+1,...,ϕn)= 0 for all i. Moreover,∈ ∩ there is an equality of ideals

(f1,...,fn k)=(xk+1 ϕk+1,...,xn ϕn). − − − Proof. Let Ψ: k[[x ,...,x ]] k[[x ,...,x ]] be the continuous homomor- 1 n → 1 n phism deﬁned by xi xi, i k, and xi fi k, i > k, and hence 7→ ≤ 7→ − Ψ(g) = g(x1,...,xk, f1,...,fn k) for all g k[[x1,...,xn]]. By the in- − verse function theorem, Ψ has an inverse, which∈ we denote by Φ. Clearly, Φ(x )= x for i k, and thus i i ≤

Φ(g)= g(x1,...,xk, Φ(xk+1),..., Φ(xn)) for all g k[[x ,...,x ]]. For i>k, we have ∈ 1 n

xi = Φ Ψ(xi)=Φ(fi k)= fi k(Φ(x1),..., Φ(xn)) ◦ − − = fi k(x1,...,xk, Φ(xk+1),..., Φ(xn)). −

Set ϕi(x1,...,xk)=Φ(xi)(x1,...,xk, 0,..., 0). Then

fi(x1,...,xk,ϕk+1,...,ϕn)=0 for all i>k as claimed. To see the ﬁnal statement, we ﬁrst show that (x ϕ ,...,x k+1− k+1 n− ϕn) (f1,...,fn k). Since ϕi(x1,...,xk) = Φ(xi)(x1,...,xk, 0,..., 0), we ⊆ − can write Φ(xi)= ϕi(x1,...,xk)+ xjgij Xj>k for some g k[[x ,...,x ]]. Then, for i>k, ij ∈ 1 n xi =Ψ Φ(xi)=Φ(xi)(x1,...,xk, f1,...,fn k)= ϕi(x1,...,xk)+ fjhij ◦ − Xj>k for some hij k[[x1,...,xn]]. Hence xi ϕi (f1,...,fn k) for all i>k, and ∈ − ∈ − thus (xk+1 ϕk+1,...,xn ϕn) (f1,...,fn k). By the previous theorem, − − ⊆ − 7.8. HENSEL’S LEMMA AND THE IMPLICIT FUNCTION THEOREM291

(f1,...,fn k) is a prime ideal of coheight k, and clearly (xk+1 ϕk+1,...,xn − − − ϕ ) is also a prime ideal of coheight k. Since (x ϕ ,...,x ϕ ) is n k+1 − k+1 n − n contained in (f1,...,fn k), the two ideals must be equal. − For n k = 1, the inverse function theorem says that, if f k[[x ,...,x ]] − ∈ 1 n and ∂f/∂xn(0) = 0, then there exist u,ϕ k[[x1,...,xn]], with u a unit, such that f = u6 (x ϕ). This is a special∈ case of the formal Weierstrass n − preparation theorem, which is discussed in the exercises.

Exercises

Exercise 7.1. Let S be a graded ring, and suppose that S is generated as an S -algebra by S , the elements of degree 1. Show that gr S = S as graded 0 1 S+ ∼ rings.

Exercise 7.2. Let S = k[x1,...,xn] and let m be the maximal ideal (x1,...,xn). Show that B (S) = k[x ,...,x ,y ,...,y ]/(x y x y , 1 i, j n), where m ∼ 1 n 1 n i j − j i ≤ ≤ the xi have degree zero and the yj have degree one.

Exercise 7.3. (Tangent cones) (i) Let g(x1,...,xn) k[x1,...,xn], g = 0, N ∈ 6 and write g = d=0 gd, where each gd is homogeneous and g0 = 0. Define the multiplicity mult0 g of g at 0 to be the smallest m such that gd = 0, and P n 6 set in g = gm. The multiplicity of g at an arbitrary point of A is defined similarly. Thus, if m is the maximal ideal corresponding to p, then multp g is the smallest nonnegative integer m such that g mm mm+1. For an ideal ∈ − I in k[x1,...,xn], define

in(I)= in h : h I . { ∈ }

Show that I is a homogeneous ideal in k[x1,...,xn]. If I =(f1,...,fk), is it always the case that in(I) = (in f1,..., in fk)? (ii) Let X = V (J) An be an aﬃne variety, let 0 X correspond to the ⊆ ∈ maximal ideal m, and let R = A(X). Show that grm R is a quotient of the graded ring k[x1,...,xn], viewed as gr(x1,...,xn) k[x1,...,xn]. More precisely, show that grm R = k[x1,...,xn]/ in(J). n 1 It follows that the projective scheme V (in(J)) P − depends only on + ⊆ the isomorphism class of the local ring A(X)m. This scheme is called the (projective) tangent cone to X at 0. The tangent cone to X at an arbitrary point p is deﬁned similarly. 292 CHAPTER 7. GRADED RINGS

