Commutative Algebra

Home , Affine variety, Algebraic independence, Commutative algebra, Completion of a ring, Depth (ring theory), Equivalence class

Commutative Algebra

Andrew Kobin

Spring 2016 / 2019 Contents Contents

Contents

1 Preliminaries 1 1.1 Radicals ...... 1 1.2 Nakayama’s Lemma and Consequences ...... 4 1.3 Localization ...... 5 1.4 Transcendence Degree ...... 10

2 Integral Dependence 14 2.1 Integral Extensions of Rings ...... 14 2.2 Integrality and Field Extensions ...... 18 2.3 Integrality, Ideals and Localization ...... 21 2.4 Normalization ...... 28 2.5 Valuation Rings ...... 32 2.6 Dimension and Transcendence Degree ...... 33

3 Noetherian and Artinian Rings 37 3.1 Ascending and Descending Chains ...... 37 3.2 Composition Series ...... 40 3.3 Noetherian Rings ...... 42 3.4 Primary Decomposition ...... 46 3.5 Artinian Rings ...... 53 3.6 Associated Primes ...... 56

4 Discrete Valuations and Dedekind Domains 60 4.1 Discrete Valuation Rings ...... 60 4.2 Dedekind Domains ...... 64 4.3 Fractional and Invertible Ideals ...... 65 4.4 The Class Group ...... 70 4.5 Dedekind Domains in Extensions ...... 72

5 Completion and Filtration 76 5.1 Topological Abelian Groups and Completion ...... 76 5.2 Inverse Limits ...... 78 5.3 Topological Rings and Module Filtrations ...... 82 5.4 Graded Rings and Modules ...... 84

6 Dimension Theory 89 6.1 Hilbert Functions ...... 89 6.2 Local Noetherian Rings ...... 94 6.3 Complete Local Rings ...... 98

7 Singularities 106 7.1 Derived Functors ...... 106 7.2 Regular Sequences and the Koszul Complex ...... 109 7.3 Projective Dimension ...... 114

i Contents Contents

7.4 Depth and Cohen-Macauley Rings ...... 118 7.5 Gorenstein Rings ...... 127

8 Algebraic Geometry 133 8.1 Aﬃne Algebraic Varieties ...... 133 8.2 Morphisms of Aﬃne Varieties ...... 142 8.3 Sheaves of Functions ...... 146 8.4 Finite Morphisms and the Fibre Theorem ...... 147 8.5 Projective Varieties ...... 150 8.6 Nonsingular Varieties ...... 153 8.7 Abstract Curves ...... 156 8.8 The Spectrum of a Ring ...... 159

ii 1 Preliminaries

1 Preliminaries

The contents of this document come from an Algebra IV course taught by Dr. Peter Abra- menko at the University of Virginia in Spring 2016. The main theory of commutative rings are covered, along with applications to algebraic geometry towards the end. Topics include: Integral dependence Noetherian, Artinian and Dedekind rings Valuations and discrete valuation rings Dimension theory Affine and projective varieties Affine algebras Hilbert’s Nullstellensatz Morphisms of varieties The commutative algebra may be found in Atiyah and MacDonald’s Introduction to Commu- tative Algebra, while the algebraic geometry generally follows the first chapter of Hartshorne’s Algebraic Geometry. Several topics in these notes also come from a course on commutative algebra taught by Dr. Craig Huneke at UVA in Spring 2019. The main topics I’ve included from this course are: associated primes, Cohen’s structure theory for complete local rings, some dimension theory and notions of singularity in rings. We will adhere to several conventions throughout these notes: All rings are commutative with unity 1. A will always denote such a ring. All subrings contain 1. In particular, proper ideals are never subrings. For any ring homomorphism f : A → B, in addition to the homomorphism axioms, we stipulate that f(1A) = 1B. In any integral domain, we assume 1 6= 0.

1.1 Radicals

There are three important types of radicals that we may define in a commutative ring A. The first two are defined below. Definition. The nil radical of A is defined as the intersection of all prime ideals of A, written \ N(A) = P. prime ideals P ⊂A We say A is reduced if N(A) = 0.

1 1.1 Radicals 1 Preliminaries

Definition. The Jacobson radical of A is the intersection of all maximal ideals of A, written \ J(A) = M. maximal ideals M⊂A Remark. Clearly N(A) ⊆ J(A). Lemma 1.1.1. For any x ∈ A, x ∈ J(A) if and only if 1 − xy is a unit in A for all y ∈ A. Proof. ( =⇒ ) Suppose A(1 − xy) 6= A for some y ∈ A. Then A(1 − xy) ⊂ M for a maximal ideal M. By definition, x ∈ J(A) ⊆ M and since J(A) is an ideal, xy ∈ M as well. This means 1 = xy + (1 − xy) ∈ M, a contradiction. Therefore A(1 − xy) = A so there is some a ∈ A such that a(1 − xy) = 1; that is, 1 − xy is a unit. ( ⇒ = ) Conversely, suppose x 6∈ M for some maximal ideal M ⊂ A. Then M + Ax = A by maximality of M, so b + xy = 1 for some b ∈ M, or 1 − xy = b ∈ M, meaning 1 − xy cannot be a unit in A. Proposition 1.1.2. N(A) = {x ∈ A | xn = 0 for some n ≥ 1}. That is, the nil radical consists of all nilpotent elements in A. Proof. (⊇) Suppose xn = 0. Then for any prime ideal P , xn = 0 ∈ P which implies x ∈ P . Thus x ∈ N(A). (⊆) If x ∈ A is not nilpotent, we will produce a prime ideal not containing x. Let Σ denote the set of ideals I ⊂ A such that xn 6∈ I for any n ≥ 1. Notice that 0 is an element of Σ since we are assuming x is not nilpotent; thus Σ is nonempty. Therefore we may apply Zorn’s Lemma to choose an ideal P ∈ Σ which is maximal among this collection of ideals. We claim that P is prime. Clearly P 6= A since A is not an element of Σ. Suppose y, z ∈ A but y, z 6∈ P . Since P is maximal in Σ, the ideals P + Ay and P + Az are both strictly larger ideals than P . Thus P + Ay, P + Az 6∈ Σ so there exist m, n ∈ N such that xm ∈ P + Ay and xn ∈ P + Az. Then xm+n ∈ (P + Ay)(P + Az) ⊆ P + Ayz, but xm+n 6∈ P by assumption, so we must have yz 6∈ P . This proves P is prime. n Definition. For an ideal I ⊂ A, the radical of I√is r(I) = {x ∈ A | x ∈ I for some n ≥ 1}. Alternate notations for the radical ideal include I and rad(I). Corollary 1.1.3. For any ideal I ⊂ A, r(I) equals the intersection of all prime ideals of A containing I. In particular, r(I) is an ideal. Proof. Consider the quotient ring A¯ = A/I. For x ∈ A we have the following equivalent statements:

n x ∈ r(I) ⇐⇒ there exists an n ∈ N such that x ∈ I n ⇐⇒ there exists an n ∈ N such that (x + I) = 0 ∈ A¯ ⇐⇒ x + I is nilpotent, i.e. x + I ⊂ N(A¯) ⇐⇒ x + I is contained in every prime of A¯ ⇐⇒ x lies in every prime ideal of A containing I. \ Therefore r(I) = P as desired. prime ideals P ⊇I

2 1.1 Radicals 1 Preliminaries

Remark. By Corollary 1.1.3, N(A) = r(0). Lemma 1.1.4. Suppose I,J ⊂ A are ideals. Then (a) I ⊆ r(I). (b) I = r(I) if and only if A/I is reduced. In particular, every prime ideal is radical. (c) r(r(I)) = r(I).

(d) r(Im) = r(I) for all m ∈ N. (e) r(IJ) ⊇ r(I)r(J). (f) r(I + J) = r(r(I) + r(J)). (g) r(I) = A if and only if I = A. Proof. (a) is clear from the definition. (b) By Corollary 1.1.3, r(I) equals the intersection of all primes lying over I. Let π : A → A/I be the quotient map. Then N(A/I) = 0 ⇐⇒ π(r(I)) = 0 ⇐⇒ r(I) ⊆ ker π = I. By part (a), we always have I ⊆ r(I) so this shows that N(A/I) = 0 ⇐⇒ r(I) = I. (c) By part (a), it suffices to show r(r(I)) ⊆ r(I). If x ∈ r(r(I)) then xn ∈ r(I) for some n ≥ 1, but then (xn)m = xnm ∈ I for some m ≥ 1. Thus x ∈ r(I). (d) If x ∈ r(Im) then xn ∈ Im for some n ≥ 1, but Im ⊆ I so this shows x ∈ r(I). On the other hand, if x ∈ r(I) then xn ∈ I for some n ≥ 1. Letting k = max{n, m}, we see that xk ∈ Im. Hence r(I) = r(Im). (e) It suffices to prove this for elements of the form xy ∈ r(I)r(J), where x ∈ r(I) and y ∈ r(J). Then xn ∈ I and ym ∈ J for some n, m ≥ 1. Without loss of generality, assume n ≥ m. Then (xy)n = xnyn ∈ IJ so xy ∈ r(IJ). (f) First take z ∈ r(I + J), so that zn ∈ I + J for some n ≥ 1. Then zn = x + y for x ∈ I, y ∈ J. Since I ⊆ r(I) and J ⊆ r(J) by part (a), we have zn = x + y ∈ I + J ⊆ r(I) + r(J). Therefore z ∈ r(r(I) + r(J)). On the other hand, if w ∈ r(r(I) + r(J)) then wm ∈ r(I) + r(J) for some m ≥ 1. This means wm = x + y for some x ∈ r(I), y ∈ r(J). In turn this says that xk ∈ I and y` ∈ J for some k, ` ≥ 1. Using the binomial formula, write k + ` wm(k+`) = (x + y)k+` = xk+` + (k + `)xk+`−1y + ... + xk+`−qyq + ... + yk+`. q Note that q ranges from 0 to k + `. When 0 ≤ q ≤ `, the qth term (the first term in the sum being the 0th term) is divisible by xk and so it lies in I. Otherwise when 1 + ` ≤ q ≤ k + `, the qth term is divisible by y` and so it lies in J. Therefore wm(k+`) is of the form a + b where a ∈ I and b ∈ J. Hence w ∈ r(I + J). (g) Suppose r(I) = A. Then in particular 1 ∈ r(I) so for some n ≥ 1, 1 = 1n ∈ I. Thus I = A. On the other hand, if I = A then part (a) says A ⊆ r(A), which implies r(A) = A.

Example 1.1.5. Let A = Z and I = J = 2Z. Then IJ = 4Z so r(IJ) = 2Z, but I and J are each prime and thus radical, so r(I)r(J) = IJ = 4Z. Hence r(IJ) ) r(I)r(J), showing that the containment in (e) may be strict.

3 1.2 Nakayama’s Lemma and Consequences 1 Preliminaries

Example 1.1.6. Let k be an algebraically closed field and consider A = k[x, y]. Take I = (y) and J = (y − x2). Then I = r(I) since A/I = k[x] which is a PID and hence reduced. Likewise, k[x, y]/(y − x2) ∼= k[x] by the first isomorphism theorem applied to the map k[x, y] → k[x], f(x, y) 7→ f(x, x2). Thus J = r(J), so r(I) + r(J) = I + J. However, I + J = (y) + (y − x2) = (y, x2) and the radical of this ideal is r((y, x2)) = (y, x), a strictly larger ideal than I + J. This shows the containment r(I + J) ⊂ r(I) + r(J) is sometimes strict, so the second radical on the right side of property (f) is necessary for equality. Tn Tn Lemma 1.1.7. Let J1,...,Jn be ideals of a ring A and J = i=1 Ji. Then r(J) = i=1 r(Ji). n Proof. It suffices to prove the case when n = 2. Suppose x ∈ r(J1 ∩J2). Then x ∈ J1 ∩J2 for n n some n ∈ N, so in particular x ∈ J1, implying x ∈ r(J1), and x ∈ J2, implying x ∈ r(J1). Hence x ∈ r(J1) ∩ r(J2). k On the other hand, if y ∈ r(J1) ∩ r(J2), then y ∈ r(J1) so y ∈ J1 for some k ∈ N. ` m Similarly, y ∈ r(J2) implies y ∈ J2 for some ` ∈ N. Set m = max(k, `). Then y = m−k k m m−` ` m y y ∈ J1 and y = y y ∈ J2 by the setup, so y ∈ J1 ∩ J2. Hence y ∈ r(J1 ∩ J2). This proves r(J1 ∩ J2) = r(J1) ∩ r(J2). Example 1.1.8. The conclusion of Lemma 1.1.7 does not hold for arbitrary intersections n of ideals. Let k be a field and A = k[t] and consider the family of ideals Jn = (t ) for n ∈ . Then T J = 0 since if t` | f(t) for all ` ≥ 1 then necessarily f = 0. So N n∈N n r T J = r(0) = N(A) = 0 since A is a PID. However, for each n ∈ , r((tn)) = (t) so n∈N n N T r(J ) = (t) 6= 0. n∈N n Lemma 1.1.9. If f : A → B is a ring homomorphism and J is an ideal of B, then f −1(r(J)) = r(f −1(J)). Proof. Let x ∈ A. Then x ∈ f −1(r(J)) ⇐⇒ f(x) ∈ r(J) n ⇐⇒ f(x) ∈ J for some n ∈ N n ⇐⇒ f(x ) ∈ J for some n ∈ N since f is a homomorphism n −1 ⇐⇒ x ∈ f (J) for some n ∈ N ⇐⇒ x ∈ r(f −1(J)). Hence f −1(r(J)) = r(f −1(J)).

1.2 Nakayama’s Lemma and Consequences

A classic result in ring theory is Nakayama’s Lemma, which will be useful later on. Theorem 1.2.1 (Nakayama’s Lemma). If I ⊂ A is an ideal such that I ⊆ J(A), and M is a finitely generated A-module such that IM = M, then M = 0. Proof. Assume IM = M but M 6= 0. Since M is finitely generated, pick some minimal finite set of generators {m1, . . . , mn} of M. Then mn ∈ M = IM = Im1 + ... + Imn so there exist elements a1, . . . , an ∈ I such that mn = a1m1 + ... + anmn. This can be written

(1 − an)mn = a1m1 + ... + an−1mn−1.

4 1.3 Localization 1 Preliminaries

Since an ∈ I ⊆ J(A), Lemma 1.1.1 shows 1−an is a unit in A. Then mn ∈ Am1+...+Amn−1. This shows that {m1, . . . , mn−1} generates M as an A-module, contradicting minimality. Therefore we must have M = 0. One can ﬁnd examples of ideals not lying in the Jacobson radical for which IM = M for nontrivial modules M.

Example 1.2.2. Let A be a local ring which is an integral domain but not a field. As an a example, for any prime p, the ring of p-adic integers Z(p) = b : a, b ∈ Z, p - b is a local subring of Q which is not a field. Let M be the unique maximal ideal of A and take F to be its field of fractions (see below for details). Since A is not a field, M 6= 0. In fact, in the local ring situation, M = J(A). Then we have MF = F for the A-module F , but F 6= 0. This provides a counterexample to Nakayama’s Lemma when the module is not finitely generated, as is the case with the residue field of a local ring.

The following is another version of Nakayama’s Lemma which may be useful in some contexts.

Corollary 1.2.3. Let M be a ﬁnitely generated A-module and take N ⊆ M to be a submodule. If I ⊂ A is an ideal such that I ⊆ J(A) and M = IM + N, then M = N.

Proof. Consider the ﬁnitely generated A-module M = M/N. Then

IM = (IM + N)/N = M/N = M.

Applying Nakayama’s Lemma (1.2.1) shows that M = 0, that is, M = N. Let A be a local ring with maximal ideal M = J(A). We call the quotient field k = A/M the residue field of A. If N is a finitely generated A-module then N/MN is a finite dimensional k-vector space.

Proposition 1.2.4. If x1, . . . , xn are elements of N such that x¯1,..., x¯n ∈ N/MN span N/MN as a k-vector space, then N is generated by the xi as an A-module.

0 Pn Proof. Set N = i=1 Axi, which is a submodule of N. Consider the canonical projection π : N → N/MN. By assumption, π(N 0) = N/MN. Thus N = N 0 + MN but since M = J(A) and N is ﬁnitely generated as an A[x]-module, the conditions of Corollary 1.2.3 are satisﬁed. Hence N 0 = N.

1.3 Localization

One of the most important constructions in commutative algebra is the localization of a ring at a multiplicatively closed subset.

Deﬁnition. A subset S ⊆ A is said to be multiplicatively closed provided that 1A ∈ S and for all x, y ∈ S, xy ∈ S as well.

Examples. Two of the most important examples of multiplicatively closed subsets are:

5 1.3 Localization 1 Preliminaries

1 For any prime ideal P ⊂ A, the set S = A r P is multiplicatively closed. 2 For a fixed element f ∈ A (usually a function in a ring of functions), the set S = {f n | 0 n ≥ 0} is multiplicatively closed, where by convention f = 1A. Definition. Let S ⊆ A be multiplicatively closed and suppose M is a finitely generated A- module. The module of fractions S−1M is defined as S−1M = M × S/ ∼ where ∼ is the equivalence relation (m, s) ∼ (n, t) iff u(tm − sn) = 0 for some u ∈ S. It is common to write −1 m −1 elements of S M as s instead of (m, s). Addition and the A-module action of S M are given by m n tm + sn m am + = and a = . s t st s s Lemma 1.3.1. The above operations make S−1M into an A-module. −1 a b Proposition 1.3.2. For the A-module A, S A is a ring under the multiplication law s · t = ab 1A , with unity 1S−1A = . st 1A Proof. We will verify that multiplication is well-defined. The other ring axioms are easily checked. If (a, s) ∼ (a0, s0) then u(s0a − sa0) = 0 for some u ∈ S. Thus we have u(s0tab − sta0b) = tbu(s0a − sa0) = 0 so (a0b, s0t) ∼ (ab, st). Example. 3 If A is an integral domain then S = A r {0} is multiplicatively closed. The ring S−1A is called the field of fractions of A. −1 a Lemma 1.3.3. There is a homomorphism j : A → S A defined by a 7→ 1 which is one-to- one if and only if S contains no zero divisors. Proposition 1.3.4 (Universal Property of Localization). Let S ⊆ A be a multiplicatively closed subset. For any ring homomorphism ϕ : A → B such that ϕ(S) ⊆ B∗, there exists a unique ring homomorphism ϕ0 : S−1A → B making the following diagram commute: ϕ A B

j ϕ0

S−1A

a −1 0 0 a 0 a 0 1 Proof. First, if s ∈ S A and such a ϕ is defined, we must have ϕ s = ϕ 1 ϕ s = −1 0 0 a −1 ϕ(a)ϕ(s) . This implies uniqueness. Now define ϕ in this way: ϕ s = ϕ(a)ϕ(s) for 0 a b −1 any a ∈ A, s ∈ S. To see that ϕ is well-defined, suppose s = t in S A. Then there is an element u ∈ S such that u(at − bs) = 0. Applying ϕ, we have ϕ(u)ϕ(at − bs) = 0. Since ϕ(u) is a unit in B, it follows that ϕ(at − bs) = 0 =⇒ ϕ(a)ϕ(t) = ϕ(b)ϕ(s) =⇒ ϕ(a)ϕ(s)−1 = ϕ(b)ϕ(t)−1. Hence ϕ0 is well-defined. The rest follows easily from this definition of ϕ0.

6 1.3 Localization 1 Preliminaries

Examples.

−1 4 If S = ArP for a prime ideal P ⊂ A, the localization of A at P is the ring AP := S A.

5 If S = {f n | n ≥ 0} for some element f ∈ A then the localization of A at f is −1 Af := S A. ∼ Proposition 1.3.5. If f, g ∈ A are elements such that r((f)) = r((g)) then Af = Ag. Proof. The assumption r((f)) = r((g)) means that f n = xg and gm = yf for some n, m ≥ 1 and x, y ∈ A. Let jf : A → Af and jg : A → Ag be the canonical homomorphisms sending a g n a 7→ 1 in each localization. Then jf (g) = 1 is invertible in Af , since f − xg = 0 and so g x 1 f 1 · f n = 1 . Symmetrically, jg(f) = 1 is invertible in Ag. Therefore by the universal property of localization, there are unique homomorphisms ϕ1 : Ag → Af and ϕ2 : Af → Ag such that ϕ1 ◦ jg = jf and ϕ2 ◦ jf = jg: A

jf jg

ϕ2 Af Ag ϕ1 ∼ Then it is clear that ϕ1 and ϕ2 are inverses of each other, so Af = Ag. Proposition 1.3.6. If M is an A-module and S ⊆ A is a multiplicatively closed subset then there exists a well-deﬁned isomorphism of S−1A-modules (or of A-modules)

−1 −1 f : S A ⊗A M −→ S M a am s ⊗ m 7−→ s . Proof. Deﬁne a map

S−1A × M −→ S−1M a am s , m 7−→ s . It is easy to check that is A-bilinear, so by the universal property of tensor products, we get a −1 −1 well-deﬁned homomorphism of A-modules f : S A ⊗A M → S M. Clearly f is surjective. P ai −1 Q Q For an arbitrary element ⊗ mi ∈ S A ⊗ M, set s = si ∈ S and ti = sj ∈ S. i si i j6=i Then X ai X aiti X 1 1 X ⊗ mi = ⊗ mi = ⊗ aitimi = ⊗ aitimi. si s s s i i i i −1 1 Therefore every element of S A ⊗ M has the form s ⊗ m for s ∈ S and m ∈ M. Now 1 m suppose f s ⊗ m = 0. Then s = 0 by deﬁnition of the map, so tm = 0 for some t ∈ S. 1 t 1 1 Thus s ⊗ m = st ⊗ m = st ⊗ tm = st ⊗ 0 = 0. This proves f is injective and thus we have −1 ∼ −1 an isomorphism S A ⊗A M = S M as A-modules. Finally, it’s easy to see that the same map is also an isomorphism of S−1A-modules.

7 1.3 Localization 1 Preliminaries

Given an ideal I ⊂ A, we can deﬁne the subset S−1I ⊆ S−1A by viewing I as an −1 a A-module. Speciﬁcally, S I = s : a ∈ I, s ∈ S . Lemma 1.3.7. Let I ⊂ A be an ideal and let S ⊂ A be a multiplicatively closed subset. Then

(a) S−1I ⊆ S−1A is an ideal.

(b) S−1I = S−1A if and only if S ∩ I 6= ∅.

a −1 b −1 a b ab Proof. (a) Let s ∈ S A and t ∈ S I, meaning s, t ∈ S, a ∈ A and b ∈ I. Then s · t = st lies in S−1I since ab ∈ I by absorption and st ∈ S by the multiplicatively closed property. Hence S−1I is an ideal of S−1A. s 1 −1 −1 −1 (b) On one hand, if s ∈ S ∩ I then s = 1 ∈ S I so S I = S A. On the other hand, −1 −1 1 −1 a 1 S I = S A implies 1 ∈ S I so there exist elements a ∈ I and s ∈ S such that s = 1 . Then there is a t ∈ S such that t(a − s) = 0, that is, ts = ta ∈ S ∩ I. Thus S ∩ I is nonempty.

Proposition 1.3.8. Let S be a multiplicatively closed subset of A. Then

(a) Every proper ideal J ⊂ S−1A is of the form S−1I for some ideal I not intersecting S.

(b) For any prime ideal P ⊂ A not intersecting S, j−1(S−1P ) = P .

−1 {P ⊂ A | P is a prime ideal,S ∩ P = ∅} −→ {prime ideals of S A} P 7−→ S−1P

is an inclusion-preserving bijection.

−1 −1 a −1 Proof. (a) Let J ⊂ S A be an ideal. Then I = j (J) is an ideal of A. If s ∈ S I for a a a 1 a a a s a ∈ I, s ∈ S, then 1 ∈ J so s = 1 · s ∈ J. On the other hand, if s ∈ J then 1 = s · 1 ∈ J. −1 a −1 −1 Then a ∈ I since I = j (J). Hence s ∈ S I, which proves that S I = J. Also, S ∩I = ∅ since S−1I = J ( S−1A. (b) By deﬁnition of the map j, P ⊆ j−1(S−1P ). For the other containment, suppose −1 −1 a −1 a b a ∈ j (S P ). Then 1 ∈ S P so there exist b ∈ P and s ∈ S such that 1 = s . Thus for some t ∈ S we have t(sa − b) = 0, or tsa = tb which lies in P since b ∈ P . Since s, t ∈ S and S ∩ P = ∅, s, t 6∈ P but then tsa ∈ P implies a ∈ P since P is prime. Thus P = j−1(S−1P ) as claimed. x y −1 (c) Suppose P ⊂ A is a prime ideal with S ∩ P = ∅. Let s , t ∈ S A such that x y xy −1 xy a s · t = st ∈ S P . Then there are elements a ∈ P and u ∈ S such that st = u . So for some v ∈ S, v(uxy − sta) = 0; that is, vuxy = vsta which lies in P since a ∈ P . Now x −1 y −1 v, u ∈ S so v, u 6∈ P , meaning x ∈ P or y ∈ P . Thus s ∈ S P or t ∈ S P . This proves that S−1P is a prime ideal of S−1A. Finally, denote the map P 7→ S−1P by Φ. By (b), Φ is injective. For surjectivity, assume J ⊂ S−1A is prime. Then P = j−1(J) is also prime. By (a), J = S−1P = Φ(P ). Also, S ∩ P = ∅ since J is prime and therefore J 6= S−1A. Hence Φ is surjective, so it is a bijection.

8 1.3 Localization 1 Preliminaries

Corollary 1.3.9. If P ⊂ A is prime then AP is a local ring with unique maximal ideal S−1P , where S = A r P . Moreover, there is an inclusion-preserving bijection

{Q ⊂ A | Q is a prime ideal,Q ⊆ P } ←→ {prime ideals of AP }.

Proof. We know that S−1P ⊂ S−1A is a prime ideal. Observe that

−1 −1 a a ∗ S A r S P = s : a ∈ A r P, s ∈ S = s : a ∈ S, s ∈ S = AP ,

−1 −1 the set of units in AP . This implies S P is the unique maximal ideal of S A. The bijection follows from (c) of Proposition 1.3.8, using the fact that S ∩Q = ∅ if and only if Q ⊆ P . Lemma 1.3.10. Let S be a multiplicatively closed subset and I,J ideals of the ring A. Then

(a) S−1(IJ) = (S−1I)(S−1J).

(b) S−1(I ∩ J) = (S−1I) ∩ (S−1J).

x −1 Pn Proof. (a) Let s ∈ S (IJ), with x ∈ IJ and s ∈ S. Then x = k=1 akbk for ak ∈ I and bk ∈ J. Moreover, we assume 1 ∈ S by convention so

n n x X akbk X ak bk = = · . s s s 1 k=1 k=1

−1 −1 x Then each summand is a product of an element in S I and an element in S J, so s ∈ −1 −1 −1 −1 −1 y −1 −1 (S I)(S J). Thus S (IJ) ⊆ (S I)(S J). On the other hand, if t ∈ (S I)(S J) then there are elements ak ∈ I, bk ∈ J, sk, tk ∈ S such that

n Pn y X akbk s1t1 ··· sˆktˆk ··· sntnakbk = = k=1 . t sktk s1t1 ··· sntn k=1

The numerator lies in IJ since ak ∈ I, bk ∈ J so each summand is a product of an element of I and an element of J. The denominator is an element of S since this set is multiplicative. y −1 −1 −1 −1 Therefore t ∈ S (IJ) by deﬁnition. Hence (S I)(S J) ⊆ S (IJ). x −1 (b) Observe that for any s ∈ S A, x ∈ S−1(I ∩ J) ⇐⇒ x ∈ I ∩ J and s ∈ S s x x ⇐⇒ ∈ S−1I and ∈ S−1J s s x ⇐⇒ ∈ (S−1I) ∩ (S−1J). s So S−1(I ∩ J) = (S−1I) ∩ (S−1J). (c) If I ∩ S 6= ∅, I ⊆ r(I) implies r(I) ∩ S 6= ∅ as well so we have S−1r(I) = S−1A = r(S−1A) = r(S−1I).

9 1.4 Transcendence Degree 1 Preliminaries

Now assume I ∩ S = ∅. Since S is multiplicative, if s were in r(I) for any s ∈ S then sn ∈ I ∩ S for some n ∈ N, which is impossible. Therefore r(I) ∩ S = ∅. By Corollary 1.1.3, T r(I) = p where p runs over all prime ideals of A that contain I. Then by Proposition 1.3.8, we have ! \ \ \ S−1r(I) = S−1 p = S−1p = q = r(S−1I) p⊇I p⊇I q⊇S−1I where p runs over the primes of A containing I and q runs over the primes of S−1A containing S−1I. Hence we have S−1r(I) = r(S−1I) in all cases.

−1 Lemma 1.3.11. For a prime ideal P ⊂ A, the quotient ﬁeld AP /S P is isomorphic to the ﬁeld of fractions of A/P .

Proof. Let ϕ : A −→π A/P −→i Frac(A/P ) be the canonical quotient map composed with a¯ i :a ¯ 7→ 1 . Then P ⊆ ker ϕ, so S = ArP gets mapped to nonzero elements in Frac(A/P ), but these are all units since Frac(A/P ) is a ﬁeld. Therefore the universal property of localization gives us a unique homomorphism f : AP → Frac(A/P ) such that f ◦ j = ϕ. Notice that if a −1 a −1 s ∈ S P then a ∈ P so ϕ(a) = 0 and therefore f s = 0. So S P ⊆ ker f. Clearly f is −1 ∼ onto, so by the ﬁrst isomorphism theorem, we have AP /S P = Frac(A/P ). Example.

6 Let A = Z. Since Z is an integral domain, (0) is a prime ideal and its localization is A(0) = Q, the ﬁeld of fractions of Z. For any nonzero prime p ∈ Z, the localization a Z(p) = s : a, s ∈ Z, p - s is called the ring of p-adic integers. Each Z(p) contains only −1 the prime ideals (0) and pZ(p) := S (p).

1.4 Transcendence Degree

Let K/k be an extension of ﬁelds and let E ⊆ K be any subset.

Deﬁnition. We say E is algebraically independent over k if for any ﬁnite subset {x1, . . . , xn} ⊆ E and any function F ∈ k[t1, . . . , tn] for which F (x1, . . . , xn) = 0, it must be that F = 0. When E = {x}, we say the element x is transcendental (or transcendent) over k.

Clearly x ∈ K is transcendental over k if and only if it is not algebraic over k.

Deﬁnition. A set E ⊆ K is a transcendence basis for K/k if E is a maximal algebraically independent subset of K. In the case that E is a transcendence basis for K/k and K = k(E), we say the extension is purely transcendental.

Note that the empty set E = ∅ is a transcendence basis if and only if K/k is an algebraic extension.

Example 1.4.1. Let K = k(t1, . . . , tn) be the ﬁeld of fractions of the polynomial ring k[t1, . . . , tn]. Then K is a purely transcendental extension of K, with transcendence basis E = {t1, . . . , tn}.

10 1.4 Transcendence Degree 1 Preliminaries

∼ Lemma 1.4.2. If {x1, . . . , xn} ⊂ K is algebraically independent then k(x1, . . . , xn) = k(t1, . . . , tn). Proof. Consider the surjection

ϕ : k[t1, . . . , tn] −→ k[x1, . . . , xn]

ti 7−→ xi.

Then ϕ is injective because {x1, . . . , xn} is algebraically independent. Moreover, ϕ is a k- algebra homomorphism so this extends to an isomorphism k(t1, . . . , tn) → k(x1, . . . , xn). Lemma 1.4.3. Let K/k be a ﬁeld extension and suppose E ⊆ K with |E| = n. Then

(a) If E is algebraically independent over k then there exists a transcendence basis E0 of K/k such that E0 ⊇ E.

(b) If E is ﬁnite and K = k(E) then the elements of E may be numbered such that k(x1, . . . , xr)/k is purely transcendental for some 0 ≤ r ≤ n and K/k(x1, . . . , xr) is algebraic.

(c) E is a transcendence basis for K/k if and only if K/k(E) is algebraic and K/k(E1) is not algebraic for any proper subset E1 ( E. Proof. (a) Apply Zorn’s Lemma (exercise). (b) Let E be numbered so that {x1, . . . , xr} is a maximal algebraically independent subset of E. Clearly {x1, . . . , xr} is a transcendence basis for k(x1, . . . , xr) so k(x1, . . . , xr)/k is purely transcendental. We will show that for each r + 1 ≤ j ≤ n, xj is algebraic over k(x1, . . . , xr). We know for each j, {x1, . . . , xr, xj} is algebraically dependent over k, so there exists a nonzero polynomial F ∈ k[t1, . . . , tr+1] such that F (x1, . . . , xr, xj) = 0. Write

d F = Fd(t1, . . . , tr)tr+1 + ... + F1(t1, . . . , tr)tr+1 + F0(t1, . . . , tr)

where Fi ∈ k[t1, . . . , tr] and Fd 6= 0. Then

d 0 = Fd(x1, . . . , xr)xj + ... + F1(x1, . . . , xr)xj + F0(x1, . . . , xr)

and Fd(x1, . . . , xr) 6= 0 since {x1, . . . , xr} is algebraically independent. We must therefore have d ≥ 1. This shows xj is algebraic over k(x1, . . . , xr) for each r + 1 ≤ j ≤ n, as we had claimed. Finally, it follows that K/k(x1, . . . , xr) is algebraic. (c) Pick any x ∈ K r E. Then since E is a maximal algebraically independent subset of K, E ∪ {x} is algebraically dependent over k. This means there exist x1, . . . , xr ∈ E such that {x1, . . . , xr, x} is algebraically dependent over k. Hence x must be algebraic over k(x1, . . . , xr) as in (b). In particular, x is algebraic over k(E). To show the second property, we may suppose without loss of generality that E1 = {x1, . . . , xr−1} so that K/k(x1, . . . , xr−1) is algebraic. Consider the tower K ⊃ k(E) ⊃ k(x1, . . . , xr−1). Then k(E) = k(x1, . . . , xr) is algebraic over k(x1, . . . , xr−1), which implies xr is algebraic over k. This clearly contradicts the algebraic independence of {x1, . . . , xr}. Hence K/k(E1) cannot be algebraic for any proper subset E1 ( E.

11 1.4 Transcendence Degree 1 Preliminaries

Going the other direction, suppose E satisﬁes the properties that K/k(E) is algebraic and K/k(E1) is not algebraic for any E1 ( E. By (b), we may number E = {x1, . . . , xr, . . . , xn} so that k(x1, . . . , xr)/k is purely transcendental and K/k(x1, . . . , xr) is algebraic. By assumption {x1, . . . , xr} cannot be a proper subset of E, so we get r = n and k(E)/k is purely transcendental. This implies E is algebraically independent over k. Finally, since K/k(E) is algebraic, for any x ∈ K r k(E), x is algebraic. So E ∪ {x} is not algebraically independent. This proves maximality of E, hence E is a transcendence basis.

Proposition 1.4.4. Assume E1,E2 ⊆ K are ﬁnite subsets with K/k(E1) algebraic and E2 algebraically independent over k. Then |E1| ≥ |E2|.

Proof. Let E1 = {x1, . . . , xm} and E2 = {y1, . . . , yn}, so that |E1| = m and |E2| = n. We claim that the elements of E1 can be numbered so that K/k(y1, . . . , y`, x`+1, . . . , xm) is algebraic for all 0 ≤ ` ≤ min{m, n}. We prove this by induction on `. For ` = 0, K/k(E1) is already algebraic by assumption. Now suppose we can number E1 so that K/k(y1, . . . , y`−1, x`, . . . , xm) is algebraic; set K`−1 = k(y1, . . . , y`−1, x`, . . . , xm). Now y` is algebraic over K`−1 since y` ∈ K. This implies there exist polynomials G0,...,Ge in m variables such that

e Ge(y1, . . . , y`−1, x`, . . . , xm)y` + ... + G0(y1, . . . , y`−1, x`, . . . , xm) = 0 and Ge(y1, . . . , y`−1, x`, . . . , xm) 6= 0. In particular, Ge must be nonzero. It follows that there exists a nonzero polynomial F in m + 1 variables such that

F (y1, . . . , y`, x`, . . . , xm) = 0 (∗)

We may assume F is such a polynomial of minimal degree. Now {y1, . . . , y`} is algebraically independent over k so one of x`, . . . , xm must occur in (∗). After renumbering, we may assume x` occurs in (∗), which then becomes

d Fd(y1, . . . , y`, x`+1, . . . , xm)x` + ... + F0(y1, . . . , y`, x`+1, . . . , xm) = 0 for polynomials Fi in m variables over k with Fd 6= 0. Then Fd(y1, . . . , y`, x`+1, . . . , xm) 6= 0 due to the minimality of deg F . This also implies d ≥ 1. Then deg F ≥ deg Fd + d. The above equation shows that x` is algebraic over K` := k(y1, . . . , y`, x`+1, . . . , xm). By induction, K/K`−1 is algebraic but since K`−1 ⊆ K`, we have now shown that K/K`(x`) and K`(x`)/K` are both algebraic extensions. By the tower property of algebraic extensions, K/K` is algebraic, proving the claim. If min{m, n} = n then n ≤ m and we’re done. On the other hand, if min{m, n} = m then we can use the above induction to conclude that K/k(y1, . . . , ym) is algebraic. But then we must have n = m since if n > m, ym+1 will be algebraic over k(y1, . . . , ym), in which case there is a nontrivial polynomial equation H(y1, . . . , ym+1) = 0. This however contradicts {y1, . . . , ym+1} being algebraically independent. In every case we have n ≤ m as desired.

Corollary 1.4.5. If E1 and E2 are transcendence bases for K/k then |E1| = |E2| or they are both inﬁnite. Proof. Apply Proposition 1.4.4 and (c) of Lemma 1.4.3.

12 1.4 Transcendence Degree 1 Preliminaries

This allows us to deﬁne the following:

Definition. The transcendence degree of a field extension K/k is defined as tr degk K = |E| if K/k admits a finite transcendence basis E, and tr degk K = ∞ otherwise.

Remark. tr degk K = 0 if and only if K/k is algebraic.

Corollary 1.4.6. The number r in Lemma 1.4.3(b) such that k(x1, . . . , xr)/k is purely transcendental and K/k(x1, . . . , xr) is algebraic is equal to tr degk K.

Proof. We proved that {x1, . . . , xr} is algebraically independent over k and {x1, . . . , xr, x} is algebraically dependent for any x ∈ K r {x1, . . . , xr}. Hence {x1, . . . , xr} is a maximal algebraically independent set in K, and thus a transcendence basis for K/k. Therefore

r = #{x1, . . . , xr} = tr degk K.

Example 1.4.7. For the purely transcendental extension k(t1, . . . , tn), its transcendence degree is tr degk k(t1, . . . , tn) = n. Proposition 1.4.8. For a tower of ﬁelds L ⊃ K ⊃ k in which each extension is ﬁnitely generated, we have

tr degk L = tr degK L + tr degk K.

Proof. Suppose E1 = {x1, . . . , xn} is a transcendence basis for K/k and E2 = {y1, . . . , ym} is a transcendence basis for L/K. Set E = E1 ∪ E2 = {x1, . . . , xn, y1, . . . , ym}. We ﬁrst show E is algebraically independent over k. Suppose f ∈ k[t1, . . . , tn+m] such that

f(x1, . . . , xn, y1, . . . , ym) = 0.

Then we may write X an+1 an+m f = fi1,...,im (t1, . . . , tn)tn+1 ··· tn+m i1,...,im where each fi1,...,im ∈ k[t1, . . . , tn], aj ≥ 0 and the sum is over all tuples (i1, . . . , im) ∈ m N . Notice that f(E1, tn+1, . . . , tn+m) ∈ K[tn+1, . . . , tn+m] so because E2 is algebraically independent over K and f(E1,E2) = 0, we must have f(E1, tn+1, . . . , tn+m) ≡ 0. Therefore each fi1,...,im (E1) = 0 but because E1 is algebraically independent over k, each fi1,...,im ≡ 0. Hence f ≡ 0, showing E is algebraically independent over k. To ﬁnish the proof, note that K/k(E1) and L/K(E2) are algebraic by Lemma 1.4.3(c), and we have a tower k(E) ⊂ K(E2) ⊂ L in which each intermediate extension is algebraic, so L/k(E) is algebraic by transitivity. Hence E is a transcendence basis for L/k. It follows that tr degk L = |E| = |E1| + |E2| = tr degk K + tr degK L.

Proposition 1.4.9. If k is a perfect ﬁeld, K = k(E) and tr degk K = r, then the elements of E can be numbered such that k(x1, . . . , xr)/k is purely transcendental and K/k(x1, . . . , xr) is a separable algebraic extension.

Proof. If char k = 0 then this follows from Corollary 1.4.6 and (b) of Lemma 1.4.3. If char k = p > 0 then k = kp and the result follows from Proposition 4.1 in Lang.

13 2 Integral Dependence

2 Integral Dependence

A critical concept in algebraic number theory and algebraic geometry is the idea of integrality: the property that an element (or a collection of elements) satisfies a polynomial identity with coefficients in a ring. In this chapter, we define integrality and explore the basic properties of integral extensions of rings, eventually proving the ‘going up’ and ‘going down’ theorems (Section 2.3) and Noether’s Normalization Lemma (Section 2.4). These results have vital consequences in later chapters.

2.1 Integral Extensions of Rings

In this section we deﬁne and outline the main results about integrality over a ring A. Let A ⊆ B be a ring extension, i.e. A is a subring of B.

Lemma 2.1.1. If M is a B-module which is ﬁnitely generated as an A-module, then for any x ∈ B, there exists a monic polynomial f(t) ∈ A[t] of degree at least 1 such that f(x)M = 0. Pn Proof. Let M = i=1 Ami. Since M is a B-module, there exist elements aij ∈ A for 1 ≤ i, j ≤ n, such that x satisﬁes the equations

n X xmi = aijmj for each 1 ≤ i ≤ n. j=1

Pn On the other hand, x also satisﬁes xmi = j=1 xδijmj for each i, so subtracting these equations gives us a system

n X (xδij − aij)mj = 0 for each 1 ≤ i ≤ n. j=1

Set cij = xδij − aij ∈ B and consider the matrix C = (cij) ∈ Mn(B). Then the system of equations above can be written     m1 0  .  . C  .  = . (∗) mn 0

From linear algebra, there exists an adjoint matrix adj(C) ∈ Mn(B) such that adj(C)C = (det C)In. Multiplying through (∗) on the left by adj(C) gives us     m1 0  .  . (det C)  .  = . mn 0

14 2.1 Integral Extensions of Rings 2 Integral Dependence

Since the mi generated M, this proves that (det C)M = 0. Now, expanding det C using the Leibniz formula for a determinant yields X det C = sgn(σ)cσ(1)1 ··· cσ(n)n

σ∈Sn n Y 0 n−2 0 0 = (x − aii) + an−2x + ... + a0 for some ai ∈ A i=1 n n−1 = x + an−1x + ... + a0 for some ai ∈ A, n n−1 after combining coefficients. Set f(t) = t + an−1t + ... + a0. Then f(x)M = 0. Definition. An element x ∈ B is integral over A if there exists a monic polynomial f(t) ∈ A[t] of degree at least 1 such that f(x) = 0. If every element in B is integral over A then we say B is integral over A, or B is an integral extension of A. Lemma 2.1.2. For x ∈ B the following are equivalent: (i) x is integral over A. (ii) A[x] is a finitely generated A-module. (iii) There exists a subring C ⊆ B containing A[x] such that C is finitely generated as an A-module. (iv) There exists a faithful A[x]-module M which is finitely generated as an A-module. n n−1 n Pn−1 i Proof. (i) =⇒ (ii) If x +an−1x +...+a1x+a0 = 0 for ai ∈ A then x ∈ M := i=0 Ax which is a finitely generated A-module. By induction, for all m ≥ n, xm ∈ M which proves A[x] = M. Hence A[x] is finitely generated. (ii) =⇒ (iii) Let C = A[x]. Then C is finitely generated as an A-module by assumption. (iii) =⇒ (iv) Let M = C. Then since C is a subring, 1A ∈ C and hence C is a faithful A-module. (iv) =⇒ (i) Lemma 2.1.1 implies that f(x)M = 0 for some monic polynomial f(t) ∈ A[t], but if M is faithful, f(x) = 0. Hence x is integral over A.

Corollary 2.1.3. Let B/A be a ring extension and suppose x1, . . . , xn ∈ B are integral over A. Then A[x1, . . . , xn] is a ﬁnitely generated A-module. Proof. We prove this by induction on n. The base case for n = 1 follows from (ii) of Lemma 2.1.2. For n ≥ 2, suppose C := A[x1, . . . , xn] is ﬁnitely generated over A. Then there exist elements y1, . . . , y` ∈ C such that ` X C = A[x1, . . . , xn−1] = Ayi. i=1 Pm−1 j By hypothesis, xn is integral over A and therefore over C as well. Then C[xn] = j=0 Cxn for some m ∈ N. Thus m−1 ` X X j A[x1, . . . , xn] = C[xn] = Ayixn, j=0 i=1

so in particular, A[x1, . . . , xn] is a ﬁnitely generated A-module.

15 2.1 Integral Extensions of Rings 2 Integral Dependence

Deﬁnition. For a ring extension B/A, the set A0 of elements x ∈ B that are integral over A is called the integral closure of A in B. If A0 = A, we say A is integrally closed in B. Further, if A0 = B then B/A is called an integral extension.

Example 2.1.4. By Corollary 2.1.3, A[x1, . . . , xn] is an integral extension of A for any set of algebraic integers x1, . . . , xn over A. Lemma 2.1.5. Let B/A be a ring extension. Then A0, the integral closure of A in B, is a subring of B which contains A. Proof. It is trivial that A ⊆ A0, for every a ∈ A is a root of t − a ∈ A[t]. Suppose x, y ∈ A0. Then by Corollary 2.1.3, A[x, y] is a ﬁnitely generated A-module which is a subring of B. Clearly x + y, xy ∈ A[x, y] so applying Lemma 2.1.2(iii) shows that x + y and xy are integral over A. Proposition 2.1.6. If B/A and C/B are integral extensions of rings, then C/A is also an integral extension. Proof. Take x ∈ C. We must show that x is integral over A. First, x is integral over B so there exist elements b0, . . . , bn−1 ∈ B such that

n n−1 x + bn−1x + ... + b0 = 0.

0 0 Then since B/A is integral, x is integral over B := A[b0, . . . , bn−1]. By Corollary 2.1.3, B is ﬁnitely generated as an A-module since all the bi are integral over A. Then there are 0 0 0 0 elements y1, . . . , y` ∈ B which generate B as an A-module. Also, B [x] is a B -module which is ﬁnitely generated by xj, 0 ≤ j ≤ m − 1. Then we have

m−1 m−1 ` 0 X 0 j X X j A[x] ⊆ B [x] = B x = Ayix . j=0 j=0 i=1 Thus by Lemma 2.1.2(iii), x is integral over A. This proves C/A is an integral extension. Corollary 2.1.7. If A0 is the integral closure of A in B then A0 is integrally closed in B. Proof. Take x ∈ B which is integral over A0. Then A0[x]/A0 is integral by Corollary 2.1.3 and A0/A is integral by definition of A0. By Proposition 2.1.6, this implies A0[x]/A is integral, so in particular x is integral over A. Thus x ∈ A0. Definition. An integral domain A is called integrally closed (or alternatively, normal) if A is integrally closed in its field of fractions. Proposition 2.1.8. Every UFD is integrally closed. Proof. Let A be a UFD and denote the field of fractions of A by K. Take x ∈ K and write a x = b for a, b ∈ A and b 6= 0; we may assume a and b are relatively prime, that is, there is no prime p ∈ A that divides both a and b. Suppose x is integral over A. Then there are elements a0, . . . , an−1 ∈ A such that an an−1 + a + ... + a = 0. b n−1 b 0

16 2.1 Integral Extensions of Rings 2 Integral Dependence

Multiplying through by bn gives us

n n a + an−1b + ... + a0b = 0.

Subtracting an to the other side shows that b | an in A, but since a and b are relatively prime, and therefore an and b are also relatively prime, this is only possible when b is a unit. −1 a −1 Hence b exists in A, so x = b = ab ∈ A, proving A is integrally closed. Example 2.1.9. Both the integers Z and the Gaussian integers Z[i] are Euclidean domains, so in particular they are UFDs and therefore integrally closed by Proposition 2.1.8.

Example 2.1.10. Let K be√ a quadratic extension of Q, i.e. a ﬁeld extension such that [K : Q] = 2. Then K = Q( d) for some squarefree integer d ∈ Z r {0, 1}, meaning d has prime factorization d = ±p1p2 ··· pr for primes pi. We want to compute the integral closure of Z in K. Since K/Q is Galois, there is one nontrivial Q-automorphism σ ∈ Gal(K/Q): σ : K −→ K √ √ a + b d 7−→ a − b d.

Using the norm and trace maps for K/Q, we have that

α ∈ K is integral over Z ⇐⇒ N(α) = ασ(α) and Tr(α) = α + σ(α) ∈ Z.

To verify this, suppose f(α) = 0 for a monic polynomial f(t) ∈ Z[t]. Then 0 = σ(f(α)) = f(σ(α)), so σ(α) is integral over Z. Since the integral elements form a ring, N(α) = ασ(α) and Tr(α) = α + σ(α) are also integral over Z. But the norm and trace maps take values in Q, so N(α), Tr(α) ∈ Q and they’re integral, so both lie in Z. Conversely, if ασ(α), α + σ(α) ∈ Z then α is a root of the monic polynomial

2 (t − α)(t − σ(α)) = t − (α + σ(α))t + ασ(α) ∈ Z[t]. Therefore α is integral. For any number ﬁeld K/Q, we let OK denote the integral closure of Z in K, called the ring of integers of K. In the quadratic case, our work above shows that √ 2 2 OK = {a + b d | a, b ∈ Q and a − b d, 2a ∈ Z} √ ( [ d] if d ≡ 2, 3 (mod 4) Z √ = h 1+ d i Z 2 if d ≡ 1 (mod 4).

√ 1+ d To see how the second case arises, suppose d ≡ 1 (mod 4). Then ω = 2 satisﬁes the 2 1−d integral expression ω − ω + 4 = 0.

17 2.2 Integrality and Field Extensions 2 Integral Dependence

2.2 Integrality and Field Extensions

In this section, let A and B be integral domains and let K be the ﬁeld of fractions of A.

Lemma 2.2.1. If B/A is an integral extension then A is a ﬁeld if and only if B is a ﬁeld.

Proof. ( =⇒ ) Assume A is a ﬁeld and choose x ∈ B r {0}. Then x is integral over A so n n−1 there exist elements a0, . . . , an−1 ∈ A such that x + an−1x + ... + a0 = 0. This can be rewritten as n−1 n−2 x(x + an−1x + ... + a1) = −a0 (∗) Choose such an expression for x with n minimal. By assumption, x 6= 0 and since n was n−1 n−2 minimal, x + an−1x + ... + a1 6= 0. Then since B is an integral domain, (∗) implies −1 a0 6= 0, and since A is a ﬁeld, a0 ∈ A exists. Divide through (∗) by −a0 to obtain

−1 −1 n−1 x := −a0 (x + ... + a1).

This shows x is invertible in B. ( ⇒ = ) Conversely, suppose B is a ﬁeld and take a ∈ A r {0}. Then a−1 ∈ B exists. Since B/A is integral, there exist elements ai ∈ A such that

−n −(n−1) −1 n−1 a + an−1a + ... + a0 = 0 ⇐⇒ a = −(an−1 + ... + a0a ) ∈ A.

This shows that a is invertible in A.

Lemma 2.2.2. Let A be integrally closed in its fraction ﬁeld K, let L/K be a ﬁnite extension and suppose α ∈ L is algebraic over K. Then α is integral over A if and only if α has a minimal polynomial f(t) ∈ A[t].

Proof. ( ⇒ = ) is the deﬁnition of integrality. ( =⇒ ) Take a splitting ﬁeld E of f over K. Then f splits into linear factors:

n Y f(t) = (t − αi) for αi ∈ E, α1 = α. i=1 Since α is integral, there are K-isomorphisms

σi : K(α) −→ K(αi), α 7→ αi, ∼ using the fact that K(αi) = K[t]/(f(t)) for each root αi. By assumption, there exists a monic polynomial g(t) ∈ A[t] such that g(α) = 0. Applying σi to this equation gives

0 = σi(g(α)) = g(σi(α)) = g(αi).

Qn Thus αi is integral over A for all 1 ≤ i ≤ n. It follows that f(t) = i=1(t − αi) ∈ K[t] has coeﬃcients which are integral over A. Finally, since A is integrally closed, this proves f ∈ A[t].

18 2.2 Integrality and Field Extensions 2 Integral Dependence

Remark. The most important rings in algebraic number theory are rings of integers: for a finite extension K/Q, OK denotes the integral closure of Z in K. By Lemma 2.2.2, we can write OK = {α ∈ K | fα ∈ Z[t]} where fα denotes the minimal polynomial of α over Q. Definition. For any finite field extension L/K of degree n = [L : K], there is a K-algebra homomorphism ∼ ϕ : L,→ EndK (L) = Mn(K)

α 7→ (ϕα : x 7→ αx).

The trace form of the extension L/K is deﬁned as

Tr = TrL/K : L −→ K

α 7−→ tr(ϕα).

By properties of the matrix trace tr, the trace form TrL/K is K-linear.

Lemma 2.2.3. A ﬁnite extension L/K is separable if and only if TrL/K 6= 0. Proof. We will only show the forward direction. Suppose L/K is separable. Then L = K(α) for some α ∈ L by the primitive element theorem. We claim that for some 0 ≤ i ≤ n − 1, Tr(αi) 6= 0. Let n Y f(t) = (t − αj) j=1

be the minimal polynomial of α over K. By separability, f has distinct roots α = α1, . . . , αn in some splitting ﬁeld E of f over K. By the Cayley-Hamilton theorem, the characteristic polynomial χ of ϕα satisﬁes 0 = χ(ϕ(α)) = ϕ(χ(α)). Thus χ(α) = 0 by injectivity of ϕ. Hence f(t) | χ(t) in K[t] but deg f = n = deg χ and both polynomials are monic, so we have f = χ. This implies that α1, . . . , αn are the distinct eigenvalues of ϕα in E. Let Mα be the Pn matrix representing ϕα in Mn(K). Then Tr(α) = tr(Mα) = j=1 αj. Taking powers of Mα, we have n i i X i Tr(α ) = tr(Mα) = αj. j=1 i Consider the n × n matrix C = (αj) where 0 ≤ i ≤ n − 1 and 1 ≤ j ≤ n. Then C ∈ Mn(E) and its determinant is a Vandermonde determinant:

Y 2 det C = (αj − α`) `

which is nonzero because the αj are distinct. In particular,

0 1  1 + ... + 1   Tr(1)  . .  .   .  . 6= C . =  .  =  .  . n−1 n−1 n−1 0 1 α1 + ... + αn Tr(α )

So Tr(αi) 6= 0 for some i. Therefore the trace form on L/K is nondegenerate.

19 2.2 Integrality and Field Extensions 2 Integral Dependence

When Tr 6= 0, one may deﬁne a nondegenerate, symmetric, K-bilinear form b : L×L → K by b(x, y) = TrL/K (xy).

Lemma 2.2.4. For any β ∈ L, the characteristic polynomial χ of ϕβ satisﬁes χ(t) = (f(t))[L:K(β)] where f(t) is the minimal polynomial of β over K. Therefore TrL/K (β) = [L : K(β)] TrK(β)/K (β).

Proof. Let ϕβ be the map L → L given by ϕβ(x) = βx. Set m = [L : K(β)] and n = [K(β): K]. If m = 1, then L = K(β) so deg f(t) = [L : K] = n. By Cayley-Hamilton, f(t) divides χ(t), but both are by deﬁnition monic polynomials and deg χ(t) = n, so we have χ(t) = f(t). In the general case, let {x1, . . . , xr} be a basis of K(β)/K and let {y1, . . . , ym} be a basis of L/K(β). We know the set {xiyj : 1 ≤ i ≤ r, 1 ≤ j ≤ m} is a basis for L/K. Let B = (bk`) Pr be the standard matrix for ϕβ in the extension K(β)/K. Then βxi = k=1 bkixk and thus r X β(xiyj) = bki(xkyj). k=1

If we write the basis {xiyj} as {x1y1, x2y1, . . . , xry1, x1y2, . . . , xrym} then ϕβ has the following standard matrix in L/K:   B 0  B     ..   0 .  B There are m blocks, each of which is B, so χ(t) = [det(B − tI)]m but by the m = 1 case, det(B − tI) = f(t). Therefore χ(t) = (f(t))m as claimed. Proposition 2.2.5. If A is an integrally closed integral domain with K its fraction ﬁeld, and L/K is a ﬁnite separable extension, then for the integral closure B of A in L, there Pn exists a K-basis {v1, . . . , vn} of L such that B ⊆ i=1 Avi. Proof. If x ∈ L then we claim there is some a ∈ A r {0} such that ax ∈ B. Since x is algebraic, there exist q0, . . . , qn−1 ∈ K such that

n n−1 x + qn−1x + ... + q0 = 0.

Then a may be chosen to be the common denominator of the qi, for which we have

n n−1 n (ax) + aqn−1(ax) + ... + a q0 = 0. Each coeﬃcient of this expression now lies in A, so ax is integral over A and thus lies in B. Now this implies that there is a K-basis {u1, . . . , un} of L with ui ∈ B for each 1 ≤ i ≤ n. Since L/K is separable, Lemma 2.2.3 says that TrL/K 6= 0 and so b(·, ·) is nondegenerate. Then there exists a dual basis {v1, . . . , vn} of L with respect to b(·, ·). Let x ∈ B. Then we Pn can write x = j=1 cjvj for cj ∈ K. We ﬁnish by showing each cj ∈ A. Consider

n ! n X X Tr(uix) = b(ui, x) = b ui, cjvj = cjb(ui, vj) = ci. j=1 j=1

20 2.3 Integrality, Ideals and Localization 2 Integral Dependence

Now each ui, x ∈ B so uix ∈ B and thus uix is integral over A. Up to sign, Tr(uix) is a coefficient of the characteristic polynomial χ of ϕ(uix). By Lemma 2.2.4, χ is a power of the minimal polynomial of uix, which according to Lemma 2.2.2 has coefficients in A. Hence Pn χ ∈ A[t] so Tr(uix) = ci ∈ A. This shows that x ∈ i=1 Avi as desired. Corollary 2.2.6. If A is a PID and L/K is a finite separable extension, where K is the fraction field of A, then the integral closure B of A in L is a finitely generated free A-module of rank n = [L : K].

Proof. First, by Proposition 2.1.8, every PID is integrally closed so we may apply Proposi- tion 2.2.5 and its proof to produce K-bases {u1, . . . , un} and {v1, . . . , vn} of L such that

n n X X Aui ⊆ B ⊆ Avi. i=1 i=1

The vi form a K-basis, so in particular they are linearly independent over A and thus Pn i=1 Avi is a free A-module of rank n. By the theory of modules over a PID, we have that B is free of rank at most n over A, but reversing the roles of the ui and vi, we see that the rank of B equals n.

Example 2.2.7. If K/Q is ﬁnite, then there exists a Q-basis of K, say {α1, . . . , αn}, which is a Z-basis of OK . Such a basis is called an integral basis of K (or of OK ). In particular, Corollary 2.2.6 shows that OK is a free abelian group of rank√ n = [K : Q]. In the special case of a quadratic extension, K = Q( d), the ring of integers is OK = ⊕ ω, with Z Z (√ d if d ≡ 2, 3 (mod 4) √ ω = 1+ d 2 if d ≡ 1 (mod 4).

Thus an integral basis of OK is {1, ω}. Remark. Proposition 2.2.5 and Corollary 2.2.6 are not true in general if L/K is not a separable extension, as we will show in Section 4.5.

2.3 Integrality, Ideals and Localization

In this section we develop the theory of ideals in integral extensions and explore in particular how these notions interact with localization. We will prove the famous ‘going up’ and ‘going down’ theorems.

Lemma 2.3.1. Assume B/A is a ring extension with I ⊂ A and J ⊂ B ideals such that J ∩ A = I. Then

(a) B/J is integral over A/I.

(b) If I and J are prime ideals then I is maximal in A if and only if J is maximal in B.

21 2.3 Integrality, Ideals and Localization 2 Integral Dependence

Proof. (a) We have a sequence A,→ B B/J where the kernel of A → B/J is A ∩ J = I. Therefore A/I injects into B/J so the ring extension (B/J)/(A/I) makes sense. Take x¯ ∈ B/J withx ¯ = x + J for some x ∈ B. Then there exist ai ∈ A such that

n n−1 x + an−1x + ... + a0 = 0 mod J n n−1 =⇒ x¯ +a ¯n−1x¯ + ... +a ¯0 = 0. The coefficients of the second equation are in A/I, so we see thatx ¯ is integral over A/I. (b) If I and J are prime ideals then A/I and B/J are integral domains. By (a), B/J is integral over A/I, and by Lemma 2.2.1, B/J is a field exactly when A/I is a field. Hence J is maximal in B if and only if I is maximal in A. Lemma 2.3.2. Let B/A be a ring extension and S ⊆ A be a multiplicatively closed subset. (a) If B/A is integral then S−1B/S−1A is integral.

(b) If C is the integral closure of A in B then S−1C is the integral closure of S−1A in S−1B.

−1 −1 a a Proof. First, there is a canonical map S A → S B, s 7→ s which is a well-deﬁned, one- to-one ring homomorphism. As a consequence, we can regard S−1A as a subring of S−1B. x −1 (a) Let s ∈ S B where x ∈ B and s ∈ S. Given that x is integral over A, there exist a0, . . . , an−1 ∈ A such that

n n−1 x + an−1x + ... + a0 = 0 in B. Applying the map j : B → S−1B to this expression and dividing by sn gives us xn a xn−1 a + n−1 + ... + 0 = 0 in S−1B. s s s sn −1 x −1 Since each coeﬃcient lies in S A, this shows that s is integral over S A and thus the extension S−1B/S−1A is integral. (b) We have A ⊆ C ⊆ B so by the preliminary remark, we may consider the tower of ring extensions S−1A ⊆ S−1C ⊆ S−1B. Since C/A is integral, part (a) implies S−1C/S−1A is integral. If x ∈ S−1B is integral over S−1A, there exist fractions a0 ,..., an−1 ∈ S−1A such s s0 sn−1 that xn a xn−1 a + n−1 + ... + 0 = 0. s sn−1 s s0 Multiplying through by sn and the common denominator, we can write

0 n 0 n−1 0 −1 t x + an−1x + ... + a0 = 0 in S B

0 0 00 where t ∈ S and ai ∈ A. Then there is some t ∈ S such that

00 0 n 0 n−1 0 t (t x + an−1x + ... + a0) = 0 in B. Setting t = t0t00 ∈ S, we see that

n 00 n−1 00 00 tx + an−1x + ... + a0 = 0 in B, where ai ∈ A.

22 2.3 Integrality, Ideals and Localization 2 Integral Dependence

Multiplying through by tn−1 gives us

n 00 n−1 00 n−1 (tx) + an−1(tx) + ... + a0t = 0.

x tx −1 Therefore tx is integral over A, so tx ∈ C. This means s = ts ∈ S C, so we conclude that S−1C is the integral closure of S−1A in S−1B. Note that Lemma 2.3.2(b) does not imply that if B/A is an integral extension of rings with Q ⊂ B prime and Q ∩ B = p, then BQ/Ap is integral. Lemma 2.3.3. Suppose B/A is an integral extension of rings. If Q, Q0 ⊂ B are prime ideals with Q ⊆ Q0 and Q ∩ A = Q0 ∩ A, then Q = Q0.

Proof. Set p = Q ∩ A = Q0 ∩ A. Then p is a prime ideal of A. If S = A r p then by −1 −1 Lemma 2.3.2(a), S B/Ap is an integral extension. By Corollary 1.3.9, S p is the unique 0 0 maximal ideal of the local ring Ap. Notice that S ∩ Q = S ∩ A ∩ Q = S ∩ p = ∅, and likewise S ∩Q = ∅. Therefore S−1Q ⊆ S−1Q0 are prime ideals of S−1B by Proposition 1.3.8. 1 −1 0 −1 0 −1 −1 −1 0 −1 −1 Also, 1 6∈ S Q so S Q ∩ S A 6= S A. Thus S Q ∩ S A = S p, and likewise S−1Q ∩ S−1A = S−1p. Since S−1p is maximal in S−1A, Lemma 2.3.1(b) implies S−1Q and S−1Q0 are both maximal ideals in S−1B. Therefore S−1Q = S−1Q0 so Proposition 1.3.8(c) shows that we must have Q = Q0.

Lemma 2.3.4. Suppose B/A is an integral extension of rings and p ⊂ A is a prime ideal. Then there exists a prime ideal Q ⊂ B with Q ∩ A = p.

Proof. Let S = A r p and consider the commutative diagram A B

i j

−1 −1 Ap = S A S B

Choose any maximal ideal m ⊂ S−1B and set Q = j−1(m) which is a prime ideal of B by Proposition 1.3.8. We claim that Q ∩ A = p. Since B is integral over A, Lemma 2.3.2(a) −1 −1 −1 implies S B is integral over S A = Ap. Then by Lemma 2.3.1(b), m maximal in S B −1 implies m ∩ Ap is maximal in Ap. But Ap is a local ring, so m ∩ Ap = S p. Since the ﬁrst diagram commutes, we have another commutative diagram: Q ∩ A Q

i j

−1 S p = m ∩ Ap m

−1 −1 −1 −1 Therefore Q ∩ A = j (m) ∩ A = i (m ∩ Ap) = i (S p). Finally, Proposition 1.3.8(b) shows that Q ∩ A = p as desired.

23 2.3 Integrality, Ideals and Localization 2 Integral Dependence

Lemmas 2.3.3 and 2.3.4 together show that in an integral extension B/A, any prime ideal p ⊂ A lifted to B is contained in a prime ideal of B which is unique in any chain of ideals of B. This generalizes in an important way, in what is known as the going up theorem. Theorem 2.3.5 (Going Up). Suppose B/A is an integral extension of rings and

p0 ( p1 ( ··· ( pn is a chain of prime ideals in A, in which each inclusion is strict. Then for any chain of prime ideals Q0 ( Q1 ( ··· ( Qm−1 in B such that 0 ≤ m ≤ n and Qi ∩ A = pi for each 0 ≤ i ≤ m − 1, there exists a prime ideal Qm ⊂ B such that Qm−1 ( Qm and Qm ∩ A = pm.

Proof. When m = 0, Lemma 2.3.4 says there exists a prime ideal Q0 ⊂ B with Q0 ∩ A = p0. Now to induct, suppose the theorem holds for some m, 1 ≤ m ≤ n. Then Qm−1 ∩ A = pm−1 so consider the extension of quotient rings ¯ ¯ A = A/pm−1 ,−→ B = B/Qm−1. ¯ ¯ By Lemma 2.3.1(a), B/A is an integral extension. By the correspondence theorem, p¯m := ¯ ¯ pm/pm−1 is a prime ideal of A. Then by Lemma 2.3.4, there exists a prime ideal Qm ⊂ B ¯ −1 ¯ such that Qm ∩A = p¯m. Let Qm be the preimage π (Qm) where π : B → B is the canonical projection. Clearly Qm ) Qm−1 since p¯m 6= 0. Finally, the above work implies Qm ∩A = pm. Hence the statement holds by induction. Deﬁnition. If A is a ring and p ⊂ A is prime, the height of p is deﬁned as

ht(p) = sup{` ≥ 0 | there is a chain of prime ideals p0 ( p1 ( ··· ( p`−1 ( p}. The Krull dimension of A is then deﬁned by

dim A = sup{ht(p) | p ⊂ A is prime}.

Remark. If p is not maximal then p ( m for some maximal ideal m ⊂ A, so we can reformulate the Krull dimension deﬁnition in one of the following ways:

dim A = sup{ht(m) | m ⊂ A is maximal}

= sup{` ≥ 0 | p0 ( p1 ( ··· ( p` is a chain of prime ideals}. Examples. 1 If K is a ﬁeld then dim K = 0 since (0) is the only prime ideal.

2 If A is an integral domain, then dim A = 0 if and only if A is a ﬁeld. However, there exist non-domains which have dimension 0.

3 If A is a PID but not a ﬁeld, then all nonzero prime ideals are of the form (π) for a prime element π ∈ A. This shows that dim A = 1.

24 2.3 Integrality, Ideals and Localization 2 Integral Dependence

Lemma 2.3.6. Suppose p ⊂ A is a prime ideal. Then

(a) ht(p) = dim Ap. (b) ht(p) + dim A/p ≤ dim A.

Proof. (a) By Proposition 1.3.8(c), prime ideals of Ap are in bijective, inclusion-preserving correspondence with primes Q ⊂ A intersecting S = A r p trivially, that is, primes Q ⊆ p. Therefore any chain p0 ( p1 ( ··· ( pn−1 ( p of prime ideals in A corresponds to a −1 −1 −1 −1 −1 chain S p0 ( S p1 ( ··· ( S pn−1 ( S p of prime ideals in Ap = S p; this proves −1 dim Ap ≥ ht(p). Conversely, any chain Q0 ( Q1 ( ··· ( Q`−1 ( S p of prime ideals in −1 −1 −1 Ap corresponds to a chain j (Q0) ( j (Q1) ( ··· ( j (Q`−1) ( p in A; this establishes ht(p) ≥ dim Ap. Hence ht(p) = dim Ap. (b) Suppose dim A/p = ` and let Q0 ( Q1 ( ··· ( Q` be a maximal chain of prime ideals in A/p. By the correspondence theorem, this lifts to a chain of prime ideals

Q0 ( Q1 ( ··· ( Q` (∗)

in A such that each Qi ⊇ p. Of course we can always add p to the bottom of such a chain, so if ` is maximal, we must have Q0 = p. Now for any chain of prime ideals p0 ( p1 ( ··· ( pn−1 ( p in A, we can extend (∗) by this new chain, giving

p0 ( p1 ( ··· ( pn−1 ( p = Q0 ( Q1 ( ··· ( Q`, a chain of prime ideals in A. This shows dim A ≥ n + `. In particular, this holds for any such n ≤ ht(p), so we have

dim A ≥ ht(p) + ` = ht(p) + dim A/p.

Theorem 2.3.7. If B is an integral extension of A then dim A = dim B.

Proof. If Q0 ( Q1 ( ··· ( Q` is a chain of prime ideals in B, then

Q0 ∩ A ( Q1 ∩ A ( ··· ( Q` ∩ A is a chain of prime ideals in A – the strictness of each containment comes from Lemma 2.3.3. This implies dim B ≤ dim A. On the other hand, if p0 ( p1 ( ··· ( p` is a chain of prime ideals in A, then by the going up theorem, we get a chain of prime ideals

Q0 ( Q1 ( ··· ( Q`

in B, with Qi ∩ A = pi for each i. Hence dim A ≤ dim B so we have dim A = dim B. There is an analog of the going up theorem for descending chains of prime ideals, but it requires additional conditions. First we prove two important lemmas. Lemma 2.3.8. For a ring extension B/A and any prime ideal p ⊂ A, there exists a prime ideal Q ⊂ B with the property that Q ∩ A = p if and only if pB ∩ A ⊆ p.

25 2.3 Integrality, Ideals and Localization 2 Integral Dependence

Proof. ( =⇒ ) If Q ⊂ B exists and satisﬁes Q ∩ A = p, then pB ∩ A = (Q ∩ A)B ∩ A ⊆ Q ∩ A = p. ( ⇒ = ) Set T = A r p. By assumption, pB ∩ A ⊆ p implies (pB ∩ A) ∩ T = ∅. Therefore T −1(pB) 6= T −1B so there is a maximal ideal m ⊂ T −1B such that T −1(pB) ⊆ m. By Proposition 1.3.8, there is a prime ideal Q ⊂ B such that Q ∩ T = ∅ and m = T −1Q. Then p ⊆ pB ∩ A ⊆ Q ∩ A = Q ∩ p ⊆ p so it follows that p = Q ∩ A as required.

Lemma 2.3.9. Suppose A and B are integral domains with B/A an integral extension of rings, K the ﬁeld of fractions of A and A integrally closed (in K). If p ⊂ A is a prime ideal of A and α ∈ pB is any element, then the nonleading coeﬃcients of the minimal polynomial f of α over K lie in p.

Proof. By Lemma 2.2.2, the coeﬃcients of f must lie in A. Let α = b1p1 + ... + bkpk for bi ∈ B, pi ∈ p. Replacing B with A[b1, . . . , bk], we may assume B is a ﬁnitely generated Pn A-module generated by x1, . . . , xn. For each 1 ≤ i ≤ n, write αxi = j=1 aijxj with aij ∈ p. Pn Then j=1(δijαxi −aij) = 0. Multiplying by the classical adjoint of the matrix (δijαxi −aij), we get det(δijαxi − aij)xj = 0 for all 1 ≤ j ≤ n.

Therefore αxi satisfies a monic polynomial with coefficients in p. Since this holds for every generator xi, it follows that α itself satisfies such a polynomial, say g ∈ p[t]. If f is the minimal polynomial of α over K, f must divide g, say g = fh for h ∈ A[t]. Reducing modulo p, we get tdeg g =g ¯ = f¯h¯ ∈ (A/p)[t]. Since p is prime, A/p is a domain, so f¯ and h¯ must also be powers of t. Thus all nonleading terms of f lie in p.

Theorem 2.3.10 (Going Down). Suppose B/A is an integral extension of integral domains with A integrally closed. If p0 ) p1 ) ··· ) pn is a chain of prime ideals in A and

Q0 ) Q1 ) ··· ) Qm−1

is a chain of prime ideals in B such that 0 ≤ m ≤ n and Qi ∩ A = p for each i ≤ m − 1, then there is a prime ideal Qm ⊂ B such that Qm ( Qm−1 and Qm ∩ A = pm. Proof. If m = 0, this is given to us by Lemma 2.3.4. Otherwise assume 1 ≤ m ≤ n. Then 0 −1 since B is an integral domain, the map B → BQm−1 is injective. Set B = BQm−1 = S B, 0 where S = B r Qm−1. Then we get a tower of ring extensions A ⊆ B ⊆ B . It suﬃces to 0 0 0 show there exists a nonzero prime ideal Q ⊂ B such that Q ∩A = pm, for then we will have 0 0 pm = Q ∩ A = (Q ∩ B) ∩ A = Qm ∩ A for some prime ideal Qm ⊂ B, by Proposition 1.3.8. 0 0 0 Set p = pm. By Lemma 2.3.8, showing the existence of Q ⊂ B satisfying Q ∩ A = p is equivalent to showing pB0 ∩ A ⊆ p. If pB0 ∩ A = 0, we are done. Otherwise, take a nonzero element a ∈ pB0 and write a = αs for some α ∈ pB and s ∈ S. Also let f be the minimal

26 2.3 Integrality, Ideals and Localization 2 Integral Dependence

n n−1 polynomial of α over K. Then Lemma 2.3.9 implies f = t + cn−1t + ... + c0 for ci ∈ p. Since a ∈ A, we may write 1 c c f(at) = tn + n−1 tn−1 + ... + 0 = tn + c0 tn−1 + ... + c0 , an a an n−1 0

0 ci 1 with ci = an−i . Then an f(at) is also irreducible, and plugging in t = s gives 1 1 f(as) = f(α) = 0. an an

1 0 It follows that an f(at) is the minimal polynomial of s over K. Thus each ci lies in A for 0 i 0 0 ≤ i ≤ n − 1. Suppose that a 6∈ p. Then since p is prime, cn−ia = cn−i ∈ p implies cn−i ∈ p for each 0 ≤ i ≤ n − 1. Since s ∈ B, we get

n 0 n−1 0 0 s = −cn−1s − ... − c0 ∈ pB ⊆ pm−1B ⊆ Qm−1 by hypothesis. Then s ∈ Qm−1 since Qm−1 is a prime ideal, but this contradicts s ∈ S = 0 B r Qm−1. Hence a ∈ p, implying pB ∩ A ⊆ p after all. For the remainder of the section, let A be an integral domain with ﬁeld of fractions K. Whenever S ⊆ Ar{0}, we can consider the local ring S−1A as a subring of K. In particular, Ap ⊆ K for every prime p ⊂ A. \ Lemma 2.3.11. Any integral domain A can be written A = Am. maximal ideals m⊂A

Proof. Let D be the intersection of Am over all maximal ideals m ⊂ A. Then D ⊆ K. For a ﬁxed x ∈ D, consider the ideal I = {a ∈ A | ax ∈ A} ⊆ A. If m is a ﬁxed maximal ideal b of A, by assumption x ∈ Am so there exist elements a, b ∈ A with a 6∈ m such that x = a . It follows that ax = b ∈ A, so a ∈ I. Since we took a 6∈ m, this implies I 6⊂ m. Now m was arbitrary, so we must have I = A. Thus 1 ∈ I, implying x = 1 · x ∈ A and therefore A = D as required.

Proposition 2.3.12. For an integral domain A, the following are equivalent:

(i) A is integrally closed.

(ii) Ap is integrally closed for every prime ideal p ⊂ A.

(iii) Am is integrally closed for every maximal ideal m ⊂ A.

Proof. (i) =⇒ (ii) Take p ⊂ A to be a prime ideal and set S = A r p. By assumption, A −1 is the integral closure of A in K, and by Lemma 2.3.2(b), S A = Ap is integrally closed in K, which is also the ﬁeld of fractions of Ap. (ii) =⇒ (iii) is trivial. (iii) =⇒ (i) If m ⊂ A is a maximal ideal, we have A ⊆ Am ⊆ K. Then the assumption that Am is integrally closed implies that the integral closure of A in K lies in Am for any such m. Therefore by Lemma 2.3.11, A is integrally closed.

27 2.4 Normalization 2 Integral Dependence

2.4 Normalization

Let k be a ﬁeld and A a ﬁnitely generated k-algebra, but not necessarily an integral domain.

Deﬁnition. Elements a1, . . . , an ∈ A are said to be algebraically dependent over k if there exists a nonzero polynomial f ∈ k[t1, . . . , tn] such that f(a1, . . . , an) = 0. Otherwise the ai are said to be algebraically independent over k.

If a1, . . . , an ∈ A are algebraically independent, then the k-algebra homomorphism

ϕ : k[t1, . . . , tn] −→ A

ti 7−→ ai ∼ is injective. As a consequence, k[a1, . . . , an] = k[t1, . . . , tn] as k-algebras. Theorem 2.4.1 (Noether’s Normalization Lemma). If A is a ﬁnitely generated k-algebra with A = k[x1, . . . , xn] for xi ∈ A, then there exist algebraically independent elements y1, . . . , yr ∈ A, 0 ≤ r ≤ n, such that A/k[y1, . . . , yr] is an integral extension. Moreover, Pn if k is inﬁnite, we may choose the yi in j=1 kxj. Proof. We prove the theorem by induction on n. Starting with n = 1, there are two possibilities:

x1 is transcendental over k, in which case we may take r = 1 and y1 = x1.

x1 is algebraic over k, in which case we may take r = 0, so that k[x1]/k is integral.

Now suppose n ≥ 2. If x1, . . . , xn are already algebraically independent, take r = n and yi = xi for each 1 ≤ i ≤ n. If not, we claim that there exist elements z1, . . . , zn−1 ∈ A (and Pn zi ∈ j=1 kxj if k is inﬁnite) such that the following hold:

(i) k[z1, . . . , zn−1, xn] = A.

(ii) xn is integral over k[z1, . . . , zn−1].

Once we prove the claim, the proof finishes as follows. Set B = k[z1, . . . , zn−1] ⊆ A. Then A = B[xn] and A/B is integral by (i) and (ii). By induction, there exist y1, . . . , yr ∈ B (and Pn−1 Pn if k is infinite, yi ∈ j=1 kzj ⊆ j=1 kxj) such that the yi are algebraically independent over k and B is integral over k[y1, . . . , yr]. It will then follow that A/k[y1, . . . , yr] is an integral extension by Proposition 2.1.6. To prove the existence of z1, . . . , zn−1 satisfying (i) and (ii), assume k is infinite. By assumption there is a nonzero polynomial f ∈ k[t1, . . . , tn] such that f(x1, . . . , xn) = 0. Pd Write f = i=0 fi(t1, . . . , tn) where each fi is homogeneous of (total) degree i and fd 6= 0. (Since f0 ∈ k, we know d ≥ 1.) We can write fd in two ways:

X j1 jn fd(t1, . . . , tn) = cJ t1 ··· tn where cJ ∈ k n J=(j1,...,jn)∈N0 j1+...+jn=d d X d−j fd(t1, . . . , tn) = Gj(t1, . . . , tn−1)tn where each Gj is homogeneous of degree i. j=0

28 2.4 Normalization 2 Integral Dependence

Since fd 6= 0, not all of the Gj are 0. This implies fd(t1, . . . , tn−1, 1) 6= 0. Since k is inﬁnite, there exist scalars λ1, . . . , λn−1 ∈ k such that fd(λ1, . . . , λn−1, 1) 6= 0. Plugging these into the top expression for fd, we get

X j1 jn−1 j1+...+jn fd(λ1xn, . . . , λn−1xn, xn) = cJ λ1 ··· λn−1 xn J

X j1 jn−1 d d = cJ λ1 ··· λn−1 xn = fd(λ1, . . . , λn−1, 1)xn. J

Now set zi = xi − λixn for each 1 ≤ i ≤ n − 1. Then k[z1, . . . , zn−1, xn] = k[x1, . . . , xn] and

d X 0 = f(x1, . . . , xn) = fj(x1, . . . , xn) j=0 d X = fj(z1 + λ1xn, . . . , zn−1 + λn−1xn, xn) j=0

= fd(z1 + λ1xn, . . . , zn−1 + λn−1xn, xn) + lower degree terms

X j1 jn−1 jn = cJ (z1 + λ1xn) ··· (zn−1 + λn−1xn) xn + lower degree terms J

= fd(λ1xn, . . . , λn−1xn, xn) + lower terms in xn d−1 d X j = fd(λ1, . . . , λn−1, 1)xn + Hj(z1, . . . , zn−1)xn j=1

d × where fd(λ1, . . . , λn−1, 1)xn ∈ k and each Hj ∈ k[t1, . . . , tn−1]. This shows xn satisfies a nontrivial monic polynomial equation with coefficients in B = k[z1, . . . , zn−1], since in particular we can divide out by fd(λ1, . . . , λn−1, 1) 6= 0. Thus xn is integral over B. If k is finite, we need a slightly different argument. By the assumption on x1, . . . , xn, there is a nonzero polynomial F ∈ k[t1, . . . , tn] such that F (x1, . . . , xn) = 0. Write

X j1 jn F (t1, . . . , tn) = cjt1 ··· tn n j∈N0

n for cj ∈ k for all j = (j1, . . . , jn) ∈ N0 . Set J = {j | cj 6= 0}, which is a ﬁnite set since F is a n polynomial. We claim there exists a tuple m = (m1, . . . , mn) ∈ N0 such that for any distinct j, j0 ∈ J, jm 6= j0m. To justify such a claim, let N ∈ N be greater than any component n−1 n−2 ji of any tuple j = (j1, . . . , jn) ∈ J. Then m = (N ,N ,...,N, 1) satisﬁes the desired conclusion, since for N large enough, every base N expansion of a tuple j ∈ J is unique. mi For each 1 ≤ i ≤ n − 1, set zi = xi − xn . Then k[z1, . . . , zn−1, xn] = k[x1, . . . , xn] so (i)

29 2.4 Normalization 2 Integral Dependence

is satisﬁed. Moreover, we have

X j1 jn−1 jn 0 = F (x1, . . . , xn) = cjx1 ··· xn−1 xn j∈J

X m1 j1 mn−1 jn−1 jn = cj(z1 + xn ) ··· (zn−1 + xn ) xn j∈J

X m1j1 mn−1jn−1 jn = cjxn ··· xn xn + lower terms in xn j∈J

X m1j1+...+mn−1jn−1+mnjn = cjxn + lower terms in xn j∈J X jm = cjxn + lower terms, j∈J where the lower terms at the end have total degree strictly bounded by m. By our choice jm of m, we have that the powers xn over all j ∈ J are distinct, so in particular there is a ∗ ∗ ∗ maximum exponent j m, j ∈ J and for this j , cj∗ 6= 0. This proves xn satisﬁes a nonzero polynomial in k[z1, . . . , zn−1][t] which may be made monic by dividing out by cj∗ 6= 0. Hence xn is integral over k[z1, . . . , zn−1].

Remark. If A is an integral domain with field of fractions K = k(x1, . . . , xn), the r in the normalization lemma equals tr degk K. We use Noether’s normalization lemma to prove the fundamental theorem in algebraic geometry, Hilbert’s Nullstellensatz. This first version is stated in terms of algebraic extensions of fields; we will prove other versions later. Theorem 2.4.2 (Hilbert’s Nullstellensatz, Algebraic Version). If K/k is a field extension such that K is a finitely generated k-algebra, then K/k is a finite algebraic extension.

Proof. Noether’s normalization lemma says that there are yi ∈ K such that K/k[y1, . . . , yr] is integral. Then by Lemma 2.2.1, K a field implies k[y1, . . . , yr] is also a field. However, ∼ k[y1, . . . , yr] = k[t1, . . . , tr] which is not a field unless r = 0. Thus no such yi exist, so K/k is an integral extension, which is obviously equivalent to K/k being algebraic. Finally, the extension is of finite degree because K is finitely generated as a k-algebra.

n Deﬁnition. For k a ﬁeld and a = (a1, . . . , an) ∈ k , the ideal associated to a is

ma := (t1 − a1, . . . , tn − an) ⊂ k[t1, . . . , tn].

Remark. Notice that for each 1 ≤ i ≤ n, ti ≡ ai (mod ma), so for any polynomial F ∈ k[t1, . . . , tn], F (t1, . . . , tn) ≡ F (a1, . . . , an) (mod ma).

Therefore k[t1, . . . , tn] = k + ma. This deﬁnes an evaluation homomorphism

ϕa : k[t1, . . . , tn] −→ k

F (t1, . . . , tn) 7−→ F (a1, . . . , an).

30 2.4 Normalization 2 Integral Dependence

One can check that ϕa is a k-algebra homomorphism. Clearly ma ⊆ ker ϕa, and by k[t1, . . . , tn] = n k + ma, we get ma = ker ϕa and this is a maximal ideal of k[t1, . . . , tn]. Also, if a 6= b ∈ k then ma 6= mb. This deﬁnes an injection n k ,−→ MaxSpec(k[t1, . . . , tn])

a 7−→ ma where MaxSpec(A) = {maximal ideals of A} for any ring A. This mapping is not surjective in general, but in the case that k is algebraically closed, the Nullstellensatz (2.4.2) implies that it is surjective. Corollary 2.4.3. If k is algebraically closed, then the mapping n k −→ MaxSpec(k[t1, . . . , tn]) is bijective.

Proof. We saw that if a 6= b then ma 6= mb, so it remains to show surjectivity. Let m be a maximal ideal of A := k[t1, . . . , tn]. Then the quotient ring K = A/m is a ﬁeld which is ﬁnitely generated as a k-algebra by images of the monomials t¯i = ti + m in A/m. By the Nullstellensatz (2.4.2), K/k is algebraic, but since k was assumed to be algebraically closed, this implies K = k. Therefore for all 1 ≤ i ≤ n, there exists an ai ∈ k such that t¯i = ai =a ¯i in K. Equivalently, ti ≡ ai (mod m) so ti − ai ∈ m for all i. This proves ma ⊆ m for a = (a1, . . . , an), but we saw that ma is always maximal. Thus ma = m. We conclude that kn → A is both surjective and injective.

By Corollary 2.4.3, if k is algebraically closed, every proper ideal I ⊂ k[t1, . . . , tn] has n a common zero in k , called a Nullstelle in German. Explicitly, I ⊆ ma for some maximal n ideal ma deﬁned by a point a ∈ k . Lemma 2.4.4. If A is a ring in which every prime ideal P which is not maximal is an intersection of prime ideals properly containing P , then every prime ideal is an intersection of maximal ideals. Proof. Suppose to the contrary that p ⊂ A is a prime ideal that is not an intersection of maximal ideals. In particular, p is not maximal so by hypothesis, \ p = {Q ⊂ A | Q prime, Q ) p}. Consider the integral domain B = A/p. Then J(B) 6= 0, where J(B) denotes the Jacobson radical of B, so there is a nonzero element x ∈ J(B). Let S = {xn | n ≥ 0} be the −1 multiplicatively closed subset of powers of x in B, and consider the localization Bx = S B. Since p is prime, A/p is reduced, i.e. N(A/p) = 0, but this means x 6∈ N(B). In particular, x is not nilpotent, so Bx is nonzero. Let m ⊂ Bx be a maximal ideal and let Q be its corresponding prime ideal in B. If Q were itself maximal, we would have x ∈ Q but then Q ∩ S 6= ∅. Therefore Q is not maximal, but for any prime ideal Q0 ) Q, there is no 0 0 corresponding prime in Bx, so Q ∩S must be nonempty. In particular, x ∈ Q for every prime ideal Q0 ) Q. However, the assumption on p implies every non-maximal prime in B = A/p is T 0 0 an intersection of prime ideals properly containing it. Thus Q = {Q ⊂ B prime | Q ) Q}, but then x ∈ Q, a contradiction. Hence no such p exists in A, so every prime ideal in A is an intersection of maximal ideals.

31 2.5 Valuation Rings 2 Integral Dependence

Corollary 2.4.5. If A is a finitely generated k-algebra, for k a field, every prime ideal of A is an intersection of maximal ideals. Proof. By Lemma 2.4.4, it suffices to show that for every prime ideal p ⊂ A which is not T maximal, p = {Q ⊂ A | Q prime, Q ) p}. Suppose to the contrary that there exists such T a prime p with p ( {Q ⊂ A | Q prime, Q ) p}. Let B = A/p. Then B is not a field, and by assumption the intersection of all nonzero prime ideals in B is nonzero. In particular this intersection contains a nonzero element f, which as in Lemma 2.4.4 is not nilpotent. n b o Consider the (nonzero) localization Bf = f n : b ∈ B, n ≥ 0 . Then Bf is finitely generated as a k-algebra since A and B are. Suppose Bf is a field. Then by Nullstellensatz (2.4.2), Bf is an algebraic extension of k. In particular, Bf /k is integral, so Bf is integral over B as well. Then Lemma 2.2.1 implies Bf and B are fields simultaneously, but this contradicts the fact that B is not a field. Therefore Bf cannot be a field, so it has a nonzero prime ideal 0 0 m ⊂ Bf . This pulls back along j : B → Bf to a prime ideal m ⊂ B. Then f 6∈ m but clearly this is impossible, since f was chosen to lie in the intersection of all prime ideals of B. Thus no such prime p ⊂ A can exist, so every non-maximal prime ideal of A is equal to the intersection of all prime ideals properly containing it.

2.5 Valuation Rings

Deﬁnition. An integral domain A with ﬁeld of fractions K is called a valuation ring if x ∈ A or x−1 ∈ A for every x ∈ K×.

Example 2.5.1. If A is a UFD and p ∈ A is a prime element, then the localization A(p) = a c × b : a, b ∈ A, p - b is a valuation ring. To see this, take d ∈ K with c, d ∈ A r {0}. By c unique factorization, we may assume d is in lowest terms, i.e. gcd(c, d) = 1. Then if p - d, c c −1 d we have d ∈ A(p). On the other hand, if p | d, we must have p - c so d = c ∈ A(p). Theorem 2.5.2. Every valuation ring A is a local ring and is integrally closed.

Proof. We ﬁrst prove A is a local ring. Let A× be the set of units in A and let m = A r A×. One can prove that m is an ideal of A, and since every element outside of m is a unit, m must be the unique maximal ideal of A. To prove A is integrally closed, assume x ∈ K× is integral over A. Then there exist n n−1 −1 a0, . . . , an−1 ∈ A such that x + an−1x + ... + a0 = 0. If x ∈ A, we can multiply the polynomial expression through by x−(n−1) to obtain

−1 −(n−1) x = −(an−1 + an−2x + ... + a0x ). Since x−1 ∈ A, the right side lies in A and thus x ∈ A. This shows A is integrally closed. Theorem 2.5.3. For K a ﬁeld and A ⊆ K a nonzero subring, the integral closure of A in K is equal to \ A = B. valuation rings B A⊆B⊆K We will see in Section 4.1 how valuation rings arise naturally as subrings of a ﬁeld K.

32 2.6 Dimension and Transcendence Degree 2 Integral Dependence

2.6 Dimension and Transcendence Degree

Question. How does one determine the dimension of a ﬁnitely generated k-algebra?

Example 2.6.1. Consider A = k[t1, . . . , tn]. We would like to conclude that dim A = n, but this is not obvious from the deﬁnition of Krull dimension. Clearly dim A ≥ n, since

0 ( (t1) ( (t1, t2) ( ··· ( (t1, . . . , tn) is a chain of prime ideals in A. We will show the opposite inequality shortly. Assume C and D are integral domains, with D/C a ring extension and K = Frac(C) and L = Frac(D) their fields of fractions. Then L/K is a field extension. Define the transcendence degree of D over C to be tr degC D := tr degK L. If C = k is a field, we have

tr degk D = tr degk L = sup{|E| : E ⊂ D,E is algebraically independent over k}. Lemma 2.6.2. Assume A and B are ﬁnitely generated k-algebras which are integral domains, and there is a surjective k-algebra homomorphism π : B → A with ker π 6= 0. Then tr degk A < tr degk B.

Proof. Set r = tr degk B and s = tr degk A. Since A and B are ﬁnitely generated, r and s are each ﬁnite. If a1, . . . , as ∈ A are algebraically independent over k, choose b1, . . . , bs ∈ B such that π(bi) = ai for each i. Then for any polynomial F ∈ k[t1, . . . , ts] vanishing at (b1, . . . , bs), we have 0 = π(F (b1, . . . , bs)) = F (π(b1), . . . , π(bs)) = F (a1, . . . , as).

Since the ai are algebraically independent over k, we must have F = 0, so s ≤ r. To show s < r strictly, we demonstrate that any set of r elements in A is algebraically dependent over k. Let a1, . . . , ar ∈ A be given. Again choose bi ∈ B such that π(bi) = ai for 1 ≤ i ≤ r and take b0 ∈ ker π r {0}. Since tr degk B = r, there is a nonzero polynomial F ∈ k[t0, t1, . . . , tr] such that F (b0, b1, . . . , br) = 0. Choose F of minimal degree in t0. Then

d X i F (t0, t1, . . . , tr) = Fi(t1, . . . , tr)t0 for Fi ∈ k[t1, . . . , tr]. i=1

Pd i−1 We claim F0 6= 0; otherwise, F = t0 i=1 Fi(t1, . . . , tr)t0 and evaluating in the integral domain B implies

d d X i−1 X i−1 0 = b0 Fi(b1, . . . , br)b0 =⇒ Fi(b1 . . . , br)b0 = 0. i=1 i=1

This is a nonzero polynomial of strictly smaller degree in t0 that vanishes on (b0, . . . , br), so we in fact must have F0 6= 0. Now plug in the bj and apply π to obtain

d ! d X i X i 0 = π Fi(b1, . . . , br)b0 = Fi(a1, . . . , ar)π(b0) = F0(a1, . . . , ar). i=1 i=1

Since F0 6= 0, we see that a1, . . . , ar are algebraically dependent over k.

33 2.6 Dimension and Transcendence Degree 2 Integral Dependence

Corollary 2.6.3. For any n ≥ 1, dim k[t1, . . . , tn] = n.

Proof. Set A = k[t1, . . . , tn]. We showed in Example 2.6.1 that dim A ≥ n. Let

p0 ( p1 ( ··· ( p` be a chain of prime ideals in A. Then we have a well-deﬁned sequence of surjections

A/p0 A/p1 ··· A/p`

with all kernels nonzero since pi ( pi+1 for all i. By Lemma 2.6.2,

n ≥ tr degk A/p0 > tr degk A/p1 > ··· > tr degk A/p` ≥ 0.

In this sequence, there are ` strict inequalities, so ` ≤ n. Further, ` was arbitrary so dim A = n.

Theorem 2.6.4. If A is a ﬁnitely generated k-algebra which is an integral domain, then dim A = tr degk A.

Proof. By Noether’s normalization lemma, we can pick y1, . . . , yr ∈ A which are algebraically independent over k such that A/k[y1, . . . , yr] is an integral extension of rings. Then dim A = dim k[y1, . . . , yr] = r by Theorem 2.3.7. Now for all x ∈ A, x is integral over k[y1, . . . , yr]. If K is the field of fractions of A, then every z ∈ K is algebraic over the field k(y1, . . . , yr), but then Lemma 1.4.3(c) shows that {y1, . . . , yr} is a transcendence basis of K/k. Hence tr degk A = tr degk K = r = dim A. Definition. We say a ring A of finite Krull dimension has the strong dimension property (SDP) if whenever p0 ( ··· ( p` is a maximal (non-refinable) chain of prime ideals in A, we have ` = dim A.

Lemma 2.6.5. Assume A has SDP and n = dim A. Then

(1) ht(p) + dim A/p = dim A for any prime ideal p ⊂ A.

(2) ht(m) = n for all maximal ideals m ⊂ A.

(3) If p0 ( p1 ( ··· ( p` is a chain of prime ideals in A which cannot be reﬁned, then ` = ht(p`) − ht(p0). (4) If p ⊂ A is a prime ideal then A/p has SDP.

Proof. (1) Given a prime ideal p ⊂ A with ht(p) = m, choose a chain

p0 ( ··· ( pm−1 ( p of prime ideals in A. If dim A/p = d then there is a maximal chain

0 = Q0 ( Q1 ( ··· ( Qd

34 2.6 Dimension and Transcendence Degree 2 Integral Dependence

of prime ideals in A/p. Then we may form a chain

p0 ( ··· ( pm−1 ( p = Q0 ( Q1 ( ··· ( Qd

where Qi ⊂ A are the prime ideals such that Qi = Qi/p. This is a maximal, non-refinable chain of primes in A, so by SDP, dim A = m + d = ht(p) + dim A/p. (2) follows immediately from (1) since A/m is a field and dim F = 0 for any field F . (3) follows from a similar argument to the proof of (1). Given such a chain of primes

p0 ( ··· ( p` we can extend this to a maximal chain of primes in A:

Q0 ( ··· ( Qj = p0 ( ··· ( P` ( p`+1 ( ··· ( pm.

Then SDP implies n = j + m. When such a chain is chosen so that j = ht(p0) and m − ` = dim A/p`, we must have

n = j + m = j + ` + (m − `) = ht(p0) + ` + dim A/p`.

However by (1), n = ht(p`) + dim A/p` so we see that ht(p`) = ht(p0) + `. (4) Any maximal prime ideal chain in A/p is of the form

0 ( p1/p ( ··· ( p`/p

where p ( p1 ( ··· ( p` is a non-reﬁnable prime ideal chain in A. By (3), ` = ht(p`) − ht(p) but p` must be maximal, so by (2), ht(p`) = n. Finally, applying (1) gives ` = n − ht(p) = dim A/p. Hence SDP holds for A/p. An important example of a ring with the strong dimension property is the polynomial algebra k[t1, . . . , tn]. Before proving this has SDP, we need a preliminary result which relies on the going down theorem.

Proposition 2.6.6. Let A be a ﬁnitely generated k-algebra that is an integral domain. Then for any nonzero prime ideal p ⊂ A such that ht(p) = 1, we have dim A/p = dim A − 1.

Proof. By Noether’s normalization lemma, there exist elements y1, . . . , yn ∈ A which are algebraically independent over k and have the property that A/k[y1, . . . , yn] is an integral extension. Then by Theorems 2.3.7 and 2.6.4, dim A = dim k[y1, . . . , yn] = n. Since ht(p) = 1, there is no nonzero prime ideal of A properly contained in p. Set q = p ∩ k[y1, . . . , yn]. If q = 0, then we would have 0 = p ∩ k[y1, . . . , yn] = 0 ∩ k[y1, . . . , yn], which would imply p = 0 by Lemma 2.3.3, but this contradicts ht(p) = 1. Suppose there exists a nonzero 0 0 prime ideal q ⊂ k[y1, . . . , yn] such that q ( q. Then the going down theorem would give us a nonzero prime ideal p0 ( p, contradicting ht(p) = 1. This shows that ht(q) = 1 also. Now by Lemma 2.3.1(a), A/p is integral over k[y1, . . . , yn]/q, so it suﬃces to prove dim k[y1, . . . , yn]/q = n − 1. Set B := k[y1, . . . , yn]. In light of Lemma 1.4.2 and Corollary 2.6.3, we have that dim B = n. We ﬁrst show that q is a principal ideal. Since q 6= 0, there exists a nonzero element

35 2.6 Dimension and Transcendence Degree 2 Integral Dependence

f 0 ∈ q. Then because B is a UFD, f 0 is divisible by some irreducible element f ∈ q, but then f is prime and 0 6= (f) ⊆ (f 0) ⊆ q. Since q is a minimal prime, we must have (f) = q. Finally, B/q = B/(f) is an integral extension of k[z2, . . . , zn] where {1, z2, . . . , zn} is a change of basis of {y1, . . . , yn} such that f(1, z2, . . . , zn) is monic. It follows from Theorem 2.3.7 and Corollary 2.6.3 that dim B/q = dim k[z2, . . . , zn] = n − 1. Theorem 2.6.7. If A is a ﬁnitely generated k-algebra which is an integral domain, then A has the strong dimension property.

Proof. We induct on n = dim A. If n = 0, A is a ﬁeld so the property holds trivially. Now assume that every k-algebra which is a domain and has dimension less than n has the strong dimension property. Take a maximal chain of prime ideals p0 ( ··· ( p` in A. Since A is a domain, we must have p0 = 0 and p` maximal. Consider the integral domain B = A/p1. By the correspondence theorem,

0 = p1/p1 ( p2/p1 ( ··· ( p`/p1

is a maximal chain of prime ideals in B of length ` − 1. Moreover, ht(p1) = 1 so by Proposition 2.6.6, dim B = dim A/p1 = dim A − 1 = n − 1. Thus by induction, n − 1 = ` − 1 and so we get n = ` as required.

Corollary 2.6.8. For any n ≥ 1, k[t1, . . . , tn] has the strong dimension property. In particular, dim k[t1, . . . , tn]/p = n − ht(p) for any prime ideal p ⊂ k[t1, . . . , tn].

36 3 Noetherian and Artinian Rings

3 Noetherian and Artinian Rings

3.1 Ascending and Descending Chains

Definition. A partially ordered set (Σ, ≤) satisfies the ascending chain condition (ACC) if any ascending chain x1 ≤ x2 ≤ · · · in Σ is stationary, i.e. there is some N ∈ N such that xn = xN for all n ≥ N. Definition. A partially ordered set (Σ, ≤) satisfies the descending chain condition (DCC) if any descending chain x1 ≥ x2 ≥ · · · in Σ is stationary, i.e. there is some N ∈ N such that xn = xN for all n ≥ N.

In ring theory, we study chain conditions on the collection ΣM of submodules of a ﬁxed A-module M, partially ordered by inclusion. The modules M that satisfy the above chain conditions have special names.

Definition. An A-module M is noetherian if the collection of submodules of M satisfies the ascending chain condition. On the other hand, M is artinian if the collection of submodules of M satisfies the descending chain condition.

Lemma 3.1.1. Let (Σ, ≤) be a poset. Then

(i) Σ satisﬁes ACC if and only if every nonempty subset of Σ has a maximal element.

(ii) Σ satisﬁes DCC if and only if every nonempty subset of Σ has a minimal element.

In particular, M is noetherian iﬀ every nonempty collection of submodules has a maximal element, and M is artinian iﬀ every nonempty collection of submodules has a minimal element.

Proposition 3.1.2. An A-module M is noetherian if and only if every submodule of M is ﬁnitely generated (as an A-module).

f g Lemma 3.1.3. Let 0 → M 0 −→ M −→ M 00 → 0 be a short exact sequence of A-modules. Then

(a) M is noetherian if and only if M 0 and M 00 are noetherian.

(b) M is artinian if and only if M 0 and M 00 are artinian.

Proof. ( =⇒ ) Consider the maps induced by f and g:

∗ ∗ f :ΣM 0 −→ ΣM g :ΣM 00 −→ ΣM N 0 7−→ f(N 0) N 00 7−→ g−1(N 00).

∗ ∗ Then f and g are both order-preserving and injective maps, so the posets ΣM 0 and ΣM 00 inherit the chain conditions from ΣM .

37 3.1 Ascending and Descending Chains 3 Noetherian and Artinian Rings

( ⇒ = ) We prove this for the artinian property; the proof for noetherian modules is similar. Let C : M1 ⊇ M2 ⊇ M3 ⊇ · · · be a descending chain of submodules Mi ⊂ M. These correspond to chains of submodules

−1 −1 −1 −1 0 f (C):f (M1) ⊇ f (M2) ⊇ f (M3) ⊇ · · · in M 00 g(C):g(M1) ⊆ g(M2) ⊇ g(M3) ⊇ · · · in M .

−1 −1 By the DCC on ΣM 0 and ΣM 00 , there exists a single N ∈ N such that f (Mn) = f (MN ) and g(Mn) = g(MN ) for all n ≥ N. For all n ≥ N, we already have Mn ⊇ MN . On the other hand, given x ∈ MN , there exists a y ∈ Mn such that g(x) = g(y). This implies g(x) − g(y) = g(x − y) = 0 by A-linearity. So x − y ∈ ker g which equals im f by exactness. 0 −1 −1 Thus there is some z ∈ M such that f(z) = x − y ∈ MN . Then z ∈ f (MN ) = f (Mn) so x − y = f(z) ∈ Mn. Hence x = y + f(z) ∈ Mn so MN ⊆ Mn. This proves the DCC on ΣM , so M is artinian. Ln Corollary 3.1.4. If M1,...,Mn are noetherian (resp. artinian) A-modules then i=1 Mi is also noetherian (resp. artinian).