Exercise 7.4. (i) Let X Pn be a projective variety, and let C(X) An+1 ⊆ ⊆ be the affine cone over X. Let R = R(X) be the affine coordinate ring of C(X) and let m be the maximal ideal corresponding to 0 C(X). Show ∈ that the multiplicity of the local ring R(X)m is equal to deg X. (ii) Let X An be an affine variety, let p X correspond to the maximal ⊆ ∈ ideal m, and let R = A(X)m. Show that the degree of the tangent cone to X n 1 at p (viewed as a subscheme of P − ) is equal to the multiplicity of the local ring R.

Exercise 7.5. Let R be a regular local ring with maximal ideal m and residue field k, with the property that, for all prime ideals p of R, the ring Rp is also regular. (In fact, this is automatically satisfied by every regular local ring.) Show that, for every prime ideal q of R[x], the ring R[x]q is regular. Hence by induction, for every prime ideal q of R[x1,...,xn], the local ring R[x1,...,xn]q is regular. In particular, if k is a field, not necessarily algebraically closed, or a Dedekind domain, and q is a prime ideal of k[x1,...,xn], not necessarily maximal, the local ring k[x1,...,xn]q is regular. (Let dim R = d. We can assume by the hypothesis that q R = m and hence that mR[x] q. Thus either q = mR[x] and is generated∩ by d ⊆ ≤ elements or q is the inverse image in R[x] of a nonzero ideal in k[x], and hence is generated by d + 1 elements. Now dim R[x]q = ht q. Use Corollary 7.5.6 to see that ht q ≤= d if q = mR[x] and that ht q = d + 1 otherwise.)

Exercise 7.6. Let k be a ﬁeld of positive characteristic p = 2, let a k kp, 6 ∈ − and let R = k[x, y]/(y2 (xp a)). − − (i) Show that m =(y, xp a) is a maximal ideal of k[x, y] and thus induces − a maximal ideal of R, which we also denote by m.

(ii) Show that the local ring Rm is regular.

1 1 (iii) Describe ΩR/k. In particular, is ΩR/k a projective R-module? Exercise 7.7. Let R be a Noetherian ring, I an ideal of R, M a ﬁnitely generated R-module and N a submodule of M. Show that, in the I-adic n topology, the closure of N in M is the submodule N = n 0(N + I M). Show that ≥ T N = m M : there exists r I such that (1 r)m N . { ∈ ∈ − ∈ } 7.8. HENSEL’S LEMMA AND THE IMPLICIT FUNCTION THEOREM293

Exercise 7.8. Let R be a ring (not necessarily Noetherian) and let m be a maximal ideal of R. Show that the m-adic completion Rˆ of R is a local ring. ˆ (Begin by noting that R/mˆ ∼= R/m and hence that mˆ is a maximal ideal in Rˆ. Show that if r Rˆ mˆ , then r is a unit, by ﬁrst considering the case ∈ − r 1 mod mˆ .) ≡ Conclude that Rˆ is isomorphic to the completion of Rm at the ideal mRm.

Exercise 7.9. Let k be a ﬁeld, and consider the group of units (k[[x]])∗ in the power series ring k[[x]].

(i) Show that (k[[x]])∗ = k∗ U, where U is the subgroup of (k[[x]])∗ ∼ × deﬁned by

∞ U = a xi k[[x]] : a =1 = f k[[x]] : f 1 mod(x) . { i ∈ 0 } { ∈ ≡ } i=0 X (ii) Suppose that k has characteristic zero. Show that the multiplicative group U is isomorphic to the additive group k[[x]]. (Use the exponen- tial function to deﬁne a homomorphism from (k[[x]], +) to (U, ) with · inverse equal to the logarithm.)

(iii) If k has characteristic p > 0, show that U is not isomorphic to the additive group k[[x]].

Exercise 7.10. Let (R, m)and (S, n) be two complete local rings (in particular, Hausdorﬀ), and let f : R S be a local homomorphism (i.e. f(m) n). → ⊆ Show that f is continuous for the respective topologies. Suppose further that the induced homomorphism R/m S/n is an isomorphism. Then f is surjective if and only if the induced homomorphism→ m/m2 n/n2 is surjective. →