Proof. This is proved by induction on n, considering short exact sequences

n n−1 M M 0 → Mn → Mi → Mi → 0 i=1 i=1 and applying Lemma 3.1.3. We distinguish the special case of A as a module over itself.

Deﬁnition. A (commutative) ring A is a noetherian ring if it is noetherian as a module over itself. Similarly, A is an artinian ring if it is artinian as a module over itself.

Let A be a ring. Then the A-submodules of A are exactly the ideals of A. Then by Proposition 3.1.2, A is noetherian if and only if all ideals of A are ﬁnitely generated.

Corollary 3.1.5. If A is noetherian (resp. artinian), then any ﬁnitely generated A-module M is also noetherian (resp. artinian).

Proof. Write M = Am1 + ... + Amn for mi ∈ M. Consider the projection

An −→ M n X (a1, . . . , an) 7−→ aimi. i=1 Then An is noetherian (resp. artinian) by Corollary 3.1.4 so M is noetherian (resp. artinian) by Lemma 3.1.3.

38 3.1 Ascending and Descending Chains 3 Noetherian and Artinian Rings

Lemma 3.1.6. For a ﬁeld k and a k-vector space V , the following are equivalent:

(i) dimk V < ∞. (ii) V is a noetherian k-module.

(iii) V is an artinian k-module.

Proof. (i) =⇒ (ii), (iii) follow from the linear algebra fact that every subspace of a finite dimensional vector space is finite dimensional, and thus has a finite basis. (ii), (iii) =⇒ (i) If dimk V = ∞ then there is a countable linear independent subset ∞ {vj}j=1 ⊂ V . Let Vi = Span{vj | j ≤ i} for each i ∈ N. Then V1 ( V2 ( ··· is an ascending chain in which each subspace is finitely generated, so it cannot stabilize. Likewise, let Wi = Span{vj | j ≥ i} for each i ∈ N. Then W1 ) W2 ) ··· is a descending chain which doesn’t stabilize. Hence dimk V = ∞ implies is V is neither noetherian nor artinian.

Proposition 3.1.7. Suppose A has maximal ideals m1,..., mn, not necessarily distinct, with m1 ··· mn = 0. Then A is noetherian if and only if A is artinian. Proof. Say N is an A-module and m ⊂ A is a maximal ideal. Then N/mN becomes a k- vector space for k = A/m via (a + m)(n + mN) = an + mN. Then A-submodules of N/mN are the same as k-subspaces of N/mN. Thus N/mN is noetherian (resp. artinian) as an A-module if and only if N/mN is noetherian (resp. artinian) as a k-vector space; this in turn is equivalent to dimk N/mN < ∞ by Lemma 3.1.6. Hence N/mN is noetherian if and only if N/mN is artinian. Now let m1,..., mn ⊂ A be maximal ideals satisfying m1 ··· mn = 0. Set Ai = A/m1 ··· mi for each 1 ≤ i ≤ n. In particular, An = A/{0} = A by hypothesis. If A is noetherian (resp. artinian) then each Ai is noetherian (resp. artinian) by Lemma 3.1.3. Note that A1 = A/m1 is a ﬁeld, so A1 is both noetherian and artinian. We claim that Ai is noetherian for each 1 ≤ i ≤ n if and only if Ai is artinian for each 1 ≤ i ≤ n. Indeed, A1 is the base case so for n ≥ 2, consider the short exact sequence

0 → m1 ··· mi−1/m1 ··· mi → Ai → Ai−1 → 0.

Set Ni = m1 ··· mi−1/m1 ··· mi. Then by Lemma 3.1.3,

Ai is noetherian ⇐⇒ Ni and Ai−1 are noetherian

⇐⇒ Ni is noetherian and Ai−1 is artinian (by induction)

⇐⇒ Ni is artinian and Ai−1 is artinian (by preliminary remarks)

⇐⇒ Ai is artinian.

Finally, since A = An, this implies that A is noetherian if and only if A is artinian.

39 3.2 Composition Series 3 Noetherian and Artinian Rings

3.2 Composition Series

Let M be an A-module.

Deﬁnition. A chain in M is a strictly descending chain of submodules of M of the form

M = M0 ) M1 ) ··· ) Mn = 0. The integer n is called the length of the chain.

Definition. A composition series of M is a chain in M which cannot be refined, or equivalently, the quotient Mi/Mi+1 is simple for all 0 ≤ i ≤ n − 1. Definition. The length of a module M, denoted `(M), is by convention `(M) = ∞ if M does not have a composition series. Otherwise,

`(M) = min{n ∈ N | M has a composition series of length n}. One can view the length function `(·) as a generalization of the dimension of a vector space.

Proposition 3.2.1. Assume M is an A-module that has a composition series. Then

(a) For every submodule N ⊂ M, `(N) ≤ `(M) and if N is a proper submodule, `(N) < `(M).

(b) Any chain in M has length at most `(M).

(d) Any chain in M can be extended to a composition series.

Proof. (a) Choose a composition series of minimal length,

M = M0 ) M1 ) ··· ) M` = 0.

Then ` = `(M). Intersecting with N, setting Ni = N ∩ Mi, gives a sequence of containments that are not necessarily strict:

N = N0 ⊇ N1 ⊇ · · · ⊇ N` = 0.

However for each i, we have an inclusion Ni/Ni+1 ,→ Mi/Mi+1 since

Ni+1 = N ∩ Mi+1 = N ∩ Mi ∩ Mi+1 = Ni ∩ Mi+1. ∼ By assumption, each Mi/Mi+1 is simple, so either Ni/Ni+1 = 0 or Ni/Ni+1 = Mi/Mi+1, in which case Ni/Ni+1 is also simple. After removing repetitions, i.e. where Ni = Ni+1, we obtain a composition series of N of length at most `. Hence `(N) ≤ `(M). Now suppose N ( M but `(N) = `(M). Then Ni ) Ni+1 for all 0 ≤ i ≤ ` − 1 in the above chain. By construction, M`−1 is a simple A-module and 0 6= N`−1 ⊆ M`−1, so we have N`−1 = M`−1.

40 3.2 Composition Series 3 Noetherian and Artinian Rings

By induction, Ni = Mi for all 0 ≤ i ≤ ` − 1. Hence N = N0 = M0 = M, a contradiction. Therefore `(N) < `(M) for every proper submodule N. (b) Consider any chain M = M0 ) M1 ) ··· ) Mn = 0. Then by (a),

` = `(M0) > `(M1) > ··· > `(Mn) > 0. In this string we have n strict inequalities so it must be that n ≤ `. (c) By definition of `(M), any compositions series of M has length at least `(M), but by (b), every composition series of M has length at most `(M). (d) Given any chain, refine it as many times as possible. By (b), we cannot get a chain of length strictly greater than `(M), so the process will eventually terminate when the length of the chain is `(M). At this point we will have a composition series for M. Corollary 3.2.2. An A-module M has a composition series if and only if M is both noetherian and artinian. Proof. ( =⇒ ) Assume `(M) < ∞. Then by Proposition 3.2.1, all chains in M have length at most `(M). In particular, any ascending or descending chain must be finite, which implies M has satisfies ACC and DCC. ( ⇒ = ) Assume M 6= 0. Then the set of all proper submodules of M has a maximal element by Lemma 3.1.1; call it M1. If M1 = 0, then M ) M1 = 0 is a composition series for M. Otherwise, the set of proper submodules of M1 has a maximal element M2, by Lemma 3.1.1 again, using the fact that submodules of noetherian modules are noetherian. Continue in this fashion to construct a chain

M = M0 ) M1 ) M2 ) ···

By maximality in each step, Mi/Mi+1 is simple for all i ≥ 0. Since M is artinian, this process must terminate, else the DCC is violated. Then at some point we have

M = M0 ) M1 ) M2 ) ··· ) Mn ) 0 which is a composition series for M. Lemma 3.2.3. If M is an A-module that admits a composition series, then for any short exact sequence 0 → M 0 → M → M 00 → 0 of A-modules, `(M) = `(M 0) + `(M 00). That is, length is additive on short exact sequences.

f g Proof. Label the morphisms 0 → M 0 −→ M −→ M 00 → 0. First, M is both noetherian and artinian by Corollary 3.2.2. Note that M 0 and M 00 are also noetherian and artinian by Lemma 3.1.3, so each admits a composition series. Consequently, `(M 0) and `(M 00) are deﬁned. Let

0 0 0 0 00 00 00 00 M = M0 ) M1 ) ··· ) M` = 0 and M = M0 ) M1 ) ··· ) Mn = 0 be composition series of M 0 and M 00. Then by Proposition 3.2.1(c), `(M 0) = ` and `(M 00) = n. By exactness, we get a sequence of inclusions

−1 00 −1 00 0 0 0 M = g (M0 ) ⊃ · · · ⊃ g (Mn ) = ker g = im f = f(M0) ⊃ f(M1) ⊃ · · · ⊃ f(M`) = 0. (∗)

41 3.3 Noetherian Rings 3 Noetherian and Artinian Rings

−1 00 00 −1 00 Now since g is surjective, g(g (Mi )) = Mi so the inclusions are strict between the g (Mi ), −1 0 0 0 ≤ i ≤ n. Similarly, since f is injective, f (f(Mi )) = Mi so the inclusions between the 0 f(Mj), 0 ≤ j ≤ `, are strict. Therefore (∗) is a strictly descending chain of submodules of M of total length ` + n. Consider each quotient in the ﬁrst half; since g is surjective, we get an isomorphism −1 00 −1 00 ' 00 00 g (Mi )/g (Mi+1) −→ Mi /Mi+1 for 0 ≤ i ≤ n. −1 00 −1 00 Then each g (Mi )/g (Mi+1) is simple. Similarly, injectivity of f gives us an isomorphism

0 0 ' 0 0 Mi /Mi+1 −→ f(Mi )/f(Mi+1) for 0 ≤ j ≤ `,

0 0 which shows the f(Mi )/f(Mi+1) are all simple. Hence (∗) is a composition series for M, so by Proposition 3.2.1(c), `(M) = ` + n = `(M 0) + `(M 00).

3.3 Noetherian Rings

In this section we present the key details regarding noetherian rings and their modules in commutative algebra.

Lemma 3.3.1. If A is a noetherian ring, then

(a) For any ideal I ⊂ A, A/I is a noetherian ring.

(b) For any multiplicatively closed subset S ⊆ A, S−1A is a noetherian ring. In particular, Ap is a local noetherian ring for any prime ideal p ⊂ A. Proof. (a) By Lemma 3.1.3, we know A/I is noetherian as an A-module. It follows from the isomorphism theorems that the A-submodules of A/I are exactly the ideals of A/I, so it follows that A/I is noetherian as a ring. −1 −1 (b) If J1 ⊆ J2 ⊆ · · · is a chain of ideals in S A then applying j gives a chain of −1 −1 ideals j (J1) ⊆ j (J2) ⊆ · · · in A. Since A is noetherian, the chain stabilizes, i.e. there −1 −1 is some n ∈ N such that j (Ji) = j (Jn) for all i ≥ n. Applying the localization functor −1 −1 −1 S to these modules returns the original Ji, since S (j (Ji)) = Ji for each i ≥ 1 by Proposition 1.3.8: J1 ⊆ J2 ⊆ · · · ⊆ Jn = Jn+1 = ··· . −1 −1 Hence the ACC holds for S A, so S A is noetherian. The ﬁnal remark about Ap follows from Corollary 1.3.9.

Remark. A similar argument as in Lemma 3.3.1(b) shows that for every noetherian A- module M, any localization S−1M is also a noetherian A-module.

One of the fundamental results in the theory of noetherian rings is known as Hilbert’s basis theorem. We will see that it implies every ﬁnitely generated ring extension of a noetherian ring is also noetherian.

Theorem 3.3.2 (Hilbert’s Basis Theorem). If A is a noetherian ring, then A[t] is also noetherian.

42 3.3 Noetherian Rings 3 Noetherian and Artinian Rings

Proof. Let I ⊂ A[t] be an ideal and let L denote the set of all leading coeﬃcients of elements of I. That is,

( n ) X i L = α ∈ A : there is some ait ∈ I with an = α . i=0 We claim L is an ideal. Clearly 0 ∈ L since I contains the zero polynomial. Suppose f, g ∈ I n m such that f = ant + ... + a0 and g = bmt + ... + b0. We must show an + bm ∈ L. If n = m then the leading coeﬃcient of f + g is an + bm, and since f + g ∈ I, we have an + bn ∈ L. If n 6= m, without loss of generality assume n < m. Then

m−n m m−n m t f + g = (ant + ... + a0t ) + (bmt + ... + b0) m m−n = (an + bm)t + ... + (a0 + bm−n)t + ... + b0 which lies in I, showing an+bm ∈ L. Finally, if c ∈ A then can is the leading term of cf ∈ I, so L is an ideal of A. Now, since A is noetherian, L is finitely generated, say by a1, . . . , an ∈ A. For each 1 ≤ i ≤ n, let fi ∈ I be a polynomial having ai as its leading coefficient. For each ei of these, write fi = ait + ... so that deg fi = ei. Set E = max{e1, . . . , en}. For each 0 ≤ d ≤ E − 1, let Ld be the set of all leading coefficients of polynomials in I of degree d, together with 0:

( d ) X i Ld = α ∈ A : there is some ait ∈ I with ad = α . i=0

The proof that each Ld is an ideal of A is similar to the proof above for L. Then each

Ld is finitely generated, say by bd1, . . . , bdnd ∈ A. Let fd1, . . . , fdnd ∈ I be polynomials such that the leading coefficient of fdr is bdr, 1 ≤ r ≤ nd. We claim that f1, . . . , fn and fdr, 0 ≤ d ≤ E − 1, 1 ≤ r ≤ nd, are a generating set for I. Let Q denote the ideal generated by these polynomials. By construction, Q ⊆ I since the generators of Q are elements of I. If Q ( I, choose f ∈ I r Q of minimum degree d with leading coefficient a. First suppose d ≥ E. Then a ∈ L so there exist elements c1, . . . , cn ∈ A such that d−e1 d−en a = c1a1 + ... + cnan. Observe that g = c1t f1 + ... + cnt fn is a polynomial in Q of degree d and leading coefficient a, so f − g ∈ I has strictly smaller degree than f. By minimality, we must have f − g = 0, that is, f = g ∈ Q, a contradiction. On the other hand, if d < E then f ∈ Ld so we have f = c1bd1 + ... + cnd bdnd for some cr ∈ A. Then g = c1fd1 + ... + cnd fnd is a polynomial in Q of degree d and leading coefficient a, producing the same contradiction. It follows that Q = I, so I is finitely generated. This proves that A[t] is noetherian. Remark. The converse of Hilbert’s basis theorem is trivially true: A[t] noetherian implies A noetherian since A ∼= A[t]/(t) as rings. Corollary 3.3.3. If A is noetherian, then

(a) A[t1, . . . , tn] is noetherian for any n ≥ 1. (b) B is noetherian for any ﬁnitely generated ring extension B/A.

43 3.3 Noetherian Rings 3 Noetherian and Artinian Rings

Proof. (a) Induct on the number of generators. The base case is Hilbert’s basis theorem. (b) We can write B = A[b1, . . . , bn] for some bi ∈ B. Then there is a (unique) surjective ring homomorphism

ϕ : A[t1, . . . , tn] −→ B

ti 7−→ bi. ∼ So B = A[t1, . . . , tn]/ ker ϕ and we can apply (a) and Lemma 3.3.1. Corollary 3.3.4. For a field k, any finitely generated k-algebra is noetherian. Proof. Every field is noetherian (it admits a trivial composition series) so Lemma 3.3.3(b) may be applied. Corollary 3.3.5. Every finitely generated commutative ring is noetherian.

Proof. Let A be such a ring. There is a unique ring homomorphism f : Z → A given by f(n) = n · 1A for all n. Set A0 = f(Z). Then A0 is noetherian by Lemma 3.3.1(a) and the hypothesis that A is finitely generated as a ring means that A/A0 is a finitely generated ring extension. Then the statement follows from Corollary 3.3.3(b). Proposition 3.3.6. Suppose A is noetherian and C ⊃ B ⊃ A are extensions of rings such that C/A is a finitely generated extension and C is a finitely generated B-module. Then B/A is also finitely generated.

Proof. Choose x1, . . . , xm ∈ C and y1, . . . , yn ∈ C with y1 = 1 such that n X C = A[x1, . . . , xm] = Byj. j=1

Then there exist bij` ∈ B such that for all 1 ≤ i ≤ m, 1 ≤ j ≤ n,

n X xiyj = bij`y`. (∗) `=1

Deﬁne the subring B0 = A[bij` | 1 ≤ i ≤ m, 1 ≤ j, ` ≤ n]. Then B0 is ﬁnitely generated as an A-module by construction, so by Corollary 3.3.3, B0 is noetherian. Consider the B0-submodule n X M = B0y` ⊆ C. `=1 Pn By (∗), xiyj ∈ M for all i, j. Thus xi j=1 B0yj ⊆ M, or in other words xiM ⊆ M for each i. In particular, C = A[x1, . . . , xm] ⊆ M ⊆ C, so M = C. Now we have

A ⊆ B0 ⊆ B ⊆ C

with B0 noetherian and C a finitely generated B0-module. Thus by Corollary 3.1.5, C is a noetherian B0-module, and Proposition 3.1.2 implies that B ⊆ C is finitely generated as a B0-module. Hence B/B0 is a finitely generated ring extension, and by construction B0/A is a finitely generated ring extension, so we conclude that B/A is finitely generated as well.

44 3.3 Noetherian Rings 3 Noetherian and Artinian Rings

As a result, we obtain the following interesting fact. Theorem 3.3.7. If a field K is finitely generated as a ring then K is a finite field.

Proof. Consider the map f : Z → K. Since K is a field, f(Z) is an integral domain. We have ∼ ∼ two cases: (1) when char K = p > 0, f(Z) = Fp; and (2) when char K = 0, f(Z) = Z. In case (1), K is a finitely generated Fp-algebra. By Hilbert’s Nullstellensatz (2.4.2), K/Fp is an algebraic extension, so in fact it is an extension of finite fields. Therefore K is finite. On the other hand, in (2), K is finitely generated over Z and K is a field, so we have Z ⊂ Q ⊂ K. This means K is a finitely generated Q-algebra, so once again the Nullstellensatz says that K/Q is a finite extension. However, one sees that the conditions of Proposition 3.1.2 are satisfied, so Q is a finitely generated ring extension of Z, which is false. Therefore the only possibility is that K is a finite field.

Corollary 3.3.8. Any matrix ring GLn(K), n ≥ 2, over an infinite field K cannot be finitely generated. The proof of Hilbert’s basis theorem can be generalized to prove that power series rings over a noetherian ring are also noetherian. Theorem 3.3.9. If A is noetherian, then the power series ring A[[t]] is also noetherian. P∞ i Proof. For a power series f ∈ A[[t]], where f = i=0 ait , let j be the smallest index such that aj 6= 0. We say aj is the coefficient of lowest degree of f. If ai = 0 for all i ≥ 0, i.e. f is the zero power series, we will say its coefficient of lowest degree is 0. Let I ⊂ A[[t]] be an ideal and define L ⊂ A to be the set of all coefficients of lowest degree of elements in I:

( ∞ ) X i L = aj : ait ∈ I for some j ∈ N0 and ai, i > j . i=j We first show L is an ideal. Since 0 ∈ I and 0 has lowest degree coefficient 0, we have 0 ∈ L. P∞ i P∞ k Let f, g ∈ I, with f = i=j ait and g = k=` bkt . We must show aj + b` ∈ L. If j = `, the lowest degree coefficient of f + g is aj + bj, and since f + g ∈ I, we have aj + bj ∈ L. If j 6= `, without loss of generality assume j < `. Then ∞ ∞ ∞ ∞ ∞ `−j `−j X i X k X i+`−j X k X k t f + g = t ait + bkt = ait + bkt = (ak+j−` + bk)t i=j k=` i=j k=` k=`

so we see that the lowest degree coefficient is aj + b`. Therefore aj + b` ∈ L. Now, if α ∈ A, the lowest degree coefficient of αf is αai, and αf ∈ I since I is an ideal, so αai ∈ L. This proves L is an ideal of A. By the noetherian hypothesis, L is finitely generated by some α1, . . . , αn ∈ A. Now for each k = 1, . . . , n, take fk ∈ I such that the lowest degree term of fk is αk; let ∞ e be the exponent of t in f having α as its coefficient, i.e. f = α tek + P a ti. Set k k k k k i=ek+1 i E = max{ek | 1 ≤ k ≤ n}. For all 0 ≤ d ≤ E − 1, define Jd ⊂ A by

( ∞ ) X i Jd = ad : ait ∈ I for some ai, i > d ∪ {0}. i=d

45 3.4 Primary Decomposition 3 Noetherian and Artinian Rings

The proof that each Jd is an ideal of A is similar to the proof for L. In particular, since A is noetherian we get that Jd is ﬁnitely generated for each 0 ≤ d ≤ E − 1. Say Jd is generated by βd1, . . . , βdnd ∈ A. For each 0 ≤ d ≤ E − 1 and for each 1 ≤ r ≤ nd, let fdr ∈ A[[t]] be d P∞ i such that fdr = βdrt + i=d+1 ait ∈ I. Let Q be the ideal of A[[t]] ﬁnitely generated by all fk, 1 ≤ k ≤ n and fdr, 0 ≤ d ≤ E − 1, 1 ≤ r ≤ nd. We claim that Q = I. Since all of the generators of Q lie in I, it is clear that Q ⊆ I. On the other hand, take f ∈ I and suppose d the lowest degree term of f is at , for some a ∈ L and d ≥ 0. If d ≤ E − 1, then a ∈ Jd so Pnd there exist cr ∈ A such that a = r=1 crβdr, which implies

nd nd ∞ ! nd ∞ nd X X d X i X d X X i f − crfdr = f − cr βdrt + ait = f − crβdrt − crait r=1 r=1 i=d+1 r=1 i=d+1 r=1 is a power series in A[[t]] of lowest degree at least d + 1. (We can switch the sums in the last Pnd step since one is ﬁnite.) Since f − r=1 crfdr ∈ I, we may assume that the lowest degree of f is at least E. If the lowest degree of f is d > E, then

nE X d−E f − crfdrt r=1

which is a power series in A[[t]] of lowest degree at least d + 1. Notice that the ideals Jd are ascending: J0 ⊆ J1 ⊆ J2 ⊆ · · · . Since A is noetherian, this chain stabilizes, or in other words, we can always ﬁnd power series

∞ X `−E gr = c`rt for 1 ≤ r ≤ nE `=d such that f can be written n XE f = grfdr. r=1 Hence f ∈ Q so Q = I and it follows that I is ﬁnitely generated.

3.4 Primary Decomposition

For this section, we assume A is a noetherian ring. Our goal is to prove an analog of primary decomposition (of modules over a PID) for modules over A.

Lemma 3.4.1. For any ideal I ⊂ A, there exists n ∈ N such that r(I)n ⊆ I.

Proof. Since A is noetherian, r(I) is ﬁnitely generated; say x1, . . . , xm ∈ r(I) such that ei (x1, . . . , xm) = r(I). Then there exist ei ∈ N such that xi ∈ I for each 1 ≤ i ≤ m. Set j1 jm n = e1 + ... + em. Then x1 ··· xm ∈ I whenever j1 + ... + jm ≥ n, but the products j1 jm n n x1 ··· xm generate r(I) , so we see that for this choice of n, r(I) ⊆ I. We obtain the following fact about the nil radical of a noetherian ring.

Corollary 3.4.2. There exists an n ∈ N such that N(A)n = 0.

46 3.4 Primary Decomposition 3 Noetherian and Artinian Rings

Proof. For the ideal I = (0), r(0) = N(A). The statement follows from Lemma 3.4.1.

Deﬁnition. A proper ideal I ⊂ A is irreducible if for any ideals J, K ⊂ A with J ∩K = I, we have J = I or K = I.

Example 3.4.3. Prime ideals are always irreducible. Indeed, if J ∩K = p is prime but p ( J and p ( K, then there are some elements x ∈ J rp and y ∈ Krp. But xy ∈ JK ⊆ J ∩K = p which is a contradiction since p is prime. Thus p is irreducible.

Lemma 3.4.4. Any proper ideal I ⊂ A is a ﬁnite intersection of irreducible ideals.

Proof. Consider the collection

Σ = {proper ideals I ⊂ A | I is not an intersection of irreducible ideals}.

If we assume Σ is nonempty, then since A is noetherian, Lemma 3.1.1 guarantees that Σ has a maximal element I0. In particular, I0 is not irreducible. Then there exist ideals J, K ⊂ A such that I0 = J ∩ K but I0 ( J and I0 ( K. Since I0 is maximal in Σ, neither J nor K lie in Σ. Thus J and K are each a ﬁnite intersection of irreducible ideals, but then I0 = J ∩K is also a ﬁnite intersection of irreducible ideals, a contradiction. Hence Σ must be empty.

Deﬁnition. A proper ideal I ⊂ A is primary if for every x, y ∈ A, xy ∈ I implies x ∈ I or y ∈ r(I). Equivalently, I is primary if xy ∈ I implies that x ∈ I, y ∈ I or x, y ∈ r(I).

Lemma 3.4.5. If a proper ideal I ⊂ A is irreducible, then I is primary.

Proof. Consider B = A/I. Then if I is irreducible, (0) is irreducible in B, i.e. there do not exist two nonzero ideals J, K ⊂ B with J ∩ K = 0. Assume x, y ∈ B such that xy = 0 and x 6= 0. We will show that yn = 0 for some n ∈ N, i.e. y is nilpotent. Consider the ascending chain Ann(y) ⊆ Ann(y2) ⊆ Ann(y3) ⊆ · · · By the ACC, there is some n ∈ N such that Ann(yi) = Ann(yn) for all i ≥ n. Now we show (x) ∩ (yn) = 0. If b ∈ (x) ∩ (yn) then b = rx for some r ∈ B, so by = rxy = 0. Also, b = cyn for some c ∈ B, so cyn+1 = cyny = by = 0. This shows c ∈ Ann(yn+1) = Ann(yn), and thus 0 = cyn = b. Hence (x) ∩ (yn) = 0 but since (0) is irreducible and x 6= 0, we must have (yn) = 0. Finally, this implies I ⊂ A is primary, since if x0, y0 ∈ A such that x0y0 ∈ I but x0 6∈ I, we have x0 + I 6= I and (x0 + I)(y0 + I) = x0y0 + I = I. Then by the ﬁrst part of the proof, x0 + I 6= I means (y0 + I)n = (y0)n + I = I for some n ∈ N, and thus (y0)n ∈ I. Theorem 3.4.6 (Primary Decomposition). For any proper ideal I ⊂ A, there exist ﬁnitely many primary ideals Q1,...,Qn such that I = Q1 ∩ · · · ∩ Qn. Proof. Apply Lemmas 3.4.4 and 3.4.5.

Lemma 3.4.7. If Q ⊂ A is a primary ideal, then r(Q) is the smallest prime ideal of A containing Q.

47 3.4 Primary Decomposition 3 Noetherian and Artinian Rings

Proof. Assume x, y ∈ A with xy ∈ r(Q). Then (xy)n = xnyn ∈ Q for some n ∈ N, so xn ∈ Q or yn ∈ r(Q). In the ﬁrst case, we see immediately that x ∈ r(Q). Otherwise, yn ∈ r(Q) implies that y ∈ r(r(Q)) = r(Q) by Lemma 1.1.4(c). Hence r(Q) is a prime ideal. Now, by Corollary 1.1.3, r(Q) = T{p0prime | Q ⊆ p0 ⊂ A} but since r(Q) is prime, it is the smallest prime containing Q. Deﬁnition. For a prime ideal p ⊂ A, any proper ideal Q ⊂ A is called p-primary if p = r(Q). Examples. 1 If A is a UFD and p ∈ A is a prime element, then (pn) is p-primary (the usual notation for (p)-primary) for any n ∈ N. 2 In general, a p-primary ideal need not be a power of p. Let A = k[x, y] and Q = (x, y2). Then A/Q ∼= k[y](y2) and in this quotient, the zero divisors are the multiplies of y, hence nilpotent. This implies Q is primary, and clearly its radical is p = (x, y). However, p2 ( Q ( p are strict inclusions, so Q is not a power of a prime ideal. 3 If p is a prime ideal, pn need not always be primary. Let A = k[x, y, z]/(xy − z2) and letx, ¯ y,¯ z¯ ∈ A be the respective images of x, y, z in the quotient. Set p = (¯x, z¯). Then p is a prime ideal of A since A/p ∼= k[y] is an integral domain. However,x ¯y¯ =z ¯2 ∈ p2, butx ¯ 6∈ p2 andy ¯ 6∈ r(p2) = p. So p2 is not primary. The next result shows that the counterexamples in 2 and 3 do not occur when we consider maximal ideals. Lemma 3.4.8. Let A be a noetherian ring. (a) If I ⊂ A is a proper ideal such that r(I) is maximal, then I is primary.

(b) If m ⊂ A is a maximal ideal, then mn is primary for all n ∈ N. Proof. (a) Assume r(I) is maximal. Then by Corollary 1.1.3, the set of all prime ideals in A containing I just consists of r(I). Thus B = A/I is a local ring with unique maximal ideal r(I)/I. If x, y ∈ B such that xy = 0 but y 6∈ r(I)/I, then since r(I)/I is the unique maximal ideal, y ∈ B×. So xy = 0 =⇒ xyy−1 = 0 =⇒ x = 0. Now if x0, y0 ∈ A such that x0y0 ∈ I but y0 6∈ r(I), then by the previous statement, (x0 + I)(y0 + I) = I =⇒ x0 + I = I =⇒ x0 ∈ I. This shows I is primary. (b) By Lemma 1.1.4, r(mn) = r(m) = m. By (a), this shows that mn is m-primary. Corollary 3.4.9. If m ⊂ A is a maximal ideal and I ⊂ A is any proper ideal, then I is m-primary if and only if there is some n ∈ N such that mn ⊆ I ⊆ m. Proof. ( =⇒ ) By Lemma 1.1.4, I ⊆ r(I) = m. Then by Lemma 3.4.1, r(I)n = mn ⊆ I for some n. ( ⇒ = ) Conversely, by Lemma 3.4.8(a), m = r(mn) ⊆ r(I) ⊆ r(m) = m so we have r(I) = m. By deﬁnition, this means I is m-primary.

48 3.4 Primary Decomposition 3 Noetherian and Artinian Rings

Tn Proposition 3.4.10. Assume I ⊂ A is a proper ideal and I = i=1 Qi for primary ideals Qi. If p1,..., pn are (not necessarily distinct) prime ideals such that Qi is pi-primary for each 1 ≤ i ≤ n, then

(a) For every prime ideal p ⊂ A such that p ⊇ I, p ⊇ pi for some 1 ≤ i ≤ n. (b) The minimal elements of the collection {p ⊂ A prime | p ⊇ I} are exactly the minimal n elements of {pi}i=1. (c) If A is noetherian, it only has ﬁnitely many minimal prime ideals. Tn Proof. (a) If p ⊇ I = i=1 Qi then by Lemma 1.1.4,

n n \ \ p = r(p) ⊇ r(I) ⊇ r(Qi) = pi. i=1 i=1

Thus p contains one of the pi. (b) If p ⊂ A is a minimal prime ideal containing I, then by (a), p ⊇ pi for some pi. But by minimality, p = pi. (c) Apply (b) and primary decomposition (3.4.6) to the ideal I = (0).

Lemma 3.4.11. If Q1,...,Qn are p-primary ideals for a ﬁxed prime ideal p ⊂ A, then Tn i=1 Qi is also p-primary. Tn Tn Tn Proof. By assumption r(Qi) = p for all 1 ≤ i ≤ n, so r ( i=1 Qi) = i=1 r(Qi) = i=1 p = p. Tn Tn Tn It remains to show i=1 Qi is primary. Assume x, y ∈ A with xy ∈ i=1 Qi but x 6∈ i=1 Qi. Then for each i, xy ∈ Qi but there is some j such that x 6∈ Qj. Since Qj is p-primary, we Tn Tn must have y ∈ r(Qi) = p = r ( i=1). Therefore i=1 Qi is primary. Tn Deﬁnition. A primary decomposition I = Qi is called minimal (or reduced) if for T i=1 every 1 ≤ i ≤ n, Qi 6⊃ j6=i Qj and r(Qi) 6= r(Qj) whenever i 6= j. By Theorem 3.4.6 and Lemma 3.4.11, minimal primary decompositions exist for all ideals I ⊂ A.

Example 3.4.12. Let A = k[x, y, z] and consider the ideals p1 = (x, y), p2 = (x, z) and ∼ ∼ m = (x, y, z). Then A/p1 = k[z] and A/p2 = k[y] which are integral domains, so p1 and p2 are prime. Also, A/m ∼= k which is a ﬁeld, so m is a maximal ideal of A. We claim that p1 ∩ p2 = (x, yz). Clearly x and yz both lie in (x, y) ∩ (x, z), so (x, yz) ⊆ p1 ∩ p2. On the other hand, for any f ∈ p1 ∩ p2, f ∈ (x, y) implies f = gx + hy for g, h ∈ A. But f ∈ (x, z) is possible if and only if h = h0z for some h0 ∈ A, in which case we have f = gx+h0yz ∈ (x, yz). Hence p1 ∩ p2 = (x, yz). 2 Now we show that p1p2 = p1 ∩ p2 ∩ m and this is a minimal primary decomposition 2 2 of p1p2. The left can be written p1p2 = (x , xy, xz, yz). Notice that x , xy, xz and yz all 2 2 2 lie in m , and since p1p2 ⊆ p1 ∩ p2, this shows that x , xy, xz, yz ∈ p1 ∩ p2 ∩ m . Hence 2 2 2 p1p2 ⊆ p1 ∩ p2 ∩ m . Going the other way, suppose F ∈ p1 ∩ p2 ∩ m . Then F ∈ m so F = ax2 + bxy + cxz + dy2 + eyz + fz2 for some a, b, c, d, e, f ∈ A. Notice that by the 2 paragraph above, ax + bxy + cxz + eyz ∈ p1p2 so to show that F ∈ p1p2, it will suﬃce to

49 3.4 Primary Decomposition 3 Noetherian and Artinian Rings

2 2 2 show that dy + fz ∈ p1p2. By the previous statement, ax + bxy + cxz + eyz ∈ p1 ∩ p2 z 2 2 so because F ∈ p1 ∩ p2, we must have dy + fz ∈ p1 ∩ p2. Observe that fz ∈ p2 so it 2 2 follows that dy ∈ p2. But y 6∈ p2 and this ideal is prime, so we must have d ∈ p2. A similar 2 2 2 2 argument shows that f ∈ p1. Thus we have dy ∈ p2p1 and fz ∈ p1p2, so dy + fz ∈ p1p2. This proves F ∈ p1p2, so p1 ∩ p2 ∩ m ⊆ p1p2. 2 Since p1 and p2 are prime, they are primary (3.4.5). Also, since m is maximal, m is primary (3.4.8(b)). Each term in the primary decomposition has a distinct radical, so we need only check that none of them contains the intersection of the remaining two. Notice:

2 x ∈ p1 ∩ p2 but x 6∈ m , 2 2 2 y ∈ p1 ∩ m but y 6∈ p2, 2 2 2 and z ∈ p2 ∩ m but z 6∈ p1.

2 This shows p1p2 = p1 ∩ p2 ∩ m is a minimal primary decomposition. Definition. Let I ⊂ A be an ideal and take x ∈ A. The colon ideal, or ideal quotient of I by x is defined as (I : x) = {y ∈ A | xy ∈ I}. If J ⊂ A is another ideal, we can define the ideal quotient of I by J as

(I : J) = {y ∈ A | yJ ⊆ I}.

The following lemma collects some basic facts about ideal quotients.

Lemma 3.4.13. Let I,J ⊂ A be ideals and x ∈ A. Then

(a) (I : J) is an ideal of A and I ⊆ (I : J).

(b) x ∈ I if and only if (I : x) = A. T T (c) For a collection of ideals {Ij} of A, j Ij : x = j(Ij : x).

(d) AnnA(x) = (0 : x).

(e) Let D = {z ∈ A | yz = 0 for some y ∈ A r {0}} be the set of zero divisors of A. Then [ [ [ D = (0 : x) = AnnA(x) = r(AnnA(x)). x6=0 x6=0 x6=0

Proof. (a) Since IJ ⊆ I, I ⊆ (I : J) is clear. To prove (I : J) is an ideal, let y, z ∈ (I : J) and r ∈ A. Then for any j ∈ J, yj, zj ∈ I and so (ry − z)j = ryj − zj = r(yj) − zj ∈ I. Hence (I : J) is an ideal. (b) This is the deﬁnition of an ideal I.

50 3.4 Primary Decomposition 3 Noetherian and Artinian Rings

⇐⇒ xy ∈ Ij for each Ij

⇐⇒ y ∈ (Ij : x) for each j \ ⇐⇒ y ∈ (Ij : x). j

(d) This is the deﬁnition of the annihilator. (e) The ﬁrst equality is trivial and the second is immediate from (d). To prove the third, we have AnnA(x) ⊆ r(AnnA(x)) by Lemma 1.1.4(a). On the other hand, for any n n y ∈ r(AnnA(x)), there is some n ≥ 1 such that y ∈ AnnA(x), i.e. xy = 0. Then y ∈ D so all equalities hold.

Lemma 3.4.14. If Q ⊂ A is a p-primary ideal and x ∈ ArQ then (Q : x) is also p-primary.

Proof. Fix x ∈ A r Q. Then for any y ∈ (Q : x), xy ∈ Q and since Q is primary, this implies y ∈ r(Q) = p. So (Q : x) ⊆ p. By Lemma 3.4.13(a), we have

p = r(Q) ⊆ r((Q : x)) ⊆ r(p) = p.

Therefore r((Q : x)) = p. It remains to show (Q : x) is primary. Suppose y, z ∈ A such that yz ∈ (Q : x) but z 6∈ r((Q : x)) = p. Then xyz ∈ Q but z 6∈ r(Q) = p, so xy ∈ Q since Q is primary. By deﬁnition this implies y ∈ (Q : x). Hence (Q : x) is p-primary. Tn Proposition 3.4.15. If i=1 Qi = 0 is a minimal primary decomposition of the zero ideal, Sn with Qi a pi-primary ideal for each 1 ≤ i ≤ n, then i=1 pi ⊆ D. T Proof. By minimality, for each 1 ≤ i ≤ n there is some xi ∈ j6=i Qj such that xi 6∈ Qi. Then by Lemma 3.4.13(c) and (d),

n ! n \ \ AnnA(xi) = (0 : xi) = Qj : xi = (Qj : xi) = (Qi : xi). j=1 j=1

By Lemma 3.4.14, (Qi : xi) is pi-primary for each 1 ≤ i ≤ n. Moreover, pi = r((Qi : xi)) = Sn r(AnnA(xi)) ⊆ D by Lemma 3.4.13(e), so i=1 pi ⊆ D as claimed. Corollary 3.4.16. If A is noetherian and p ⊂ A is a minimal prime ideal, then p ⊆ D.

Proof. By Theorem 3.4.6, (0) has a primary decomposition and we may assume it is minimal Tn by Lemma 3.4.11. Set 0 = i=1 Qi where Qi ⊂ A is a pi-primary ideal for some prime ideal pi. By Proposition 3.4.10, the minimal prime ideals of A are elements of {pi | 1 ≤ i ≤ n} and by Proposition 3.4.15, each pi ⊆ D. Thus all minimal prime ideals of A are contained in D.

51 3.4 Primary Decomposition 3 Noetherian and Artinian Rings

Lemma 3.4.17. If Qi is a pi-primary ideal for 1 ≤ i ≤ n and pi + pj = A whenever i 6= j, then n n Y \ Qi = Qi. i=1 i=1 Qn Tn Proof. We always have i=1 Qi ⊆ i=1 Qi so it suﬃces to show the reverse inclusion. First note that by Lemma 1.1.4(f), whenever j 6= i, we have

r(Qi + Qj) = r(r(Qi) + r(Qj)) = r(pi + pj) = r(A) = A.

So 1 ∈ r(Qi + Qj) which implies Qi + Qj = A. Now if n = 2 and x ∈ Q1 ∩ Q2 then the previous statement tells us that there exist q1 ∈ Q1, q2 ∈ Q2 such that

x = x · 1 = x(q1 + q2) = xq1 + xq2 ∈ Q1Q2.

Therefore the statement holds when n = 2. To induct, we need to show that Q1 ··· Qn−1 + Qn = A. By hypothesis, for each 1 ≤ i ≤ n − 1 there are elements xi ∈ Qi, yi ∈ Qn−1 such that xi + yi = 1. Then

n−1 Y 1 = (xi + yi) ∈ x1 ··· xn−1 + Qn ⊆ Q1 ··· Qn−1 + Qn. i=1

Therefore Q1 ··· Qn−1 + Qn = A. Finally, using the base case and inductive hypothesis, we have n n−1 ! n−1 ! n−1 ! \ \ \ Y Qi = Qi ∩ Qn = Qi ∩ Qn = Qi Qn. i=1 i=1 i=1 i=1 Hence the statement holds for any n. In the next proposition, we generalize the correspondence of Proposition 1.3.8 to a bijection on primary ideals.

Proposition 3.4.18. Let S ⊆ A be a multiplicatively closed subset and suppose Q ⊂ A is a p-primary ideal. Then

(a) If S ∩ p 6= ∅ then S−1Q = S−1p.

(b) If S ∩ p = ∅ then j−1(S−1Q) = Q and S−1Q is S−1p-primary. (c) The correspondence

Q0 ⊂ A is primary and Φ: ←→ {primary ideals of S−1A} r(Q0) ∩ S = ∅

given by Q 7→ S−1Q is bijective and inclusion-preserving.

52 3.5 Artinian Rings 3 Noetherian and Artinian Rings

n sn −1 Proof. (a) Take s ∈ S ∩ p. Then for some n ∈ N, s ∈ Q ∩ S which implies that 1 ∈ S Q, sn −1 −1 −1 but 1 is a unit in S A, so we must have S Q = S A. −1 −1 q −1 −1 −1 (b) The inclusion Q ⊆ j (S Q) is trivial: if q ∈ Q then 1 ∈ S Q, so q ∈ j (S Q) by −1 −1 x −1 deﬁnition of the map j. For the reverse inclusion, suppose x ∈ j (S Q). Then 1 ∈ S Q x q so there exist elements q ∈ Q, s ∈ S such that 1 = s . Hence for some t ∈ S, t(xs − q) = 0, i.e. stx = tq ∈ Q. Now stx ∈ Q but st ∈ S, so since S ∩ p = ∅, we must have st 6∈ p. Since Q is p-primary, this implies x ∈ Q. This proves Q = j−1(S−1Q). Next, r(Q) = p by assumption and r(S−1Q) = S−1(r(Q)) = S−1p, so it remains to −1 x y −1 xy −1 show S Q is a primary ideal. Suppose s , t ∈ S A such that st ∈ S Q. Then there are xy q elements u ∈ S, q ∈ Q such that st = u . Thus for some v ∈ S, v(uxy − stq) = 0, that is, vuxy = stq ∈ Q. Since vu ∈ S by multiplicative closure, vu 6∈ p as before, so xy ∈ Q because x −1 y −1 −1 Q is primary. Thus x ∈ Q or y ∈ r(Q), which implies s ∈ S Q or t ∈ S (r(Q)) = S p. Hence S−1Q is S−1p-primary. (c) By part (b), Φ is injective and for any primary ideal Q0 ⊂ A, S−1Q0 is primary in S−1A. Inclusion-preserving is as in Proposition 1.3.8. Suppose J ⊂ S−1A is a primary ideal. To show Φ is surjective, it’s enough to show I = j−1(J) is a primary ideal of A, since then Proposition 1.3.8(a) says that S−1I = S−1(j−1(J)) = J. Take x, y ∈ A such that xy ∈ I xy −1 x y but x 6∈ I. Then 1 ∈ S I = J so since J is primary, 1 ∈ J or 1 ∈ r(J). The former x −1 y is impossible, since 1 ∈ J = S I would imply x ∈ I. So we must have 1 ∈ r(J). Then yn n for some n ≥ 1, 1 ∈ J, and thus y ∈ I. Thus I is primary. We conclude that Φ is an inclusion-preserving bijection.

3.5 Artinian Rings

Recall from Section 3.1 that a (commutative) ring A is artinian if the descending chain condition holds on the set of ideals of A. We have seen many examples of noetherian rings, but it turns out that the artinian condition is much more restrictive. Assume that A 6= 0. Example 3.5.1. If k is a field and A is a finite dimensional k-algebra, then A is artinian. Proposition 3.5.2. Let A be artinian. Then (a) If A is an integral domain then A is a field. (b) dim A = 0. (c) J(A) = N(A). Proof. (a) Let x ∈ A, x 6= 0, be given. Then the ideals generated by powers of x form a descending chain in A: (x) ⊇ (x2) ⊇ (x3) ⊇ · · · By the descending chain condition, there is some n ∈ N such that (xn) = (xn+1). So there is a y ∈ A such that xn = xn+1y, that is, xn(1 − xy) = 0. Since A is a domain, xn 6= 0 so we must have 1 − xy = 0. Hence x is a unit. (b) If A is any artinian ring with a prime ideal p ⊂ A, then A/p is also artinian by Lemma 3.1.3. Since A/p is a domain, by (a) it is a field, so p is maximal. Thus all prime ideals in A are maximal, and thus also minimal, so dim A = 0.

53 3.5 Artinian Rings 3 Noetherian and Artinian Rings

(c) follows directly from the deﬁnitions of J(A) and N(A) as intersections of maximal and prime ideals. Lemma 3.5.3. If A is artinian then Spec(A) is ﬁnite.

Proof. Consider the collection Σ = {p1 ∩ · · · ∩ pn | pi ⊂ is a prime ideal and n ≥ 1}. Since A 6= 0, Σ is nonempty. By Lemma 3.1.1, Σ has a minimal element, say I = p1 ∩ · · · pn. For any prime ideal p ∈ Spec(A), I ∩ p ∈ Σ but by minimality, I = I ∩ p, so p ⊇ I. Since p is prime, we must have p ⊇ pi for some 1 ≤ i ≤ n. Now all prime ideals are maximal by Proposition 3.5.2(b), so p = pi. Hence Spec(A) = {p1,..., pn}.

Proposition 3.5.4. If A is artinian, then N(A)n = 0 for some n ∈ N. Proof. Consider the descending chain

N(A) ⊇ N(A)2 ⊇ N(A)3 ⊇ · · ·

By DCC, there is an n ∈ N such that N(A)m = N(A)n for all m ≥ n. Set I = N(A)n and assume I 6= 0. Then the collection Σ = {ideals J ⊂ A | IJ 6= 0} is nonempty. By Lemma 3.1.1, Σ contains a minimal element J0. Then IJ0 6= 0 so there exists an element x ∈ J0 so that xI 6= 0. This means (x) ∈ Σ, but by minimality of J0, we must have (x) = J0. Now I2 = N(A)2n = N(A) = I so we have 0 6= xI = xI2 = (xI)I. Thus xI ∈ Σ and xI ⊆ (x) = J0. By minimality, this implies xI = J0 = (x), and thus there is some y ∈ I so that x = xy. By induction, x = xym for all m ∈ N, but y ∈ I ⊆ N(A) so y is nilpotent. Thus for some m, ym = 0, giving us x = xym = 0, contradicting the initial assumption that x 6= 0. Hence I = N(A)n = 0. Theorem 3.5.5 (Hopkins). For a (commutative) ring A, the following are equivalent: (i) A is artinian.

(ii) A is noetherian and dim A = 0. Proof. By Proposition 3.5.2(b), A artinian directly implies dim A = 0. Assuming (i), Lemma 3.5.3 implies that A has ﬁnitely many maximal (and prime) ideals m1,..., mn. Then

N(A) = J(A) = m1 ∩ · · · mn ⊇ m1 ··· mn.

k k k By Proposition 3.5.4, there is some k ∈ N such that N(A) = 0, but (m1 ··· mn) ⊆ N(A) = 0 so by Proposition 3.1.7, A is noetherian. On the other hand, assuming (ii) implies that A has ﬁnitely many minimal prime ideals, by Proposition 3.4.10. But if dim A = 0, each minimal prime ideal is maximal as well. Thus by Corollary 3.4.2, N(A)k = 0 for some k ∈ N. Then once again the hypothesis of Proposition 3.1.7 is satisﬁed, so A is artinian. Theorem 3.5.6. Let A be a local noetherian ring with unique maximal ideal m. Then one of the following conditions holds:

(i) mn 6= mn+1 for all n ∈ N, and thus A is not artinian.

(ii) mn = 0 for some n ∈ N, in which case A is artinian.

54 3.5 Artinian Rings 3 Noetherian and Artinian Rings

Proof. Consider the descending chain

m ⊇ m2 ⊇ m3 ⊇

If mn 6= mn+1 for all n ∈ N, the chain is not stationary so A is not artinian. Clearly in this case mn 6= 0 for any n ∈ N, or else the chain would stabilize. Alternatively, if mn = mn+1 for some n, the locality condition implies mn+1 = J(A)mn. Since A is noetherian, mn is a ﬁnitely generated submodule of A, so Nakayama’s Lemma (1.2.1) implies mn = 0. Now if p ⊂ A is a prime ideal, 0 = mn ⊆ p so we must have m ⊆ p; otherwise, if x ∈ m r p then xn ∈ p, which is impossible. By maximality, m = p so m is in fact the unique prime ideal of A. Hence dim A = 0, so by Theorem 3.5.5, A is artinian. Examples. 1 If k is a ﬁeld, A = k[[t]] is noetherian by Theorem 3.3.9. Also, A is local with maximal ideal m = (t). In this ring, (0) is prime so dim A = 1. Therefore A cannot be artinian, so by Theorem 3.5.6, mn = (tn) 6= 0 for all n ∈ N.

2 Choose a prime p ∈ Z, an integer n ∈ N and set A = Z/pnZ. Since A is finite, it is local artinian with maximal ideal pZ/pnZ. 3 Let k be a field and consider the polynomial ring B = k[t]. If f ∈ B is a prime element, i.e. an irreducible polynomial in t, then A = B/(f n) is a local artinian ring with maximal ideal (f)/(f n). Proposition 3.5.7. Let A be a local artinian ring with maximal ideal m and residue field k = A/m. Then the following are equivalent: (a) Every ideal of A is principal. (b) m is principal.

2 2 (c) dimk m/m ≤ 1, where m/m is viewed as a k-vector space. Proof. (a) =⇒ (b) is trivial and (b) =⇒ (c) follows from the fact that if m = Ax, then m/m2 = k(x + m2) which has dimension at most 1. 2 To prove (c) =⇒ (a), note that if dimk m/m ≤ 1 then there is some x ∈ m such that m/m2 = k(x+m2). By Theorem 3.5.5, A is noetherian so in particular m is ﬁnitely generated as an A-module. Thus by Nakayama’s Lemma (1.2.1), with M = N = J(A) = m, we get N/MN = m/m2 =⇒ m = (x). Now Proposition 3.5.4 implies there exists an n ∈ N such that mn = N(A)n = 0. Take a nonzero ideal I ⊂ A. Then I ⊆ m and there is an ` ∈ N such that I ⊆ m` = (x`) but I 6⊆ m`+1 = (x`+1). So there is a y ∈ I such that y = ax` for some a ∈ A r m. Since A is local, this a must be a unit in A. Thus we have (y) ⊆ I ⊆ m` = (x`) = (y),

so I = (y) and I = m` as claimed. Proposition 3.5.8. Let k be a field and A a finitely generated k-algebra which is also a local artinian ring. Then dimk A is finite.

55 3.6 Associated Primes 3 Noetherian and Artinian Rings

Proof. For any local ring A which is finitely generated as a k-algebra, we have surjections k[t1, . . . , tr] → A → A/m so that A/m is also a finitely generated k-algebra. By the algebraic Nullstellensatz (2.4.2), A/m is a finite (algebraic) extension. Since A is artinian, A is also noetherian by Theorem 3.5.5. Then Theorem 3.5.6 shows that mn = 0 for some n ∈ N, in which case the chain of submodules m` ⊇ m`+1 eventually stabilizes: A ⊇ m ⊇ m2 ⊇ · · · ⊇ mn−1 ⊇ mn = 0. For each 0 ≤ ` ≤ n − 1, m` is a finitely generated A-module (since A is noetherian), so each quotient m`/m`+1 is a finite dimensional A/m-vector space. By the first paragraph and transitivity of dimension, we then have that

` `+1 ` `+1 dimk m /m = (dimk A/m)(dimA/m m /m ) < ∞ for each 0 ≤ ` ≤ n − 1. In particular, taking ` = n − 1 shows that mn−1/mn = mn−1 is finite dimensional over A/m. From linear algebra, (∗) if W ⊆ V is a vector subspace over a field F such that W and V/W are both finite dimensional F -vector spaces, then V is finite dimensional. Taking V = mn−2 and W = mn−1, we now have that mn−2 is a finite dimensional A/m-vector space. By induction, m` is finite dimensional for all 1 ≤ ` ≤ n−1. In particular, m1 = m is finite dimensional over A/m, and of course A/m is finite dimensional over itself, so by (∗), A is a finite dimensional A/m-vector space. This proves the statement. The structure of commutative artinian rings is completely described in the next theorem.

Theorem 3.5.9. Any artinian ring A is the ﬁnite direct product of local artinian rings, and the factors are uniquely determined by A up to isomorphism.

Corollary 3.5.10. If A is a ﬁnitely generated k-algebra which is artinian, then dimk A is ﬁnite.

Proof. By Theorem 3.5.9, A is the direct product of ﬁnitely many local artinian rings, and each of these has ﬁnite dimension by Proposition 3.5.8.

3.6 Associated Primes

Deﬁnition. A prime ﬁltration of an A-module M is a sequence of submodules

0 = M0 ⊆ M1 ⊆ · · · ⊆ Mn = M ∼ such that for each 0 ≤ i ≤ n − 1, Mi+1/Mi = R/pi+1 for some prime ideals p1,..., pn ⊂ A. Deﬁnition. Let M be an A-module. A prime ideal p ⊂ A is an associated prime of M if there is an embedding of A-modules A/p ,→ M. Let Ass(M) denote the set of all associated primes of M.

Lemma 3.6.1. For any A-module M,

Ass(M) = {p ⊂ A prime | there is some nonzero x ∈ M with AnnA(x) = p}.

56 3.6 Associated Primes 3 Noetherian and Artinian Rings

Proof. If p ∈ Ass(M) then x = 1 + p ∈ A/p ,→ M is a nonzero element of M and by definition AnnA(x) = p. Conversely, if p = AnnA(x) for some x ∈ M, then the natural map ∼ ϕx : A → M, a 7→ ax has kernel p so it factors through A/p = im ϕx ⊆ M. We will prove that every A-module has a prime filtration, but first we need a lemma. Lemma 3.6.2. For all nonzero A-modules M, Ass(M) is nonempty. Proof. Consider the set

Λ = {I = AnnA(x) | x ∈ M, x 6= 0}.

Then since M 6= 0, Λ is nonempty. Let I = AnnA(x) be a maximal element of Λ; we claim I is a prime ideal of A. Suppose a, b ∈ A such that ab ∈ I, so that abx = 0. If a, b 6∈ I then ax 6= 0, but then b ∈ AnnA(ax) which shows I ( AnnA(ax). This contradicts maximality of I, so either a or b lies in I, showing I is prime. Theorem 3.6.3. If A is noetherian, then every finitely generated A-module has a prime filtration. Proof. Let M be a finitely generated A-module and consider the collection of submodules

Γ = {N ⊆ M | N has a prime ﬁltration}.

By Lemma 3.6.2, there exists a prime p ∈ Ass(M) so the submodule A/p ,→ M has trivial prime ﬁltration 0 ⊆ A/p. Thus Γ is nonempty. Let N be a maximal element of Γ, say with prime ﬁltration 0 ⊆ N1 ⊆ · · · ⊆ Nk = N.

If N 6= M, use Lemma 3.6.2 again to choose p ∈ Ass(M/N) with p = AnnA(x + N) for some x ∈ M r N. Then N + Ax has a prime ﬁltration:

0 ⊆ N1 ⊆ · · · ⊆ Nk = N ( N + Ax ∼ with (N + Ax)/N = AnnA(x + N) = p, contradicting maximality of N. Hence M = N. Lemma 3.6.4. For every short exact sequence of A-modules 0 → K → M → N → 0,

Ass(M) ⊆ Ass(N) ∪ Ass(K).

Proof. Suppose p ∈ Ass(M) so that by Lemma 3.6.1, p = AnnA(y) for some nonzero y ∈ M. If y ∈ N then p ∈ Ass(N) already. Otherwise, u = y + N ∈ M/N is a nonzero element of K. Suppose p 6= AnnA(u). Then there is some s ∈ AnnA(u) such that s 6∈ p and 0 = su = sy+N in K, i.e. sy ∈ N. Then p = AnnA(y) ⊆ AnnA(sy). On the other hand, if t ∈ AnnA(sy) then sty = 0, so st ∈ p but since s 6∈ p and p is prime, this means t ∈ p. Hence p = AnnA(y) or AnnA(sy), so p lies in either Ass(K) or Ass(N). Corollary 3.6.5. For any A-module M, Ass(M n) = Ass(M) for all n ≥ 1. Example 3.6.6. If A is an integral domain, then Ass(A) = {0} (by Lemma 3.6.1 for example).

57 3.6 Associated Primes 3 Noetherian and Artinian Rings

Theorem 3.6.7. Let A be a noetherian ring and M a ﬁnitely generated A-module. Then

(1) Let p ⊂ A be a prime ideal. Then p ∈ Ass(M) if and only if pAp ∈ Ass(Mp). (2) | Ass(M)| < ∞.

(3) If p is minimal among prime ideals containing AnnA(M) then p ∈ Ass(M). Proof. (1) On one hand, if p ∈ Ass(M) then A/p ,→ M by deﬁnition, but since localization is exact, (A/p)p ,→ Mp as well so pAp ∈ Ass(Mp). Conversely, suppose pAp ∈ Ass(Mp). By

Lemma 3.6.1, pAp = AnnAp (α) for some nonzero α ∈ Mp. Write p = (x1, . . . , xn) so that xiα = 0 for all 1 ≤ i ≤ n. Then there exist elements si ∈ A r p such that sixiα = 0. Set s = s1 ··· sn. Since p is prime, s 6∈ p and we have sα ∈ M and xi(sα) = sxiα = 0 for each i. Thus p ⊆ AnnA(sα). If t ∈ AnnA(sα) r p then we would have (ts)α = 0 but this implies α = 0 in Mp, a contradiction. Hence p = AnnA(sα) so p ∈ Ass(M). (2) By Theorem 3.6.3, M has a prime ﬁltration

0 = M0 ⊆ M1 ⊆ · · · ⊆ Mn = M ∼ ∼ with Mi+1/Mi = A/pi+1 for some primes p1,..., pn ⊂ A. If n = 1, notice that M1 = A/p1 is a domain so Ass(M1) = {p1} by Example 3.6.6. For n ≥ 2, we have a short exact sequence

0 → Mn−1 → M → M/Mn−1 → 0

so Lemma 3.6.4 says that Ass(M) ⊆ Ass(Mn−1) ∪ Ass(M/Mn−1). By induction, Ass(Mn−1) is finite, and Ass(M/Mn−1) = Ass(A/pn) = {pn} so Ass(M) is finite as well. (3) Suppose p is minimal among all prime ideals of A containing AnnA(M). Then Mp 6= 0 and Mp is an (A/ AnnA(M))p-module, so we may assume AnnA(M) = 0. Now p is minimal, so Spec(Ap) = {pAp}. By (1), we may replace A by Ap and M by Mp. Consider the filtration

M ⊇ pM ⊇ p2M ⊇ · · ·

Since A is noetherian and Spec(A) = {p}, Theorem 3.5.5 shows that A is artinian and thus pn+1M = pnM for some smallest n ≥ 1. Then p(pnM) = pnM so Nakayama’s lemma (1.2.1) n n−1 says that p M = 0. Take y ∈ p M, y 6= 0. Then p ⊆ AnnA(y) but since Spec(A) = {p}, p is a maximal ideal and hence p = AnnA(y). So by Lemma 3.6.1, p ∈ Ass(M).

Remark. The proof of (2) shows that if p1,..., pn are the primes corresponding to a prime ﬁltration of M, then Ass(M) ⊆ {p1,..., pn}. Example 3.6.8. Let A = k[x, y](x2 − y3) and deﬁne an A-module M by the free resolution

 2x  −3y2 0 → A −−−−−−→ A2 → M → 0.

(M is called the module of K¨ahlerdiﬀerentials of A). We claim the only associated primes of M are 0 and (x, y). Let p be a prime ideal of A and consider the localized sequence

2 0 → Ap → Ap → Mp → 0

58 3.6 Associated Primes 3 Noetherian and Artinian Rings

which remains exact since localization is exact. If p 6= (x, y) then x 6= 0 or y 6= 0, so 2x 1 0 the matrix can be put into the form or after a change of basis. Thus −3y2 0 1 ∼ Mp = Ap so Ass(Mp) = {0} and thus by Theorem 3.6.7(1), p 6∈ Ass(M). On the other hand, if p = (x, y) then consider the element

2x3 m = ∈ M. −3y5

Notice that 2x4 2x4 2x2 xm = = = x2 = 0 −3xy5 −3x2y2 −3y2 2x3y 2x3y 2x while ym = = = x2y = 0. −3y6 −3x2y3 −3y2

Hence AnnA(m) = (x, y) so (x, y) is an associated prime of M. Proposition 3.6.9. For a noetherian ring A and a ﬁnitely generated A-module M, [ p = {a ∈ A | ax = 0 for some nonzero x ∈ M}. p∈Ass(M)

Proof. Let Z be the set of all a ∈ A for which ax = 0 for some x 6= 0 in M. It is clear that any p ∈ Ass(M) is contained in Z, so one inclusion is true. On the other hand, suppose a ∈ Z with ax = 0 for some nonzero x ∈ M. The set

Λ = {AnnA(bx) | b ∈ A, b 6= 0}

is nonempty since AnnA(ax) ∈ Λ, so it has a maximal element, say q = AnnA(bx) for some b 6= 0. If we can show q is prime, then we are done by Lemma 3.6.1. Suppose uv ∈ q but u 6∈ q. Then ubx 6= 0 and q ⊆ AnnA(ubx), so by maximality, q = AnnA(ubx). Hence v ∈ AnnA(ubx) = q which shows q is prime as desired.

59 4 Discrete Valuations and Dedekind Domains

4 Discrete Valuations and Dedekind Domains

4.1 Discrete Valuation Rings

Deﬁnition. A discrete valuation on a ﬁeld K is a surjective group homomorphism v : K× → Z such that v(x + y) ≥ min(v(x), v(y)) for all x, y ∈ K×. We often extend v to all of K by setting v(0) = ∞.

0 Lemma 4.1.1. Let A be an integral domain with ﬁeld of fractions K. If v : A r {0} → N0 is a function satisfying

(i) v0(xy) = v0(x) + v0(y)

(ii) v0(x + y) ≥ min(v0(x), v0(y)) for all x, y ∈ A r {0} and v0(0) = ∞ when necessary, then there exists a unique discrete × 0 0 valuation v : K → Z which extends v , i.e. v|A = v .

× a 0 0 Proof. Define v : K → Z by v b = v (a)−v (b). Then one can check that v is well-defined, a surjective homomorphism and satisfies v(x + y) ≥ min(v(x), v(y)).

Corollary 4.1.2. If A is a UFD with ﬁeld of fractions K and p ∈ A is a prime element, × n there is a discrete valuation vp : K → Z such that for any a ∈ A, vp(a) = n if p | a and pn+1 - a.

Proof. It’s enough to show that vp satisfies conditions (i) and (ii) of Lemma 4.1.1. Take x, y ∈ A r {0} and write x = pnx0 and y = pmy0 where n, m ≥ 0 and p - x0, y0. Then n+m 0 0 0 0 xy = p x y and since p is prime, p - x y . Thus vp(xy) = n + m, so (i) is satisfied. On the other hand, suppose without loss of generality that n ≥ m, so that min(v(x), v(y)) = m. Then x + y = pnx0 + pmy0 = pm(pn−mx0 + y0) but since p | pn−mx0 and p - y0, we have n−m 0 0 p - (p x + y ). Thus vp(x + y) = m and condition (ii) is satisfied.

× Deﬁnition. For a prime element p in a UFD A, the valuation vp : K → Z is called the p-adic valuation on A (or on K).

Examples.

1 For A = Z and K = Q, and p a prime integer, the classical p-adic valuation is defined a n by vp b = vp(a) − vp(b), where vp(a) = n if a = p m, p - m. 2 Let k be a field and consider the field of rational polynomials K = k(t). The so- × called “infinite place” of K is a discrete valuation v∞ : K → Z defined by the degree f f × function: v∞(f) = − deg f for f ∈ k[t] and v∞ g = deg g − deg f for g ∈ K . For A = k[t−1], the prime element t−1 induces the same valuation on Frac(A) = k(t): vt−1 = v∞.

60 4.1 Discrete Valuation Rings 4 Discrete Valuations and Dedekind Domains

3 For k a ﬁeld, consider the ﬁeld of rational power series

−n K = k((t)) = {t f | n ∈ N0, f ∈ k[[t]]}.

P∞ i Deﬁne a function v on K by v(x) = m ∈ Z if x = i=m ait for ai ∈ k and am 6= 0. Formally let v(0) = ∞ so that v : K → Z is a discrete valuation. As in 2 , the prime element t in the UFD A[[t]] induces this valuation.

Deﬁnition. If v : K× → Z is a discrete valuation, then O(v) = {v ∈ K | v(x) ≥ 0} is called the valuation ring of v.

Lemma 4.1.3. For any discrete valuation v : K× → Z, O(v) is a subring of K. Proof. By convention, v(0) = ∞ so 0 ∈ O(v). If x, y ∈ O(v) then by deﬁnition v(xy) = v(x)+v(y) ≥ 0 so xy ∈ O(v). Likewise, v(x+y) ≥ min(v(x), v(y)) ≥ 0 shows that xy ∈ O(v). Hence O(v) ⊆ K is a subring.

Proposition 4.1.4. An integral domain A is a DVR if there exists a discrete valuation v on K = Frac(A) such that O(v) = A.

Proof. Suppose v : K× → Z is a discrete valuation with A = O(v). By Lemma 4.1.3, A is a ring so it is enough to show that A is a DVR. If x ∈ K× such that x−1 6∈ A, then v(x−1) < 0. Note that the deﬁnition of a valuation forces v(1) = 0 since v(1) = v(1 · 1) = v(1) + v(1), so for such an x we have

0 = v(1) = v(xx−1) = v(x) + v(x−1) < v(x) + 0

and so v(x) > 0. Thus x ∈ A, so A is a DVR. Examples, continued.

1 For a prime p and the p-adic valuation vp : Q → Z, the valuation ring O(vp) = a b : a, b ∈ Z, b 6= 0, p - b coincides with Z(p), the p-adic integers described in Exam- ple 1.2.2.

2 For v∞ on k(t), the valuation ring is similar to the p-adic integers:

f O(v ) = : f, g ∈ k[t], g 6= 0, deg g ≥ deg f . ∞ g

3 For the valuation v on k((t)), O(v) = k[[t]].

Proposition 4.1.5. Let v : K× → Z be a discrete valuation with valuation ring O. Then (1) O× = {x ∈ K | v(x) = 0}.

(2) O is a local ring with unique maximal ideal m = O r O× = {x ∈ K | v(x) > 0}. (3) O is a DVR.

61 4.1 Discrete Valuation Rings 4 Discrete Valuations and Dedekind Domains

(4) If π ∈ O satisﬁes v(π) = 1, then m = (π).

(5) Every proper ideal I ⊂ O is of the form I = (πm) = {x ∈ K | v(x) ≥ m}, where m = min(v(x) | x ∈ I).

(6) O is a PID with unique prime elements of the form uπ, where u is a unit.

(7) Spec(O) = {(0), m} and hence dim O = 1.

Proof. Exercise.

Deﬁnition. For a valuation ring O = O(v), an element π ∈ O such that m = (π) is the unique maximal ideal of O is called a uniformizer of O.

Remark. Suppose A is a local noetherian domain.

(1) If dim A = 1 then the fact that A is local and a domain implies Spec(A) = {0, m}, where m is the maximal ideal of A.

(2) Further, if I ⊂ A is a nonzero proper ideal then Spec(A/I) = {m/I} by the correspondence theorem. Thus dim A/I = 0. By Lemma 3.1.3, A noetherian implies A/I is noetherian as well. Thus by Theorem 3.5.5, A/I is also artinian.

(3) Now m/I = N(A/I) so by Proposition 3.5.4, (m/I)` = 0 for some ` ∈ N, and hence m` ⊆ I ⊆ m. This shows that r(I) = m, i.e. that I is m-primary by Lemma 3.4.8.

(4) Finally, since dim A = 1, Theorem 3.5.5 says that A is not artinian, so by Theorem 3.5.6, ` `+1 m 6= m for any ` ∈ N0. The next result completely characterizes local noetherian DVRs.

Theorem 4.1.6. Let A be a local noetherian domain with maximal ideal m and residue ﬁeld k = A/m, such that dim A = 1. Then the following are equivalent:

(i) A is a DVR.

(ii) A is integrally closed.

(iii) m is a principal ideal.

2 (iv) dimk m/m = 1.

(v) For all nonzero proper ideals I ⊂ A, there exists an n ∈ N such that I = mn. (vi) There exists an x ∈ A such that for any nonzero proper ideal I ⊂ A, I = (xn) for some n ∈ N.

62 4.1 Discrete Valuation Rings 4 Discrete Valuations and Dedekind Domains

Proof. (i) =⇒ (ii) If A is a DVR then A is a PID by (6) of Proposition 4.1.5, so A is a UFD. By Proposition 2.1.8, this proves A is integrally closed. (ii) =⇒ (iii) Choose a nonzero element a ∈ m. By Remark (3), there is some ` ∈ N with m` ⊆ (a) and m`−1 6⊂ (a). Choose b ∈ m`−1 with b 6∈ (a) and set x = ab ∈ k. Then −1 b −1 x = a 6∈ A since b 6∈ (a). Since A is integrally closed, this means x is not integral over A. Suppose x−1m ⊂ m. Then m is an A[x−1]-module, m is faithful since m 6= 0 and A[x−1] ⊆ k, and m is ﬁnitely generated as an A-module because A is noetherian. Therefore the conditions of Lemma 2.1.2(iv) are satisﬁed, showing x−1 is integral over A, a contradiction. Thus we must have x−1m 6⊂ m. Now b 1 1 1 x−1m = m = (bm) ⊆ m` ⊆ (a) = A, a a a a showing x−1m is an ideal of A. But x−1m 6⊂ m so by maximality we must have x−1m = A, i.e. m = (x). 2 2 2 (iii) =⇒ (iv) If m = (x) then m/m = k(x + m ) as a k-vector space, so dimk m/m . By 2 2 (4) of the Remark, m 6= m , so we must have dimk m/m = 1. (iv) =⇒ (v) By (3) of the Remark, there exists an ` ∈ N such that m` ⊆ I ⊆ m. Without loss of generality, assume ` ≥ 2 and consider the quotient A/m`. Then by (2) of the Remark, A/m` is local artinian. Thus

` 2 ` 2 dimk(m/m )/(m /m ) = dimk m/m = 1 by the isomorphism theorems. By Proposition 3.5.7, this implies I/m` = (m/m`)n = mn/m` for some n ≤ `. However, I ⊇ m` and mn ⊇ m` so we must have I = mn. (v) =⇒ (vi) By Remark (4), m 6= m2 so we can find some element x ∈ m r m2. By assumption, there is some n ∈ N such that (x) = mn. Since x 6∈ m2, n has to equal 1, so m = (x). Hence for any nonzero ideal I ⊂ A, I = m` = (x`) for some ` ∈ N. (vi) =⇒ (i) Suppose m = (xn) for some x ∈ A and n ∈ N. Since m is maximal, x ∈ m so we have (x) ⊆ m = (xn) ⊆ (x) and thus m = (x), that is, m is principal. By Remark (4), ` `+1 (x ) 6= (x ) for any ` ∈ N0. This means for any nonzero element a ∈ A, there is a unique ` ` ∈ N0 such that (a) = (x ) by assumption. Define v : A r {0} → N0 by v(a) = ` where ` is ` chosen as described. This map is surjective because x 7→ ` for any ` ∈ N0. If a, b ∈ A such that (a) = (x`) and (b) = (xm), then (ab) = (x`)(xm) = (x`+m) =⇒ v(ab) = v(a) + v(b). Moreover, suppose v(a) = ` and v(b) = m. Then (a) = (x`) and (b) = (xm) so there exist u, w ∈ A× such that a = ux` and b = wxm. Without loss of generality assume ` ≤ m. Then a + b = ux` + wxm = x`(u + vxm−`) and u + vxm−` ∈ A, so this shows a + b ∈ (x`) and hence (a + b) ⊆ (x`). If (a + b) = (xr), we have (xr) ⊆ (x`) so in particular r ≥ ` and thus v(a + b) ≥ ` = v(a). Now by Lemma 4.1.1, v extends uniquely to a discrete valuation v : K× → Z. It remains to verify that O(v) = A and we will be finished. By construction A ⊆ O(v) since a v|Ar{0} : A r {0} → N0. On the other hand, if y ∈ O(v) is nonzero then y = b for some a, b ∈ A r {0}. Then 0 ≤ v(y) = v(a) − v(b), that is, v(a) ≥ v(b). If v(a) = ` and v(b) = m as above, then (a) = (x`) and (b) = (xm), and since ` ≥ m, we have (a) ⊆ (b). Hence a ∈ (b) a so y = b ∈ A. This completes the proof that A = O(v), i.e. A is a valuation ring.

63 4.2 Dedekind Domains 4 Discrete Valuations and Dedekind Domains

4.2 Dedekind Domains

Definition. An integral domain A is called a Dedekind domain if (i) A is noetherian. (ii) dim A = 1, i.e. every prime ideal is maximal. (iii) A is integrally closed. Notice that the Krull dimension requirement means that a field is not a Dedekind domain. Example 4.2.1. Every PID that is not a field is a Dedekind domain. Dedekind domains are the central object in algebraic number theory, as one can construct a canonical Dedekind domain OK , called the ring of integers, for any number field K ⊇ Q. In this way the axioms and properties of Dedekind domains were devised in an attempt to generalize the algebraic properties of Z. Proposition 4.2.2. If A is a noetherian domain of Krull dimension 1, the following are equivalent: (i) A is integrally closed, and hence a Dedekind domain.

(ii) Ap is a DVR for every nonzero prime ideal p ⊂ A. (iii) Every primary ideal in A is a prime power.

Proof. (i) =⇒ (ii) It follows from Proposition 2.3.12 that Ap is integrally closed for every prime ideal p in A. Also, by Lemma 2.3.6, dim Ap = ht(p) = 1 and by Lemma 3.3.1, Ap is noetherian, so Ap is a DVR by Lemma 3.3.1. (ii) =⇒ (i) If Ap is a DVR then it is a PID, hence a UFD, hence integrally closed by Proposition 2.1.8. Since this holds for all nonzero prime ideals p, Proposition 2.3.12 implies that A is also integrally closed. (ii) =⇒ (iii) Take a p-primary ideal 0 6= Q ⊂ A so that p = r(Q). Then p is maximal −1 −1 since dim A = 1. For S = Arp, S Q is a proper ideal in S A = Ap. By Proposition 3.4.18, −1 −1 −1 −1 n S Q is S p-primary in Ap. Now Ap is a DVR, so S Q = (S p) for some n ∈ N according to Theorem 4.1.6. Since p is maximal, Lemma 3.4.8 says that pn is p-primary, and by Lemma 1.3.10, (S−1p)n = S−1pn, so Proposition 3.4.18 implies Q = pn. (iii) =⇒ (ii) We are assuming A is a noetherian domain, so Ap is a local noetherian domain for all nonzero prime ideals p ∈ Spec(A) = MaxSpec(A) by Lemma 3.3.1. Then dim A = 1 implies dim Ap = 1. We want to show Ap is a DVR, so by Theorem 4.1.6, it’s −1 enough to show that all nonzero ideals in Ap are powers of S p, where S = A r p. Take −1 some nonzero ideal I ⊂ Ap. By Remark (3) preceding Theorem 4.1.6, I is S p-primary so Proposition 3.4.18 says there exists a primary ideal Q ⊂ A, Q ∩ S = ∅, with S−1Q = I. Since Q ∩ S = ∅, Q ⊆ p and thus r(Q) = p since p is the only prime ideal of A containing Q. By assumption, Q = pn for some n ∈ N, and thus Lemma 1.3.10 gives us I = S−1Q = S−1pn = (S−1p)n.

−1 Hence I is a power of S p and we conclude that Ap is a DVR.

64 4.3 Fractional and Invertible Ideals 4 Discrete Valuations and Dedekind Domains

Corollary 4.2.3. If A is a Dedekind domain, then any nonzero ideal of A is a product of prime ideals in A. T` Proof. Suppose I ⊂ A is a nonzero ideal. Since A is noetherian, I = i=1 Qi for primary ideals Qi ⊂ A with r(Qi) = pi prime, by Theorem 3.4.6. By Lemma 3.4.11, we may take this to be a minimal primary decomposition of I, so pi 6= pj for i 6= j. Since dim A = 1, each T` Q` pi is maximal so pi + pj = A for all i 6= j. Hence I = i=1 Qi = i=1 Qi by Lemma 3.4.17. Now by Proposition 4.2.2(iii), each primary ideal Qi is a power of a prime ideal, namely its Q` ni radical r(Qi) = pi. So there exist n1, . . . , n` ∈ N such that I = i=1 pi . We will prove in the next section that nonzero ideals in a Dedekind domain factor into a unique product of prime ideals. This is the most important feature of Dedekind domains, as it allows us to generalize the unique factorization properties of Z to similar objects in number theory which turn out to be Dedekind domains. For example, taking the integral closure of Z in a finite separable extension of Q preserves the Dedekind property: Theorem 4.2.4. If A is a Dedekind domain with field of fractions K and L/K is a finite separable field extension, then the integral closure B of A in L is a Dedekind domain.

Proof. First, B is integrally closed by Corollary 2.1.7. Moreover, since B/A is an integral extension, dim B = dim A = 1 by Theorem 2.3.7. Finally, since L/K is separable, Proposi- Pn tion 2.2.5 says that there is a K-basis {v1, . . . , vn} of L such that B ⊆ i=1 Avi. Since A is noetherian, we get that B is noetherian as well. Hence B is a Dedekind domain.

4.3 Fractional and Invertible Ideals

In this section we prove that every nonzero ideal in a Dedekind domain factors uniquely into a product of prime ideals. In order to prove this, we need to expand the notion of an ideal to include other submodules of the fraction ﬁeld. Let A be an integral domain with ﬁeld of fractions K. Then A-ideals are also A-submodules of K.

Deﬁnition. For two A-submodules I,J ⊆ K, the generalized ideal quotient of I by J with respect to K is (I :K J) = {x ∈ K | xJ ⊆ I}.

Lemma 4.3.1. For every pair of submodules I,J ⊆ K, (I :K J) is an A-submodule of K. Lemma 4.3.2. Let I,J ⊆ K be submodules. If S is any multiplicatively closed subset of A, −1 −1 −1 then S (I :K J) = (S I :K S J) if J is a ﬁnitely generated A-module. Deﬁnition. A nonzero A-submodule I ⊆ K is called a fractional ideal of A if there is some x ∈ K× such that xI ⊆ A. Equivalently, I is fractional if there is an ideal J ⊂ A such that I = yJ for some y ∈ K×.

Deﬁnition. A nonzero A-submodule I ⊆ K is an invertible ideal of A if there is an A-submodule J ⊆ K such that IJ = A.

Proposition 4.3.3. Let I ⊆ K be a nonzero A-submodule. Then

65 4.3 Fractional and Invertible Ideals 4 Discrete Valuations and Dedekind Domains

(a) If I is ﬁnitely generated then I is a fractional ideal.

(b) If A is noetherian and I is fractional, then I is ﬁnitely generated.

(c) If I is invertible with IJ = A, then J = (A :K I) and J is unique. (d) If I is invertible then I is ﬁnitely generated and therefore fractional.

Pn ai Proof. (a) Let I = Axi where xi = for ai, bi ∈ A {0}. Then b1 ··· bnI ⊆ A so I is i=1 bi r fractional. (b) If A is noetherian, then any A-ideal J is finitely generated, so if I = yJ for J ⊂ A and y ∈ K× then I is clearly finitely generated as an A-module. (c) Suppose IJ = A for some submodule J ⊆ K. Then J ⊆ (A :K I) by definition of the generalized ideal quotient. On the other hand,

(A :K I) = (A :K I)A = (A :K I)IJ ⊆ AJ = J.

Therefore we have J = (A :K I) and this implies uniqueness since J was arbitrary. (d) Suppose IJ = A for some submodules J ⊆ A. Since 1 ∈ A = IJ, there are xi ∈ I Pn and yi ∈ J such that 1 = i=1 xiyi. For each x ∈ I, we have

n n X X x = x · 1 = x xiyi = xi(xyi). i=1 i=1 Pn Pn Since each xyi ∈ IJ = A, we have i=1 xi(xyi) ∈ i=1 Axi ⊆ I. Therefore x ∈ I, so Pn I = i=1 Axi, meaning I is ﬁnitely generated. Finally, (a) implies that I is a fractional A-ideal.

−1 If I is an invertible ideal, we will write I = (A :K I). When I ⊂ A is an ideal, we will call it an integral ideal to distinguish from fractional ideals that are not submodules of A. −1 Take a nonzero prime ideal p ∈ Spec(A) and consider the local ring Ap = S A, where −1 S = A r p. If I ⊆ K is an A-submodule, we will write Ip = S I. Proposition 4.3.4. For an A-submodule I ⊆ K, the following are equivalent:

(i) I is an invertible ideal of A.

(ii) I is ﬁnitely generated as an A-module and Ip is an invertible Ap-ideal for all prime ideals p ∈ Spec(A).

(iii) I is a ﬁnitely generated A-module and Im is an invertible Am-ideal for all maximal ideals m ∈ MaxSpec(A).

Proof. (i) =⇒ (ii) By Proposition 4.3.3(d), I is ﬁnitely generated as an A-module. If IJ = A for a submodule J ⊆ K, then by Lemma 1.3.10(a), we have Ap = IpJp so in particular Ip is an invertible ideal of Ap. (ii) =⇒ (iii) is trivial.

66 4.3 Fractional and Invertible Ideals 4 Discrete Valuations and Dedekind Domains

(iii) =⇒ (i)) Set L = I(A :K I). Then L ⊆ A is an ideal. For any maximal ideal m ⊂ A, set S = A r m. Then

−1 −1 −1 Lm = ImS (A :K I) = Im(S A :K S I) by Lemma 4.3.2 −1 = Im(Am :K Im) = ImIm = Am.

−1 In particular Lm is not contained in the maximal ideal S m, so either L ∩ S 6= ∅ or L 6⊂ m. However, if Lm ⊂ Am is an ideal, then L ∩ S = ∅, so we must have L 6⊂ m. Since m is maximal, it follows that L = A, i.e. I is invertible.

Proposition 4.3.5. For a local domain A which is not a ﬁeld, A is a DVR if and only if every fractional ideal of A is invertible.

Proof. (i) =⇒ (ii) Suppose A is a DVR. By Proposition 4.1.5, A is a PID so every fractional ideal of A is of the form Ax for some x ∈ K×. The inverse of Ax is Ax−1, so all fractional ideals of A are also invertible. (ii) =⇒ (i) Every nonzero (integral) ideal of A is fractional, so by assumption they are all invertible. Then Proposition 4.3.3(d) implies every ideal of A is ﬁnitely generated, so by Proposition 3.1.2, A is noetherian. Let m be the unique maximal ideal of A. Then there exists a fractional ideal m ⊆ K such that mm−1 = A. We claim that every ideal of A is of the form mn for some n ∈ N. Let J ⊂ A be a nonzero integral ideal. Then J ⊆ m so Jm−1 ⊆ mm−1 = A, which shows that Jm−1 is an integral ideal of A. Moreover, A ( m implies J ( Jm−1, for if J = Jm−1 then we would have Jm = JA = J, and thus m = 0 by Nakayama’s Lemma (1.2.1). Now starting with any ideal J ⊂ A, we get a strictly ascending chain of ideals −1 −2 J ( Jm ( Jm ( Since A is noetherian, the chain stabilizes, i.e. Im−n = A for some n ∈ N. Therefore I = mn for this n. This implies Spec(A) = {0, m} so dim A = 1. Finally, Theorem 4.1.6 shows that A is a DVR. Our results in the previous two sections combine to give us the following characterization of Dedekind domains.

Theorem 4.3.6. For an integral domain A which is not a ﬁeld, the following are equivalent:

(i) A is a Dedekind domain.

(ii) Every fractional ideal of A is invertible.

Proof. (i) =⇒ (ii) Suppose A is Dedekind. Then by Proposition 4.2.2, Ap is a DVR for every prime ideal p ∈ Spec(A) = MaxSpec(A). Thus Proposition 4.3.5 implies every fractional ideal Ip is invertible (as an Ap-module) for any such prime ideal p. Finally, since A is noetherian, every ideal I is ﬁnitely generated (3.1.2) and hence Proposition 4.3.4 implies I is an invertible ideal. (ii) =⇒ (i) If every fractional ideal of A is invertible, then each is ﬁnitely generated by Proposition 4.3.3(d), so A is noetherian. We want to show Ap is a DVR for all maximal ideals m ∈ MaxSpec(A). By Proposition 4.3.5, it’s enough to show that every fractional

67 4.3 Fractional and Invertible Ideals 4 Discrete Valuations and Dedekind Domains

0 0 −1 ideal of Am is invertible. Let I ⊂ Am be a nonzero ideal. Then I = S I = Im for some nonzero ideal I ⊂ A. By assumption I is invertible, so Im is invertible. It follows that every fractional ideal of Am is invertible, so we have that ht(m) = dim Am = 1 by Lemma 2.3.6(a) and this holds for any maximal ideal m of A. Hence dim A = 1 so Proposition 4.2.2 tells us that A is Dedekind. Dedekind domains have the important property that every nonzero ideal factors uniquely into a product of prime ideals.

Theorem 4.3.7. If A is a Dedekind domain then every nonzero ideal I ⊂ A factors uniquely Qm into a product of prime ideals I = i=1 pi, with repetitions allowed. Qm Proof. By Corollary 4.2.3, I factors as I = i=1 pi for nonzero prime ideals pi ⊂ A that Qn need not be distinct. Thus it remains to check uniqueness. Suppose that I = j=1 qj for prime ideals qj ⊂ A. Then q1 ··· qn ⊆ p1 so there exists a 1 ≤ j ≤ n such that qj ⊆ p1. Since primes are maximal in a Dedekind domain, qj = p1. We may relabel the qj so that q1 = p1. −1 By Theorem 4.3.6, p1 exists so we have

p1p2 ··· pm = p1q2 ··· qn −1 −1 =⇒ p1 p1p2 ··· pm = p1 q2 ··· qn

=⇒ p2 ··· pm = q2 ··· qn.

By induction, m = n and after relabelling, qi = pi for all 1 ≤ i ≤ m. Hence prime factorizations are unique up to reordering.

Qm ei Qm fi Proposition 4.3.8. Take nonzero ideals I,J ⊂ A with I = i=1 pi and J = i=1 pi for distinct primes p1,..., pm ⊂ A and exponents ei, fi ≥ 0. Prove the following:

(a) I ⊆ J if and only if ei ≥ fi for all i.

Qm max(ei,fi) (b) I ∩ J = i=1 pi .

Qm min(ei,fi) ei fj (c) I + J = i=1 pi . In particular, pi + pj = A for all ei and fj when i 6= j.

ei fi Proof. (a) One direction is clear: if ei ≥ fi for all i then pi ⊆ pi for all i, and hence I ⊆ J. Qm fi f1 On the other hand, suppose I ⊆ J. Then I ⊆ i=1 pi ⊆ p1 . By Theorem 4.3.6, p1 is −1 invertible, so we may multiply by p1 to obtain:

f1 e1 e2 em f1 I ⊆ p1 =⇒ p1 p2 ··· pm ⊆ p1

−e1 e1 e2 em −e1 f1 =⇒ p1 p1 p2 ··· pm ⊆ p1 p1

e2 em f1−e1 =⇒ p2 ··· pm ⊆ p1 .

f1−e1 If f1 − e1 > 0 then p1 ⊆ p1. Since p1 is prime, the above implies p1 ⊇ pj for some 2 ≤ j ≤ m. But prime ideals are maximal in a Dedekind domain, so p1 = pj, contradicting the fact that p1,..., pm are distinct primes. Therefore f1 − e1 ≤ 0. Repeating the proof for each pi, 2 ≤ i ≤ m shows that ei ≥ fi for all i.

68 4.3 Fractional and Invertible Ideals 4 Discrete Valuations and Dedekind Domains

(b) Using unique factorization of ideals, write

n n n n Y ei Y fi Y gi Y hi I = pi ,J = pi ,I ∩ J = pi ,I + J = pi i=1 i=1 i=1 i=1

for distinct prime ideals p1,..., pm, pm+1,..., pn, n ≥ m, and unique exponents ei, fi, gi, hi ≥ 0, with ei = fi = 0 for m < i ≤ n. Then by (a), we get the following implications:

I ∩ J ⊆ I =⇒ gi ≥ ei for all 1 ≤ i ≤ n and

I ∩ J ⊆ J =⇒ gi ≥ fi for all 1 ≤ i ≤ n.

So gi ≥ max(ei, fi) for all 1 ≤ i ≤ n. This shows gi ≥ max(ei, fi) ≥ ei and gi ≥ max(ei, fi) ≥ fi for all 1 ≤ i ≤ n, so because I ∩ J is the largest ideal of A which is contained in both I and J, part (a) implies

n n Y max(ei,fi) Y max(ei,fi) I ∩ J ⊆ pi ⊆ I ∩ J =⇒ I ∩ J = pi . i=1 i=1 (c) As in part (b), we have:

I ⊆ I + J =⇒ ei ≥ hi for all 1 ≤ i ≤ n and

J ⊆ I + J =⇒ fi ≥ hi for all 1 ≤ i ≤ n.

So min(ei, fi) ≥ hi for all 1 ≤ i ≤ n, meaning ei ≥ min(ei, fi) ≥ hi and fi ≥ min(ei, fi) ≥ hi. Since I + J is the smallest ideal of A containing both I and J, part (a) means that

n n Y min(ei,fi) Y min(ei,fi) I ⊆ pi ⊆ I + J and J ⊆ pi ⊆ I + J i=1 i=1 n Y min(ei,fi) imply I + J ⊆ pi ⊆ I + J i=1 n Y min(ei,fi) so I + J = pi . i=1

Qm ei Theorem 4.3.9 (Chinese Remainder Theorem). If I = i=1 pi for distinct prime ideals pi ⊂ A and exponents ei ≥ 0, then there is a canonical isomorphism m Y ei A/I −→ A/pi . i=1 Remark. Let A be a Dedekind domain and take a nonzero prime (maximal) ideal p ∈ Qm ei Spec(A), so that Ap is a DVR. For 0 6= x ∈ A, we have a prime factorization (x) = i=1 pi with p1 = p and ei ≥ 0. The ei are determined by (x), so deﬁne a function

vp : A r {0} −→ N0 x 7−→ e1.

69 4.4 The Class Group 4 Discrete Valuations and Dedekind Domains

Then vp extends to a valuation on K = Frac(A) such that O(vp) = Ap. As this generalizes the construction of the p-adic valuation on Q, the function vp is called the p-adic valuation on K. Proposition 4.3.10. If A is a Dedekind domain with Spec(A) ﬁnite, then A is a PID.

Proof. Let p1,..., pn be the prime ideals of A. By the Chinese remainder theorem, 2 ∼ 2 A/p1p2 ··· pn = A/p1 × A/p2 × · · · × A/pn.

2 2 2 In particular, for any ﬁxed x1 ∈ p1 r p1, there exists y ∈ A that satisﬁes y + p1 = x1 + p1 2 and y + pi = 1 + pi for all 2 ≤ i ≤ n. Thus y − x1 ∈ p1 ⊂ p1 so it follows that y ∈ p1. We 2 have (y) ⊆ p1 but (y) 6⊂ p1 and (y) 6⊂ pi for any 2 ≤ i ≤ n. Since A is a Dedekind domain, (y) has a prime factorization n Y ei (y) = pi for ei ≥ 0. i=1

By Proposition 4.3.8(a) and the above work, we conclude that 1 ≤ e1 < 2 and 0 ≤ ei < 1 for each 2 ≤ i ≤ n. In other words, e1 = 1 and e2 = ... = en = 0, so (y) = p1. Repeating the proof for each pi, 1 ≤ i ≤ n shows that every prime ideal of A is principal. Suppose pi = (yi). Then any ideal I ⊂ A has a prime factorization

n n Y ai Y ai a1 an I = pi = (yi ) = (y1 ··· yn ). i=1 i=1 Therefore I is principal. Corollary 4.3.11. Let I be a nonzero ideal in a Dedekind domain A. Then for any nonzero a ∈ I, there is some b ∈ I such that I = (a, b). Proof. We first prove A/I is a PID. By the correspondence theorem, every prime ideal of A/I pulls back along the quotient map A → A/I to a unique prime ideal of A which contains I. Qr aj Write I = j=1 pj for prime ideals pj ⊂ A and integers aj ≥ 1. Then Proposition 3.4.10(a) and the fact that dim A = 1 imply that the only prime ideals containing I are p1,..., pr. It follows that A/I has finitely many prime ideals. Therefore by Proposition 4.3.10, A/I is a PID. Now take 0 6= a ∈ I. Then by the first paragraph, A/(a) is principal and therefore the ideal I/(a) can be generated by a single element b+(a), b ∈ I. (If b ∈ (a), this means I = (a) so I was principal to begin with.) On one hand, it’s immediate that (a, b) ⊆ I. On the other hand, every y ∈ Ir(a) has reduction y+(a) ⊆ I/(a) in A/(a), so y+(a) = r(b+(a)) = rb+(a) for some r ∈ A, and thus y − rb ∈ (a), so y = rb + sa ∈ (a, b). This shows I = (a, b).

4.4 The Class Group

Deﬁnition. For a Dedekind domain A with ﬁeld of fractions K, the ideal group of A is the free abelian group

IA = {I ⊂ K | I is a fractional ideal of A} = {I ⊆ K | I is an invertible ideal}.

70 4.4 The Class Group 4 Discrete Valuations and Dedekind Domains

× The subgroup PA = {Ax | x ∈ K } ≤ IA is called the subgroup of principal fractional ideals of A. The class group of A is then deﬁned as the quotient group

CA = IA/PA. The order of this group, denoted h(A), is called the class number of A. In general, class numbers can be arbitrarily large, but a highly important fact for number ﬁelds is that their rings of integers have ﬁnite class number:

Theorem 4.4.1. For an algebraic extension K/Q, with ring of integers OK , the class number hK := h(OK ) is ﬁnite. Lemma 4.4.2. For a Dedekind domain A, the following are equivalent: (i) h(A) = 1. (ii) A is a PID. (iii) A is a UFD. Proof. (i) ⇐⇒ (ii) Suppose h(A) = 1. Then for any ideal I ⊂ A, I is a fractional ideal (e.g. by Proposition 4.3.3), but since the class group of A is trivial, I must be principal. Hence A is a PID. Conversely, suppose every ideal I ⊂ A is principal. Any fractional ideal Qr ei J ∈ IA may be written as a formal product of prime ideals J = i=1 pi with ei ∈ Z for each i. Then each prime ideal is principal: pi = (xi) for an element xi ∈ A, 1 ≤ i ≤ r, so we have r r Y ei Y ei e1 er J = pi = (xi) = Ax1 ··· xr . i=1 i=1

Therefore J is a principal fractional ideal, and it follows that PA = IA, that is, h(A) = |IA/PA| = 1. (ii) ⇐⇒ (iii) It is well-known that every PID is a UFD. For the converse, suppose p ⊂ A is a (proper) nonzero prime ideal. Choose 0 6= x ∈ p (which is not a unit since p 6= A) and e1 er write x = p1 ··· pr for distinct irreducible elements pi ∈ A and exponents ei ∈ N. Since p is a prime ideal, pi ∈ p for some 1 ≤ i ≤ r. Then (pi) ⊆ p. Since pi is irreducible, (pi) is a prime ideal, so the fact that dim A = 1 implies (pi) = p. Thus every nonzero prime ideal of A is principal. This implies the result by unique factorization of ideals. √ Example√ 4.4.3. Let K = Q( −5) be the quadratic number ﬁeld with ring of integers OK = Z[ −5]. Then OK is a Dedekind domain, but it is not a UFD, since for example √ √ 6 = 2 · 3 = (1 + −5)(1 − −5) and each of the elements in the two factorizations is irreducible but not prime. However, since OK is Dedekind, the ideal (6) factors uniquely into a product of prime ideals: 2 (6) = p p2p3, 1 √ √ where p1 = (2, 1 + −5) = (2, 1 − −5) √ p2 = (3, 1 + −5) √ p3 = (3, 1 − −5).

71 4.5 Dedekind Domains in Extensions 4 Discrete Valuations and Dedekind Domains

√ √ 2 In particular,√ (2) = p1, (3)√ = p2p3, (1 +√ −5) = p1p2 and√ (1 − −5) = p1p3. We have (2) ( p1 (Z[ −5] and [Z[ −5] : 2Z[ −5]] = 4, so [Z[ −5] : p1] = 2 and hence p1 is a maximal ideal. Similar arguments show that p2 and p3 are also√ maximal (thus prime) ideals. Moreover, notice that each of p1, p2, p3 is not principal, so h(Z[ −5]) > 1 (in fact, the class number is 2 in this case).

4.5 Dedekind Domains in Extensions

Theorem 4.2.4 says that extending a Dedekind domain in a finite separable extension of its fraction field (i.e. taking its integral closure) again yields a Dedekind domain. The critical step in the proof of this result is to prove the integral closure is noetherian, which requires Proposition 2.2.5. The issue in generalizing Theorem 4.2.4 is that Proposition 2.2.5 does not hold if L/K is not separable – in fact, we will demonstrate in a moment several counterexamples to 2.2.5 in the inseparable case, even when A and B are DVRs. Recall the following definition:

Deﬁnition. A ﬁeld extension L/K is purely inseparable if for each α ∈ L, the minimal polynomial of α over K has a single distinct root.

Lemma 4.5.1. For a ﬁnite extension L/K, the following are equivalent:

(i) L/K is purely inseparable.

(ii) If α ∈ L is separable over K then α ∈ K.

n (iii) There is some n ∈ N such that for all α ∈ L, αp ∈ K. Lemma 4.5.2. Let L/K be any algebraic extension of ﬁelds. Then there is a unique subﬁeld K ⊆ Ksep ⊆ L such that Ksep/K is separable and L/Ksep is purely inseparable.

We now prove the general version of 4.2.4.

Theorem 4.5.3. Let A be a Dedekind domain with ﬁeld of fractions K, L/K a ﬁnite extension and B the integral closure of A in L. Then B is a Dedekind domain.

Proof. It remains to prove the theorem when L/K is inseparable. Assume char K = p > 0. By Lemma 4.5.2, there is a unique subextension K ⊂ L0 ⊂ L such that L0/K is separable and L/L0 is purely inseparable. Let B0 denote the integral closure of A in L0. Since L0/K is separable, Theorem 4.2.4 shows B0 is Dedekind. Thus it suﬃces to consider the case when K = L0 and A = B0. After this reduction, L/K is purely inseparable, so by Lemma 4.5.1, there is some n ∈ N so that for every α ∈ L, αpn ∈ K. Set q = pn. Let M be the splitting ﬁeld of all polynomials of the form tq − α, with α ∈ K. Alternatively,

M = {α1/q ∈ K | α ∈ K} where K is an algebraic closure of K. Let C be the integral closure of A in M.

72 4.5 Dedekind Domains in Extensions 4 Discrete Valuations and Dedekind Domains

purely insep. K L M

A B C

Deﬁne the Frobenius map ϕ : M → K by ϕ(x) = xq = xpn . Notice that ϕ is well-deﬁned by Lemma 2; injective since char K = p; and surjective by the construction of M. It is clear that ϕ is a homomorphism, so it is an isomorphism (of rings) M → K. Moreover, we claim A = ϕ(C). Indeed, if y ∈ A then any root of tq − y lies in C and maps to y via ϕ. Conversely, if x ∈ C then xq − α = 0 for some α ∈ K, i.e. xq ∈ K, but xq ∈ C as well, so ϕ(x) = xq ∈ C ∩ K = A. We have thus established that A ∼= C as rings, so C is in particular a Dedekind domain. By Theorem 4.3.6, B is a Dedekind domain if and only if every ideal of B is invertible. For an ideal I ⊂ B, let J = IC be the corresponding ideal of C. Then J is invertible Pm −1 since C is Dedekind, so i=1 aibi = 1 for some elements ai ∈ I, bi ∈ J . Since we are in Pm q q q q−1 q characteristic p, i=1 ai bi = 1 and bi ∈ K. For each 1 ≤ i ≤ m, set ci = ai bi so that Pm q−1 −q q −q q −q i=1 aici = 1 and ci ∈ I J ⊆ L. Now ciI ⊆ I J ⊆ J J = C but the previous statement shows that ciI ⊆ L. So ciI ⊆ C ∩ L = B, and thus ci ∈ (B : I). In summary, Pm we have i=1 aici = 1 for elements ai ∈ I and ci ∈ (B : I), so it follows that I(B : I) = B. Hence every ideal of B is invertible, so B is a Dedekind domain. In fact, Theorem 4.5.3 can be obtained as a corollary to the much more powerful Krull- Akizuki Theorem, which we don’t quite have the tools to prove yet.

Theorem (Krull-Akizuki). Let A be a noetherian integral domain and suppose dim A = 1,K = Frac(A), L/K is a ﬁnite algebraic extension and B ⊂ L is any subring containing A. Then B is a noetherian domain with dim B ≤ 1.

For the remainder of the section, we discuss counterexamples to Proposition 2.2.5 and discuss a generalization of this result for aﬃne algebras.

Example 4.5.4. (Hochster) Consider a perfect field k of characteristic p > 0 (e.g. k = Fp p if you like) and let K = k(tn | n ∈ N). Also, for each n ∈ N, set Kn = k(t1, . . . , tn) and S∞ p S∞ L = n=1 Kn = k(tn | n ∈ N) so that L = K. Define a ring A = n=1 Kn[[x]] where x is an indeterminate. It is a fact that A is a local noetherian domain with maximal ideal (x), and any ideal of A is of the form I = (xm) for some m ∈ N. In particular, Theorem 4.1.6 P∞ i shows that A is a DVR. Observe that K[[x]] ⊆ A ⊆ L[[x]]. Set u = i=1 tix ∈ L[[x]] so that up ∈ A. If F = Frac(A), the extension F (u)/F is purely inseparable, but the integral closure of A in F (u) is not a finitely generated A-module.

Example 4.5.5. (Kaplansky) Let k be a ﬁeld of characteristic 2 and set C = k[[t]]. Take some u ∈ C r k and let K = k(t, u2) and L = k(t, u). If u is chosen so that u and t are algebraically independent over k, for example, then we will have [L : K] = 2. Consider the rings A = C ∩ K,B = C ∩ L and the ﬁeld of fractions M = Frac(C) = k((t)).

73 4.5 Dedekind Domains in Extensions 4 Discrete Valuations and Dedekind Domains

k K L M

A B C

Since C is a DVR (Example 3 in Section 4.1), A and B are also DVRs with Frac(A) = K and Frac(B) = L. In particular, B is the integral closure of A in L – if B0 is this integral closure, then B ⊆ B0 because B is integral over A, but on the other hand B is integrally closed (using the fact that char k = 2), so it must be that B = B0. Suppose B were a ﬁnitely generated A-module. Then we would have B ⊆ Aα1 + ... + Aαr + Aβ1u + ... + Aβsu for m some αi, βj ∈ K. By clearing denominators, there is some m ∈ N such that t αi ∈ A and m m P∞ ` t βj ∈ A for all 1 ≤ i ≤ r, 1 ≤ j ≤ s. So t B ⊆ A[u]. Write u = `=0 a`t with a` ∈ k. Set

∞ m −m−1 −m−1 X ` v = (u − a0 − a1t − ... − amt )t = t a`t . `=m+1 By the first description, v ∈ L and by the second, v ∈ C, so v ∈ C ∩L = B ⊆ A[u]. However, the second expression for v shows the coefficient of u in the expansion of tmv must be t−1, contradicting the fact that v ∈ A[u]. Thus B cannot be finitely generated over A. To close, we generalize Proposition 2.2.5 to the context of affine algebras. Theorem 4.5.6 (Noether). Let k be a field and suppose A is a finitely generated k-algebra that is also an integral domain with field of fractions K. If L/K is any finite extension and B is the integral closure of A in L, then B is a finitely generated A-module.

Proof. (Eisenbud) By Noether normalization (2.4.1), there exist y1, . . . , yr ∈ A such that A is integral over the polynomial ring k[y1, . . . , yr]. Then the result will follow if we prove B/k[y1, . . . , yr] is a finite extension of rings, so we may replace A with k[y1, . . . , yr], and K 00 00 with k(y1, . . . , yr), to begin with. Moreover, if L is the normal closure of L over K and B is the integral closure of A in L00, then demonstrating B00/A is finite also implies the result. So we make the reduction L = L00, so that L/K is a normal extension. Set G = Gal(L/K) and let L0 = LG be the subfield fixed by G. It is a general fact from Galois theory that L/L0 is Galois and L0/K is a purely inseparable extension. purely insep. Galois K L0 L

A B0 B

Since L/L0 is finite and separable, Proposition 2.2.5 gives us that B is a finitely generated B0-module, so it’s enough to prove B0 is a finitely generated A-module to obtain the result. If L0 = K, we’re done so assume L0 6= K. This a nontrivial purely inseparable extension, so char K = p > 0 and by Lemma 4.5.1, there is some n ∈ N with q = pn and αq ∈ K for every

74 4.5 Dedekind Domains in Extensions 4 Discrete Valuations and Dedekind Domains

0 0 0 α ∈ L . In particular, if L = K(z1, . . . , zn) for z1, . . . , zn ∈ L , then for each 1 ≤ i ≤ n, q q fi z ∈ K, so there exist fi, gi ∈ A such that z = . Since A = k[y1, . . . , yr], we have i i gi

X j1 jr X j1 jr fi = cJ,iy1 ··· yr and gi = dJ,iy1 ··· yr r r J=(j1,...,jr)∈N0 J=(j1,...,jr)∈N0

for cJ,i, dJ,i ∈ k, finitely many of which are nonzero. Let S be the finite set of all nonzero r cJ,i and dJ,i, where J ranges over all tuples (j1, . . . , jr) ∈ N0 and 1 ≤ i ≤ n. Then the field 0 1/q 0 1/q 1/q k := k(s | s ∈ S) is a finite extension of k, and consequently Le := k (y1 , . . . , yr ) is a finite extension of L0. If Be is the integral closure of A in Le, then Be contains B0 so it’s enough to prove the result when L0 = Le and B0 = Be. 0 1/q 1/q 0 Now consider C := k [y1 , . . . , yr ] ⊆ L . Then C is integrally closed (e.g. it’s a UFD) 0 0 1/q 0 1/q 1/q 0 and Frac(C) = L , so B ⊆ C. But each yj is integral over A, so C = k [y1 , . . . , yr ] ⊆ B since B0 is the integral closure of A in L0. Hence B0 = C, which is finitely generated over A by definition. This finishes the proof.

75 5 Completion and Filtration

5 Completion and Filtration

5.1 Topological Abelian Groups and Completion

Definition. A topological abelian group is a topological space G with an abelian group structure such that the addition and inversion maps α : G × G −→ G η : G −→ G (x, y) 7−→ x + y x 7−→ −x are continuous, where G × G is endowed with the product topology. Lemma 5.1.1. The set {0} is closed in G if and only if G is Hausdorff as a topological space. Proof. Since addition is continuous, {0} is closed in G if and only if the diagonal ∆ = {(x, x) | x ∈ G} is closed in G × G. It is well known that the latter condition is equivalent to G being Hausdorff.

Lemma 5.1.2. For a ﬁxed a ∈ G, the translation map ta : G → G, x 7→ x + a, is continuous with inverse t−a, and hence a homeomorphism. Lemma 5.1.3. For any subgroup N ≤ G of a topological abelian group G, N is a topological abelian group with the subspace topology and G/N is a topological abelian group with the quotient topology.

If U = U(0) is a neighborhood of 0, then U + a = ta(U) is also open, and thus a neighborhood of a. Since ta is a homeomorphism, every neighborhood of any point of G is of the form U + a for some U = U(0). Thus the topology of G is uniquely determined by the collection {U(0)} of neighborhoods of the identity. T Lemma 5.1.4. Let H = U(0)30 U(0) be the intersection of all neighborhoods of 0. Then (i) H is a subgroup of G. (ii) H = {0}. (iii) G/H is Hausdorff. (iv) G is Hausdorff if and only if H = {0}. Proof. (i) Take x, y ∈ H. Then x, y ∈ U(0) for every neighborhood U(0) of the identity. In particular, 0 ∈ −x + U(0) and this is an open set since t−x is a homeomorphism. Hence H ⊆ −x + U(0) so y ∈ −x + U(0). Thus x + y ∈ U(0). Since U(0) was arbitrary, we have that x + y ∈ H. This argument also shows that −x ∈ U(0) for any neighborhood U(0), so −x ∈ H. Hence H is a subgroup. (ii) For any x ∈ G, x ∈ H ⇐⇒ −x ∈ H ⇐⇒ −x ∈ U(0) for all neighborhoods U(0) ⇐⇒ 0 ∈ x + U(0) for all U(0) ⇐⇒ x is a limit point of {0}. Therefore H = {0}. (iii) For a fixed a ∈ G, a + H = ta(H) = ta({0}) is closed in G/H since ta is continuous. Thus cosets, aka the ‘points’ in G/H, are closed, so G/H is Hausdorff. (iv) If H = {0} then (iii) implies G is Hausdorff. On the other hand, for all x 6= 0, there is a neighborhood Ux of x such that 0 6∈ Ux. Thus x 6∈ H, which implies H = {0}.

76 5.1 Topological Abelian Groups and Completion 5 Completion and Filtration

Assume 0 ∈ G has a countable basis of subgroups U(0) ⊆ G.

∞ Deﬁnition. A Cauchy sequence in G is a sequence of elements (xn)n=0, where xn ∈ G for all n ∈ N0, such that for all neighborhoods U(0), there is some N = N(U(0)) ∈ N such that xn − xm ∈ U(0) for all n, m ≥ N.

Deﬁnition. We say two Cauchy sequences (xn) and (yn) are equivalent, written (xn) ∼ (yn), if for any neighborhood U(0), there is some N = N(U(0)) ∈ N such that xn −yn ∈ U(0) for all n ≥ N. Lemma 5.1.5. ∼ is an equivalence relation on the set of Cauchy sequences in G.

Proof. Let (xn), (yn) and (zn) be Cauchy sequences in G. Then for any U(0), xn − xn = 0 ∈ U(0) for every n ∈ N, so (xn) ∼ (xn). If (xn) ∼ (yn) then there is an N ∈ N such that for all n ≥ N, xn − yn ∈ U(0). Since U(0) is assumed to be a subgroup of G, yn − xn = −(xn − yn) ∈ U(0) as well, so the same N shows that (yn) ∼ (xn). Finally, if (xn) ∼ (yn) and (yn ∼ zn) and U(0) is a neighborhood of 0, there are N1,N2 ∈ N such that for all n ≥ N1, xn − yn ∈ U(0) and for all n ≥ N2, yn − zn ∈ U(0). Thus for all n ≥ max{N1,N2}, we have xn − zn = xn − yn + yn − zn ∈ U(0) since U(0) is a subgroup. Therefore (xn) ∼ (zn), so ∼ is an equivalence relation.

Deﬁnition. The completion of a topological group G is the quotient space Gb of all equivalence classes under ∼ of Cauchy sequences in G. For a Cauchy sequence (xn), we let [(xn)] denote its equivalence class in Gb. Proposition 5.1.6. For any topological group G, Gb is a topological group via pointwise addition and inversion: [(xn)] + [(yn)] = [(xn + yn)] and −[(xn)] = [(−xn)].

Proof. Consider Cauchy sequences (xn) and (yn) in G. We first prove pointwise addition 0 and inversion of Cauchy sequences are invariant on equivalences classes. If (xn) ∼ (xn) and 0 0 (yn) ∼ (yn) then for all U(0), there are numbers N1,N2 ∈ N such that xn − xn ∈ U(0) for 0 all n ≥ N1 and yn − yn ∈ U(0) for all n ≥ N2. Taking N = max{N1,N2}, we have that for 0 0 0 0 all n ≥ N,(xn + yn) − (xn + yn) = (xn − xn) + (yn − yn) ∈ U(0) since U(0) is a subgroup. 0 0 0 Thus (xn + yn) ∼ (xn + yn), so pointwise addition is well-defined. Similarly, (−xn) ∼ (−xn) so inversion is well-defined. To prove that Gb is a group under these actions, consider two Cauchy sequences (xn) and (yn) in G. Then for a neighborhood U(0), there are N1,N2 ∈ N such that xm − xn ∈ U(0) for all m, n ≥ N1 and ym − yn ∈ U(0) for all m, n ≥ N2. Then for all n ≥ N = max{N1,N2}, (xm + ym) − (xn − yn) = (xm − xn) + (ym − yn) ∈ U(0) since U(0) is a subgroup. Hence (xn + yn) is a Cauchy sequence. The zero sequence (0) represents the identity class in Gb and it’s easy to check that [(xn)] + [(−xn)] = [(0)] for all Cauchy sequences (xn), so we have that Gb is an abelian group. Finally, for each neighborhood U(0) in G, define a subgroup Ub(0) in Gb by

Ub(0) = {[(xn)] : there exists an N ∈ N such that xn ∈ U(0) for all n ≥ N}.

Then {Ub(0) : U(0) is a neighborhood of 0 in G} forms a basis for a topology on Gb which is compatible with the group structure on Gb. Hence Gb is a topological group.

77 5.2 Inverse Limits 5 Completion and Filtration

Proposition 5.1.7. There is a homomorphism ϕ : G → Gb which is injective precisely when G is Hausdorff. T Proof. Define ϕ(x) = [(x, x, x, . . .)] = [(x)]. Then ker ϕ = H where H = U(0) U(0) so ϕ is injective if and only if H = 0, which by Lemma 5.1.4 is equivalent to G being Hausdorff.

Deﬁnition. A homomorphism of topological abelian groups is a continuous homomorphism f : G → H. Such a map f is an isomorphism of topological abelian groups if f is a homeomorphism and an isomorphism, that is, f is a continuous homomorphism with a continuous inverse f −1 : H → G which is also a homomorphism.

Lemma 5.1.8. Given a homomorphism of topological abelian groups f : G → H, if (xn) is a Cauchy sequence in G then (f(xn)) is a Cauchy sequence in H. Proof. Suppose V (0) is a neighborhood of 0 ∈ H. Since f is continuous, U(0) := f −1(V (0)) is open in G, and since f is a homomorphism, f(0) = 0 and hence U(0) is a neighborhood of 0 ∈ G. Thus there is some N ∈ N such that for all m, n ≥ N, xm − xn ∈ U(0). Applying f and using the fact that it is a homomorphism, we get f(xm) − f(xn) = f(xm − fn) ∈ f(U(0)) = V (0) for all m, n ≥ N. Therefore (f(xn)) is Cauchy. Corollary 5.1.9. Every homomorphism of topological abelian groups f : G → H induces a ˆ ˆ homomorphism f : Gb → Hb such that f[(xn)] = [f(xn)].

In particular, this says that the completion Gb a solution to a universal mapping property for G. In addition, the completion operation (c·) is a covariant functor on the category TopAbGps of topological abelian groups:

f g Proposition 5.1.10. For every sequence G −→ H −→ K of homomorphisms of topological abelian groups, we have (\g ◦ f) =g ˆ ◦ fˆ.

5.2 Inverse Limits

In this section we formulate an alternative construction of the completion Gb from Section 5.1 using inverse limits.

Deﬁnition. An inverse system is a family of groups (Am)m∈I , for some directed index set I, together with homomorphisms θm` : Am → A` for every m ≥ ` such that when m ≥ ` ≥ k, the following diagram commutes:

θm` Am A`

θmk θ`k

78 5.2 Inverse Limits 5 Completion and Filtration

Deﬁnition. Given an inverse system (Am, θm`) over a directed set I, we deﬁne a coherent Q sequence to be a sequence (ym) ∈ m∈I Am such that θm`(ym) = y` for all m ≥ `. The set of all coherent sequences of (Am, θm`) is called the inverse limit of the inverse system: ( ) Y lim Am = (ym) ∈ Am : θm`(ym) = y` for all m ≥ ` . ←− m∈I

Now let G be a topological group possessing a countable basis of open subgroups Gn = Gn(0) such that G = G0 ⊇ G1 ⊇ G2 ⊇ · · ·

Lemma 5.2.1. Each Gn is both open and closed.

Proof. For all a ∈ G, a + Gn = ta(Gn) is open since translation is a homeomorphism. Since the a + G partition the complement of G in G, we have G G = S (a + G ) which n n r n a6∈Gn n is a union of open sets and therefore open. Hence Gn is closed.

For each n ≥ 1, Gn ⊆ Gn−1 so we have projections θn : G/Gn → G/Gn−1.

Lemma 5.2.2. The system (G/Gn, θn)n∈N is an inverse system.

Lemma 5.2.3. lim G/Gn is a topological group. ←− Q Proof. Consider G/Gn with the product topology. Then lim G/Gn has the subspace n∈N ←− topology from Q G/G . Indeed, if p : Q G/G → G/G is the mth projection map, n≥0 n m n∈N n m −1 then the sets S(x, m) := pm (x + Gm) = {(yn) | ym = x + Gm} over all m form a subbasis Q for the topology on G/Gn, so the sets Se(x, m) := S(x, m) ∩ lim G/Gn form a subbasis n∈N ←− for the (subspace) topology on lim G/Gn. In fact, Se(0, m) form a basis of this topology ←− because the collection of Se(0, m) are closed under intersections to begin with. The axioms of a topological abelian group are then easily checked for lim G/Gn. ←−

When (G/Gn, θn)n∈N is the inverse system of quotients in G, the inverse limit coincides with the completion of G from Section 5.1.

Proposition 5.2.4. For a topological abelian group G with inverse system (G/Gn, θn)n∈N, ∼ there is a canonical isomorphism of topological abelian groups Gb = lim G/Gn. ←− Proof. For each n ∈ N, deﬁne a map

ζn : Gb −→ G/Gn

[(xk)] 7−→ xkn + Gn where kn is the smallest N (or any N) such that xk + Gn = xN + Gn for all k ≥ N. Such a choice is possible because (xk) is Cauchy. Clearly the ζn are homomorphisms, and one can verify that they are well-deﬁned on equivalence classes of Cauchy sequences. Now deﬁne a map into the inverse limit by

ζ : Gb −→ lim G/Gn ←−

[(xk)] 7−→ (ζn([(xk)]))n∈N.

79 5.2 Inverse Limits 5 Completion and Filtration

The image of ζ consists of coherent sequences: for all n ∈ N, we have

θn+1(xkn+1 + Gn+1) = xkn+1 + Gn = xkn + Gn

since kn+1 ≥ kn, and therefore θn+1(ζn+1[(xk)]) = ζn[(xk)]. Now given (yn) ∈ lim G/Gn, ←− construct a Cauchy sequence mapping to (yn) by taking a coset representative xn for each n ∈ N such that yn = xn + Gn. Since (yn) is a coherent sequence, for each n ∈ N we have

θn+1(yn+1) = yn =⇒ θn+1(xn+1 + Gn+1) = xn+1 + Gn = xn + Gn.

So xn+1 − xn ∈ Gn and hence (xn) is a Cauchy sequence. Clearly ζ([(xn)]) = (yn) so ζ is surjective. To see that ζ is continuous, suppose [(xk)] ∈ Gb. Then using the notation from the proof of Lemma 5.2.3, for each m ∈ N,

−1 [(xk)] ∈ ζ (Se(0, m)) ⇐⇒ xkm + Gm = 0 + Gm in G/Gm ⇐⇒ [(xk)] ∈ Gbm.

Since Gbm is open for all m, we have that ζ is continuous. Finally, to show ζ is an isomorphism, we deﬁne an inverse function by

η : lim G/Gn −→ Gb ←−

xn + Gn 7−→ [(xn)].

We showed above that this map is well-deﬁned and a homomorphism of topological abelian groups, and clearly ζ ◦ η = id by surjectivity of ζ. Finally, for any [(xk)] ∈ Gb, we lim G/Gn ←− have η ◦ ζ([(xk)]) = η(xkn + Gn) = [(xkn )] = [(xk)] since subsequences of Cauchy sequences are equivalent to their parent Cauchy sequence. This proves that ζ is an isomorphism of topological abelian groups.

Definition. When (Gn)n∈N is the system of all subgroups of finite index in G, the inverse limit Gb = lim G/Gn is called the profinite completion of G. ←−

Deﬁnition. A surjective inverse system is one in which every map θm` is surjective.

Deﬁnition. Let (Am, αm`) and (Bm, βm`) be two inverse systems indexed over the same set I.A morphism of inverse systems is a set of maps (ϕ•) such that ϕm : Am → Bm is a homomorphism for each m ∈ I that is compatible with the αm` and βm`, i.e. the following diagrams commute for all m ≥ ` ∈ I:

ϕm Am Bm

αm` βm`

A` B` ϕ`

80 5.2 Inverse Limits 5 Completion and Filtration

Deﬁnition. Let (Am, αm`), (Bm, βm`) and (Cm, γm`) be inverse systems with the same index set I.A short exact sequence of inverse systems is a sequence of maps of inverse (im) (pm) im pm systems 0 → (Am) −−→ (Bm) −−→ (Cm) → 0 such that 0 → Am −→ Bm −→ Cm → 0 is an exact sequence for all m ∈ I.

Proposition 5.2.5. For any short exact sequence of inverse systems 0 → (Am) → (Bm) → (Cm) → 0, there is an exact sequence of topological abelian groups

0 → lim Am → lim Bm → lim Cm. ←− ←− ←− In other words, inverse limit is a left exact functor from the category of inverse systems to the category of topological abelian groups. Furthermore, if (Am) is a surjective inverse system then 0 may be added on the right, i.e. lim(−) is an exact functor. ←− Q A Proof. Let A = Am and deﬁne a map d : A → A by (am) 7→ (am −θm+1,m(am+1)). Deﬁne B C A B B, d ,C and d similarly for (Bm) and (Cm). Then ker d = lim Am, ker d = lim Bm and ←− ←− C kerd = lim Cm. We have a commutative diagram with exact rows: ←− 0 A B C 0

dA dB dC

0 A B C 0

so by the Snake Lemma, we get an exact sequence

0 → ker dA → ker dB → ker dC → coker dA → coker dB → coker dC → 0.

The first three terms of this are 0 → lim Am → lim Bm → lim Cm so we have our exact ←− ←− ←− sequence. For the final statement, suppose (Am, αm`) is a surjective system. We need to prove A A coker d = 0, i.e. for each (am) ∈ A, we want to find (xm) ∈ A such that d (xm) = (am). Let x0 = 0. Then a0 = x0 − θ1(x1) so θ1(x1) = −a0 and since θ1 is surjective, this allows us to choose some x1 ∈ A1 so that this holds. Proceed to inductively define (xm). This shows dA is surjective, so we are done. p Corollary 5.2.6. Let 0 → G0 → G −→ G00 → 0 be a short exact sequence of topological 0 00 0 0 abelian groups having systems of open subgroups {Gn}, {Gn} and {Gn}, where Gn = G ∩ Gn 00 and Gn = p(Gn) for each n ∈ N. Then the sequence of completions

0 → Gb0 → Gb → Gb00 → 0 is exact. Proof. This is done by considering the short exact sequence of surjective inverse systems 0 0 00 00 0 → (G /Gn) → (G/Gn) → (G /Gn) → 0 and applying Proposition 5.2.5.

Corollary 5.2.7. For a topological abelian group G with system of open subgroups Gn and ∼ completion Gb, for each n ∈ N, Gbn is a topological subgroup of Gb and G/b Gbn = G/Gn.

81 5.3 Topological Rings and Module Filtrations 5 Completion and Filtration

Proof. Apply Corollary 5.4.9 to the sequence 0 → Gn → G → G/Gn → 0. Notice that ∼ ∼ G/Gn = G/G\n = G/b Gbn because each has the discrete topology.

Corollary 5.2.8. For any topological abelian group G, Gb ∼= Gb.

Deﬁnition. A topological abelian group G is complete if the canonical map ϕ : G → G,b x 7→ [(x)], is an isomorphism.

In this language, Corollary 5.2.8 says that completions are complete. T∞ Remark. Notice that if G is complete, ϕ : G → Gb is an isomorphism, so n=1 Gn = ker ϕ = 0 and hence by Lemma 5.1.4, G is Hausdorﬀ.

5.3 Topological Rings and Module Filtrations

One of the most important concepts in commutative algebra is the completion of a ring with respect to a certain topology induced by one of its ideals. We ﬁrst deﬁne the notion of a topological ring.

Definition. A topological ring is a ring A which is a topological abelian group such that the ring multiplication A × A → A, (a, b) 7→ ab, is a continuous map. Topological fields are defined analagously.

Given an ideal J ⊂ A, one can always give A the structure of a topological ring:

Deﬁnition. Let A be a ring and J ⊂ A an ideal. Then the chain

A = J 0 ⊇ J 1 ⊇ J 2 ⊇ J 3 ⊇ · · · forms a basis of additive subgroups of (A, +) as a topological abelian group. This topology, called the J-adic topology, makes A into a topological ring.

Lemma 5.3.1. The completion Ab of a topological ring A is a topological ring. Examples.

1 Let A = k[t] be the polynomial ring in a single variable and take J = (t). Then Ab = k[[t]], the power series ring.

2 When A = Z and J = (p) for a prime integer p, the completion Zb is called the p-adic n completion of , denoted p. Alternatively, p = lim /p . Z Z Z ←− Z Z Deﬁnition. Let A be a topological ring. An A-module M is called a topological A-module if M is a topological abeliain group and the A-action map A × M → M, (a, m) 7→ am, is a continuous map.

82 5.3 Topological Rings and Module Filtrations 5 Completion and Filtration

If J ⊂ A is an ideal of A, so that A is a topological ring with the J-adic topology, n for an A-module M one can define open sets Mn = J M that make M into a topological A-module. Then the topological completion Mc of M with respect to this topology is a topological Ab-module. Rather than using J nM to define the J-adic topology on M, we instead use systems of additive subgroups of M called filtrations. Definition. A filtration of an A-module M is an infinite chain of submodules

M = M0 ⊇ M1 ⊇ M2 ⊇ · · ·

If J ⊂ A is an ideal, a J-filtration of M is a filtration such that JMn ⊆ Mn+1 for all n ≥ 0. If in addition JMn = Mn+1 for all n larger than some n0 ∈ N, then the filtration is called a stable J-filtration.

0 Lemma 5.3.2. If (Mn) and (Mn) are stable J-ﬁltrations of M, then there exists an n0 ∈ N 0 0 such that Mn+n0 ⊆ Mn and Mn+n0 ⊆ Mn for all n ≥ 0. 0 0 Proof. Let n0 ∈ N such that JMn = Mn+1 and JMn = Mn+1 for all n ≥ n0. Then 0 0 n n 0 0 JMn0 = Mn0+1 and JMn0 = Mn0+1. By induction, J Mn0 = Mn0+n and J Mn0 = Mn0+n. Then we have

J nM ⊆ J n−1M ⊆ · · · ⊆ JM ⊆ M n 0 n−1 0 0 0 and J M ⊆ J M ⊆ · · · ⊆ JM ⊆ M for all n ≥ n0.

n n 0 0 From this we obtain Mn+n0 = J Mn0 ⊆ J M ⊆ Mn. Reversing the roles of the Mn and Mn gives the other containment.

Remark. The completion Ab along an ideal J can also be described in terms of a metric, as T∞ n long as n=1 J = 0: if x, y ∈ A, deﬁne 1 d(x, y) = where v(x, y) = sup{k ≥ 1 | x − y ∈ J k}. 2v(x,y) Formally, we set d(x, y) = 0 if v(x, y) = ∞. Completion of the induced metric topology on A, that is, by formally adding in limits of Cauchy sequences in this topology, yields a topological ring isomorphic to Ab. Moreover, this description makes it easier to see that the ring operations + and · on A are continuous with respect to the metric topology and extend to continuous operations on Ab, and that the canonical embedding A,→ Ab has dense image. T∞ n Proposition 5.3.3. Suppose A is a ring and J ⊂ A is an ideal such that n=1 J = 0. Let Ab be the completion of A along J. Then JAb ⊆ J(Ab), the Jacobson radical of Ab.

Proof. By Lemma 1.1.1 it suﬃces to show 1 − x is a unit in Ab for all x ∈ J. Consider the formal expansion 1 = 1 + x + x2 + ... 1 − x n n+1 n+1 Set sn = 1 + x + ... + x ∈ A. Then for each n ≥ 1, sn − sn+1 = −x ∈ J so (sn) is a 1 Cauchy sequence with respect to the J-adic metric and hence its limit 1−x exists in Ab.

83 5.4 Graded Rings and Modules 5 Completion and Filtration

T∞ n n ∼ n Lemma 5.3.4. If J ⊂ A is an ideal with n=1 J = 0, then A/Jb Ab = A/J for all n ≥ 1. In particular, if m ⊂ A is a maximal ideal, then Ab is a local ring with maximal ideal mb = mAb.

Example 5.3.5. The completion of a polynomial ring A = k[t1, . . . , tn] along the maximal ideal m = (t1, . . . , tn) is a power series ring Ab = k[[t1, . . . , tn]], which is local with maximal ideal (t1, . . . , tn). More generally:

T∞ n Proposition 5.3.6. If J = (a1, . . . , an) is a ﬁnitely generated ideal of A with n=1 J = 0, then Ab = A[[t1, . . . , tn]]/(t1 − a1, . . . , tn − an).

Corollary 5.3.7. If A is noetherian then any completion Ab is also noetherian. Example 5.3.8. Be aware that localization and completion do not commute. For example, consider the ideal J = (x) in A = k[x, y]. Then A(x) = (k(y)[x])(x) and the completion of this ring along (x) is k(y)[[x]]. Meanwhile, the completion Ab of A along (x) is k[[x]][y], whose localization at (x)Ab is the proper subring (k[[x]][y])(x) ( k(y)[[x]].

5.4 Graded Rings and Modules

L∞ Deﬁnition. A ring A is said to be graded if as an abelian group, A = n=0 An for subgroups An ≤ A with the property that AmAn ≤ Am+n for all m, n ∈ N0.

Notice that A0 is always a subring of A, so each An is an A0-module. We also deﬁne L∞ A+ = n=1 An, which is an ideal of A. Deﬁnition. If A is a graded ring, an abelian group M is a graded A-module if M is an L∞ A-module and M = n=0 Mn for subgroups Mn ≤ M such that AmMn ⊆ Mm+n for all m, n ∈ N0.

As above, each Mn is not just an A-module but also an A0-module. L∞ Deﬁnition. Given a graded A-module M = n=0 Mn, an element x ∈ Mn is said to be homogeneous of degree n in M.

Deﬁnition. Let A and B be graded rings. A homomorphism of graded rings is a ring map f : A → B such that f(An) ⊆ Bn for all n ∈ N0. Likewise, if M and N are two graded A-modules, a homomorphism of graded modules is an A-module map f : M → N such that f(Mn) ⊆ Nn for all n ∈ N0.

Example 5.4.1. If A = k[t1, . . . , tn] then A is a graded ring with respect to total degree:

∞ M X i1 ir A = An where A0 = k and An = kt1 ··· tr r n=0 (i1,...,ir)∈N0 i1+...+ir=n

This is called the standard grading on the polynomial ring A. The ideal A+ here is (t1, . . . , tn).

84 5.4 Graded Rings and Modules 5 Completion and Filtration

Lemma 5.4.2. For a graded ring A, the following are equivalent:

(i) A is noetherian.

(ii) A0 is noetherian and A/A0 is a ﬁnitely generated ring extension. Proof. (ii) =⇒ (i) follows directly from Corollary 3.3.3. ∼ (i) =⇒ (ii) If A is noetherian, then A0 = A/A+ is noetherian and A+ is ﬁnitely generated.

Let A+ = (x1, . . . , xs) for xi ∈ A. We may assume each xi ∈ Aki is homogeneous of degree ki. We will show An ⊆ A0[x1, . . . , xs] for all n ∈ N. The n = 0 case is trivial so assume Pn n ≥ 1. Then An ⊂ A+ so for any x ∈ An there are elements ci ∈ A such that x = i=1 cixi.

Since x is homogeneous of degree n, we may assume ci ∈ An−ki where by convention we treat A` = 0 when ` < 0. Since all ki > 0, then ci ∈ A0[x1, . . . , xs] by induction, and therefore x ∈ A0[x1, . . . , xs]. It follows that A ⊆ A0[x1, . . . , xs], so A = A0[x1, . . . , xs] and A/A0 is a ﬁnitely generated extension of rings.

Remark. If A is a ring and I ⊂ A is an ideal, there are two canonical graded rings we can associate to I: ∞ ∞ ∗ M n M n n+1 A := I and GI (A) := I /I . n=0 n=0 0 0 ∗ ∗ By convention, we let I = A. Notice that I = A,→ A , so that A0 = A. If A is noetherian, then I is ﬁnitely generated as a submodule, that is, I = (x1, . . . , xr) for some xi ∈ I. Then powers of I are generated by homogeneous products of the generators of I:

r ! n Y ei I = xi : ei ∈ N0, e1 + ... + er = n . i=1

∗ ∗ In particular, this shows that A = A[x1, . . . , xs], so A is a noetherian ring by Corol- lary 3.3.3(b).

Suppose I is an ideal of a ring A. Let M be an A-module with I-ﬁltration (Mn)n∈N0 ; recall that this means there is a chain of submodules

M = M0 ⊇ M1 ⊇ M2 ⊇ · · ·

n such that IMn ⊆ Mn+1 for all n ∈ N0. The standard I-filtration on M is when Mn = I M, which defines the I-topology (or I-adic topology) on M. For any filtration (Mn)n∈N0 , we define a graded module from M by

∞ ∗ M M = Mn. n=0 Lemma 5.4.3. M ∗ is a graded A∗-module.

Lemma 5.4.4. Let A be a noetherian ring, I ⊂ A and ideal and M a ﬁnitely generated

A-module. Then for an I-ﬁltration (Mn)n∈N0 of M, the following are equivalent:

85 5.4 Graded Rings and Modules 5 Completion and Filtration

∗ L∞ ∗ (i) M = n=0 Mn is a ﬁnitely generated A -module.

(ii) (Mn)n∈N0 is a stable I-filtration. Proof. Note that by Corollary 3.1.5, the assumptions that A is noetherian and M is finitely generated imply that M is a noetherian A-module. Thus each submodule Mn ⊆ M is finitely generated, so the modules n M Qn := Mi n=0

are ﬁnitely generated for each n ∈ N, by Corollary 3.1.4. For each n, Qn generates the following A∗-submodule of M ∗:

∞ ∗ M m Qn := Qn ⊕ I Mn. m=1

∗ ∗ m (That Qn is an A -submodule follows from the fact that I Mn ⊆ Mm+n for all m ∈ N0.) ∗ ∗ Moreover, Qn is a finitely generated A -module generated by the A-module generators of ∗ ∗ m m−1 Qn. By construction, Qn ⊆ Qn+1 for all n, since I Mn ⊆ I Mn+1 for all m ≥ 1. Also, S∞ ∗ ∗ n=0 Qn = M . With these observations, we now proceed to the main proof. (i) =⇒ (ii) If M ∗ is a finitely generated A∗-module then it follows (3.1.5) that M ∗ is noetherian as an A∗-module, i.e. M ∗ satisfies the ACC on A∗-submodules. In particular, ∗ ∗ ∗ ∗ the ascending chain Qn ⊆ Qn+1 stabilizes, so there is some n0 ∈ N0 such that Qn = Qn0 for ∗ S∞ ∗ ∗ ∗ ∗ all n ≥ n0. By the above, M = n=0 Qn = Qn0 so Qn0 is a finitely generated A -module. Then ∞ n0 ∞ M M M m m Mn = Mn ⊕ I Mn0 =⇒ Mn0+m = I Mn0 for all m ∈ N0. n=0 n=0 m=n0

This is equivalent by Lemma 5.3.2 to (Mn) ∈n∈N0 being a stable I-filtration. m (ii) =⇒ (i) If (Mn) is a stable filtration, there is some n0 ∈ N0 such that Mn0+m = I Mn0 ∗ ∗ ∗ for all m ∈ N0. This implies M = Qn0 so by the preliminary remarks, M is a finitely generated A∗-module. Proposition 5.4.5. Assume A is a noetherian ring, I ⊂ A is an ideal, M is a finitely 0 generated A-module with stable I-filtration (Mn)n∈N0 and M ⊆ M is a submodule. Then the 0 0 0 filtration Mn = M ∩ Mn induced on M is a stable I-filtration.

0 Proof. First, (Mn) is an I ﬁltration, since for each n,

0 0 0 0 0 IMn = I(M ∩ Mn) ⊆ IM ∩ IMn ⊆ M ∩ Mn+1 ⊆ M ∩ Mn.

∗ L∞ ∗ If (Mn) is stable then by Lemma 5.4.4, M = n=0 Mn is a finitely generated A -module, and thus by Lemma 5.4.2 a noetherian A∗-module. Then (M 0)∗ exists and is an A∗-submodule of M ∗. Since M ∗ is noetherian, (M 0)∗ is finitely generated as an A∗-submodule. Finally, 0 Lemma 5.4.4 implies that (Mn) is a stable I-filtration. We now obtain the useful Artin-Rees Lemma, which has a number of important consequences that follow.

86 5.4 Graded Rings and Modules 5 Completion and Filtration

Corollary 5.4.6 (Artin-Rees Lemma). Assume A is a noetherian ring, I ⊂ A is an ideal, M 0 is a finitely generated A-module and M ⊆ M is a submodule. Then there exists an n0 ∈ N0 0 n+n0 n 0 n0 such that M ∩ I M = I (M ∩ I M) for all n ∈ N0. n Proof. Apply Proposition 5.4.5 to the stable I-filtration Mn = I M of M. Corollary 5.4.7. Assume A is a noetherian ring, I ⊂ A is an ideal, M is a finitely generated A-module and M 0 ⊆ M is a submodule. Then the subspace topology induced on M 0 by the I-topology on M is precisely the I-topology on M 0. Corollary 5.4.8 (Krull’s Intersection Theorem). Suppose A is a noetherian ring. Then (a) If I ⊂ A is an ideal such that I ⊆ J(A) and M is a finitely generated A-module, then T∞ n n=0 I M = 0. In particular, the I-topology on M is Hausdorff. T∞ n (b) If A is an integral domain and I is a proper ideal of A, then n=0 I = 0. 0 T∞ n 0 Proof. (a) Set M = n=0 I M so that M ⊆ M is a finitely generated A-submodule of M (A is noetherian and M is finitely generated). By the Artin-Rees lemma, there exists an n0 ∈ N0 0 n+n0 n 0 n0 0 n+n0 0 so that for all n ∈ N0, M ∩ I M = I (M ∩ I M). But notice that M ∩ I M = M and M 0 ∩ In0 M = M 0, so we have M 0 = InM 0. Since I ⊆ J(A), Nakayama’s lemma (1.2.1) implies M 0 = 0. T∞ n (b) Now suppose A is an integral domain. Set J = n=0 I and suppose x ∈ J. Then k k x ∈ I for all k ∈ N0, so in particular (x) ⊆ I for every k. By the Artin-Rees lemma, there n+n0 n n0 k is an n0 ∈ N0 such that for all n ∈ N0,(x) ∩ I = I ((x) ∩ I ). Since (x) ⊆ I for all k, this becomes (x) = In(x) for all n, even (x) = I(x). Thus there is some a ∈ I such that x = ax, which can be written x(1 − a) = 0. If a were 1, I would be the unit ideal, but this contradicts the assumption that I is proper. So a 6= 1, and since A is a domain, it follows that x = 0. Corollary 5.4.9. Suppose A is a noetherian ring and 0 → M 0 → M → M 00 → 0 is a short exact sequence of A-modules. For a fixed ideal I ⊂ A, consider the I-topologies on M 0,M,M 00 and the corresponding completions Mc0, M,c Mc00. Then the sequence

0 → Mc0 → Mc → Mc00 → 0 is also exact. Proof. This follows from Corollary 5.4.7 and Lemma 5.3.2. Now recall the other canonical graded ring associated to an ideal I ⊂ A, namely

∞ M n n+1 GI (A) = I /I . n=0 The multiplication law is given by

Im/Im+1 × In/In+1 −→ Im+n/Im+n+1 (x + Im+1, y + In+1) 7−→ xy + Im+n+1.

87 5.4 Graded Rings and Modules 5 Completion and Filtration

This is well-deﬁned since In ⊇ In+1 is a chain. If M is an A-module with I-ﬁltration

(Mn)n∈N0 , we can deﬁne a GI (A)-module by

∞ M GI (M) = Mn/Mn+1. n=0

Then GI (M) is a graded GI (A)-module via the multiplication law

m m+1 I /I × Mn/Mn+1 −→ Mm+n/Mm+n+1 m+1 (x + I , z + Mn+1) 7−→ xz + Mm+n+1

m which is well-deﬁned since xz ∈ I Mn ⊆ Mm+n. Lemma 5.4.10. If A is a noetherian ring, then for every ideal I ⊂ A,

(a) GI (A) is noetherian.

(b) For any finitely generated A-module M with stable I-filtration (Mn)n∈N0 , GI (M) is finitely generated as a GI (A)-module.

Proof. (a) Since A is noetherian, I is ﬁnitely generated, say by x1, . . . , xr ∈ I. Setx ¯i = 2 2 n n+1 xi + I ∈ I/I . Then as before, it can be checked that each I /I is generated by products of powers of thex ¯i. By this, GI (A) = A/I[¯x1,..., x¯r] which is noetherian by Corollary 3.3.3(b). n (b) If (Mn) is a stable I-ﬁltration of M, there is some n0 ∈ N0 such that Mn0+n = I Mn0 n n+1 for all n0 ∈ N0. This implies Mn0+n/Mn0+n+1 = (I /I )Mn0 /Mn0+1 for all n ∈ N0, and thus n M0 GI = G(A) Mi/Mi+1. i=0

Now each Mi/Mi+1 is a ﬁnitely generated A-module and even an A/I-module, and A/I ⊆ GI (A). It follows that GI (M) is a ﬁnitely generated GI (A)-module.

88 6 Dimension Theory

6 Dimension Theory

6.1 Hilbert Functions

L∞ Let A = n=0 An be a graded noetherian ring. Recall from Lemma 5.4.2 that A0 is noethe- L∞ rian and A = A0[x1, . . . , xs] for some xi ∈ Aki , ki ≥ 1, 1 ≤ i ≤ s. Also let M = n=0 Mn be a ﬁnitely generated graded A-module, generated by homogeneous elements m1, . . . , mt such

that mj ∈ Mrj for each 1 ≤ j ≤ t.

Lemma 6.1.1. Each Mn is a ﬁnitely generated A0-module.

Proof. It is clear that each Mn is an A0-module by the deﬁnition of a graded A-module. Pt Take z ∈ Mn. Then there exist aj ∈ A for 1 ≤ j ≤ t such that z = j=1 ajmj. Further, we can choose polynomials fj ∈ A0[t1, . . . , ts] such that aj = fj(x1, . . . , xs) for each 1 ≤ j ≤ t. e1 es Note that monomials in the xi are homogeneous: x1 ··· xs ∈ Ak1e1+...+kses . Since z ∈ Mn, we can write t X z = gj(x1, . . . , xs)mj j=1

e1 es where for each j, gj(x1, . . . , xs) is an A0-linear combination of monomials x1 ··· xs such that k1e1 + ... + kses = n − rj. This shows Mn is ﬁnitely generated as an A0-module: the e1 es explicit generators are all elements of the form x1 ··· xs mj for 1 ≤ j ≤ t and ei ≥ 0 such that k1e1 + ... + kses = n − rj.

Let C be the class of ﬁnitely generated A0-modules.

Deﬁnition. A function λ : C → Z is said to be additive if for any short exact sequence 0 → M 0 → M → M 00 → 0 in C, we have

λ(M) = λ(M 0) + λ(M 00).

Examples.

1 If A0 is noetherian, the length function `(M) is additive by Lemma 3.2.3.

2 In the special case that A0 = k is a ﬁeld, C is the class of all ﬁnite dimensional k-vector spaces. Here, the dimension function λ(M) = dimk M is an additive function.

f1 f2 fn fn+1 Lemma 6.1.2. Suppose 0 → M0 −→ M1 −→· · · −→ Mn −−→ 0 is an exact sequence in C. Then for any additive function λ : C → Z, n X i (−1) λ(Mi) = 0. i=0

Proof. For each 1 ≤ i ≤ n + 1, set Ni = im fi which also equals ker fi+1 by exactness. Then we have short exact sequences

fi+1 0 → Ni → Mi −−→ Ni+1 → 0

89 6.1 Hilbert Functions 6 Dimension Theory

for every 1 ≤ i ≤ n + 1. By additivity, λ(Ni) − λ(Mi) + λ(Ni+1) = 0 for all i, so we get a telescoping sum:

n n X i X i (−1) λ(Mi) = (−1) (λ(Ni) + λ(Ni+1)) = λ(N0) ± λ(Nn+1). i=0 i=0

Finally, additivity implies λ(0) = 0 so we have λ(N0) = λ(Nn+1) = 0. The result then follows.

Definition. Let A be a graded ring, C the class of finitely generated A0-modules and λ an additive function on C. Then for any module M ∈ C, we define the Poincaréseries of M with respect to λ as ∞ X n PM (t) = λ(Mn)t ∈ Z[[t]]. n=0

Formally, the Poincaréseries of any module is a power series with coefficients in Z.A surprising result is that for finitely generated A0-module M, PM (t) is in fact a rational function.

Theorem 6.1.3 (Hilbert, Serre). For any M ∈ C and additive function λ, there is some polynomial f(t) ∈ Z[t] such that f(t) PM (t) = s ∈ (t). Q ki Q i=1(1 − t )

Proof. We prove this by inducting on s, the number of generators of A0-generators of A. If s = 0, A = A0 and M is a finitely generated A-module, so Mn = 0 for all n > max{rj | 1 ≤ j ≤ t}. Thus PM (t) is a polynomial in this case, so f(t) = PM (T ) ∈ Z[t] works. Now assume the result holds for all modules in C generated as an A0-module by s − 1 elements or fewer. Consider the A-module homomorphism f : M → M, z 7→ xsz. This gives us A-modules K = ker f and L = coker f = M/xsM. Since A is noetherian, K and L are finitely generated (by Corollary 3.1.5 and Proposition 3.1.2). Note that xsK = f(K) = 0 and ¯ xsL = xsM/xsM = 0, so K and L are in fact graded modules over the quotient A = A/xsA: L∞ L∞ K = n=0 Kn and L = n=0 Ln. Also notice that Lr = Mr whenever r < ks. Moreover, by Lemma 3.1.3, A¯ is noetherian so K and L are finitely generated as A¯-modules. We can view A¯ as a graded noetherian ring:

∞ ¯ M ¯ ¯ A = An = A0[¯x1,..., x¯s] n=0 ¯ where An = An/xsAn−ks andx ¯i = xi + xsAki−ks for all n, i. Since ki ≥ 1 for all i, we actually ¯ ∼ ¯ have A0 = A0, so the ﬁnitely generated modules over A0 are precisely the elements of C. In a moment, this will allow us to apply the induction hypothesis to K and L. For n ∈ N0, let fn denote the restriction of f to Mn. We get the following exact sequences:

fn 0 → Kn → Mn −→ Mn+ks → Ln+ks → 0 for n ∈ N0.

90 6.1 Hilbert Functions 6 Dimension Theory

n+ks By Lemma 6.1.2, λ(Kn) − λ(Mn) + λ(Mn+ks ) − λ(Ln+ks ) = 0. Multiplying through by t and rearranging terms yields

ks n n n+ks t (t λ(Kn) − t λ(Mn)) = t (λ(Ln+ks ) − λ(Mn+ks ))

ks =⇒ t (PK (t) − PM (t))n+ks = (PL(t) − PM (t))n+ks for all n ∈ N0 ks =⇒ t (PK (t) − PM (t)) = PL(t) − PM (t) since both power series have 0 coeﬃcients for indices less than ks, and the second line implies that the power series agree in every remaining term. This can be rearranged to read

ks ks (1 − t )PM (t) = PL(t) − t PK (t).

By the induction hypothesis, there exist polynomials g(t), h(t) ∈ Z[t] such that

s−1 s−1 Y ki Y ki (1 − t )PL(t) = g(t) and (1 − t )PK (t) = h(t). i=1 i=1 It follows that s Y ki ks (1 − t )PM (t) = g(t) − t h(t) ∈ Z[t]. i=1

Thus the result holds for PM (t) for all M ∈ C by induction. Definition. The dimension of a graded A-module M is defined to be the order of the pole of PM (t) at t = 1, denoted d(M). Formally, we set d(M) = 0 if there is no pole at t = 1. For a graded A-module M of dimension d = d(M), Theorem 6.1.3 allows us to write the Poincaréseries M as g(t) P (t) = M h(t)(1 − td) where h(1) 6= 0, and if d ≥ 1, g(1) 6= 0.

Corollary 6.1.4. If d(M) ≥ 1 and x ∈ Ak such that xz 6= 0 for all z ∈ M r {0}, then d(M/xM) = d(M) − 1. Proof. Consider the map f : M → M, z 7→ xz. As in Theorem 6.1.3, we have modules K = ker f and L = coker f = M/xM. Since xz 6= 0 for any z ∈ M r {0}, K = 0. Then the k proof of Theorem 6.1.3 shows that (1−t )PM (t) = PL(t). It follows that d(L) = d(M)−1.

Lemma 6.1.5. For all n, s ∈ N0 with n ≥ s, n + s − 1 #{(i , . . . , i ) ∈ s : i + ... + i = n} = . 1 s N0 1 s s − 1

Example 6.1.6. Let k be a ﬁeld and consider the polynomial ring A = k[t1, . . . , ts]. Then by Example 5.4.1, A is a graded ring with

M i1 is A0 = k and An = kt1 ··· ts for each n ≥ 1. ij ∈N0 i1+...+is=n

91 6.1 Hilbert Functions 6 Dimension Theory

Let λ(M) = dimk M be the additive dimension function for k-vector spaces M. Notice that n+s−1 by Lemma 6.1.5, dimk An = s−1 for each n ≥ 1, so the Poincar´eseries for A as a module over itself is ∞ ∞ X X n + s − 1 1 P (t) = (dim A )tn = tn = . A k n s − 1 (1 − t)s n=0 n=0 Hence d(A) = s so the notion of vector space dimension coincides with the dimension of a graded ring. Notice that in this example, k1 = k2 = ··· = ks, i.e. the generators t1, . . . , ts are all homogeneous of degree 1. The following proposition generalizes this and introduces the Hilbert polynomial of an A-module.

Proposition 6.1.7. Suppose A = A0[x1, . . . , xs] is a graded noetherian ring such that xi ∈ A1 for all 1 ≤ i ≤ s. Then for any additive function λ and graded A-module M, there exists n0 ∈ N0 and a polynomial g(t) ∈ Q[t] with deg g = d(M) − 1 such that λ(Mn) = g(n) for all n ≥ n0.

n+k 1 Proof. For ﬁxed k, ` ∈ N, ` = `! (n + k)(n + k − 1) ··· (n + k − ` + 1) is a polynomial in n of degree ` with coeﬃcients in Q. The assumption that each xi ∈ A1 means k1 = . . . ks, so Theorem 6.1.3 gives us a function f(t) ∈ Z[t] such that

∞ f(t) X n + s − 1 P (t) = = f(t) tn M (1 − t)s s − 1 n=0

Pn0 k by Lemma 6.1.5. Write f(t) = k=0 akt , with ak ∈ Z and n0 ∈ N. If d(M) = 0, then PM (t) ∈ Z[t] and so λ(Mn) = 0 for all n > deg f. In this case, g = 0 works; formally we let deg 0 = −1. Now if d(M) 6= 0, we may assume s = d = d(M) and f(1) 6= 0. Then in Pn0 particular k=0 ak = f(1) 6= 0 so we have

n ! ∞ ! ∞ X0 X m + d − 1 X P (t) = a tk tm = λ(M )tn. M k d − 1 n k=0 m=0 n=0

This shows that for all n ≥ n0,

n X0 n − k + d − 1 λ(M ) = a . n k d − 1 k=0

Pn0 t−k+d−1 Set g(t) = k=0 ak d−1 . Then by the preliminary comment, g is a polynomial of degree at most d − 1 in Q[t], and g(n) = λ(Mn) for all n ≥ n0. Finally, notice that

n n ! X0 n − k + d − 1 X0 nd−1 λ(M ) = a = a + lower terms n k d − 1 k (d − 1)! k=0 k=0

Pn0 and k=0 ak 6= 0, so the degree of g is exactly d − 1. Deﬁnition. The function g is called the Hilbert polynomial of M with respect to the additive function λ.

92 6.1 Hilbert Functions 6 Dimension Theory

Proposition 6.1.8. Given a noetherian ring A with maximal ideal m ∈ MaxSpec(A), an m- ∞ primary ideal Q ⊂ A and a ﬁnitely generated A-module M with stable Q-ﬁltration (Mn)n=0, the following hold:

(a) `(M/Mn) < ∞ for all n ∈ N0.

(b) If Q is generated by s elements, there exist n0 ∈ N and a polynomial g(t) ∈ Q[t] of degree at most s with the property that `(M/Mn) = g(n) for all n ≥ n0. (c) The degree and leading coeﬃcient of g(t) are independent of the choice of Q-ﬁltration.

Proof. (a) Recall the graded noetherian ring

∞ M n n+1 G(A) = Gn(A), where G0(A) = A/Q and Gn(A) = Q /Q . n=0

Notice that G0(A) is noetherian and dim A/Q = 0 since r(Q) = m is the only prime ideal of A containing Q. Thus by Theorem 3.5.5, G0(A) is artinian. The associated graded L∞ module G(M) = n=0 Mn/Mn+1 is a graded G(A)-module. By Lemma 5.4.10(b), G(M) is finitely generated as a G(A)-module. Moreover, for each j ∈ N0, Gj(M) = Mj/Mj+1 is a finitely generated G0(A)-module, hence artinian and noetherian by Corollary 3.1.5. Hence by Corollary 3.2.2, each quotient Mj/Mj+1 has a composition series, so `(M/Mn) = Pn−1 j=0 `(Mj/Mj+1) is finite. 2 (b) We have that Q = (x1, . . . , xs) as an ideal of A. For each xi, definex ¯i = xi + Q ∈ 2 Q/Q = G1(A). Then G(A) = G0(A)[¯x1,..., x¯s]. Note that k1 = ... = ks = 1 in the notation of Theorem 6.1.3. Consider the Poincaréseries

∞ X n+1 PG(M)(t) = `(Mn/Mn+1)t . n=0

By Proposition 6.1.7, there exist n0 ∈ N and a polynomial f(t) ∈ Q[t] of degree d−1 ≤ s−1, where d = d(G(M)), such that `(Gj(M)) = f(j) for each j ≥ n0. Then by (a), for all n ≥ n0 we have n−1 n −1 n−1 X X0 X `(M/Mn) = `(Mj/Mj+1) = `(Mj/Mj+1) + f(j).

j=0 j=0 j=n0 Now we show that Pn−1 f(j) is a polynomial in n of degree d ≤ s with rational coeﬃcients. j=n0 It is a fact that if f ∈ Q[t] with deg f = d − 1, then there existsg ˜ ∈ Q[t] with degg ˜ = d Pn P` r such that j=1 f(j) =g ˜n for all n. To see this explicitly, write f(t) = r=0 art so that Pn P` Pn r Pn r j=1 f(j) = r=0 ar j=1 j . For all r ∈ N0, there exists an hr ∈ Q[t] satisfying j=1 j = hr(n); indeed, these arise as certain Bernoulli polynomials:

If r = 0, h0(t) = t. 1 1 2 If r = 1, h1(t) = 2 t + 2 t . 1 1 2 1 3 If r = 2, h2(t) = 6 t + 2 t + 3 t , etc.

93 6.2 Local Noetherian Rings 6 Dimension Theory

Pn P` In any case, we have j=1 f(j) = r=0 arhr(n) which is now a polynomial in n as claimed. This proves (b). 0 (c) If (Mn) and (Mn) are two stable Q-ﬁltrations of M, then by Lemma 5.3.2, there is 0 0 some m0 ∈ N0 such that Mn+m0 ⊆ Mn and Mn+m0 ⊆ Mn for all n ∈ N0. By (b), we get 0 0 0 integers n0, n0 ∈ N0 and polynomials g, g ∈ Q[t] such that for all n ≥ max{n0, n0}, 0 0 g(n + m0) = `(M/Mn+m0 ) ≥ `(M/Mn) = g (n) 0 0 and g (n + m0) = `(M/Mn+m0 ) ≥ `(M/Mn) = g(n). g(n) It follows that lim = 1 so g and g0 must have the same degree and leading coeﬃcients. n→∞ g0(n)

In the next section, we will prove that when A is local noetherian with maximal ideal m, the dimension of A may be computed by

dim A = min{s ∈ N0 | there exists an m-primary ideal Q ⊂ A generated by s elements}. ∞ n M With A, m,M,Q and (Mn)n=0, where Mn = Q M, as in Proposition 6.1.8, let χQ (t) ∈ M Q[t] denote the polynomial (g in the proposition) such that `(M/Mn) = χQ (n) for all large enough n.

A Deﬁnition. When M = A, the polynomial χQ(t) := χQ(t) is called the characteristic polynomial of Q.

Corollary 6.1.9. For a maximal ideal m ⊂ A and an m-primary ideal Q ⊂ A, χQ(t) is a polynomial of degree at most min{s ∈ N0 | Q is generated by s elements}. Proposition 6.1.10. For a maximal ideal m ⊂ A and any m-primary ideal Q ⊂ A, deg χQ(t) = deg χm(t). Proof. Since A is assumed to be noetherian, mr ⊆ Q ⊆ m for some r ∈ N by Lemma 3.4.1. Then mnr ⊆ Qn ⊆ mn for all n ∈ N, so `(A/mn) ≤ `(A/Qn) ≤ `(A/mnr). Thus for sufficiently large n, χm(n) ≤ χQ(n) ≤ χm(nr), so taking the limit as n → ∞ implies the statement. Remark. Let A be a graded noetherian ring and m ⊂ A a maximal ideal. By Proposi- tions 6.1.7 and 6.1.8, deg χm(t) = d(Gm(A)) where Gm(A) is considered as a graded Gm(A)- module. By definition, d(Gm(A)) equals the order of the pole at t = 1 of the Poincaréseries

PGm(A)(t), which equals deg g(Gm(A))+1 when g is the Hilbert function of Gm(A). Therefore we can write d(A) = deg χm(t) = d(Gm(A)).

6.2 Local Noetherian Rings

For the entirety of this section, assume A is a local noetherian ring with maximal ideal m and set

δ(A) = min{s ∈ N0 | there exists an m-primary ideal Q ⊂ A generated by s elements}.

Our goal is to show that δ(A) = d(A) = deg χm(t) = dim A.

94 6.2 Local Noetherian Rings 6 Dimension Theory

Lemma 6.2.1. δ(A) ≥ d(A). Proof. This follows immediately from Corollary 6.1.9 and Proposition 6.1.10. Lemma 6.2.2. If Q is an m-primary ideal and M is a ﬁnitely generated A-module, then for any x ∈ A satisfying xz 6= 0 for all z ∈ m r {0}, M 0 M deg χQ (t) ≤ deg χQ (t) − 1 where M 0 = M/xM. Proof. Set N = xM ⊆ M. We have exact sequences p 0 → N → M −→ M → 0 (E) m n pn 0 n 0 0 → N/N ∩ Q → M/Q M −→ M /Q M → 0 (En) n n ∞ n 0 ∞ ∞ for each n ∈ N. Set Nn = N ∩ Q M. Now (Q M)n=0,(Q M )n=0 and (Nn)n=0 are 0 stable Q-ﬁltrations of M, M and N, respectively (for (Nn), use Proposition 5.4.5). The n length function ` is additive, so by the short exact sequence (En), `(N/Nn) − `(M/Q M) + 0 n 0 `(M /Q M ) = 0. Thus for some n0 ∈ N0, N M M 0 χQ (n) − χQ (n) + χQ (n) = 0

for all n ≥ n0. By assumption, M → N, z 7→ xz is an isomorphism of A-modules. So the stable Q-filtration (Nn) of N corresponds to a stable Q-filtration of M. By Proposition 6.1.8(c), N M χQ (t) and χQ (t) have the same degree and leading coefficient. Taking n → ∞, this implies M 0 M N M that det χQ (t) = deg(χQ (t) − χQ (t)) ≤ deg χQ (t) − 1. Corollary 6.2.3. If x ∈ A is not a zero divisor or a unit then d(A/(x)) ≤ d(A) − 1.

0 A/(x) Proof. Let M = A, M = A/(x) and apply Lemma 6.2.2 to get deg χm (t) ≤ deg χm(t)−1 = A/(x) d(A) − 1. By deﬁnition, d(A/(x)) = deg χm/(x)(t) = deg χm (t) so the result follows. Proposition 6.2.4. d(A) ≥ dim A. Proof. We prove this by induction on d = d(A). If d = 0, `(A/mn) is stationary for n n+1 large enough n, so there exists an n ∈ N0 for which `(A/m ) = `(A/m ). This implies mn = mn+1 = J(A)mn, so by Nakayama’s Lemma (1.2.1), mn = 0. Thus A is artinian by Theorem 3.5.6. In particular, dim A = 0 by Theorem 3.5.5. Now suppose d = d(A) ≥ 1. Let p0 ( ··· ( pr be a chain of prime ideals in A. We 0 must show r ≤ d for all such r. Assuming r ≥ 1, pick x ∈ p1 r p0 and set A = A/p0 0 0 0 0 0 and x = x + p0 ∈ A . Since A is an integral domain, x is not a zero divisor and x 6= 0 0 0 0 0 because x 6∈ p0. Thus by Corollary 6.2.3, d(A /(x )) ≤ d(A ) − 1. Notice that m = m/p0 is the maximal ideal of A0. Since ` is additive, the projections A/mn → A0/(m0)n imply n 0 0 n A A0 0 `(A/m ) ≥ `(A /(m ) ) for all n ∈ N0. So deg χm(t) ≥ deg χm0 (t) which means d(A) ≥ d(A ). 0 0 0 This gives us d(A /(x )) ≤ d(A) − 1. Set pi = pi/p0 for 1 ≤ i ≤ r, each of which is a prime 0 0 0 ideal of A . Then x ∈ p1 so 0 0 0 0 0 0 p1/(x ) ( p2/(x ) ( ··· ( pr/(x ) is a chain of prime ideals in A0/(x0). Since d(A0/(x0)) ≤ d(A) − 1, and dim A0/(x0) ≤ d(A0/(x0)) ≤ d−1 by the inductive hypothesis, we get r−1 ≤ dim A0/(x0) ≤ d−1. Therefore r ≤ d. Taking the supremum over all such chains of prime ideals, we get dim A ≤ d.

95 6.2 Local Noetherian Rings 6 Dimension Theory

Corollary 6.2.5. Let A be any noetherian ring. Then (a) If A is local, dim A < ∞. (b) For any prime ideal p ∈ Spec(A), ht(p) < ∞.

Proof. (a) follows immediately from Proposition 6.2.4. For (b), note that Ap is local noetherian and ht(p) = dim Ap by Lemma 2.3.6(a). Corollary 6.2.6. For a local noetherian ring, δ(A) ≥ d(A) ≥ dim A.

2 Example 6.2.7. Let (A, m, k) be a local noetherian ring. Then dimk m/m ≥ dim A. Indeed, 2 2 if x1, . . . , xs ∈ m such thatx ¯1,..., x¯s ∈ m/m form a k-basis of m/m , Proposition 1.2.4 says that m is generated as an ideal by x1, . . . , xs. Thus s ≥ δ(A) ≥ dim A.

For any m-primary ideal Q ⊂ A, set δ(Q) = min{s ∈ N0 | Q is generated by s elements}, so that δ(A) = min{δ(Q) | Q is m-primary}. Clearly δ(m) ≥ δ(A). Further, for any x1, . . . , xs ∈ m, s 2 X 2 m = (x1, . . . , xs) ⇐⇒ m/m = Ax¯i wherex ¯i = xi + m , by Proposition 1.2.4 i=1 s 2 X ⇐⇒ m/m = kx¯i. i=1

2 Taking the minimum over s on both sides, it follows that δ(m) = dimk m/m .

2 Deﬁnition. A local noetherian ring A is said to be a regular ring if dim A = dimk m/m .

By the above, A is regular if and only if m = (x1, . . . , xd), where d = dim A. Theorem 6.2.8. Let (A, m, k) be a local noetherian ring of dimension d = dim A. Then the following are equivalent: ∼ (a) Gm(A) = k[t1, . . . , td] for indeterminates ti.

2 (b) A is regular, i.e. dimk m/m = d.

(c) m = (x1, . . . , xd) for some x1, . . . , xd ∈ A. Lemma 6.2.9. Let A be a regular ring. Then A is an integral domain which is integrally closed.

Proof. We prove A is an integral domain. Take x, y ∈ A r {0}. By Krull’s intersection T∞ n r r+1 s theorem (5.4.8), n=1 m = 0 so there exist r, s ∈ N0 such that x ∈ m , x 6∈ m , y ∈ m s+1 and y 6∈ m . Letx ¯ be the image of x in Gr(A) andy ¯ the image of y in Gs(A). Thenx ¯ 6= 0 ∼ andy ¯ 6= 0, sox ¯ · y¯ 6= 0 in Gm(A), since by Theorem 6.2.8, Gm(A) = k[t1, . . . , td] which is an integral domain. Thus xy 6= 0 in A.

Theorem 6.2.10 (Krull’s Height Theorem). Suppose A is noetherian and x1, . . . , xs ∈ A. Then for any prime ideal p ⊂ A which is minimal among prime ideals containing (x1, . . . , xs), ht(p) ≤ s.

96 6.2 Local Noetherian Rings 6 Dimension Theory

Proof. Assume I = (x1, . . . , xs) ⊆ p and p is minimal among primes containing I. Consider −1 the localization Ap = S A, where S = A r p. Then Ap is local noetherian with maximal ideal m = S−1p. If p0 ⊂ A is any other prime ideal containing I and p0 ⊆ p, then by 0 −1 minimality, p = p. Therefore m is the unique prime ideal of Ap containing S A, so by −1 −1 x1 xs Lemma 3.4.8(a), S is m-primary. Now S I = 1 ,..., 1 so by deﬁnition, Corollary 6.2.6 and Lemma 2.3.6(a), s ≥ δ(Ap) ≥ dim Ap = ht(p). Theorem 6.2.11 (Krull’s Principal Ideal Theorem). Assume A is noetherian, x ∈ A is a nonunit and p ∈ Spec(A) is a prime ideal minimal among prime ideals containing (x). Then (a) ht(p) ≤ 1. (b) If x is not a zero divisor of A, then ht(p) = 1. Proof. (a) is immediate from Krull’s height theorem. For (b), suppose (x) ⊆ p and ht(p) = 0. Then all elements of p are zero divisors by Corollary 3.4.16, so it must be that ht(p) = 1. Theorem 6.2.12. Let A be a local noetherian ring. Then dim A = δ(A). Proof. We have already proved dim A ≤ d := δ(A) in Corollary 6.2.6. For the other inequality, we construct a sequence of elements x1, . . . , xd ∈ m satisfying the following condition: For each 1 ≤ j ≤ d, every prime p ∈ Spec(A) (∗) containing (x1, . . . , xj) has height ht(p) ≥ j.

Given such x1, . . . , xd, let Ij = (x1, . . . , xj) for each 1 ≤ j ≤ d. Then if p ⊇ Id, ht(p) ≥ d = ht(m) but this is only possible if p = m. Hence Id is m-primary by Lemma 3.4.8(a), so we get dim A = d ≥ δ(Id) ≥ δ(A). Hence the result follows once we prove (∗). We verify (∗) inductively. For 1 ≤ j ≤ d, let x1, . . . , xj−1 be constructed. If p ∈ Spec(A) such that p ⊇ Ij−1, then ht(p) ≥ j − 1 by the inductive hypothesis. By Proposition 3.4.10, there are ﬁnitely many minimal elements of the set C = {p ∈ Spec(A) | p ⊇ Ij−1}. Suppose p1,..., pn are the elements of C with height j − 1 < d. Then pi 6= m for all 1 ≤ i ≤ n, so Sn Sn i=1 pi ( m. Choose xj ∈ m r i=1 pi. It now follows that (∗) holds for these x1, . . . , xd, and the theorem is proven. Corollary 6.2.13. If A is a local noetherian ring, then δ(A) = d(A) = dim A. We ﬁnish the section by proving a partial converse to Theorem 6.2.10. First we need a technical lemma.

Lemma 6.2.14 (Prime Avoidance). Suppose I1,...,In ⊂ A are ideals of which at least n−2 Sn are prime. Then for any ideal J ⊂ A, if J ⊆ k=1 Ik then J ⊆ Ik for some 1 ≤ k ≤ n.

Proof. Suppose J 6⊂ Ik for any 1 ≤ k ≤ n. If n = 2, choose x1 ∈ J such that x1 6∈ I1 and x2 ∈ J such that x2 6∈ I2. Then x1 + x2 ∈ J but x1 + x2 6∈ I1 ∪ I2, so J 6⊂ I1 ∪ I2. Inducting S on n > 2, we have J 6⊂ Ik for each 1 ≤ ` ≤ n. Without loss of generality, assume I1 k6=` S is prime. For each 1 ≤ ` ≤ n, choose x` ∈ J not lying in k6=` Ik, so in particular x1 ∈ I1, and set x = x1 + x2x3 ··· xn. If x ∈ I` for some ` ≥ 2, then x1 = x − x2x3 ··· xn ∈ I`, contradicting the base case. On the other hand, if x ∈ I1 then x2x3 ··· xn ∈ I1, so x` ∈ I1 Sn for some ` ≥ 2 by primality, also a contradiction. Hence J is not contained in k=1 Ik.

97 6.3 Complete Local Rings 6 Dimension Theory

Theorem 6.2.15. Let A be a local noetherian ring of dimension d, with maximal ideal m. Then there exist d elements x1, . . . , xd ∈ m such that r(x1, . . . , xd) = m.

Proof. If d = 0, we take (x1, . . . , xd) = 0. In this case, Theorems 3.5.5 and 3.5.6 say that A is artinian and mn = 0 for some n ≥ 1. Thus r(0) = m by deﬁnition. Now induct on d > 0. Since A is noetherian, the set of minimal primes of A is ﬁnite (Proposition 3.4.10), say {p1,..., pk}. Clearly m 6⊂ pi for any 1 ≤ i ≤ k, or else we would be in the base case. By the prime avoidance lemma, m 6⊂ p1 ∪ · · · pk. Choose x1 ∈ m such that x1 6∈ p1 ∪ · · · ∪ pk. Then dim(A/(x1)) ≤ d − 1 so by induction, there exist d − 1 elementsx ¯2,..., x¯d ∈ A/(x1) such that r(¯x2,..., x¯d) = m¯ in A/(x1). Lifting back to A, we see that r(x1, . . . , xd) = m as desired.

Corollary 6.2.16. If A is local noetherian of dimension d then for any nonzero x ∈ m, dim(A/(x)) = d − 1.

Proof. If dim(A/(x)) < d−1 then we could choose a smaller number of elements to generate m up to radical, but this would contradict Krull’s height theorem.

6.3 Complete Local Rings

We say a ring A is complete if A is a completion of some ring along an ideal. The structure of complete local rings is fully described in this section.

T∞ n Theorem 6.3.1 (Hensel’s Lemma). Let (A, m, k) be a complete local ring with n=1 m = 0. Suppose F (t) ∈ A[t] is a monic polynomial which factors as

F (t) ≡ g(t)h(t) mod m

for some monic, relatively prime polynomials g, h ∈ k[t]. Then there exist polynomials monic G, H ∈ A[t] of the same degrees as g and h, respectively, such that F (t) = G(t)H(t) in A[t] and G(t) = g(t) and H(t) = h(t) in k[t].

Proof. We inductively construct polynomials Gn,Hn ∈ A[t] satisfying

n (1) Gn = g, Hn = h and Gn − Gn+1,Hn − Hn+1 ∈ m A[t] for all n ≥ 1.

n (2) F − GnHn ∈ m A[t] for all n ≥ 1.

For n = 1, choose any monic polynomials G1,H1 ∈ A[t] lifting g and h, i.e. so that G1 = g and H1 = h. Then F − G1H1 = F − gh ≡ 0 mod m

so F − G1H1 ∈ mA[t]. Now assume Gn,Hn ∈ A[t] have been constructed and satisfy (1) and n (2). Then by (2), there are some y1, . . . , y` ∈ m and L1,...,L` ∈ A[t] such that

` X F − GnHn = yiLi. i=1

98 6.3 Complete Local Rings 6 Dimension Theory

Since g and h are relatively prime, for each 1 ≤ i ≤ ` we can write

Li = aig + bih for some ai, bi ∈ k[t] with deg ai < deg h by the division algorithm; note that this implies deg bi < deg g as well. Lift each ai to Ai ∈ A[t] and bi to Bi ∈ A[t], so that Li−AiGn−BiHn ∈ mA[t] for each i. Then

` X n+1 F ≡ GnHn + yi(AiGn + BiHn) mod m . i=1

P` P` n Set Gn+1 = Gn + i=1 yiBi and Hn+1 = Hn + i=1 yiAi. Then since each yi ∈ m , we see n that Gn − Gn+1,Hn − Hn+1 ∈ m A[t]. Moreover,

` X X F − Gn+1Hn+1 = F − GnHn − yi(AiGn + BiHn) − yiyjBiAj i=1 i,j X 2n = − yiyjBiAj ∈ m A[t] i,j

n+1 which implies F − Gn+1Hn+1 ∈ m A[t] since 2n ≥ n + 1. Hence (Gn) and (Hn) are Cauchy sequences in A[t] so they converge to some G, H ∈ A[t]. By construction, F − GH ∈ mnA[t] for all n ≥ 1, so by hypothesis, F = GH. T∞ Corollary 6.3.2. If (A, m, k) is a complete local ring with n=1 m = 0 and f(t) is a monic polynomial such that f¯(t) has a simple root α ∈ k, then there exists an element a ∈ A with a¯ = α and f(a) = 0.

Proof. This is the case where g(t) = t − α.

Definition. A coefficient field for a local ring (A, m, k) is a field L ⊆ A such that the composition L,→ A → A/m = k is an isomorphism.

There are three possibilities for the situation of a local ring having a coeﬃcient ﬁeld:

char A = 0 and Q ⊆ A, in which case A is called equicharacteristic of characteristic 0; char A = p > 0 and Fp ⊆ A, in which case A is called equicharacteristic of characteristic p;

A does not contain a ﬁeld, which is only possible if char A = 0 and char k = p > 0, and in this case A is called mixed characteristic.

Example 6.3.3. The power series rings C[[t]] and R[[t]] are examples of equicharacteristic complete local rings in characteristic 0.

Example 6.3.4. For a prime p, Fp[[t]] is a complete local ring that is equicharacteristic of characteristic p.

99 6.3 Complete Local Rings 6 Dimension Theory

Example 6.3.5. The p-adic integers Zp are the prototypical example of a mixed character- ∼ istic ring; they do not contain a field and have residue field Zp/pZp = Fp. Theorem 6.3.6 (Cohen’s First Structure Theorem). If (A, m, k) is a complete local ring T∞ n containing a field and n=1 m = 0, then A has a coefficient field.

Proof. First assume char A = 0, so Q ⊆ A. Use Zorn’s Lemma to choose a maximal subfield L ⊆ A. If L,→ A −→π k is already an isomorphism, we’re done. Otherwise, L still embeds into k since both are fields, so choose some α ∈ k r π(L). If α is transcendental over π(L), lift α to a ∈ A. Then for all nonzero f ∈ L[t], π(f(a)) = f(α) 6= 0 since α is transcendental, so f(a) 6∈ m. Thus L(a) ⊆ A, contradicting maximality of L. So α must be algebraic, say with minimal polynomial g(t) ∈ π(L)[t]. Since char L = 0, α is a simple root of g so by Corollary 6.3.2, α lifts to a root a ∈ A of g ∈ π(L)[t] = L[t]. Then a is algebraic, so L[a] is a field and it is contained in A, again contradicting maximality of L. So we must have π(L) = k. Now assume char A = p > 0. By Proposition 5.2.4, Ab ∼= lim A/mn so to prove the ←− theorem, we will exhibit a coefficient for each A/mn that are compatible with the maps n+1 n πn+1 : A/m → A/m of the inverse system. We induct on n; for n = 1, k = A/m itself is a coefficient field. For n = 2, set R = A/m2, which is a local ring with maximal ideal 2 mR = m/m , and consider the ring

Rp = {xp | x ∈ R} of pth powers in R – note that it is a ring by the ‘freshman’s dream’: (x + y)p = xp + yp. p p 2 Suppose x ∈ R is nonzero. Then since mR = 0, we must have x 6∈ mR and thus x is a unit in R. If x−1 ∈ R is an inverse of x, then (x−1)p ∈ Rp is an inverse of xp so we see that Rp is a subfield of R. Let L be a maximal subfield of R containing Rp and consider the projection π : R → R/mR = k. If π(L) 6= k, choose some α ∈ k r π(L) and lift it to x ∈ R. Then xp ∈ Rp ⊆ L is nonzero, so by the above it is a unit. This shows L[x] ) L ⊇ Rp which contradicts maximality of L, so we must have π(L) = k. Thus R has a coefficient field. n To induct, suppose we have constructed a coefficient field Ln for A/m . Set B = −1 n+1 n+1 n n+1 πn+1(Ln) ⊆ A/m . Then B is a subring of A/m containing m /m . If b ∈ B is n+1 not a unit, then b ∈ A/m and πn+1(b) ∈ Ln, so πn+1(b) = 0 since Ln is a field. Thus n n+1 n n+1 b ∈ m /m . This shows B is a local ring with maximal ideal mB = m /m . Moreover, 2n n+1 2 m ⊆ m implies mB = 0 so we are in the n = 2 case and there exists a coefficient field n+1 Ln+1 ⊆ B ⊆ A/m . By construction, πn+1(Ln+1) = Ln so taking L = lim Ln, we obtain a ←− coefficient field of Ab. Corollary 6.3.7. Let (A, m, k) be a complete local ring of characteristic p > 0 such that T∞ n n=1 m = 0 and k is perfect. Then

∞ \ e L = Ap e=1 is the unique coeﬃcient ﬁeld for A.

100 6.3 Complete Local Rings 6 Dimension Theory

Proof. We first prove L is a field. Let a ∈ L be nonzero. For each e ≥ 1, we can write a = xpe T∞ pe for some x ∈ A. If a ∈ m, then x ∈ m since m is prime. This shows a ∈ e=1 m = 0, contradicting a 6= 0. Therefore a 6∈ m, so a is a unit and L is a field. Now let F be a coefficient field for A. Then F ∼= k so F pe = F for all e ≥ 1. This implies

∞ ∞ \ e \ e F = F p ⊆ Ap = L e=1 e=1 so F ⊆ L. Reversing F and L gives an equality, so L is in fact unique.

Lemma 6.3.8. Let A = Ab be the completion along an ideal I ⊆ A and suppose M is an T∞ n A-module such that n=1 I M = 0. If x1, . . . , xr ∈ M such that M/IM is generated by x¯i = xi + IM, 1 ≤ i ≤ r, then M is generated by the xi. Pr Proof. Set N = i=1 Axi ⊆ M. For each m0 ∈ M, we can write

r X m0 = ai0xi + m1 i=1

2 for some ai0 ∈ A and m1 ∈ IM. By hypothesis M = N + IM, so IM = IN + I M and by extension, InM = InN + In+1M for all n ≥ 1. Therefore we can write

r X m1 = ai1xi + m2 i=1

2 for ai1 ∈ A and m2 ∈ I M. Repeating this argument, we can write

r n ! X X m0 = aijxi + mn+1 i=1 j=0

j n+1 Pn for any n ≥ 1, where aij ∈ I and mn+1 ∈ I M. But for each i, j=0 aij is a Cauchy sequence in A, so it converges to some ai ∈ A and

r X n m0 − aixi ∈ I M i=1 for all n. Hence m0 ∈ N. Theorem 6.3.9 (Cohen’s Second Structure Theorem). Let (A, m, k) be a complete local ring T∞ n containing a ﬁeld with n=1 m = 0 and suppose m is ﬁnitely generated, say by x1, . . . , xr. Then there is a surjective ring map

k[[t1, . . . , tr]] −→ A

sending ti 7→ xi.

101 6.3 Complete Local Rings 6 Dimension Theory

Proof. Deﬁne π : k[[t1, . . . , tr]] → A by sending ti 7→ xi as suggested. This means for an P n1 nr arbitrary power series αn1,...,nr t1 ··· tr in k[[t1, . . . , tr]], X n1 nr X n1 nr π αn1,...,nr t1 ··· tr = αn1,...,nr x1 ··· xr and the partial sums of this series form a Cauchy sequence in A, so they converge. Hence π is well-deﬁned and is a homomorphism by construction. It remains to show π is surjective. But A is a module over k[[t1, . . . , tr]] and A/(t1, . . . , tr)A = A/m is generated by 1, so by Lemma 6.3.8, A is generated by 1 as a k[[t1, . . . , tr]]-module. This implies π is surjective.

T∞ n Corollary 6.3.10. If A is a complete local ring containing a ﬁeld with n=1 m = 0 and m is ﬁnitely generated, then A is noetherian.

Proof. Apply Theorem 3.3.9 and Lemma 3.3.1(a).

Deﬁnition. For a local ring A with maximal ideal m, a system of parameters for A is a set of elements x1, . . . , xn ∈ m such that r((x1, . . . , xn)) = m.

Recall (Theorem 6.2.12) that if x1, . . . , xn is a system of parameters for A, then n = dim A.

Theorem 6.3.11 (Cohen’s Third Structure Theorem). Let (A, m, k) be a complete local noetherian ring of dimension d containing a ﬁeld and suppose x1, . . . , xd ∈ m is a sytem of parameters for A. Then the map

ϕ : k[[t1, . . . , td]] −→ A

ti 7−→ xi ∼ identifies k[[t1, . . . , td]] = im ϕ ⊆ A and the resulting ring extension A/k[[t1, . . . , td]] is finite. Proof. As in the proof of the second structure theorem, ϕ is a well-defined homomorphism since A is complete. Set I = (x1, . . . , xd), so that r(I) = m. Notice that A/I is a local noetherian ring with maximal ideal m/I, and by Lemma 3.4.1, mn ⊆ I for some n ≥ 1. Then Theorem 3.5.6 implies A/I is artinian. In particular, dimk A/I is finite by Proposition 3.5.8. Choose a k-basis y1, . . . , ym of A/I. Setting S = k[[t1, . . . , td]], we have k = S/(t1, . . . , td) so

A = Sy1 + ... + Sym + (t1, . . . , td)A.

Now Lemma 6.3.8 implies A = Sy1+. . . Sym as an S-module. Let B = im ϕ ⊆ A. Then A is a ﬁnite extension of B ∼= S/ ker ϕ so d = dim A = dim B by Theorem 2.3.7. On the other hand, by Krull’s height theorem (6.2.10), dim S ≤ d. If ker ϕ 6= 0, then Corollary 6.2.16 implies ∼ dim B = dim(S/ ker ϕ) ≤ d − 1, a contradiction. Hence ker ϕ = 0 so k[[t1, . . . , td]] = B. Remark. One may regard Cohen’s third structure theorem as a complete version of Noether’s normalization lemma (Theorem 2.4.1).

102 6.3 Complete Local Rings 6 Dimension Theory

Example 6.3.12. Consider the 1-dimensional local ring A = R[x](x2+1) with maximal ideal m = (x2 + 1)A. Then the residue field of A is C but C does not embed into A. By Cohen’s first structure theorem (6.3.6), C does embed into the completion Ab but it’s not immediately obvious how this embedding is defined. Note that understanding C ,→ Ab comes down to finding an element u ∈ Ab such that u2 = −1. Since x2 + 1 ∈ m, 1 − (x2 + 1) is a unit and we have x2 = −1. 1 − (x2 + 1) Applying Hensel’s lemma (6.3.1) produces a square root of 1 − (x2 + 1), so set x u = . p1 − (x2 + 1)

Explicitly, u can be identiﬁed with the Cauchy sequence 1 3 5 u = x + x(x2 + 1) + x(x2 + 1)2 + x(x2 + 1)3 + ... 2 8 16 Recall the following deﬁnition from homological algebra.

Definition. An A-module M is flat if − ⊗A M is an exact functor, i.e. takes short exact sequences to short exact sequences. A flat morphism of rings is a homomorphism f : A → B such that B is flat as an A-module.

T∞ n Theorem 6.3.13. Suppose A is a noetherian ring, I ⊂ A is an ideal with n=1 I = 0 and ∼ M is a ﬁnitely generated A-module. Then Mc = M ⊗A Ab. n n ∼ n Proof. Notice that for a free module A , we have Ab = A ⊗A Ab. Then taking a presentation of M, say Am → An → M → 0, and applying Corollary 5.4.9, we get a diagram

m n A ⊗A Ab A ⊗A Ab M ⊗A Ab 0

∼ ∼ α

Abm Abn Mc 0 with exact rows. Then by the ﬁve lemma, there exists a map α which ﬁlls in the diagram and is an isomorphism.

Corollary 6.3.14. Completion is flat, i.e. for any local noetherian ring A, the natural map A → Ab is flat. The following is an analogue of the going down theorem (2.3.10) for flat morphisms.

103 6.3 Complete Local Rings 6 Dimension Theory

Theorem 6.3.15. Let ϕ : A → B be a ﬂat morphism of noetherian rings. Suppose q ⊂ B is a prime ideal, p = ϕ−1(q) and there is a chain of prime ideals

p ) p1 ) p2 ) ··· ) pn in A. Then there exists a chain of prime ideals

q ) q1 ) q2 ) ··· ) qn

−1 in B such that ϕ (qi) = pi for all 1 ≤ i ≤ n.

Proof. If n = 1, choose q1 ⊆ q which is minimal among primes containing p1B. Suppose −1 −1 ϕ (q1) ) p1. Then there exists some x ∈ ϕ (q1) r p1. Consider the short exact sequence x 0 → A/p1 −→ A/p1 −→ A/(p1, x) → 0.

(The left map is injective since p1 is prime and x 6∈ p1.) Since B is a ﬂat A-module, tensoring with B yields another short exact sequence

x 0 → B/p1B −→ B/p1B −→ B/(p1, x)B → 0.

Since q1 is minimal over p1B, it is an associated prime of B/p1B (Theorem 3.6.7), so Propo- sition 3.6.9 implies x is a zero divisor in B/p1B which contradicts exactness of the second −1 sequence on the left. Hence we must have ϕ (q1) = p1. The rest of the chain is constructed inductively using the same argument. Theorem 6.3.16. Let ϕ :(A, m) → (B, n) be a local morphism of local noetherian rings. Then (1) dim A + dim B/mB ≥ dim B.

(2) If ϕ is ﬂat, then dim A + dim B/mB = dim B.

Proof. (1) Set d = dim A, e = dim B/mB and choose a system of parameters x1, . . . , xd for A and elements y1, . . . , ye ∈ B whose imagesy ¯1,..., y¯e form a system of parameters in B/mB. Set I = (ϕ(x1), . . . , ϕ(xd), y1, . . . , ye). We claim r(I) = n. It will then follow by Krull’s height theorem (6.2.10) that dim B ≤ d + e. To prove the claim, set J = (ϕ(x1), . . . , ϕ(xd)) ⊆ I and write r(I) r(r(J), y , . . . , y ) = 1 e r(J) r(J) r(mB, y , . . . , y ) = 1 e r(mB) r(r(mB), y¯ ,..., y¯ ) n = 1 e = . r(mB) r(mB) This proves r(I) = n by the correspondence theorem for prime ideals. (2) We must show dim B ≥ d + e. Choose a chain of prime ideals

¯q0 ( ¯q1 ( ··· ( ¯qe = n/mB

104 6.3 Complete Local Rings 6 Dimension Theory

in B/mB and use the correspondence theorem to lift it to a chain

q0 ( q1 ( ··· ( qe = n

in B. Then q0 must contract to m, so if

p0 ( p1 ( ··· ( pn = m is a chain of primes in A, since ϕ is ﬂat, Theorem 6.3.15 provides a chain of prime ideals

0 0 0 q0 ( q1 ··· ( qd−1 ( q0

lying over the pi. This proves dim B ≥ d + e as required.

Corollary 6.3.17. If A is a local noetherian ring, then dim A = dim Ab.

Proof. This follows from the fact that A → Ab is ﬂat and A/b mAb ∼= A/m which is a ﬁeld. We also obtain a generalization of Corollary 2.6.3.

Corollary 6.3.18. If A is a noetherian ring, then dim A[x1, . . . , xn] = dim A + n for every n ≥ 1.

105 7 Singularities

7 Singularities

In this chapter we study several types of singularities in commutative algebra. The strongest type of nonsingularity is regularity: as deﬁned in Section 6.2, a local ring (A, m, k) is regular if its Krull dimension equals the dimension of the k-vector space m/m2. By Theorem 6.2.8, this is equivalent to m being generated by dim A elements. The dimension theory of Section 6.2 can be weakened in several ways to yield diﬀerent notions of singularity, including the Cohen- Macauley and Gorenstein conditions.

7.1 Derived Functors

The cleanest way to characterize most singularities is in the language of homological algebra. In this section we brieﬂy review the derived functors Tor and Ext. For full details, one should consult Weibel’s or Rotman’s books, both entitled An Introduction to Homological Algebra. For an A-module M, the functor − ⊗A M : ModA → Ab is a right exact functor, but is not left exact in general.

Definition. In the category A-Mod of left A-modules, the left derived functors of − ⊗A M are called Tor: A Torn (M,N) := Ln(− ⊗A M)(N) = Hn(P• ⊗A N), where P• → M is a projective resolution of M. Likewise for a right A-module M, the left derived functors of M ⊗A − are called tor: A torn (N,M) := Ln(M ⊗A −)(N) = Hn(M ⊗A P•). A ∼ ∼ By definition, for all right A-modules M and left A-modules N, Tor0 (M,N) = M ⊗A N = A tor0 (M,N). The Comparison Theorem says that unique chain maps exist between projective resolutions of M and N when M → N is a module homomorphism, so we need not worry when defining LnT about which projective resolution we choose. This follows from a more general statement:

Theorem 7.1.1. Let P• : · · · → P2 → P1 → P0 be a projective chain complex and suppose C• : ··· C2 → C1 → C0 is an acyclic chain complex. Then for any homomorphism ϕ : H0(P•) → H0(C•), there is a chain map f : P• → C• whose induced map on H0 is ϕ, and ϕ is unique up to chain homotopy. Corollary 7.1.2 (Comparison Theorem). Let g : M → N be A-linear and pick projective resolutions P• and Q• of M and N, respectively. Then there exists a chain map f : P• → Q• such that H0(f) = g and f is unique up to chain homotopy.

Since Tor is deﬁned as the derived functors of − ⊗A M, for every short exact sequence 0 → B0 → B → B00 → 0 A in ModA, Torn (−,M) induces an exact sequence A 00 A 0 A A 00 · · · → Torn+1(B ,M) → Torn (B ,M) → Torn (B,M) → Torn (B ,M) → · · · A 00 0 00 · · · → Tor1 (B ,M) → B ⊗A M → B ⊗A M → B ⊗A M → 0.

106 7.1 Derived Functors 7 Singularities

Theorem 7.1.3. For all projective resolutions P• of M and Q• of N and for all n ≥ 0, we have isomorphisms ∼ A ∼ Hn(P• ⊗A N) = Torn (M,N) = Hn(M ⊗A Q•) which are natural in each variable. Corollary 7.1.4. Let A be a commutative ring and suppose M and N are A-modules. Then A ∼ A for all n ≥ 0, Torn (M,N) = torn (M,N). A The functors Torn can actually be axiomatized, as the following theorem shows. A Theorem 7.1.5. For each n ≥ 0, there exists a functor Torn : A−Mod × A−Mod → A−Mod which satisﬁes A (1) Tor0 (M,N) = M ⊗ N. (2) For any short exact sequence 0 → M 0 → M → M 00 → 0 and any A-module N, there is a long exact sequence

0 00 0 → Torn(M ,N) → Torn(M,N) → Torn(M ,N) → Torn−1(M ,N) → which is natural in N.

A (3) For any free module F , Torn (F,N) = 0 for all n > 0. A Moreover, any functor satisfying these three properties is naturally isomorphic to Torn . The Tor functors give an important characterization of ﬂatness for modules. Theorem 7.1.6. For an A-module M, the following are equivalent: (1) M is ﬂat.

A (2) Tor1 (N,M) = 0 for all ﬁnitely generated A-modules N. A (3) Tor1 (A/p,M) = 0 for all prime ideals p ⊂ A.

(4) For all prime ideals p ⊂ A, the multiplication map p ⊗A M → pM is an isomorphism. Proof. (1) ⇐⇒ (2), (2) =⇒ (3) and (3) ⇐⇒ (4) are all immediate from the deﬁnitions n of Tori and the long exact sequence (Theorem 7.1.5). For (3) =⇒ (2), use Theorem 3.6.3 to get a prime ﬁltration of N:

0 ⊆ N1 ⊆ N2 ⊆ · · · ⊆ Nk = N ∼ ∼ with Ni+1/Ni = A/pi for primes pi ⊂ A. If k = 1, then N = A/p1 and (3) implies A Tor1 (N,M) = 0. Inducting on k, the short exact sequence

0 → Nk−1 → N → A/pk → 0 A induces a long exact sequence in Tori , part of which reads: A A A 0 = Tor1 (Nk−1,M) → Tor1 (N,M) → Tor1 (A/pk,M) = 0. A So Tor1 (N,M) = 0.

107 7.1 Derived Functors 7 Singularities

Theorem 7.1.7 (Local Criterion for Flatness). Suppose ϕ :(A, m) → (B, n) is a local homomorphism of local noetherian rings and N is a finitely generated B-module. Then N is A a flat A-module if and only if Tor1 (A/m,N) = 0. Proof. ( =⇒ ) follows from Theorem 7.1.6. ( ⇒ = ) By Theorem 7.1.6, it’s enough to show that for all finitely generated A-modules, Tor1(N,M) = 0. We induct on d = dim(A/ AnnA(N)). If d = 0, let

N0 ⊆ · · · ⊆ N` = N be a prime ﬁltration of N. Then since r(Ann(N)) = m, the only prime appearing as a quotient is m. There is a short exact sequence

0 → k = A/m → N → N 0 → 0

0 so by induction on `, Tor1(N ,M) = 0 and by hypothesis, Tor1(k, M) = 0. Therefore Tor1(N,M) = 0. Now let d > 0; we will show Tor1(A/p,M) = 0 for all primes p in a prime ﬁltration of N (keep the same notation as above). Notice that dim(A/p) ≤ d for any such p. If p = m, then by hypothesis Tor1(A/m,M) = 0. Otherwise, choose x ∈ m r p and consider the short exact sequence

0 → A/p −→x A/p −→ A/(p, x) → 0.

Then dim(A/(p, x)) < dim(A/p) ≤ d so by induction, Tor1(A/(p, x),M) = 0. Thus the long exact sequence in Tor for the above sequence contains

x Tor1(A/p,M) −→ Tor1(A/p,M) → Tor1(A/(p, x),M) = 0.

That is, Tor1(A/p,M) = x Tor1(A/p,M) but since this is a ﬁnitely generated A-module, Nakayama’s lemma (Theorem 1.2.1) implies Tor1(A/p,M) = 0. Next we shift our focus to the functor Ext. This is in some ways the more interesting of the two derived functors, since in any module category it naturally admits an A-algebra structure. For any A-module M, the functor HomA(M, −): ModA → Ab is left exact, but not right exact in general. Similarly, the contravariant functor HomA(−,N) is left exact but not right exact in general.

Deﬁnition. In A-Mod, the right derived functors of HomA(M, −) are called Ext:

n n ExtA(M,N) := R (HomA(M, −))(N) = Hn(HomA(M,E•)), where N → E• is an injective resolution of N. Likewise, the right derived of HomA(−,N) are denoted n n ExtA(M,N) := R (HomA(−,N))(M) = Hn(HomA(P•,N)) where P• → M is a projective resolution of M.

108 7.2 Regular Sequences and the Koszul Complex 7 Singularities

0 ∼ By deﬁnition, ExtA(M,N) = HomA(M,N) for all A-modules M and N. The comparison theorem (7.1.1) implies that ∼ Hn(HomA(M,E•)) = Hn(HomA(P•,N)) for any injective resolution N → E• and projective resolution P• → M, so the notational abuse above is only temporary. As with Tor, any short exact sequence

0 → B0 → B → B00 → 0 in ModA determines two long exact sequences, corresponding to the covariant and contravariant versions of HomA:

0 00 1 0 0 → HomA(M,B ) → HomA(M,B) → HomA(M,B ) → ExtA(M,B ) → · · · 00 0 1 00 and 0 → HomA(B ,N) → HomA(B,N) → HomA(B ,N) → ExtA(B ,N) → · · ·

As with Tor, the Extn functors can be axiomatized:

n Theorem 7.1.8. For each n ≥ 0, there exists a functor ExtA : A−Mod × A−Mod → A−Mod which satisﬁes

0 (1) ExtA(M,N) = HomA(M,N). (2) For any short exact sequence 0 → M 0 → M → M 00 → 0 and any A-module N, there is a long exact sequence

→ Extn(M 00,N) → Extn(M,N) → Extn(M 0,N) → Extn+1(M 00,N) →

which is natural in N.

n (3) For any free module F , ExtA(F,N) = 0 for all n > 0.

n Moreover, any functor satisfying these three properties is naturally isomorphic to ExtA.

7.2 Regular Sequences and the Koszul Complex

Deﬁnition. An element x ∈ A is a nonzero divisor for a module M if xm = 0 implies m = 0 for all m ∈ M, or equivalently, if 0 → M −→x M is exact. A regular sequence on M is a sequence of elements (x1, . . . , xn) in A satisfying:

(1) x1 is a nonzero divisor for M and for each 2 ≤ i ≤ n, xi is a nonzero divisor for M/(x1, . . . , xi−1)M.

(2) (x1, . . . , xn)M 6= M.

Remark. If M is ﬁnitely generated and each xi lies in the Jacobson radical J(A), then condition (2) is guaranteed.

109 7.2 Regular Sequences and the Koszul Complex 7 Singularities

Deﬁnition. A local noetherian ring A is Cohen-Macauley (abbreviated CM) if there is a system of parameters x1, . . . , xd ∈ A which forms a regular sequence on A. Example 7.2.1. If (A, m) is a 0-dimensional local ring then A is trivially Cohen-Macauley. Example 7.2.2. If (A, m) is a 1-dimensional local integral domain, then every nonzero element is a nonzero divisor, so A is Cohen-Macauley.

4 4 3 3 4 4 3 3 Example 7.2.3. Let A = C[x , y , x y, xy ]m where m = (x , y , x y, xy ). Then A is a local 4 4 subring of C[x, y](x,y), so dim A = 2. Note that x , y is a system of parameters for A, since (x3y)4 = (x4)3y4 ∈ (x4, y4) and likewise (xy3)4 = x4(y4)3 ∈ (x4, y4).

However (x4, y4) is not a regular sequence on A because y4 is a zero divisor for A/(x4):

y4(x3y)2 = x4(xy3)2 = 0

but x3y 6= 0 in A/(x4). Lemma 7.2.4. An element x ∈ A is a nonzero divisor if and only if 0 → A −→x A is exact, and in this case the free resolution

0 → A −→x A −→ A/(x) → 0

is exact. Proof. Easy.

More generally, we will show that (x1, . . . , xn) is a regular sequence on A if and only if the induced free resolution

(x ··· x ) mn m1 1 n 0 → A → · · · → A −−−−−−→ A → A/(x1, . . . , xn) → 0 is exact. Example 7.2.5. For a regular sequence (x, y) on A, this free resolution is

−y  x  x y 0 −→ A −−−−→ A2 −−−−−→ A −→ A/(x, y) −→ 0

which is exact by direct calculation.

For an element x ∈ A, let K•(x; A) denote the chain complex 0 → A −→x A → 0

and for any chain complex C• of A-modules, write C•(x) = C• ⊗ K•(x; A), where ⊗ here ∼ means the tensor product of chain complexes. Then for each n ∈ Z, Cn(x) = Cn ⊕ Cn−1 and the diﬀerential on C•(x) is given by

d : Cn(x) −→ Cn−1(x) n−1 (cn, cn−1) 7−→ (dcn + (−1) xcn−1, dcn−1).

110 7.2 Regular Sequences and the Koszul Complex 7 Singularities

Proposition 7.2.6. For any x ∈ A and any chain complex of A-modules C•, there is a long exact sequence

x · · · → Hn(C•) → Hn(C•(x)) → Hn−1(C•) −→ Hn−1(C•) → Hn−1(C•(x)) → · · ·

Proof. Apply homology to the short exact sequence of complexes

0 → C• → C•(x) → C•(−1) → 0 ∼ and use Hn(C•(−1)) = Hn−1(C•).

Corollary 7.2.7. For any x ∈ A and any chain complex of A-modules C•, there is a short exact sequence

0 → Hn(C•)/xHn(C•) → Hn(C•(x)) → AnnHn−1(C•)(x) → 0.

The complex K•(x; A) is an example of the Koszul complex, which we deﬁne next.

Deﬁnition. The Koszul complex for a sequence of elements (x1, . . . , xn) in A is the chain complex K•(x1, . . . , xn; A) deﬁned inductively by

K•(x1, . . . , xn; A) = K•(x1, . . . , xn−1; A)(xn).

If M is an A-module, write K•(x1, . . . , xn; M) = K•(x1, . . . , xn; A) ⊗A M. L Remark. In general K•(x1, . . . , xn; M) and n(Kn(x1, . . . , xn; A) ⊗A M) are not the same chain complex. Keep in mind that K• ⊗A M is the tensor product of K• with the complex 0 → M → 0.

Lemma 7.2.8. For any sequence of elements (x1, . . . , xn) in A, K•(x1, . . . , xn; A) is isomorphic to the exterior algebra V An. In particular, the graded pieces of the Koszul complex for (x1, . . . , xn) are (n) Ki(x1, . . . , xn; A) = A i .

Corollary 7.2.9. For any sequence of elements (x1, . . . , xn) in A, K•(x1, . . . , xn; A) is a graded A-algebra as well as a ﬁnitely generated free resolution of K0(x1, . . . , xn; A) = A.

Theorem 7.2.10. Let x1, . . . , xn ∈ A and let M be a ﬁnitely generated A-module. Then ∼ (1) H0(K•(x1, . . . , xn; M)) = M/(x1, . . . , xn)M. ∼ (2) Hn(K•(x1, . . . , xn; M)) = AnnM (x1, . . . , xn) = {m ∈ M | xim = 0 for all 1 ≤ i ≤ n}.

(3) If (x1, . . . , xn) is a regular sequence on M, then Hi(K•(x1, . . . , xn; M)) = 0 for all i > 0.

(4) If A is noetherian and x1, . . . , xn ∈ J(A) then the converse of (3) is true: if K•(x1, . . . , xn; M) is acyclic then (x1, . . . , xn) is a regular sequence on M.

111 7.2 Regular Sequences and the Koszul Complex 7 Singularities

Proof. Write Hi(x1, . . . , xn; M) := Hi(K•(x1, . . . , xn; M)). If n = 1, then (1) – (3) are immediate for the Koszul complex

0 → M −→x1 M → 0.

Further, if H1(x1; M) = 0 then x1 is a nonzero divisor for M. Since A is noetherian and x1 ∈ J(A), Nakayama’s lemma (Theorem 1.2.1) implies M 6= x1M, assuming M 6= 0. Now assume n > 1 and induct. (1) Let C•(M) = K•(x1, . . . , xn−1; M) for any ﬁnitely generated A-module M and set C• = C•(A). Notice that K•(x1, . . . , xN−1; M) = C• ⊗A M as chain complexes, so

C•(M)(xn) = K•(x1, . . . , xn−1; M)(xn) = K•(x1, . . . , xn; A) ⊗A M = K•(x1, . . . , xn; M).

Applying Corollary 7.2.7 with n = 0, we get

0 → H0(C•(M))/xnH0(C•(M)) → H0(x1, . . . , xn; M) → AnnH−1(C•)(xn) = 0 ∼ and by induction, H0(C•(M)) = M/(x1, . . . , xn−1)M so we get ∼ H0(x1, . . . , xn; M) = M/(x1, . . . , xn; M).

(2) follows from the same argument. (3) For each i ≥ 1, induction on the short exact sequence in Corollary 7.2.7 implies ∼ Hi(x1, . . . , xn; M) = 0. Then H1(x1, . . . , xn; M) = AnnH0(C•)(xn) so by (1), H0(C•) = M/(x1, . . . , xn−1)M. Finally, since (x1, . . . , xn) is a regular sequence, we have

AnnH0(C•)(xn) = 0 as desired. (4) Again, the short exact sequence in Corollary 7.2.7 implies Hi(C•)/xnHi(C•) = 0 for all i ≥ 1 and Nakayama’s lemma then implies Hi(C•) = 0 for all i ≥ 1. By the inductive hypothesis, (x1, . . . , xn−1) is a regular sequence and we are assuming H1(x1, . . . , xn; M) =

AnnH0(C•)(xn) = 0, so it follows that xn is a nonzero divisor on M/(x1, . . . , xn−1)M and we are done.

Corollary 7.2.11. If (x1, . . . , xn) is a regular sequence on A then the Koszul complex K•(x1, . . . , xn; A) is a free resolution of A/(x1, . . . , xn). Example 7.2.12. Let (A, m, k) be a complete local noetherian ring containing a coefficient field, so that k ,→ A, and suppose A is Cohen-Macauley of dimension d. Fix a system of parameters x1, . . . , xd ∈ A that also forms a regular sequence. We know by Cohen’s third structure theorem (6.3.11) that B = k[[x1, . . . , xd]] → A is a finite ring extension. We claim that A is flat as a B-module. Notice that (x1, . . . , xd) is a regular sequence on B, so that ∼ K•(x1, . . . , xd; B) is a free resolution of B/(x1, . . . , xd) = k. Then B ∼ ∼ Tor1 (k, A) = H1(K•(x1, . . . , xd; B) ⊗B A) = H1(K•(x1, . . . , xd; A)) = 0

since (x1, . . . , xd) is a regular sequence on A. By Theorem 7.1.6, this shows A is a ﬂat B-module.

112 7.2 Regular Sequences and the Koszul Complex 7 Singularities

Conversely, if A is flat over B, then Tori(B)(k, A) = 0 for all i ≥ 1 which means Hi(x1, . . . , xd; A) = 0 for all i ≥ 1. Thus by Theorem 7.2.10(4), (x1, . . . , xd) is a regular sequence. This shows that the Cohen-Macauley property (in this context) is equivalent to being a flat B-module, where B is a power series ring. Even better, a finitely generated flat module over any local noetherian ring is free, so the Cohen-Macauley property can be detected locally by a certain freeness condition.

Example 7.2.13. Let B = k[[x4, y4]] and A = k[[x4, y4, x3y, xy3]] be the completions of the polynomial rings in Example 7.2.3. Then

A = B + (x3y)B + (xy3)B + (x6y2)B + (x2y6)B

but A is not free as a B-module, since for example y4(x6y2) − x4(x2y6) = 0. Hence A is not Cohen-Macauley.

Deﬁnition. A minimal free resolution of an A-module M is a free resolution

ϕ2 ϕ1 ϕ0 · · · → F2 −→ F1 −→ F0 −→ M → 0

in which ϕi(Fi) ⊆ mFi−1 for each i ≥ 1. Let (A, m) be a local noetherian ring and M a ﬁnitely generated A-module. We construct

a minimal free resolution of M as follows. Let b0 = dimk M/mM. If {m¯ 1,..., m¯ b0 } is a k-

basis of M/mM, lift it to a generating set m1, . . . , mb0 (by Proposition 1.2.4) and deﬁne b0 ϕ0 : A → M by ei 7→ mi. Set M1 = ker ϕ0 and repeat this procedure inductively to construct ϕ ϕ ϕ · · · → Ab2 −→2 Ab1 −→1 Ab0 −→0 M → 0.

bi This sequence is a free resolution by observation and we have built it so that ϕi(A ) ⊆ m(Abi−1 ) at each step.

Example 7.2.14. If (x1, . . . , xn) is a regular sequence on A, then K•(x1, . . . , xn; A) is a minimal free resolution of A/(x1, . . . , xn).

ϕ ϕ ϕ Deﬁnition. If · · · → Ab2 −→2 Ab1 −→1 Ab0 −→0 M → 0 is a minimal free resolution of M, then bi is called the ith Betti number of M, sometimes written βi(M).

A The Betti numbers of a module have a homological characterization: note that Tori (k, M) is the ith homology of the chain complex

ϕ ϕ · · · → kbi+1 −−→i+1 kbi −→i kbi−1 → · · ·

A ∼ bi but by minimality, each ϕi is the zero map. Hence Tori (k, m) = k .

113 7.3 Projective Dimension 7 Singularities

7.3 Projective Dimension

ϕ2 ϕ1 ϕ0 Deﬁnition. If M is an A-module and · · · → P2 −→ P1 −→ P0 −→ M → 0 is a projective resolution of M, we say M has projective dimension pdA(M) = n if there is a smallest n such that ker ϕn is projective. Otherwise M has inﬁnite projective dimension (or no projective dimension).

Example 7.3.1. pdA(M) = 0 if and only if M is itself projective. Proposition 7.3.2. For a local noetherian ring (A, m, k) and a ﬁnitely generated A-module A M, pdA(M) = sup{i | Tori (k, M) 6= 0}.

Proof. Since projective modules are flat and k ⊗A − is exact, we have automatically that A pdA(M) ≤ sup{i | Tori (k, M) 6= 0}. Conversely, set n = pdA(M) and induct on n; the base ϕ2 ϕ1 ϕ0 case follows from Nakayama’s lemma. When n ≥ 1, let · · · → P2 −→ P1 −→ P0 −→ M → 0 be a projective resolution of M and set Ki = ker ϕi. Then since P• is also a flat resolution of M, A ∼ A A we have Tori+2(k, M) = Tor1 (k, Ki). The base case shows that Ki is free, so Tori (k, M) = 0 for all i > n, proving the opposite inequality. Corollary 7.3.3. For a local noetherian ring (A, m, k) and a finitely generated A-module

M, pdA(M) = n < ∞ if and only if M has a minimal free resolution of length n. A We will need the following lemma to establish Serre’s inequality for dimk Tori (k, k). Lemma 7.3.4. Let (A, m, k) be a local noetherian ring and ϕ : F → F 0 a morphism of 0 ﬁnitely generated free A-modules. If the base change ϕ : F ⊗A k → F ⊗A k is injective, then ϕ splits. n 0 m n Proof. Write F = A ,F = A and let e1, . . . , en be the standard basis of A . By hypothesis, 0 the images ϕ(e1),..., ϕ(en) are all linearly independent in F ⊗A k, so we may complete ¯ ¯ 0 this to a basis {ϕ(e1),..., ϕ(en), fn+1,..., fm}. By Proposition 1.2.4, F is generated as a 0 free module by ϕ(e1), . . . , ϕ(en), fn+1, . . . , fm so we may deﬁne a splitting ψ : F → F by ψ(ϕ(ei)) = ei for each 1 ≤ i ≤ m and ψ(fi) = 0 for n + 1 ≤ i ≤ m. Lemma 7.3.5 (Serre). If (A, m, k) is a local noetherian ring and m = δ(A) is the minimal number of generators of m, then for all i ≥ 0, m dim TorA(k, k) ≥ . k i i

In particular, pdA(k) ≥ m.

Proof. Let F• → k be a minimal free resolution (as A-modules). Then since the image of Fi A ∼ lies in mFi−1, we can easily see that Tori (k, k) = Fi ⊗k for each i. Let m = (x1, . . . , xm) and consider the diagram below whose bottom row is the Koszul complex K•(x1, . . . , xm; A):

··· F2 F1 A k 0

ϕ2 ϕ1 id id

m m ··· A( 2 ) A A k 0

114 7.3 Projective Dimension 7 Singularities

By the comparison theorem (Corollary 7.1.2), there is a chain map ϕ• : K• := K•(x1, . . . , xm; A) → F• inducing the identity on both A and k. Notice that each ϕi splits, then rank Fi ≥ m rank Ki = i , proving the lemma. Consider the square δ Fi Fi−1

ψi ψi−1 d Ki Ki−1

Assume inductively that ψi−1 : Fi−1 → Ki−1 has been constructed so that ψi−1 ◦ ϕi−1 = 1. To show ϕi splits, by Lemma 7.3.4 it suﬃces to show that each ϕi : K• ⊗A k → F• ⊗A k is injective. Take z ∈ Ki such that ϕi(z) ∈ mFi and note that δ(Fi) ⊆ mFi−1 implies 2 2 δ ◦ ϕi(z) ∈ m Fi−1. Thus ψi−1 ◦ δ ◦ ϕi(z) ∈ m Ki−1, so we see that

2 ψi−1 ◦ ϕi−1 ◦ d(z) = d(z) ∈ m Ki−1.

Nm xi Explicitly, the graded pieces of K• = i=1(0 → A −→ A → 0) are ∼ (m) M Ki = A i = AeI I⊆{1,...,m} |I|=i

where, if I = {j1, . . . , ji}, then eI is the generator of A ⊗ Aej1 ⊗ · · · ⊗ Aeji ⊗ A with standard

ej` generators in degree 1. The diﬀerential on K• is then given by

d : Ki −→ Ki−1 X eI 7−→ εjxjeIr{j} j∈I P where εj = ±1. Write z = |I|=i zI eI ∈ Ki. Then ! X X X X 2 d(z) = εjxjzI eIr{j} = εjxjzJ∪{j} eJ ∈ m Ki−1. I⊆{1,...,m} j∈I J⊆{1,...,m} j6∈J |I|=i |J|=i−1

P 2 Since Ki−1 is a free A-module, this means each term j6∈J ±xjzJ∪{j} lies in m . However, in 2 m/m the xj form a basis so we must have zI ∈ m for all |I| = i. Hence d(z) ∈ mKi−1, so ϕi is one-to-one as required. Remark. The following is an open problem independently posed by Buchsbaum-Eisenbud and Harack. If (A, m) is a local noetherian ring of dimension d and M is a ﬁnitely generated,

artinian A-module with pdA(M) < ∞, it is not known whether the Betti numbers bi = A d dimk Tori (k, M) satisfy Serre’s lemma: bi ≥ i . However, Walker proved the following weak form of the problem: d X d bi ≥ 2 . i=0

115 7.3 Projective Dimension 7 Singularities

Auslander, Buchsbaum and Serre gave a fundamental description of regular local rings using projective dimension. Theorem 7.3.6 (Auslander-Buchsbaum-Serre). For a local noetherian ring (A, m, k) of dimension d, the following are equivalent: (1) A is a regular local ring. (2) m is generated by a regular sequence.

(3) pdA(k) < ∞.

(4) For all ﬁnitely generated A-modules M, pdA(M) < ∞.

Proof. (1) =⇒ (2) Let m = (x1, . . . , xd). By Lemma 6.2.9, A is a domain and by induction so is A/(x1, . . . , xi) for each 1 ≤ i ≤ d. Hence (x1, . . . , xd) is a regular sequence. (2) =⇒ (3) Suppose m = (x1, . . . , xt) where (x1, . . . , xt) is a regular sequence on A. Then K•(x1, . . . , xt; A) is a minimal free resolution of H0(x1, . . . , xt; A) = k, so we have pdA(k) = t < ∞. A (3) =⇒ (4) If n = pdA(k) then Tori (k, k) = 0 for all i > n, so there is a minimal free resolution of k with length n: 0 → Abn → · · · → Ab1 → A → k → 0. A Tensoring with M shows Tori (k, M) = 0 for all i > n, so pdA(M) ≤ pdA(k). (4) =⇒ (1) Assume pdA(k) = n < ∞. We claim n ≤ d = dim A. Let t be the length of the longest regular sequence in A, say (y1, . . . , yt); we will see that t is equal to the depth of A. By Proposition 3.6.9, m ∈ Ass(A/(y1, . . . , yt)). That is, R/m = k sits inside of A/(y1, . . . , yt) and in fact we have an exact sequence

0 → k → A/(y1, . . . , yt) → M → 0. If n > t, then tensoring with k gives a long exact sequence A A A · · · → Torn+1(M, k) → Torn (k, k) → Torn (A/(y1, . . . , yt), k) → · · · A but (4) implies Torn+1(M, k) = 0. On the other hand, since K•(y1, . . . , yt; A) has length t, we A A get Torn (A/(y1, . . . , yt), k) = 0. Therefore Torn (k, k) = 0, contradicting Proposition 7.3.2. This proves n ≤ t. Next, we claim t ≤ d. If t = 1, A contains a nonzero divisor y1 such that S y1 6∈ p∈Ass(A) p. In particular y1 does not lie in any minimal prime of A by Theorem 3.6.7(3). Thus d ≥ 1. Now suppose t ≥ 2 and note by Corollary 6.2.3 that dim(A/(y1)) = d − 1. Since (¯y2,..., y¯t) is a regular sequence in A/(y1), so induction on t proves t − 1 ≤ d − 1 and thus t ≤ d. Finally, let δ = δ(A) be the minimal number of generators of m. We know from Corollary 6.2.6 that d ≤ δ so to complete the proof it is enough to show δ ≤ n. But this just follows from Lemma 7.3.5. Remark. In the course of the proof of Auslander-Buchsbaum-Serre, we established that if

(A, m, k) is a local noetherian ring with pdA(k) < ∞, then

δ(A) = pdA(k) = depth(A) = dim A which strengthens the conclusion in Section 6.2. Further, this gives a new criterion for a 2 local noetherian ring to be regular: dimk m/m ≤ pdA(k).

116 7.3 Projective Dimension 7 Singularities

Corollary 7.3.7. Every regular local ring is Cohen-Macauley.

Corollary 7.3.8. If A is a regular local ring and p ⊂ A is a prime ideal, then Ap is also regular. Proof. By Theorem 7.3.6(2), the A-module M = A/p has a free resolution of ﬁnite length,

0 → Abr → · · · → Ab1 → A → M → 0,

where r = pdA(M). Tensoring with Ap yields

br b1 0 → Ap → · · · → Ap → Ap → k(p) → 0

so Corollary 7.3.3 implies pdAp (k(p)) < ∞. Hence (3) of Theorem 7.3.6 says that Ap is regular.

Corollary 7.3.9. If A is the completion of a regular local ring, then Ap is regular for all prime ideals p ⊂ A. The Auslander-Buchsbaum-Serre theorem also allows us to prove the ‘ﬁbrewise criterion for regularity’. Theorem 7.3.10. Let ϕ :(A, m) → (B, n) be a ﬂat, local homomorphism of local rings. If A and B/mB are regular, then so is A.

Proof. Set d = dim A and e = dim B/mB, so that m = (x1, . . . , xd) and n/mB = (¯y1,..., y¯e) for some yi ∈ B having imagesy ¯i ∈ B/mB. Then n = (ϕ(x1), . . . , ϕ(xd), y1, . . . , ye) so we are done by Theorem 6.3.16.

Corollary 7.3.11. If A is a regular local ring then A[[x1, . . . , xn]] is regular for any n ≥ 1.

Proof. The map A → A[[x1, . . . , xn]] is ﬂat since it is the direct limit of a system of maps to free modules. Moreover, the ﬁbre of this map is k[[x1, . . . , xn]] which is regular by Theo- rem 6.2.8(c). Hence Theorem 7.3.10 applies.

Corollary 7.3.12. If A is a regular local ring then A[x1, . . . , xn] is also regular for any n ≥ 1, meaning for all prime ideals q ⊂ A[x1, . . . , xn], the local ring A[x1, . . . , xn]q is regular. Proof. By induction, it’s enough to prove the A[x] case. Choose a prime ideal q ⊂ A[x] and set p = q ∩ A. Then the map Ap → A[x]q factors through Ap → Ap[x] and thus is ﬂat. Meanwhile, the ﬁbre is (Ap/pAp[x])q = k(p)[x]q which is regular of dimension 1 since k(p)[x] is a PID, for example, or using Noether normalization (Theorem 2.4.1). Therefore Theorem 7.3.10 implies A[x]q is regular.

Remark. Let A = k[x1, . . . , xn]. Since Ap is regular by the above, for any ﬁnitely generated

A-module M and any prime p ⊂ A, pdAp (Mp) ≤ dim Ap ≤ n. So if

0 → Q → Fn−1 → · · · → F1 → F0 → M → 0

is a resolution with Fi free, Qp will be a free Ap-module. Hence Q is projective, so any pdA(M) ≤ n. (In fact, the Quillen-Suslin theorem implies Q is even free as an A-module.)

117 7.4 Depth and Cohen-Macauley Rings 7 Singularities

Corollary 7.3.13. For a regular local ring (A, m) and any nonzero x ∈ m, the quotient A/(x) is regular if and only if x 6∈ m2.

Proof. This follows from Theorem 7.3.10 and Corollary 6.2.16.

Theorem 7.3.14 (Jacobian Criterion). Let k be a perfect ﬁeld and set A = k[x1, . . . , xn]. For a prime p = (F1,...,Fm) ⊂ A of height h = ht(p), let B = A/p, choose a prime q ⊂ B, set ξi =x ¯i ∈ B/q and deﬁne the Jacobian ∂Fi J = J(ξ1, . . . , ξn) := (ξ1, . . . , ξn) . ∂xj

Then rankk(q) J ≤ h with equality if and only if Bq is a regular local ring. Example 7.3.15. Let A = k[x, y, s, t] and p = (xy − st), which has height 1. Then for the dimension 3 ring B = A/p,  y   x  J =   −t −s so clearly (1) is satisﬁed modulo any prime q. Moreover, rank J = 1 if and only if J 6= 0 mod q, so we see that there is only an isolated singularity at q = (x, y, s, t).

Remark. The Jacobian matrix J determines a presentation of the module of K¨ahlerdiﬀer- entials of B from Example 3.6.8:

n J m 1 B −→ B → ΩA/k → 0.

7.4 Depth and Cohen-Macauley Rings

Recall from Section 7.2 that a ring A is Cohen-Macauley if there is a system of parameters x1, . . . , xd ∈ A which is a regular sequence. As projective dimension was used in the last section to characterize regularity, in this section we will give a homological characterization of the Cohen-Macauley condition using depth. As one of many consequences, we will see that if the CM condition holds for some system of parameters for A, then it holds for all systems of parameters.

Proposition 7.4.1. Let A be a noetherian ring, I ⊂ A an ideal and M a ﬁnitely generated A-module such that IM 6= M. Let (y1, . . . , yt) be any regular sequence on M which lies in I. Then i t = min{i ≥ 0 | ExtA(A/I, M) 6= 0}. In particular, the length of any regular sequence on M in I is constant.

0 Proof. Suppose HomA(A/I, M) = ExtA(A/I, M) 6= 0. Choose a nonzero map f : A/I → M and set x = f(1) ∈ M. Then for any y ∈ I, 0 = f(y) = yf(1) = yx so Ix = 0. Thus every element of I is a nonzero divisor for M and therefore every regular sequence on M

118 7.4 Depth and Cohen-Macauley Rings 7 Singularities

has length 0. Conversely, if every regular sequence has length 0, then by Proposition 3.6.9, S I ⊆ p∈Ass(M) p. By prime avoidance (Lemma 6.2.14), I ⊆ p for some p ∈ Ass(M). This means A/I → A/p ,→ M determines a nonzero element of HomA(A/I, M). So the equality holds when either side is zero. Now induct on t; let (y1, . . . , yt) be a regular sequence in I for M of maximal length. Applying HomA(A/I, −) to the short exact sequence

y1 0 → M −→ M −→ M1 → 0 yields a long exact sequence

i−1 i y1 i i · · · → ExtA (A/I, M1) → ExtA(A/I, M) −→ ExtA(A/I, M) → ExtA(A/I, M1) → · · ·

Note that (¯y2,..., y¯t) is a maximal length regular sequence on M1 = M/y1M, so by induction i t ExtA(A/I, M1) = 0 for i < t − 1 and ExtA(A/I, M1) 6= 0. Therefore the long exact sequence i implies ExtA(A/I, M) = 0 for i < t − 1 and the map

t−1 y1 t−1 ExtA (A/I, M) −→ ExtA (A/I, M) is injective. However, y1 ∈ AnnA(A/I) since y1 ∈ I, so the multiplication-by-y1 map is 0 and t−1 t ∼ t−1 hence ExtA (A/I, M) = 0. Finally, ExtA(A/I, M) = ExtA (A/I, M1) 6= 0 by induction. Deﬁnition. For an ideal I ⊂ A and a ﬁnitely generated A-module M with IM 6= M, the

I-depth of M is the minimal length of any regular sequence in I on M, written depthI (M). Deﬁnition. If (A, m) is a local noetherian ring and M is a ﬁnitely generated A-module, the

depth of M is deﬁned to be the m-depth, depth(M) = depthm(M).

Remark. For any regular sequence (y1, . . . , yt) in I on M, iterating the above argument yields an isomorphism

t ∼ ExtA(A/I, M) = HomA(A/I, M/(y1, . . . , yt)M).

Corollary 7.4.2. For any ideal I ⊂ A and any collection of ﬁnitely generated A-modules {Mj}j∈J , with IMj 6= Mj for any j ∈ J, we have ! M depthI Mj = min{depthI (Mj) | j ∈ J}. j∈J

Theorem 7.4.3 (Auslander-Buchsbaum). Let (A, m, k) be a local noetherian ring and M a

ﬁnitely generated A-module such that pdA(M) < ∞. Then depth(M)+pdA(M) = depth(A). Proof. First assume depth(A) = 0. We claim M is free – by Corollary 7.4.2, this would then

imply depth(M) = depth(A) = 0. Suppose that n = pdA(M) > 0. Given a minimal free resolution of M, say

ϕn ϕn−1 ϕ1 ϕ0 0 → Fn −→ Fn−1 −−−→· · · −→ F0 −→ M → 0,

119 7.4 Depth and Cohen-Macauley Rings 7 Singularities

we know ϕi(Fi) ⊆ mFi−1 for all i. Now depth(A) = 0 if and only if HomA(k, A) = 0, the latter of which is only possible if there is some nonzero z ∈ A with mz = 0. Consider the element z 0   u = . ∈ Fn. . 0

Then ϕn(u) = zϕn(e1) ⊆ zmFn−1 = (mz)Fn−1 = 0, but since n > 0 this contradicts ∼ injectivity of ϕn. Therefore n must be 0 in the ﬁrst place, but this of course implies M = F0 is free and as a consequence depth(M) = 0 as explained above. Now assume depth(M) = 0, so that HomA(k, M) 6= 0. Let t = depth(A) and choose a reg- A ular sequence (y1, . . . , yt) in m for A. Notice that since n = pdA(M), Tori (A/(y1, . . . , yt),M) = A 0 for all i > n and Torn (A/(y1, . . . , yt),M) 6= 0. For a minimal free resolution F• → M as A above, Torn (A/(y1, . . . , yt),M) can be computed as the nth homology of

0 → A/(y1, . . . , yt) ⊗A Fn → A/(y1, . . . , yt) ⊗A Fn−1 → · · ·

But m ∈ Ass(A/(y1, . . . , yt)) so by Proposition 3.6.9, there is some nonzero z ∈ A/(y1, . . . , yt) such that mz = 0. Then the same argument as in the paragraph above shows that

A Torn (A/(y1, . . . , yt),M) 6= 0.

On the other hand, consider the Koszul complex K•(y1, . . . , yt; A):

t (t) 0 → A → A → · · · → A 2 → A → A/(y1, . . . , yt) → 0.

Since (y1, . . . , yt) is a regular sequence, this is acyclic by Theorem 7.2.10. That is,

A Tori (A/(y1, . . . , yt),M) = 0 for all i > t. Since depth(M) = 0, there is some nonzero z ∈ M with mz = 0, so the same trick shows A Tort (A/(y1, . . . , yt),M) 6= 0. Hence t = n. The remaining cases are when d = depth(M) > 0 and t = depth(A) > 0. Note that HomA(k, M) = 0 implies m 6∈ Ass(M). By prime avoidance (Lemma 6.2.14), there is some y ∈ M lying outside the union Ass(M) ∪ Ass(A) – note that these sets are ﬁnite by Theorem 3.6.7. Then Proposition 3.6.9 implies y is a nonzero divisor for M and for A. Applying − ⊗A M to the short exact sequence y 0 → A −→ A −→ A → 0 yields another short exact sequence

y 0 → M −→ M −→ A ⊗A M → 0

A and the corresponding long exact sequence in Tor shows that Tori (A, M) = 0 for all i > 0. On the other hand, let F• → M be a minimal free resolution and apply − ⊗A A:

0 → F n → · · · → F 1 → F 0 → M ⊗A A → 0

120 7.4 Depth and Cohen-Macauley Rings 7 Singularities

where F i = Fi ⊗A A is a free A-module. Note that the Tor vanishing condition on A means this sequence is exact. We have thus shown that pdA(M) = pdA(A ⊗A M). By induction,

depth(M ⊗A A) + pdA(M ⊗A A) = depth(A)

but since A = A/(y) for a nonzero divisor y, depth(A) = depth(A)−1 and depth(M ⊗A A) = depth(M) − 1. This proves the formula. Corollary 7.4.4. A local noetherian ring A is Cohen-Macauley if and only if depth(A) = dim A. Corollary 7.4.5. Let A be a regular local ring and I ⊂ A an ideal. Then B = A/I is

Cohen-Macauley if and only if pdA(B) = dim A − dim B.

Proof. By Theorem 7.4.3, depth(B) + pdA(B) = depth(A) = dim A, which can be rewritten pdA(B) = dim A − depth(B). Therefore B is CM if and only if depth(B) = dim B if and only if pdA(B) = dim A − dim B The following is an analogue of Theorem 6.3.16 for depth, and also provides a version of Theorem 7.3.10 for the Cohen-Macauley condition. In this way it justiﬁes the inclusion of depth on our list of types of dimension in ring theory. Theorem 7.4.6. If (A, m) → (B, n) is a ﬂat, local homomorphism of local noetherian rings, then depth(A)+depth(B/mB) = depth(B). In particular, B is Cohen-Macauley if and only if A and B/mB are Cohen-Macauley. Proof. Set k = A/m, t = depth(A) and suppose t > 0. Choose a maximal regular sequence (x1, . . . , xt) in A. Consider the short exact sequence

x1 0 → A −→ A −→ A1 → 0. Since A → B is ﬂat, the sequence

x1 0 → B −→ B −→ B/x1B → 0

is also exact, so x1 is a nonzero divisor on B. Moreover, A1 → B1 = B/x1B is also ﬂat ∼ and if m1 = m/(x1) is the maximal ideal of A1, then B1/m1B1 = B/mB. At the same time, depth(A1) = depth(A) − 1 and depth(B1) = depth(B) − 1 so by induction,

depth(A)+depth(B/mB) = (depth(A1)+1)+depth(B1/m1B1) = depth(B1)+1 = depth(B). It remains to show the base case, when t = depth(A) = 0. Fix some y ∈ B whose image y¯ ∈ B/mB is a nonzero divisor. We’ll be done by induction, since dim B/mB < dim B, if we can show A → B/yB is ﬂat and y is a nonzero divisor in B. We begin with the second statement. We claim that y is a nonzero divisor on B/mnB for all n ≥ 1. Indeed, the base case is by hypothesis and in general, applying − ⊗A B to the exact sequence 0 → k` ∼= mn/mn+1 → A/mn+1 → A/mn → 0 (for some ` < ∞) yields the exact sequence in both rows of the following commutative diagram:

121 7.4 Depth and Cohen-Macauley Rings 7 Singularities

0 (B/mB)` B/mn+1B B/mnB 0

y¯ y y

0 (B/mB)` B/mn+1B B/mnB 0

Then by the Snake Lemma and induction, the kernel of the middle column is 0, so y is a nonzero divisor on B/mn+1B. Suppose yz = 0 for some z ∈ B. Theny ¯z¯ = 0 in B/mnB n T∞ n for all n ≥ 1 so by the claim,z ¯ = 0 in every B/m B. Thus z ∈ n=1 m B but by Krull’s intersection theorem (Corollary 5.4.8), this is 0 so y is a nonzero divisor in B. Finally, to A show ﬂatness, by Theorem 7.1.7 it suﬃces to prove Tor1 (k, B/yB) = 0. Since y is a nonzero divisor in B, the sequence y 0 → B −→ B −→ B/yB → 0

is exact, so applying − ⊗A k and taking Tor yields an exact sequence

A A y¯ 0 = Tor1 (k, B) → Tor1 (k, B/yB) → B/mB −→ B/mB. (The 0 on the left follows from Theorem 7.1.7.) By assumption, multiplication byy ¯ is A injective on B/mB, so exactness implies Tor1 (k, B/yB) = 0.

Corollary 7.4.7. For any local noetherian ring A, depth(A) = depth(Ab). ∼ Proof. Same as the analogous statement for dimension: A → Ab is ﬂat, A/b mb = A/m = k and all ﬁelds have depth 0.

Corollary 7.4.8. If A is Cohen-Macauley then for any n ≥ 1, A[x1, . . . , xn] is also Cohen- Macauley. Proof. It suffices to prove the case n = 1, in which case A → A[x] is flat and has regular fibres which are thus Cohen-Macauley by Corollary 7.3.7. Hence Theorem 7.4.6 applies. Definition. Let A be a noetherian ring and M a finitely generated A-module. The dimension of the module M is

dim M = sup{ht(p) | p ∈ Ass(M)}.

By Theorem 3.6.7(2), dim M < ∞ for any finitely generated A-module M. Theorem 7.4.9. Let (A, m, k) be a local noetherian ring and suppose M and N are finitely i generated A-modules. Then ExtA(M,N) = 0 for all i < depth(N) − dim M. Proof. If dim M = 0, Ass(M) = {m} so the quotients in any prime filtration

0 = M` ⊆ M`−1 ⊆ · · · ⊆ M0 = M

i of M are all isomorphic to A/m = k. By deﬁnition of depth, ExtA(k, N) = 0 for all i < depth(A), so for each short exact sequence ∼ 0 → Mj → Mj+1 → Mj+1/Mj = k → 0

122 7.4 Depth and Cohen-Macauley Rings 7 Singularities

i i the long exact sequence in Ext implies ExtA(Mj,N) = 0. In particular, ExtA(M,N) = 0. i Now induct; if dim M > 0 then by the same argument it’s enough to show ExtA(A/p,N) = 0 for all i < depth(N) − dim M and all primes p in a prime ﬁltration of M. Fix such a p. Then by deﬁnition, dim A/p ≤ dim M. Choose x ∈ m which does not lie in p and consider the short exact sequence 0 → A/p −→x A/p −→ A/(p, x) → 0.

i By Corollary 6.2.16, dim A/(p, x) = dim A/p − 1 so by induction, ExtA(A/(p, x),N) = 0 for all i < depth(N) − dim A/p + 1. Applying HomA(−,N) to the short exact sequence above yields

i i x i i 0 = ExtA(A/(p, x),N) → ExtA(A/p,N) −→ ExtA(A/p,N) → ExtA(A/(p, x),N) = 0 whenever i < depth(N) − dim A/p. Since x ∈ m r p, Nakayama’s lemma (1.2.1) implies i ExtA(A/p,N) = 0 for these i. Since p was any prime in a prime filtration of M, we are done. Corollary 7.4.10. If (A, m, k) is a local noetherian ring and M is a finitely generated A- module, then for any p ∈ Ass(M), depth(M) ≤ dim A/p. Corollary 7.4.11. If A is Cohen-Macauley, then (a) The minimal primes of A are exactly the associated primes of A. (b) For any minimal prime p ⊂ A, dim A/p = dim A. Proof. Applying Corollary 7.4.10 with M = A says that for all p ∈ Ass(A), dim A = depth(A) ≤ dim A/p ≤ dim A, so they must all be associated and the equality in (2) holds. Example 7.4.12. By Corollary 7.4.11, the ring k[x, y, z]/(x) ∩ (y, z) is not Cohen-Macauley since it has two minmal primes, (x) and (y, z), of different codimensions. Example 7.4.13. Similarly, Corollary 7.4.11 implies that k[x, y]/(x2, xy) is not Cohen- Macauley because there is an associated prime, (x, y), which is not minimal – (x) is the only minimal prime in this ring. Proposition 7.4.14. Let (A, m) → (B, n) be a local homomorphism of local noetherian rings such that A is regular and B is a finite A-module. Then B is Cohen-Macauley if and only if B is a free A-module.

Proof. Since A is regular, Theorem 7.3.6 shows pdA(B) < ∞. Then by the Auslander- Buchsbaum formula (Theorem 7.4.3), pdA(B)+depth(B) = depth(A). Next, since A → B is ﬁnite, Theorem 2.3.7 implies dim A = dim B. But A is also Cohen-Macauley (Corollary 7.3.7) so depth(A) = dim A by Corollary 7.4.4. Thus B is Cohen-Macauley ⇐⇒ depth(B) = dim B

⇐⇒ pdA(B) = dim A − dim B = 0 by Theorem 7.4.3 ⇐⇒ B is a free A-module.

123 7.4 Depth and Cohen-Macauley Rings 7 Singularities

We can now show that the deﬁnition of Cohen-Macauley is independent of the choice of regular sequence on M. Proposition 7.4.15. Let (A, m) be a d-dimensional Cohen-Macauley ring containing a ﬁeld and suppose x1, . . . , xd is a system of parameters. Then (x1, . . . , xd) is a regular sequence.

Proof. To begin, note that A is CM if and only if Ab is CM – this follows from Theorem 7.4.6. Also note that (x1, . . . , xd) is a system of parameters if and only if dim A/(x1, . . . , xd) = 0. ∼ Then A/b (x1, . . . , xd)Ab = A/(x1, . . . , xd) so because dim Ab = dim A, x1, . . . , xd is a system of parameters for Ab. Applying Theorem 7.2.10, we have:

(x1, . . . , xd) is a regular sequence on Ab ⇐⇒ K(x1, . . . , xd; Ab) is acyclic

⇐⇒ K(x1, . . . , xd; A) ⊗A Ab is acyclic (A → Ab is ﬂat)

⇐⇒ K(x1, . . . , xd; A) is acyclic

⇐⇒ (x1, . . . , xd) is a regular sequence on A. Thus we may reduce to the case when A is a complete CM ring. By Cohen’s third structure theorem (6.3.11), there is a finite ring extension B = k[[x1, . . . , xd]] → A, where k ⊆ A is a coefficient field. By Corollary 7.3.11, B is regular and because A is CM, Proposition 7.4.14 implies A is a free B-module (beware that A and B have switched roles here compared to that proposition). On the other hand, K(x1, . . . , xd; B) is acyclic by Theorem 7.2.10, but since B → A is flat, K(x1, . . . , xd; A) = K(x1, . . . , xd; B) ⊗B A is still acyclic. Hence Theorem 7.2.10 again implies that (x1, . . . , xd) is a regular sequence on A. The last result holds in general but requires a more flexible version of Cohen’s structure theorems when A does not contain a field. Proposition 7.4.16. Let A be a local Cohen-Macauley ring. Then every non-refinable chain of prime ideals p0 ( p1 ( ··· ( pk = m has length k = dim A. Proof. We induct on d = dim A. If d = 0, there’s only one prime ideal, m, so every chain of primes is trivial. If d > 0, suppose p0 ( ··· ( pk = m is a chain of primes. Since p1 is not a minimal prime, the prime avoidance lemma (6.2.14) says that [ p1 6⊂ q. q⊂A q minimal

Pick x ∈ p1 that does not lie in any minimal prime q. By Corollary 7.4.11, the minimal primes of A are exactly the associated primes, so Proposition 3.6.9 says x is not a zero divisor on A. Thus A/(x) is also CM and dim A/(x) = d − 1. Consider the chain of primes

p1 ( p2 ( ··· ( pk = m

in A/(x), where pi = pi/(x) for each 1 ≤ i ≤ k. By induction, this chain must have length k − 1 = dim A/(x) = d − 1 so k = d as desired.

124 7.4 Depth and Cohen-Macauley Rings 7 Singularities

For an ideal I in a noetherian ring A, we extend the deﬁnition of height to I by:

ht(I) := min{ht(p) | I ⊆ p is prime}.

The proof of Lemma 2.3.6(b) goes through for any ideal I, so that ht(I) + dim A/I ≤ dim A. Moreover, Proposition 7.4.16 gives us: Corollary 7.4.17. If A is a Cohen-Macauley ring, then for any ideal I ⊂ A, ht(I) + dim A/I = dim A.

Example 7.4.18. Let B = k[x, y1, . . . , yn](x,y1,...,yn), let J = (x) ∩ (y1, . . . , yn) and set A = B/I. Then the minimal primes of A are (¯x) and I = (¯y1,..., y¯n) and ht(I) = 0, but dim A/I = dim k[¯x] = 1 so ht(I) + dim A/I = 1 < dim A = n as long as n ≥ 2. Therefore A is not Cohen-Macauley. Theorem 7.4.19 (Unmixedness). Let A be a local Cohen-Macauley ring and suppose I = (x1, . . . , xn) ⊆ m such that ht(I) = n. Then (x1, . . . , xn) is a regular sequence and A/I is Cohen-Macauley. Proof. First note that in general, Krull’s height theorem (6.2.10) implies ht(I) ≤ n. Assum- ing for the moment that (x1, . . . , xn) is a regular sequence, Proposition 7.4.15 says that every maximal regular sequence in A has length depth A = dim A, so it follows that A/I is also CM. Thus it suﬃces to show (x1, . . . , xn) is a regular sequence. If n = 1, then ht((x1)) = 1 is equivalent to saying x1 is not contained in any minimal prime of A, but since A is CM, Corollary 7.4.11 and Proposition 3.6.9 show x1 is not a zero divisor. For n > 1, we induct. Suppose x1 lies in some minimal prime p ⊂ A. Then for any prime pk which is minimal over I, we have the following non-reﬁnable chain of primes in A/(x1):

p/(x1) ( p1/(x1) ( ··· ( pk/(x1).

Then k = ht(pk) = n, but pk/(x1) is minimal with respect to primes of A/(x1) containing (x2, . . . , xn)/(x1), contradicting Krull’s height theorem (6.2.10). Hence x1 cannot be contained in any minimal prime of A, but since A is CM, the argument in the base case shows x1 is a nonzero divisor on A. That is, A/(x1) is CM and ht((¯x2,..., x¯n)) = n − 1. Induction ﬁnishes the proof.

Corollary 7.4.20. If A is a local Cohen-Macauley ring then Ap is Cohen-Macauley for all primes p ⊂ A.

Proof. By Lemma 2.3.6(a), dim Ap = ht(p). Set n = ht(p). We claim there are x1, . . . , xn ∈ p such that ht((x1, . . . , xn)) = n. Assuming this is possible, the unmixedness theorem then implies (x1, . . . , xn) is a regular sequence in A, and because A → Ap is flat, (x1, . . . , xn) will also be a regular sequence in Ap. Hence dim Ap ≥ depth(Ap) ≥ n = dim Ap so by Corollary 7.4.4, Ap is CM. Finally, we can find such elements x1, . . . , xn ∈ p as follows. By the prime avoidance lemma (6.2.14), there exists an x1 ∈ p that does not lie in any minimal prime of A. Moreover, there are a finite number of minimal primes containing (x1) and by Krull’s height theorem (6.2.10), each has height 1. So choose x2 ∈ p avoiding these primes and repeat the argument to construct the rest of the sequence (x1, . . . , xn).

125 7.4 Depth and Cohen-Macauley Rings 7 Singularities

Deﬁnition. A ring A is catenary if for all primes p ⊂ q in A, every non-reﬁnable chain of prime ideals between them, p ( p1 ( ··· ( p` ( q, has the same length. Theorem 7.4.21. Suppose A = B/I for some ideal I ⊂ B. Then if B is Cohen-Macauley, A is catenary.

Proof. For any prime ideal p = q/I ⊂ A, Corollary 7.4.20 shows that Bq is still CM, but by Proposition 7.4.16, every maximal chain of primes in Bq has the same length. Therefore the same holds in Ap = Bq/Iq, so A is catenary. The next result says that the property of being Cohen-Macauley is an ‘open condition’ on the set of primes of a regular ring. This will be made precise once we define Spec(A) and the Zariski topology in Section 8.8, but for now, define V (I) := {p ⊂ A | I ⊆ p is prime} for any ideal I ⊂ A. Proposition 7.4.22. Suppose B is a regular ring and A = B/q for a prime ideal q ⊂ B. Then the set C = {p ⊂ A | p is prime and Ap is not CM} is equal to V (I) for some ideal I ⊂ A. Proof. Consider a minimal free resolution ϕ · · · → Bbc+1 −→ Bbc → · · · → Bb1 → B → A → 0 where c = ht(q) = dim B − dim A by Corollary 7.4.17. If ϕ is the zero map then pdB(A) = c so the Auslander-Buchsbaum formula (Theorem 7.4.3) and Corollary 7.4.4 show A is CM. Otherwise, it’s enough to show that after localizing at an arbitrary prime p ⊂ B, the cor- bc+1 bc responding map ϕp : Bp → Bp splits. After choosing a basis of these free Bp-modules, ϕp is split if and only if the ideal of k × k minors, where k is the rank of ϕp, is equal to the unit ideal. Let I be this minor ideal. Then we have shown that the set of primes p ⊂ A for which Ap is not CM is precisely V (I). To finish, we list some nontrivial examples and properties of Cohen-Macauley rings. Details can be found in papers by the cited authors.

Example 7.4.23. (Hochster-Eagan) Let X = (xij) be an n × m matrix of indeterminates and let It(X) be the ideal in the polynomial ring k[xij] generated by the t × t minors of X. Then k[xij]/It(X) is Cohen-Macauley for all t. More generally, Hochster-Roberts show G that if G ⊆ GLn(k) is a linearly reductive group, then the ring of invariants k[V ] is Cohen- Macauley, where V = kn. Proposition 7.4.24 (Hochster-Huneke). Suppose A and B are local noetherian ring such that A contains a ﬁeld and B is regular. If there exists a split local homomorphism A,→ B, then A is Cohen-Macauley.

Proposition 7.4.25. Let A = k[x1, . . . , xn]/I for a homogeneous ideal I ⊂ k[x1, . . . , xn].

Then A(x1,...,xn) is Cohen-Macauley if and only if Ap is Cohen-Macauley for all primes p ⊂ A.

126 7.5 Gorenstein Rings 7 Singularities

7.5 Gorenstein Rings

Recall that an ideal I ⊂ A is irreducible if it cannot be written as the intersection of two proper ideals of A.

Deﬁnition. A local noetherian ring (A, m) is Gorenstein if A is Cohen-Macauley of dimension d and there exists a system of parameters x1, . . . , xd ∈ A such that (x1, . . . , xd) is an irreducible ideal.

We will show that this definition is equivalent to the more common definition of A having finite injective dimension.

Proposition 7.5.1. A regular local ring is Gorenstein.

Proof. By Theorem 6.2.8, A is regular if and only if m = (x1, . . . , xd) for a system of parameters x1, . . . , xd, but in general a maximal ideal is irreducible, so A is Gorenstein.

Suppose (A, m) is a local noetherian ring and x1, . . . , xd is a system of parameters. Then A is Gorenstein if and only if 0 is an irreducible ideal in the 0-dimensional ring A/(x1, . . . , xd). So to begin our study of Gorenstein rings, let’s characterize when a 0-dimensional local noetherian ring is Gorenstein. Recall that a local noetherian ring has dimension 0 if and only if it’s artinian (this is Hopkins’ theorem, 3.5.5).

Example 7.5.2. Let k be a ﬁeld. Then A = k[x, y]/(x, y)2 is 0-dimensional but not Goren- stein since (x, y)2 = (x, y2) ∩ (x2, y) is not irreducible in k[x, y].

Example 7.5.3. If n ≥ 2, the ring A = k[x]/(xn) is Gorenstein. Indeed, A is a k-vector space with basis 1, x,¯ . . . , x¯n−1 and every ideal in A is of the form I = (¯xk) for some k. This means any intersection of proper ideals looks like (¯xj) ∩ (¯xk) = (¯xmax{j,k}) 6= 0. Thus 0 is not an irreducible ideal.

Example 7.5.4. For a nontrivial example, we will show later that

k[x, y, z]/(x2 − y2, y2 − z2, xy, xz, yz)

is Gorenstein.

Remark. One might hope to classify all Gorenstein rings of any dimension by the tech- niques which we will describe for local artinian rings. However, it turns out that there is no classiﬁcation of Gorenstein rings of the form k[x1, . . . , xn]/I for n ≥ 4. Recall the following deﬁnitions and basic properties from ring theory.

Deﬁnition. For an A-module M, a submodule N ⊆ M is called essential, or M is an essential extension of N, if for all nonzero submodules S ⊆ M, N ∩ S 6= 0.

Deﬁnition. The socle of a local ring (A, m) is the set soc(A) = {x ∈ A | xm = 0}.

Lemma 7.5.5. For any local ring (A, m, k),

127 7.5 Gorenstein Rings 7 Singularities

(a) soc(A) is an ideal of A.

(b) m soc(A) = 0, so in particular soc(A) is a k-vector space.

(c) If A is artinian, then A is an essential extension of soc(A). Proof. (a) and (b) are standard. For (c), take a nonzero ideal I ⊂ A. Then for a nonzero element x ∈ I, either xm = 0, so x ∈ soc(A) ∩ I, or by Theorem 3.5.6, xmn = 0 for some minimal n ≥ 2. In the latter case, xmn−1 6= 0 but xmn−1 ⊆ soc(A) ∩ I. Thus soc(A) is an essential ideal of A.

Theorem 7.5.6. A local artinian ring (A, m, k) is Gorenstein if and only if dimk soc(A) = 1.

Proof. ( =⇒ ) If dimk soc(A) ≥ 2, choose nonzero, linearly independent vectors x, y ∈ soc(A). Then (x) ∩ (y) = 0 so A is not Gorenstein. ( ⇒ = ) Assume I ∩ J = 0 for some ideals I,J 6= 0. Then I ∩ J ∩ soc(A) = 0 but by Lemma 7.5.5(c), we must have I ∩soc(A) 6= 0 and J ∩soc(A) 6= 0. However, these I ∩soc(A) and J ∩ soc(A) are distinct k-subspaces of soc(A), so we must have dimk soc(A) ≥ 2.

Let (B, m, k) be a regular local ring of dimension d and suppose m = (x1, . . . , xd). For a local artinian B-algebra of the form A = B/I, a minimal free resolution of A,

0 → Bbd → · · · → Bb1 → B → A → 0,

is necessarily of length d since pdB(A) = dim B − dim A by Theorem 7.4.19 and Corol- B ∼ lary 7.4.5. Then tensoring with k produces a resolution with all zero maps, so Tord (k, A) = kbd . On the other hand, the Koszul complex for A as a B-module is

ϕ d d 0 → A −→ A → · · · → A → A → k ⊗B A → 0

so by deﬁnition,

B Tord (k, A) = ker ϕ = ker(a 7→ (±xi)a)

= {a ∈ A | axi = 0 for all 1 ≤ i ≤ d} = soc(A).

This proves: Proposition 7.5.7. If A is a local artinian ring that is a quotient of a regular local ring of dimension d with residue ﬁeld k, then dimk soc(A) = bd, the dth Betti number of A. Corollary 7.5.8. A local artinian ring A that is a quotient of a regular local ring of dimension d is Gorenstein if and only if bd = 1.

Corollary 7.5.9. Suppose f1, . . . , fd ∈ k[x1, . . . , xd] form a system of parameters in the local

ring k[x1, . . . , xd](x1,...,xd). Then the quotient k[x1, . . . , xd](x1,...,xd)/(f1, . . . , fd) is Gorenstein.

Proof. By Corollaries 7.4.8 and 7.4.20, k[x1, . . . , xd](x1,...,xd) is Cohen-Macauley, so by Propo- sition 7.4.15, (f1, . . . , fd) is a regular sequence. Now the conclusion follows from analyzing the Koszul complex for (f1, . . . , fd).

128 7.5 Gorenstein Rings 7 Singularities

Theorem 7.5.10. Let (B, n, k) be a regular local ring of dimension n, I ⊂ B an ideal and suppose A = B/I is Cohen-Macauley of dimension d. Set c = ht(I) = n − d. Then A is B Gorenstein if and only if dimk Torc (k, A) = 1. Proof. By the Auslander-Buchsbaum formula (Theorem 7.4.3) and Corollary 7.4.4,

pdB(A) = depth(B) − depth(A) = dim B − dim A = c

B so bc = dimk Torc (k, A) is the last Betti number in a minimal free resolution of A:

0 → Bbc → Bbc−1 → · · · → Bb1 → B → A → 0.

bi Set Fi = B for each 1 ≤ i ≤ c. Suppose y1, . . . , yd ∈ B reduce to a system of parameters y¯1,..., y¯d in A. Since A is Cohen-Macauley, Proposition 7.4.15 shows (¯y1,..., y¯d) is also a regular sequence; by prime avoidance (Lemma 6.2.14), we may assume (y1, . . . , yd) is a regular B sequence in B. To prove the theorem, we will compute Tori (B/(y1, . . . , yd),A) in two ways. First, since (y1, . . . , yd) is a regular sequence, the Koszul complex K•(y1, . . . , yd; B) is acyclic by Theorem 7.2.10. But the images of the yi form a regular sequence in A, so because K•(¯y1,..., y¯d; A) = K•(y1, . . . , yd; B) ⊗B A, this reduced Koszul complex is also acyclic. B That is, Tori (B/(y1, . . . , yd),A) = 0 for i > 0. B ∼ On the other hand, Tori (B/(y1, . . . , yd),A) = Hi(K•(y1, . . . , yd; B) ⊗ F•) which implies B K•(y1, . . . , yd; B) ⊗ F• is a free resolution of Tor0 (B/(y1, . . . , yd),A) = A/(¯y1,..., y¯d). Now dim A/(¯y1,..., y¯d) = 0, by Krull’s height theorem (6.2.10) e.g., so Proposition 7.5.7 says

B dimk soc(A/(¯y1,..., y¯d)) = dimk Torn (k, A/(¯y1,..., y¯d))

equals the last Betti number in K•(y1, . . . , yd; B)⊗F•. The resolutions K•(y1, . . . , yd; B) and F• have lengths d and c, respectively, and the last free module in the tensor complex, in ∼ dimension n = d + c, is exactly Kd(y1, . . . , yd; B) ⊗B Fc = B ⊗B Fc = Fc. Therefore we have B ∼ B Torc (k, A) = Torn (k, A/(¯y1,..., y¯d)) so we conclude bc = 1 if and only if A/(¯y1,..., y¯d) is Gorenstein, by Theorem 7.5.6. However, the same argument shows that bc = 1 if and only B if dimk Torc (k, A/(z1, . . . , zd)) = 1 for any system of parameters z1, . . . , zd in A, which is equivalent to (z1, . . . , zd) being an irreducible ideal in A for any such zi.

Corollary 7.5.11. Suppose A is Gorenstein and x1, . . . , xd ∈ A is any system of parameters. Then (x1, . . . , xd) is an irreducible ideal. Example 7.5.12. (Complete intersections) A local noetherian ring A is called a complete ∼ intersection if A = B/I for a regular local ring B and an ideal I = (f1, . . . , fc) generated by a regular sequence in B. In this case we know the Koszul complex K•(f1, . . . , fc; B) is a minimal free resolution of A, so by Corollaries 7.3.3 and 7.4.5, A is Cohen-Macauley. B ∼ Further, ht(I) = c implies Torc (k, A) = k so we have inclusions regular ⊆ complete intersection ⊆ Gorenstein ⊆ Cohen-Macauley.

Example 7.5.13. Consider a graph

129 7.5 Gorenstein Rings 7 Singularities

b a

with vertices labeled with variables a, . . . , e. Let I = (ab, bc, cd, de, ea) be the edge ideal in B = k[a, b, c, d, e](a,b,c,d,e). Then ht(I) = 3 so A = B/I is a local ring of dimension 2 with free B-resolution 0 → B → B5 → B5 → B → A → 0. Hence by the Auslander-Buchsbaum formula (Theorem 7.4.3), A is Cohen-Macauley. Fur- ther, the last Betti number of A is 1 so Theorem 7.5.10 shows A is Gorenstein. However, A is not a complete intersection since I is 5-generated.

Example 7.5.14. Let k = C and choose a polynomial f ∈ C[y1, . . . , yn] such that f(0) = 0. ∂ The polynomial ring [X1,...,Xn] acts on such f by partial diﬀerentiation: Xi · f = f. C ∂yi Set ⊥ (f) := {G ∈ C[X1,...,Xn] | G · f = 0}. ⊥ Then A = C[X1,...,Xn]/(f) is a 0-dimensional Gorenstein ring and one can prove that every 0-dimensional Gorenstein ring is isomorphic to such a quotient. For example, if f = 2 2 2 ⊥ 2 2 2 2 y1 + y2 + y3 then (f) = (X1X2,X1X3,X2X3,X1 − X2 ,X1 − X3 ). On the other hand, for ⊥ 2 2 2 ⊥ g = y1y2y3 the corresponding ideal is (g) = (X1 ,X2 ,X3 ) and A = C[x1, x2, x3]/(g) is a complete intersection in this case. It is an open question which f give quotients that are complete intersections.

Theorem 7.5.15. Let (A, m, k) → (B, n, `) be a ﬂat, local homomorphism of local rings. Then B is Gorenstein if and only if A and B/mB are Gorenstein.

Proof. If dim A or dim B > 0, one can induct using Theorem 7.5.10 so we will reduce to the case that dim A = dim B = 0. By Theorem 7.5.6, we must show dim soc(B) = 1 if and only if dim soc(A) = 1 and dim soc(B/mB) = 1. First assume A and B/mB are Gorenstein. Choose a ∈ A such that soc(A) = (a) ∼= k and b ∈ B such that soc(B/mB) = (¯b) ∼= k. We claim soc(B) = (ab). On one hand, we have

abn = a(bn) ⊆ a(mB) = (am)B = 0. ¯ If ab = 0, then b ∈ AnnB(a) = AnnA(a)B = mB, so we would have b = 0 in B/mB, a contradiction. Hence ab is a nonzero element of soc(B). On the other hand, let x ∈ soc(B). Then xn = 0, so x(mB) = 0 which is equivalent to x ∈ AnnB(mB) = AnnA(m)B = aB. Thus x ∈ aB, say x = ay for some y ∈ B. Then ayn = 0 implies yn ⊆ AnnB(a) =

130 7.5 Gorenstein Rings 7 Singularities

AnnA(a)B = mB. Hence y ∈ soc(B/mB), so by hypothesis, y = zb + π for some z ∈ B and π ∈ mB. This implies x = ay = a(zb + π) = azb + aπ = zab ∈ (ab). Thus soc(B) ⊆ (ab) so we conclude soc(B) = (ab). Conversely, assume B is Gorenstein, so soc(B) = (x) for some x ∈ B. By the same argument as above, if a ∈ soc(A) and b ∈ B such that ¯b ∈ soc(B/mB) r {0}, then ab ∈ soc(B). The ﬂatness hypothesis guarantees ab 6= 0 for such a, b. If soc(A) has dimension greater than 1, then we can ﬁnd linearly independent elements a1, a2 ∈ soc(A) with × (a1b), (a2b) ⊆ soc(A). So because dim soc(B) = 1, there is some unit c ∈ B such that a2b = ca1b, or (a2 −ca1)b = 0. Thus a2 = ca1, a contradiction. A similar contradiction arises from assuming dim soc(B/mB) > 1, so we conclude dim(A) = dim(B/mB) = 1.

Let A be a ring and M an A-module. Denote by EA(M) the injective hull of M, which is the unique smallest injective A-module containing M (there are several equivalent charac- terizations, namely EA(M) is the maximal essential extension of M). If (A, m, k) is a local ˇ ring, we will write E = EA(k) and for any M, M = HomA(M,E).

Proposition 7.5.16. If (A, m, k) is a local artinian ring with E = EA(k), (1) The functor M 7→ Mˇ is exact. (2) kˇ ∼= k. (3) If λ(M) denotes the length of a prime ﬁltration of M, then λ(M) = λ(Mˇ ).

(4) For any A-module M, we have Mˇ ∼= M. Proof. (1) E is injective. ∼ (2) Note that HomA(k, E) = soc(E) as k-vector spaces. Moreover, if dim soc(E) ≥ 2, consider the essential extension k ,−→ E 1 7−→ x ∈ soc(E). Then there is a y ∈ soc(E) linearly independent from x so (y) ∩ (x) = 0, contradicting the fact that E is an essential extension of k. Hence dim soc(E) = 1. (3) Inducting on λ(M), note that AssA(M) = {m}, so there exists an embedding k ,→ M. Let 0 → k → M → M1 → 0 be the corresponding short exact sequence. Since λ is additive, λ(M) = λ(k) + λ(M1) = 1 + λ(M1) and by (1), ˇ ˇ ˇ 0 → k → M → M1 → 0 ˇ ˇ ˇ ˇ is also exact and thus λ(M) = λ(k) = λ(M1). So by induction, λ(M1) = λ(M1). Moreover, (2) implies λ(kˇ) = λ(k) so it follows that λ(Mˇ ) = λ(M). ˇ ˇ (4) Identifying M = HomA(M,E) = HomA(HomA(M,E),E) determines a map i : M,→ Mˇ which is necessarily injective. Then by (3), λ(M) = λ(Mˇ ) so coker i = 0. Hence i is an isomorphism.

131 7.5 Gorenstein Rings 7 Singularities

ϕ0 ϕ1 ϕ2 Definition. If M is an A-module and 0 → M −→ E0 −→ E1 −→· · · is an injective resolution of M, we say M has injective dimension idA(M) = n if there is a smallest n such that coker ϕn is injective. If no such n exists, M has infinite injective dimension. The next theorem presents some more common definitions of the Gorenstein property, namely (d).

Theorem 7.5.17. Let (A, m, k) be a local artinian ring and set E = EA(k). Then the following are equivalent:

(a) A is Gorenstein.

(b) A ∼= E.

(d) idA(A) < ∞. Proof. (a) =⇒ (b) By Theorem 7.5.6, soc(A) is 1-dimensional. But A is an essential extension of soc(A), so there are embeddings k ,→ A,→ E since E is the maximal essential extension of k. Applying λ to the short exact sequence

0 → A → E → E/A → 0 ˇ yields λ(E) = λ(A) = λ(E/A), but Proposition 7.5.16(4) says λ(A) = λ(A) = λ(HomA(A, E)) = λ(E) since E is injective. Thus λ(E/A) = 0, which implies E/A = 0. (b) =⇒ (c) =⇒ (d) are trivial. (d) =⇒ (a) It is well-known that over a local noetherian ring, every injective A-module is isomorphic to a direct sum of copies of E. In particular, a ﬁnite injective resolution of A over itself may be taken to be of the form

0 → A → Eb0 → Eb1 → · · · → Ebn → 0.

Applying ˇ(·) and Proposition 7.5.16 gives a free resolution

0 → Abn → · · · → Ab1 → Ab0 → E → 0

so in particular pdA(E) < ∞. Further, the Auslander-Buchsbaum formula (Theorem 7.4.3) implies pdA(E)+depth(E) = depth(A), but depth(E) and depth(A) are both 0 so pdA(E) = 0 as well. In particular, E ∼= Ab for some b ≥ 1. On the other hand, λ(E) = λ(A) by additivity of λ, so we must have b = 1. Hence A is Gorenstein.

Theorem 7.5.18. Let f1, . . . , fd be a system of parameters in the power series ring k[[x1, . . . , xd]] and set A = k[[x1, . . . , xd]]/(f1, . . . , fd). Then A is a 0-dimensional complete intersection. Pd Further, if fi = j=1 aijxj for some aij ∈ k, then soc(A) is generated by the image of det(aij) in k.

132 8 Algebraic Geometry

8 Algebraic Geometry

One of the principal mathematical offshoots of commutative algebra is the study of algebraic geometry. Many of the definitions and results in the first six chapters have geometric interpretations which give further insight into their meaning and importance. However, the beauty of algebraic geometry lies in how it transports ideas in both directions: the algebraic to the geometric and the geometric to the algebraic. This chapter only presents a brief introduction to the ideas of algebraic geometry, but the reader can learn more in such seminal texts as Hartshorne’s Algebraic Geometry or Shafarevich’s two-parts series Basic Algebraic Geometry.

8.1 Aﬃne Algebraic Varieties

In this section we introduce the fundamental object in algebraic geometry: an aﬃne variety. Let k be a ﬁeld. We will often need the assumption that k is algebraically closed in order to prove nontrivial results.

Definition. For each n ∈ N, we define affine n-space over k to be n n A = Ak = {(x1, . . . , xn) | xi ∈ k}.

As sets, An = kn, but the new notation carries with it the implication that An is viewed geometrically.

We will let A denote the polynomial ring k[t1, . . . , tn]. Deﬁnition. For a polynomial f ∈ A, deﬁne its zero set (or zero locus) to be

n Z(f) = {P ∈ A | f(P ) = 0}.

We extend the definition of zero set to sets of polynomials f1, . . . , fr ∈ A by r \ Z(f1, . . . , fr) = Z(fi). i=1 Remark. The definition of zero set can be extended to arbitrary subsets F ⊆ A by \ Z(F) = Z(f). f∈F Notice that if I = (F) is the ideal of A generated by F, then Z(F) = Z(I). By Hilbert’s basis theorem (3.3.2), there exists a finite subset {f1, . . . , fr} ⊆ F such that Z(F) = Z(f1, . . . , fr).

Deﬁnition. A subset X ⊆ An is called an algebraic set if X = Z(F) for a set F ⊆ A, that is, X is algebraic if it is the zero set of some collection of polynomials in k[t1, . . . , tn]. By the remark, it is equivalent to say X is a zero set if X = Z(I) for some ideal I ⊂ A. Thus the operation Z(·) takes a subset of a ring and assigns to it a geometric space. There is a dual notion:

133 8.1 Aﬃne Algebraic Varieties 8 Algebraic Geometry

Deﬁnition. For any subset X ⊆ An, we deﬁne the vanishing ideal of X to be J(X) = {f ∈ A | f(P ) = 0 for all P ∈ X}.

Lemma 8.1.1. For all X ⊆ An, J(X) is a radical ideal of A. Proof. Take f, g ∈ J(X) and r ∈ A. Then for any P ∈ X,(f − g)(P ) = f(P ) − g(P ) = 0 and (rf)(P ) = r(P )f(P ) = 0 so f + g, rf ∈ J(X) and thus J(X) is an ideal. Moreover, for any m ∈ N, f m(P ) = 0 if and only if f(P ) = 0 so we see that r(J(X)) = J(X). Examples.

1 ∅ = Z(A) and An = Z(0) are both algebraic sets.

n n 2 If U ⊆ A is an aﬃne subspace, i.e. U = P0 + V for a point P0 ∈ A and a linear n subspace V ⊆ k , then U = Z(L1,...,Ln−d) where d = dimk V and L1,...,Ln−d are linear polynomials in A.

n 3 For any point P = (a1, . . . , an) ∈ A , {P } = Z(t1 −a1, . . . , tn −an). Consider the maximal ideal mP = (t1 − a1, . . . , tn − an) ⊂ A (we saw in Section 2.4 that mP is maximal). Then {P } = Z(mP ). Recall that when k is algebraically closed, Corollary 2.4.3 shows that points of An are in one-to-one correspondence with the maximal ideals of A via the association P ↔ mP .

2 2 2 2 3 2 4 In A , an example of an algebraic curve is C = {(T − 1,T (T − 1))} = Z(t1 + t1 − t2):

n Lemma 8.1.2. Let X,Y ⊆ A be sets and I,I1,I2 and I` ⊂ A be ideals, with ` ∈ L some indexing set. Then

(a) If Y ⊆ X then J(Y ) ⊇ J(X).

(b) If I2 ⊆ I1 then Z(I2) ⊇ Z(I1). (c) Z(J(X)) ⊇ X.

(d) J(Z(I)) ⊇ I.

(e) Z(J(Z(I))) = Z(I).

134 8.1 Aﬃne Algebraic Varieties 8 Algebraic Geometry

(f) J(Z(J(X))) = J(X).

(g) Z(I1) ∪ Z(I2) = Z(I1 ∩ I2) = Z(I1I2). ! ! \ [ X (h) Z(I`) = Z I` = Z I` . `∈L `∈L `∈L Proof. (a) – (d) are obvious from the deﬁnitions of Z and J. (e) By (c), Z(I) ⊆ Z(J(Z(I))) so it remains to prove the reverse containment. However, we have that I ⊆ J(Z(I)) by (d) so then applying Z gives Z(I) ⊇ Z(J(Z(I))) by (b). (f) is similar to (e). Here, (d) gives us J(X) ⊆ J(Z(J(X))). On the other hand, we have X ⊆ Z(J(X)) by (c), so applying J yields J(X) ⊇ J(Z(J(X))) by (a). (g) We get Z(I1) ∪ Z(I2) ⊆ Z(I1 ∩ I2) ⊆ Z(I1I2) immediately from the containments I1 ⊇ I1 ∩ I2 ⊇ I1I2 and I2 ⊇ I1 ∩ I2 ⊇ I1I2, using (b). Suppose P ∈ Z(I1I2) and P 6∈ Z(I1). Then there is some f ∈ I1 such that f(P ) 6= 0, but for any g ∈ I2, we have (fg)(P ) = f(P )g(P ) = 0. Since A = k[t1, . . . , tn] is a domain and f(P ) 6= 0, we must have g(P ) = 0. This shows P ∈ Z(I2). Hence Z(I1) ∪ Z(I2) ⊇ Z(I1I2) so we have established all three equalities. S P S P (h) The containments I` ⊆ I` ⊆ I` give us Z(I`) ⊇ Z I` ⊇ Z I` `∈L T `∈L S `∈LP `∈L for each ` ∈ L, by (b), and therefore Z(I`) ⊇ Z I` ⊇ Z I` . Suppose T `∈L `∈L `∈L P P ∈ Z(I`). Then for every f` ∈ I`, f`(P ) = 0. In particular, for any f = f` ∈ P `∈L P T `∈L I`, f`(P ) = 0 for each ` so f(P ) = 0. Thus P ∈ Z I` . This shows Z(I`) ⊆ `∈PL `∈L `∈L Z `∈L I` , so we have all three equalities. In particular, these properties demonstrate that the algebraic subsets of An form the closed sets of a topology on An.

Deﬁnition. The topology on An having as its closed sets all algebraic subsets of An is called the Zariski topology on An. In (c) and (d), we see that Z and J are not quite inverse operations, but that they may enlarge the given object (set or ideal). The next two important results shows when Z and J are inverses: namely, ZJ is the identity on closed sets and when k is algebraically closed, JZ is the identity on radical ideals.

Lemma 8.1.3. If X ⊆ An is any subset and X is the Zariski-closure of X in An, then (a) J(X) = J(X).

(b) Z(J(X)) = X.

Proof. (a) Since X ⊆ X, we immediately get J(X) ⊇ J(X) by Lemma 8.1.2(a). On the other hand, if f ∈ J(X), f(P ) = 0 for all P ∈ X. In other words, X ⊆ Z(f) but Z(f) is closed by deﬁnition, so Z(f) ⊇ X. Thus f ∈ J(X). (b) X is algebraic by deﬁnition so there exists some ideal I ⊂ A such that Z(I) = X. Now by (a), Z(J(X)) = Z(J(X)) = Z(J(Z(I))) which by Lemma 8.1.2(e) equals Z(I) = X. So Z(J(X)) = X as required.

135 8.1 Aﬃne Algebraic Varieties 8 Algebraic Geometry

2e1 2en Example 8.1.4. Let k = R and consider f = t1 + ... + tn + 1 ∈ R[t1, . . . , tn] for some ei ∈ N. Then Z(f) = ∅ so J(Z(f)) = A and this holds for all choices of ei ∈ N. However, f does not generate the whole ring A, so this shows JZ is not always the identity on ideals of A. Hilbert’s Nullstellensatz, proven below, says that this correspondence is bijective on radical ideals when k is an algebraically closed ﬁeld.

Theorem 8.1.5 (Hilbert’s Nullstellensatz, Geometric Version). If k is algebraically closed, then J(Z(I)) = r(I) for every ideal I ⊂ A.

Proof. By Lemma 8.1.2(d), J(Z(I)) ⊇ I and by Lemma 8.1.1, J(Z(I)) is radical. Together, these imply J(Z(I)) ⊇ r(I). For the reverse containment, take f ∈ J(Z(I)). Since A is noetherian, I = (f1, . . . , fm) for a ﬁnite set of polynomials f1, . . . , fm ∈ I. We must show r Pm that there exist g1, . . . , gm ∈ A and r ∈ N so that we can write f = i=1 figi. 0 0 Consider the ideal I = (f1, . . . , fm, 1 − tn+1f) ⊂ A[tn+1] = k[t1, . . . , tn=1]. If P = n+1 n 0 0 0 (P, an+1) is a point in A , with P ∈ A , assume P ∈ Z(I ). Then fi(P ) = fi(P ) = 0 for all 1 ≤ i ≤ m. Since I = (f1, . . . , fm), this means P ∈ Z(I). Then f ∈ J(Z(I)) implies 0 0 f(P ) = 0, but 1 − tn+1f(P ) = 1 6= 0, contradicting P ∈ Z(I ). Therefore it must be that 0 0 0 n+1 Z(I ) = ∅, so I 6⊂ mP 0 for any P ∈ A , where mP 0 is the maximal ideal of A[tn+1] given by mP 0 = (t1 − a1, . . . , tn+1 − an+1). By Corollary 2.4.3, the algebraic version of Hilbert’s Nullstellensatz for algebraically closed 0 n+1 0 ﬁelds, {mP 0 | P ∈ A } = MaxSpec k[t1, . . . , tn+1]. We have therefore shown that I 0 is not contained in any maximal ideal of k[t1, . . . , tn+1], but this is only possible if I = Pm k[t1, . . . , tn+1]. In particular, 1 = i=1 fihi+(1−tn+1f)hm+1 for some h1, . . . , hm+1 ∈ A[tn+1]. 1 Substituting tn+1 = f , we obtain

m X 1 1 = fi(t1, . . . , tn)hi t1, . . . , tn, f . i=1

r 1 Then multiplying by f for suﬃciently large r clears all powers of f , yielding

m r X f = figi ∈ I for some gi ∈ k[t1, . . . , tn]. i=1 Hence f ∈ r(I). We have seen that J and Z establish a correspondence, though not always bijective, between the ideals of A and the closed subsets of An. We next explore how concepts like the noetherian property and prime ideals translate into geometry.

Definition. A topological space X is noetherian if it satisfies the descending chain condition on closed subsets, or equivalently, if it satisfies the ascending chain condition on open subsets.

Example 8.1.6. Aﬃne n-space An equipped with the Zariski topology is a noetherian space. Indeed, if X1 ⊇ X2 ⊇ X3 ⊇ · · · is a descending chain of Zariski-closed subsets

136 8.1 Aﬃne Algebraic Varieties 8 Algebraic Geometry

of An, applying the vanishing ideal operation J to each set gives an ascending chain of ideals J(X1) ⊆ J(X2) ⊆ J(X3) ⊆ · · · in A = k[t1, . . . , tn]. By Hilbert’s basis theorem (3.3.2), A is noetherian so there is some n ∈ N for which we have J(Xm) = J(Xn) for all m ≥ n. Now by Lemma 8.1.3, Z(J(Xi)) = Xi since each Xi is closed, so we get Xm = Z(J(Xm)) = Z(J(Xn)) = Xn for all m ≥ n. Hence the chain of subsets stabilizes. Lemma 8.1.7. Let X be a noetherian space. Then (a) X is quasi-compact. (b) Every subspace Y ⊆ X is noetherian, with the subspace topology. S Proof. (a) Assume X = i∈I Ui is an open cover of X for some index set I, but there is no ﬁnite subcover. Then for any sequence (ik) ⊆ I, we can construct an ascending chain

Ui1 ( Ui1 ∪ Ui2 ( Ui1 ∪ Ui2 ∪ Ui3 ( ··· which is strictly ascending because no ﬁnite union of the Ui can equal all of X. Therefore X is not noetherian. (b) Let Y1 ⊇ Y2 ⊇ Y3 ⊇ · · · be a descending chain of closed subsets of Y . Then there 0 0 Tn 0 exists a closed set Xn ⊆ X such that Yn = Xn ∩ X for each n ∈ N. Set Xn = j=1 Xj for each n. Then each Xn is closed in X and n n \ 0 \ Xn ∩ Y = Xj ∩ Y = Yj = Yn. j=1 j=1

Now X1 ⊇ X2 ⊇ X3 ⊇ · · · is a descending chain of closed subsets in X by construction, so because X is noetherian, there exists an n ∈ N such that for all m ≥ n, Xm = Xn. Thus for all m ≥ n, Ym = Xm ∩ Y = Xn ∩ Y = Yn so the chain in Y stabilizes. Hence Y is noetherian as a subspace of X. Deﬁnition. A nonempty topological space X is said to be irreducible if for any two closed subsets X1,X2 ⊆ X such that X1 ∪ X2 = X, we have X = X1 or X = X2. Lemma 8.1.8. Let Y be a subspace of a topological space X. Then Y is irreducible if and only if for any closed sets X1,X2 ⊆ X such that Y ⊆ X1 ∪ X2, we have Y ⊆ X1 or Y ⊆ X2. Proof. Obvious.

Lemma 8.1.9. A set X ⊆ An is irreducible if and only if J(X) is a prime ideal of A. Proof. ( =⇒ ) Assume f, g ∈ A such that fg ∈ J(X). Then X ⊆ X(fg) = Z(f) ∪ Z(g) by Lemma 8.1.2(g), so we can write X = (Z(f) ∩ X) ∪ (Z(g) ∩ X) – note that each of these sets is closed in X. If X is irreducible, then we must have X = Z(f) ∩ X or X = Z(g) ∩ X. In particular, X ⊆ Z(f) or X ⊆ Z(g), so f ∈ J(X) or g ∈ J(X). Hence J(X) is prime. n ( ⇒ = ) Given that J(X) is prime, suppose X ⊆ X1 ∪X2 for two closed sets X1,X2 ⊆ A . Then there exist ideals I1,I2 ⊂ A such that Z(I1) = X1 and Z(I2) = X2. By Lemma 8.1.2(g), X ⊆ Z(I1) ∪ Z(I2) = Z(I1I2) so applying J, we get J(X) ⊇ J(Z(I1I2)) ⊇ I1I2 by Lemma 8.1.2(a) and (d). Since J(X) is prime, we must have J(X) ⊇ I1 or J(X) ⊇ I2, but then X ⊆ Z(J(X)) ⊆ Z(I1) = X1 or X ⊆ Z(J(X)) ⊆ Z(I2) = X2. By Lemma 8.1.8 we are done.

137 8.1 Aﬃne Algebraic Varieties 8 Algebraic Geometry

Proposition 8.1.10. Let X be a topological space and Y ⊆ X a subspace. Then (a) X is irreducible if and only if for any nonempty open subset U ⊆ X, U = X. (b) Y is irreducible if and only if Y is irreducible. (c) If X is irreducible then X is connected. (d) If X is irreducible and f : X → Z is a continuous function, then f(X) is irreducible.

Proof. (a) For a nonempty open set U ⊆ X, X r U is a proper closed subspace and X = (X r U) ∪ U is a decomposition into closed subsets. If X is irreducible, then X 6= X r U implies X = U. Conversely, suppose X = X1 ∪X2 where X1,X2 ⊆ X are closed and X1 6= X. Then X r X1 is a nonempty open subset of X, so by assumption X r X1 = X. Then

X = X1 ∪ X2 =⇒ X r X1 ⊆ X2 =⇒ X r X1 ⊆ X2 =⇒ X ⊆ X2 =⇒ X = X2. Therefore X is irreducible. (b) By Lemma 8.1.8, Y is irreducible if and only if whenever Y ⊆ X1 ∪ X2 for closed X1,X2 ⊆ X, we have Y ⊆ X1 or Y ⊆ X2. Notice that this is equivalent to Y ⊆ X1 ∪ X2 implying Y ⊆ X1 or Y ⊆ X2. So Y is irreducible precisely when Y is irreducible. (c) Suppose X = U ∪ V for disjoint, nonempty open sets U, V ⊂ X. Then U = X r V and V = X r U are both closed, but neither U nor V is equal to the whole space X since they are disjoint and nonempty. (d) Suppose f(X) = Y1 ∪ Y2 for closed subsets Y1,Y2 ⊆ f(X). By continuity, X1 := −1 −1 f (Y1) and X2 := f (Y2) are both closed subsets of X. Then X1 ∪ X2 = X so by irreduciblity of X, X = X1 or X = X2. Suppose without loss of generality that X = X1. Then f(X) = f(X1) = Y1, so it follows that f(X) is irreducible. Deﬁnition. A subset Y of a noetherian space X is called an irreducible component of X if Y is a maximal irreducible subspace of X.

2 Example 8.1.11. Consider the aﬃne plane A . Take f = t1t2 in k[t1, t2]. Then V(f) is the 2 union of the t1 and t2 axes, each of which is an irreducible subspace of A : 2 t2 A

Example 8.1.12. Take an irreducible polynomial f ∈ k[t1, t2]. Since k[t1, t2] is a UFD, (f) is a prime ideal so C := Z(f) is irreducible by Lemma 8.1.9. C is called the (affine) algebraic curve defined by f, sometimes written f(x, y) = 0. In general, an irreducible polynomial n in k[t1, . . . , tn] corresponds to an affine variety Y = Z(f) ⊆ A , called an (affine) algebraic hypersurface.

138 8.1 Aﬃne Algebraic Varieties 8 Algebraic Geometry

Proposition 8.1.13. If X is a nonempty noetherian space, then it has finitely many irre- Sm ducible components X1,...,Xm such that X = i=1 Xi. Proof. First we show X can be written as a finite union of some closed irreducible subsets. Define collections

( ` ) [ M = ∅ 6= Y ⊆ X : Y = Yi for closed irreducible subsets Yi ⊆ X i=1 and N = {∅ 6= Y ⊆ X : Y is closed and Y 6∈ M}. Observe that M is closed under ﬁnite unions. We claim that N is empty. If Y ∈ N then Y is not irreducible (since every irreducible set is an element of M), so there exist Y 0,Y 00 ⊆ Y closed such that Y 0 ∪ Y 00 = Y and Y 0,Y 00 6= Y . Also, at least one of Y 0,Y 00 is in N in light of the observation that M is closed under unions. Without loss of generality, assume 0 0 Y ∈ N . Continuing this construction inductively, with Y1 = Y and Y2 = Y , we get a strictly descending chain Y = Y1 ) Y2 ) Y3 ) ··· contradicting the fact that X is noetherian. Therefore N must be empty. In particular, X ∈ M, so there exist closed, irreducible subsets X1,...,Xm ⊆ X such Sm that X = i=1 Xi. We may assume Xi 6⊂ Xj for any i 6= j. Now any irreducible subset Y ⊆ X is contained in at least one Xi, hence only X1,...,Xm can be irreducible components of X. In fact, each Xi is an irreducible component since if Y is another irreducible subset of X such that Xi ⊆ Y , there exists some 1 ≤ j ≤ m for which Xi ⊆ Y ⊆ Xj, but by assumption this is only possible when i = j. In this case, Xi = Y , proving Xi is an irreducible component of X for each 1 ≤ i ≤ m.

Deﬁnition. Assume X is a noetherian space. If X is irreducible, the dimension of X is deﬁned by

` ∈ | there exists a chain of closed, irreducible subsets dim X = sup N0 . Y0 ( Y1 ( ··· ( Y` with Yi ⊆ X Sm By Proposition 8.1.13, we may write X = i=1 Xi for closed, irreducible subsets Xi ⊆ X. Then we extend the deﬁnition of dimension to X by

dim X = max{dim Xi | 1 ≤ i ≤ m}.

Recall that all algebraic subsets of An are noetherian spaces (with the Zariski topology). We now deﬁne one of the fundamental objects of algebraic geometry.

Definition. An affine algebraic variety over k is an irreducible algebraic subset of An. Definition. An quasi-affine variety if a nonempty, open subset of an affine variety.

By Proposition 8.1.10, we have that if X is an aﬃne variety, then any nonempty, open subset U ⊆ X is dense in X. In particular, the Zariski topology on X is not Hausdorﬀ in general.

139 8.1 Aﬃne Algebraic Varieties 8 Algebraic Geometry

Remark. Suppose k is an algebraically closed field. Set A = k[t1, . . . , tn]. Then we have the following correspondence: n {algebraic subsets of A } ←→ {radical ideals of A} X 7−→ J(X) Z(I) 7−→ I. By Hilbert’s Nullstellensatz, this correspondence is bijective and inclusion-reversing. Fur- thermore, Lemma 8.1.9 refines this correspondence to n {affine varieties of A } ←→ Spec(A) n and {points P ∈ A } ←→ MaxSpec(A). This correspondence is preserved when restricting our attention to an algebraic subset of An. Definition. For an algebraic subset X ⊆ An, define the coordinate ring of X by k[X] := A/J(X). If k is algebraically closed, there are bijective, inclusion-reversing correspondences {algebraic subsets of X} ←→ {radical ideals of k[X]} {affine subvarieties of X} ←→ Spec(k[X]) {points of X} ←→ MaxSpec(k[X]). Lemma 8.1.14. Let X ⊆ An be an algebraic subset. Then (a) k[X] is a reduced ring. (b) k[X] is an integral domain if and only if X is irreducible. Proof. (a) By Lemma 8.1.1, J(X) = r(J(X)). Apply Lemma 1.1.4(b). (b) By Lemma 8.1.9, X being irreducible is equivalent to J(X) being a prime ideal of A = k[t1, . . . , tn]. Definition. For an affine variety X ⊆ An, we define its function field over k to be the field of fractions Frac k[X], denoted k(X). Proposition 8.1.15. Let X ⊆ An be an affine variety. Then (a) dim X ≤ dim k[X]. (b) If k is algebraically closed, then dim X = dim k[X].

Proof. (a) If Y0 ( Y1 ( ··· ( Y` is a chain of closed, irreducible subsets of X, then J(X0) ) J(Y1) ) ··· ) J(Y`) is a chain of prime ideals in k[X], by Lemma 8.1.9. (The inclusions are strict since Z(J(Yi)) = Yi for each 0 ≤ i ≤ `, by Lemma 8.1.3(b).) Thus dim X ≤ dim k[X]. (b) Assume k is algebraically closed and let p0 ( p1 ( ··· ( pm be a strictly ascending 0 0 chain of prime ideals in k[X]. For each i, pi = pi/J(X) for some prime ideal pi ⊂ A = 0 0 k[t1, . . . , tn] containing J(X). Thus by Hilbert’s Nullstellensatz, J(Z(pi)) = pi for each i, which gives us 0 0 0 Z(p0) ) Z(p1) ) ··· ) Z(pm), n 0 a strictly descending chain of aﬃne subsets of A . Since each pi contains J(X), this is a chain of closed, irreducible subsets of X. Hence dim k[X] ≤ dim X so we have equality.

140 8.1 Aﬃne Algebraic Varieties 8 Algebraic Geometry

Corollary 8.1.16. Suppose k is algebraically closed and X ⊆ An is an aﬃne variety. Then

(a) dim X = tr degk k(X).

(b) If Y ( X is an aﬃne subvariety, dim Y < dim X. Proof. (a) Apply Proposition 8.1.15 and Theorem 2.6.4. (b) This follows immediately from (a) and Lemma 2.6.2. Theorem 8.1.17. Suppose k is algebraically closed. Then

(a) If I = (f1, . . . , fs) is an ideal of A = k[t1, . . . , tn], then dim Z(I) ≥ n − s.

n (b) If X ⊆ A is any algebraic subset with irreducible components X1,...,Xm ⊆ X, then dim Xi = n − 1 for all 1 ≤ i ≤ m if and only if there exists an f ∈ A r k with X = Z(f). Proof. (a) By Proposition 3.4.10(b), the set {p ∈ Spec(A) | p ⊇ I} has only ﬁnitely many Tm minimal elements, p1,..., pm. Then Corollary 1.1.3 allows us to write r(I) = i=1 pi. By Tm Hilbert’s Nullstellensatz, J(Z(I)) = i=1 pi and so by Lemma 8.1.2,

m ! m \ [ Z(I) = Z(J(Z(I))) = Z pi = Z(pi). i=1 i=1

Lemma 8.1.9 then says that since J(Z(pi)) = pi is prime, Xi := Z(pi) is an irreducible set. We have Xi 6⊂ Xj for any i 6= j since J(Xi) = pi 6⊃ pj = J(Xj). Hence by the proof of Proposition 8.1.13, X1,...,Xm are the irreducible components of J(I) = X. Now by Krull’s height theorem (6.2.10), ht(pi) ≤ s for each 1 ≤ i ≤ m. We now exploit the fact that the polynomial ring has the strong dimension property. One has for each i,

dim Xi = dim k[Xi] = dim A/pi = dim A − ht(pi) ≥ n − s

by Corollary 2.6.8. Taking the maximum over all Xi, we get dim X ≥ n − s. (b) Starting with X = Z(f), we know A is a UFD so there exist a unit u ∈ A×, prime e1 em elements p1, . . . , pm and exponents e1, . . . , em ∈ N such that f = up1 ··· pm . Then Z(f) = Sm i=1 Z(pi) so it follows that Z(pi) are the irreducible components of Z(f). Then dim Xi = dim Z/(pi) = dim A/(pi) = n − 1 by Proposition 2.6.6. Conversely, suppose dim Xi = n − 1 for every irreducible component Xi of X. Set pi = J(Xi) to be the corresponding prime ideal in A. Then dim A/pi = dim k[Xi] = dim Xi = n − 1 which implies ht(pi) = 1 by Lemma 2.6.5(a). Since A is a UFD, there is some prime element pi ∈ A with pi = (pi). Then Xi = Z(pi) = Z(pi) so it follows that f = p1 ··· pm is the element we seek.

Example 8.1.18. Let X ⊆ An be an aﬃne variety. Then for any r ≥ 1, X × Ar is an r ∼ r aﬃne variety with coordinate ring k[X × A ] = k[X][t1, . . . , tr]. In particular, dim X × A = dim X + r. Lemma 8.1.19. If k is algebraically closed and U ⊆ X is a nonempty open subset of X ⊆ An an algebraic variety, then U is also a noetherian space with the subspace topology, U is irreducible and dim U = dim X.

141 8.2 Morphisms of Aﬃne Varieties 8 Algebraic Geometry

8.2 Morphisms of Aﬃne Varieties

In this section we define the notion of a morphism, or map, between algebraic varieties. In a basic sense, these can be thought of as polynomial maps; between affine spaces An and Am, this is well-defined but on subsets of affine space, the definition will be slightly different. n m Consider a map ϕ = (ϕ1, . . . , ϕm): A → A . We say ϕ is a morphism of varieties if there exist polynomials f1, . . . , fm ∈ A = k[t1, . . . , tn] such that fi(P ) = ϕi(P ) for all points P ∈ An. If k is infinite, notice that for all f, g ∈ A, f = g if and only if f(P ) = g(P ) for all P ∈ An. Now suppose X ⊆ An is an affine variety. A morphism from X into affine m-space is a m map ϕ = (ϕ1, . . . , ϕm): X → A such that there exist f1, . . . , fm ∈ A with ϕi(P ) = fi(P ) = ¯ ¯ ¯ fi(P ) for all P ∈ X, where fi = fi + J(X) ∈ k[X]. Formally, we identify ϕi = fi : X → k. We generalize this construction to maps between two affine varieties below.

Definition. If X ⊆ An and Y ⊆ Am are affine varieties, a morphism of affine varieties between them is a map ϕ = (ϕ1, . . . , ϕm): X → Y such that ϕ1, . . . , ϕm ∈ k[X] and for all P ∈ X, ϕ(P ) ∈ Y .

Let ϕ : X → Y be a morphism of aﬃne varieties. Since Y is closed, Y = Z(J(Y )) by Lemma 8.1.3(b) so ϕ(P ) ∈ Y is equivalent to saying g(ϕ1(P ), . . . , ϕm(P )) = 0 for all g ∈ J(Y ), or alternatively, g ◦ ϕ = 0 for all g ∈ J(Y ).

Definition. For affine varieties X ⊆ An and Y ⊆ Am, we define the ring of morphisms from X to Y as the set

M(X,Y ) = {ϕ : X → Y | ϕ is a morphism}

with ring operations given by pointwise addition and composition. A morphism ϕ ∈ M(X,Y ) is an isomorphism if ϕ is bijective and ϕ−1 is a morphism in the ring M(Y,X).

Example 8.2.1. Let k be a perfect ﬁeld of characteristic p > 0 and consider the p-power map

1 1 ϕ : A −→ A x 7−→ xp.

Then ϕ is a bijective morphism (it is explicitly given by a polynomial map and is known to be an automorphism of any ﬁeld of characteristic p) but it is not an isomorphism of varieties, since taking the pth root is not a polynomial map.

Example 8.2.2. For any affine variety X over k, k[X] = M(X, A1). Proposition 8.2.3. For an affine variety X over a field k, the points of X are in bijective correspondence with Homk(k[X], k). Proof. The correspondence is given by sending a point P ∈ X to the evaluation homomorphism evP : f 7→ f(P ), and sending a k-algebra homomorphism f : k[X] → k to the point (f(t¯1), . . . , f(t¯n)) ∈ k[X], where k[X] = k[t¯1,..., t¯n].

142 8.2 Morphisms of Aﬃne Varieties 8 Algebraic Geometry

Lemma 8.2.4. Let X and Y be aﬃne varieties. Then every morphism ϕ ∈ M(X,Y ) is continuous with respect to the Zariski topologies on X and Y .

Proof. Let ϕ : X → Y be a morphism. Take a closed set Z = Z(f1, . . . , fr) ⊆ Y , with f1, . . . , fr ∈ k[t1, . . . , tm]. Then

−1 ϕ (Z) = {P ∈ X | ϕ(P ) ∈ Z} = {P ∈ X | fi ◦ ϕ(P ) = 0}

= Z(f1 ◦ ϕ, . . . , fr ◦ ϕ).

Since each fi ◦ ϕ ∈ k[t1, . . . , tn], Z(f1 ◦ ϕ, . . . , fr ◦ ϕ) is a closed set in X. Hence ϕ is continuous.

Lemma 8.2.5. For aﬃne varieties X,Y and W and morphisms ϕ : X → Y and ψ : Y → W , the composition ψ ◦ ϕ : X → W is also a morphism.

Deﬁnition. For a morphism of varieties ϕ : X → Y , we deﬁne a comorphism ϕ∗ ∈ Homk(k[Y ], k[X]) of the k-algebras k[Y ] and k[X] by

ϕ∗ : k[Y ] −→ k[X] 1 g 7−→ (g ◦ ϕ : X → A ). Conversely, given a k-algebra homomorphism α : k[Y ] → k[X], we deﬁne its corresponding comorphism of varieties by

α∗ : X −→ Y

(t¯1,..., t¯n) 7−→ (α(t¯1), . . . , α(t¯n)).

Lemma 8.2.6. Let X,Y and W be aﬃne varieties. Then

∗ ∗ (a) (idX ) = idk[X] and (idk[X]) = idX .

ϕ ψ (b) For morphisms of varieties X −→ Y −→ W , we have (ψ ◦ ϕ)∗ = ϕ∗ ◦ ψ∗. That is, comorphisms are natural.

β (c) For k-algebra homomorphisms k[W ] −→α k[Y ] −→ k[X], we have (β ◦ α)∗ = α∗ ◦ β∗.

∗∗ ∗∗ (d) For any ϕ ∈ M(X,Y ), ϕ = ϕ. Likewise, for any α ∈ Homk(k[Y ], k[X]), α = α. Deﬁnition. A morphism ϕ : X → Y is said to be dominant if ϕ(X) = Y , that is, if it has dense image in its target. A dominant morphism is called ﬁnite if the ring extension k[X]/ϕ∗(k[Y ]) is an integral extension.

Remark. Let ϕ : X → Y be a finite morphism. Then the injection ϕ∗ : k[Y ] ,→ k[X] induces a field extension k(Y ) ,→ K(X) which is finite and algebraic. Further, if k is algebraically closed, Proposition 8.1.15 and Theorem 2.3.7 imply that dim X = dim Y .

Example 8.2.7. Let ϕ : A1 → A1 be the map sending x 7→ xn for some n ≥ 2. Then for all infinite fields k, ϕ is a finite morphism but usually not surjective.

143 8.2 Morphisms of Aﬃne Varieties 8 Algebraic Geometry

Proposition 8.2.8. If k is algebraically closed and ϕ ∈ M(X,Y ) is a ﬁnite morphism, then ϕ is surjective.

n Proof. Take a point x = (x1, . . . , xn) ∈ X ⊆ A and associate to it the maximal ideal mx = (t1 − x1, . . . , tn − xn) ⊂ A. Recall that J({x}) = mx. The ideal mx = mx/J(X) is maximal in k[X]. Similarly, for any point y ∈ Y we have my = J({y}) ∈ k[t1, . . . , tm] and an associated maximal ideal my ∈ k[Y ]. Now

ϕ(x) = y ⇐⇒ ϕ(x) ∈ {y} = Z(my)

⇐⇒ g¯(ϕ(x)) = 0 for allg ¯ ∈ my

⇐⇒ (¯g ◦ ϕ)(x) = 0 for allg ¯ ∈ my ∗ ⇐⇒ ϕ (¯g)(x) = 0 for allg ¯ ∈ my ∗ ⇐⇒ ϕ (¯g) ∈ mx for allg ¯ ∈ my ∗ ⇐⇒ ϕ (my) ⊆ mx.

Assuming ϕ is ﬁnite, the extension k[X]/ϕ∗(k[Y ]) is integral and in particular ϕ∗ : k[Y ] → ϕ∗(k[Y ]) is an isomorphism of k-algebras. For any given y ∈ Y , Lemma 2.3.4 shows there ∗ ∗ ∗ exists a prime ideal m ⊂ k[X] with m ∩ ϕ (k[Y ]) = ϕ (my). Since ϕ (my) is maximal, this m must be maximal in k[X] by Lemma 2.3.1(b). Then by Hilbert’s Nullstellensatz (algebraic ∗ version, Theorem 2.4.2), there exists x ∈ X such that m = mx. Thus ϕ (my) ⊆ m = mx. By the above equivalences, this implies ϕ(x) = y so we have proven ϕ is a surjection.

Remark. If ϕ : X → Y is a finite morphism, the fibre ϕ−1(y) is a finite set for all y ∈ Y . In particular, dim ϕ−1(y) = 0 as a variety.

Lemma 8.2.9. Let ϕ : X → Y be a morphism of aﬃne varieties. Then

(a) ϕ∗ is injective if and only if ϕ is dominant.

(b) ϕ∗ is surjective if and only if ϕ(X) is closed in Y and ϕ0 : X → ϕ(X) is an isomorphism.

Proof. (a) Notice that

∗ ∗ ϕ is injective ⇐⇒ ϕ (g) 6= 0 for all g ∈ k[Y ] r {0} ⇐⇒ g ◦ ϕ 6= 0 for all g ∈ k[Y ] r {0} ⇐⇒ g(ϕ(X)) 6= {0} for all g ∈ k[Y ] r {0} ⇐⇒ J(ϕ(X)) = J(Y ) ⊆ k[t1, . . . , tn] ⇐⇒ ϕ(X) = Y by Lemma 8.1.3.

Therefore the injectivity of ϕ∗ is equivalent to ϕ being dominant. (b) Suppose ϕ∗ is surjective. We claim ϕ(X) = Z(ker ϕ∗). On one hand, if y ∈ ϕ(X), let x ∈ X such that ϕ(x) = y. Then for any g ∈ ker ϕ∗, g(y) = g(ϕ(x)) = (g◦ϕ)(x) = ϕ∗(g)(x) =

144 8.2 Morphisms of Aﬃne Varieties 8 Algebraic Geometry

0, so ϕ(X) ⊆ Z(ker ϕ∗). This gives us a map ϕ0 : X → Z(ker ϕ∗) which corresponds on the level of k-algebras to (ϕ0)∗ : k[Y ]/ ker ϕ∗ → k[X]. Since (ϕ0)∗ is surjective, the isomorphism theorems give us k[Y ]/ ker ϕ∗ ∼= k[X] which implies, so ϕ0 : X → ϕ(X) is an isomorphism. In particular, ϕ(X) = Z(ker ϕ∗) so the image of ϕ is closed. The converse follows from Lemma 8.2.6. (c) follows from (a) and (b). Example 8.2.10. (Interpreting Noether normalization) Assume k is algebraically closed. n Take an affine variety X ⊆ A with associated affine algebra k[X] = k[t1, . . . , tn]/J(X) = ¯ ¯ k[t1,..., tn]. By Corollary 8.1.16 and Theorem 2.6.4, d := dim X = tr degk k[X] = dim k[X]. Pn ¯ Then Noether normalization (Theorem 2.4.1) prescribes elements s1, . . . , sd ∈ i=1 kti such ∼ that k[X]/k[s1, . . . , sd] is an integral extension and k[s1, . . . , sd] = k[t1, . . . , td] as k-algebras. Consider the embedding α : k[s1, . . . , sd] ,→ k[X]. Viewing k[s1, . . . , sd] = k[t1, . . . , td] = d ∗ d k[A ], we get a comorphism ϕ = α : X → A . Since k[X]/k[t1, . . . , td] is integral, ϕ is a finite morphism and hence surjective by Proposition 8.2.8. This says that any affine variety can be viewed as a finite-sheeted cover of Ad for some d. Suppose U ⊆ X is a nonempty, open subset of an affine variety X ⊆ An. Then Y = XrU is a closed set in X, so Y = Z(I) for some ideal I ⊂ k[X]. Conversely, for every ideal I ⊂ k[X] we associate to it an open set D(I) = X r Z(I). Definition. A principal open set in X is D(f) := D((f)) for some f ∈ k[X] r {0}, sometimes also denoted Xf . Explicitly, D(f) = {x ∈ X | f(x) 6= 0}. Tm If I = (f1, . . . , fm) for f1, . . . , fm ∈ k[X], then by Lemma 8.1.2(h), Z(I) = i=1 Z(fi) so Sm that D(I) = i=1 D(fi). This shows that every open subset in X can be covered by finitely many principal open subsets. For f ∈ k[X] r {0}, choose a lift F ∈ k[t1, . . . , tn] such that f = F + J(X) ∈ k[X]. Set A = k[t1, . . . , tn] and B = k[t1, . . . , tn+1].

n+1 Lemma 8.2.11. Define W ⊆ A by W = Z(L) where L is the ideal (J(X), tn+1F −1) ⊂ B. 1 −1 Let ϕ : D(f) → W be the map given by x 7→ x, f(x) with inverse ϕ : W → D(f). Then (a) ϕ is a Zariski-homeomorphism. ∼ (b) ϕ induces a k-algebra isomorphism k[W ] = k[X]f , the localization of k[X] at the powers of f. As a consequence, the principal open subsets of X can be identified with affine subvarieties of An+1, giving rise to the quasi-affine varieties mentioned previously.

Remark. If k is an arbitrary ﬁeld, k[X]f does not only depend on the principal open set Xf , but also on the choice of f. However, if k is algebraically closed, we have

Xf = Xg ⇐⇒ Z(f) = Z(g) ⇐⇒ r((f)) = r((g)) by Hilbert’s Nullstellensatz

=⇒ k[X]f = k[X]g by Proposition 1.3.5.

In this context, we denote k[X]f = k[Xf ].

145 8.3 Sheaves of Functions 8 Algebraic Geometry

8.3 Sheaves of Functions

Deﬁnition. Let X ⊆ An be an irreducible aﬃne variety. A sheaf of functions on X consists of the following data:

(a) For x ∈ X, the stalk at x is deﬁned by

n f o OX,x = g ∈ k(X): f, g ∈ k[X], g(x) 6= 0 .

Alternatively, if mx = J({x}) is the maximal ideal associated to the point x, we may

identify OX,x = k[X]mx . (b) For each nonempty, open subset U ⊆ X, we deﬁne a ring of functions on U by \ \ n o O (U) = O = h = fx ∈ k(X): f , g ∈ k[X] such that g (x) 6= 0 . X X,x gx x x x x∈U x∈U

Each h ∈ OX (U) gives a well-deﬁned, k-valued function

h : U −→ k f (x) x 7−→ x gx(x)

called a regular function on U. Formally, we set OX (∅) = 0.

Lemma 8.3.1. Let X ⊆ An be a variety and U ⊆ X a nonempty, open subset. Then

(a) For each h ∈ OX (U), h : U → k is continuous with respect to the Zariski topology.

(b) Suppose h1, h2 ∈ OX (U) and V ⊆ U is a nonempty, open subset such that h1|V = h2|V . Then h1 = h2 on U. Proof. (a) Clear from the deﬁnitions. f1 f2 (b) Fix x ∈ V and f1, g1, f2, g2 ∈ k[X] with g1(x), g2(x) 6= 0 and h1 = and h2 = . g1 g2 f1(y) f2(y) Without loss of generality we may assume V ⊆ Xg ∩ Xg . By assumption, = for 1 2 g1(y) g2(y) all y ∈ V . Thus f1g2 − f2g1 = 0 on V , so V ⊆ Z(f1g2 − f2g1), but Z(f1g2 − f2g1) is closed and V is dense, so it follows that Z(f1g2 −f2g1) = X. Thus f1g2 −f2g1 = 0 in k[X], meaning

f1 f2 h1 = = = h2 g1 g2 in k(X).

Proposition 8.3.2. Suppose k is algebraically closed and X is an aﬃne variety over k. Then

(a) OX (X) = k[X], that is, the coordinate ring of X consists of regular k-valued functions X → k.

146 8.4 Finite Morphisms and the Fibre Theorem 8 Algebraic Geometry

(b) For any f ∈ k[X] r {0}, OX (Xf ) = k[X]f .

Proof. (a) By the Nullstellensatz, the maximal ideals of k[X] are precisely mx for x ∈ X. Since k[X] is a domain, Lemma 2.3.11 allows us to write \ \ \ k[X] = k[X]m = k[X]mx = OX,x = OX (X). maximal ideals x∈X x∈X m⊂k[X]

(b) This follows from a similar proof to (a), using the fact that the maximal ideals of −1 m k[X]f are precisely those of the form S mx, where x ∈ Xf and S = {f | m ∈ N0}. Definition. Assume U is a (nonempty) quasi-affine variety. The function field of U is defined to be k(U) = Frac(OX (U)), where X = U is the Zariski-closure of U. Viewing OX (U) as a subring of k[X], we can identify k(U) = k(X) whenever X = U. Definition. Two quasi-affine varieties U, V ⊆ X are said to be birationally equivalent if k(U) ∼= k(V ) as k-algebras. Proposition 8.3.3. Suppose X and Y are affine varieties over k. Then the following are equivalent: (i) X is birationally equivalent to Y . ∼ (ii) There exist nonzero functions f ∈ k[X] and g ∈ k[Y ] such that Xf = Yg as affine varieties. ∼ (iii) There exist nonzero functions f ∈ k[X] and g ∈ k[Y ] such that k[X]f = k[Y ]g as k-algebras.

Proof. (i) ⇐⇒ (ii) By Proposition 8.1.10(a), the principal open sets Xf and Yg are dense ∼ in X and Y , respectively, so k(X) = k(Xf ) = k(Yg) = k(Y ). Thus X and Y are birationally equivalent. The converse comes from the fact that the principal open sets form bases of the Zariski topologies on X and Y . (ii) ⇐⇒ (iii) follows from the identiﬁcation k[Xf ] = k[X]f .

8.4 Finite Morphisms and the Fibre Theorem

Assume k is an algebraically closed field and let X and Y be affine varieties over k. Recall from Section 8.2 that a morphism ϕ : X → Y is finite if it is dominant – i.e. ϕ∗ : k[Y ] ,→ k[X] is injective – and k[X]/k[Y ] is integral. By Proposition 8.2.8, ϕ must also be surjective. Proposition 8.4.1. Let ϕ : X → Y be a finite morphism. Then

(a) For any closed set E ⊆ X, ϕ(E) ⊆ Y is closed and the restriction ϕ|E : E → ϕ(E) is also a ﬁnite morphism.

(b) Suppose W ⊆ Y is a closed, irreducible subset and ϕ−1(W ) ⊆ X has irreducible components E1,...,E`, with Ei ⊆ X closed for each 1 ≤ i ≤ `. Then ϕ(Ei) = W if −1 −1 and only if dim Ei = dim ϕ (W ). Furthermore, dim ϕ (W ) = dim W .

147 8.4 Finite Morphisms and the Fibre Theorem 8 Algebraic Geometry

Proof. (a) Set A = k[Y ] and B = k[X], so that B/A is an integral extension. Given a closed set E ⊆ X, set I = J(E) to be its vanishing ideal in B. By Lemma 2.3.1(a), (B/I)/(A/A∩I) is also an integral extension. From Lemma 8.1.3, we have Z(I) = Z(J(E)) = E = E. Let I0 = A ∩ I and set E0 = Z(I0). Then ϕ(E) = E0 so in particular ϕ(E) is closed. Deﬁne ψ : E → E0 by x 7→ ϕ(x). We claim ψ is ﬁnite. First, since I = J(E) is a radical ideal by Lemma 8.1.1, I0 = A ∩ I must be radical, too, e.g. by Lemma 1.1.9. Thus by the Nullstellensatz, I0 = r(I0) = J(Z(I)) = J(E0). Viewing J(E) ⊆ B, we have B/I = k[X]/J(E), but by isomorphism theorems, this can be written ∼ k[X]/J(E) = (k[t1, . . . , tn]/J(X))/(J(E)/J(X)) = k[t1, . . . , tn]/J(E) = k[E].

Similarly, A/I0 ∼= k[E0], so we get a comorphism ψ∗ : k[E0] → k[E] which is injective by the above comments. Therefore Lemma 8.2.9 shows that ψ is dominant. Moreover, k[E]/k[E0] is integral by the above, so ψ is ﬁnite. (b) Since each component Ei is closed in X, (a) says that each ϕ(Ei) is closed in Y . Further, irreducibility of Ei implies ϕ(Ei) is irreducible by Proposition 8.1.10(d). By (a),

ϕ|Ei : Ei → ϕ(Ei) is a finite morphism, so Theorem 2.6.4 implies dim Ei = dim ϕ(Ei). In particular, Corollary 8.1.16 implies that since ϕ(Ei) is a subvariety of W , ϕ(Ei) = W if and only if dim W = dim ϕ(Ei) = dim Ei. To finish, we show dim ϕ−1(W ) = dim W . Corollary 8.1.16 and the previous paragraph −1 give us dim Ei = dim ϕ (Ei) ≤ dim W , so it follows from the definition of dimension for −1 −1 ϕ (W ) that dim ϕ (W ) = max{dim Ei | 1 ≤ i ≤ `} ≤ dim W . Conversely, since ϕ is surjective we get that

−1 W = ϕ(ϕ (W )) = ϕ(E1) ∪ · · · ∪ ϕ(E`).

Since W is irreducible, we must have W = ϕ(Ej) for some 1 ≤ j ≤ `. Then

−1 dim ϕ (W ) ≥ dim Ej = dim ϕ(Ej) = dim W.

Hence the dimensions of W and ϕ−1(W ) are equal. We now prove the powerful Fibre Theorem for dominant morphisms. Recall from Exam- ple 8.1.18 that if X is an aﬃne variety, so is X × Ar, and dim X × Ar = dim X + r. Theorem 8.4.2 (Fibre Theorem). Suppose ϕ : X → Y is a dominant morphism of aﬃne varieties. Set r = dim X − dim Y . Then there exists a nonempty, open subset U ⊆ Y such that

(a) U ⊆ ϕ(X).

(b) If W ⊆ Y is closed and irreducible and W ∩ U 6= ∅, then dim(ϕ−1(W ) ∩ ϕ−1(U)) = dim W + r. In particular, dim ϕ−1(y) = r for all y ∈ U.

Proof. Suppose X ⊆ Ap and Y ⊆ Aq for p, q ≥ 1. Let B = k[X] and A = k[Y ] so that A,→ B. Set K = k(Y ) and L = k(X).

148 8.4 Finite Morphisms and the Fibre Theorem 8 Algebraic Geometry

K L

A B

For the multiplicative subset D = A r {0}, we have that K = D−1A and L = Frac(D−1B). Write B = k[t¯1,..., t¯p] = A[t¯1,..., t¯p] and note that B is a ﬁnitely generated A-module. −1 −1 −1 Then K[t¯1,..., t¯p] = D A[t¯1,..., t¯p] = D B so D B is a ﬁnitely generated K-algebra. −1 By Noether’s normalization lemma (Theorem 2.4.1), there exist elements h1, . . . , h` ∈ D B −1 which are algebraically independent over K such that D B/K[h1, . . . , h`] is an integral extension. Without loss of generality, we may assume h1, . . . , h` ∈ B. Then we have the following diagram:

K K(h1, . . . , h`) L

−1 A K[h1, . . . , h`] D B

By Proposition 1.4.8 and Corollary 8.1.16(a), we get

` = tr degK L = tr degk L − tr degk K = dim X − dim Y = r.

Note that while B/A[h1, . . . , hr] may not necessarily be integral, we can ﬁnd an element f ∈ A such that Bf /Af [h1, . . . , hr] is integral. Then h1, . . . , hr are algebraically independent over ∼ Af , so Af [h1, . . . , hr] is isomorphic to a polynomial ring in r variables, say Af [h1, . . . , hr] = Af [u1, . . . , ur]. It follows that we have maps

ψ r p Xf −→ Yf × A −→ Yf where p is the coordinate projection and ψ is the ﬁnite morphism induced as a comorphism by the k-algebra maps r ∼ ∼ k[Yf × A ] = k[Yf ][u1, . . . , ur] = k[Yf ][h1, . . . , hr] = Af [h1, . . . , hr] ,→ Bf .

Set U = Yf , the principal open subset of Y generated by f. By construction, U ⊆ ϕ(X). Take W ⊆ Y to be a closed, irreducible subset with W ∩ U 6= ∅. Set W 0 = W ∩ U; then 0 0 r r W is a closed, irreducible subset of U = Yf so W × A is closed and irreducible in U × A . Hence W 0 × Ar is an affine variety. Moreover, we have dim(ϕ−1(W ) ∩ ϕ−1(U)) = dim ϕ−1(W ∩ U) = dim ϕ−1(W 0) −1 0 r = dim ψ (W × A ) since ψ is a finite morphism 0 r = dim W × A by Proposition 8.4.1(b) = dim W 0 + r by Example 8.1.18 = dim W + r by Lemma 8.1.19. The final statement follows by applying this to the irreducible subset W = {y} ⊆ U.

149 8.5 Projective Varieties 8 Algebraic Geometry

8.5 Projective Varieties

In this section we construct a projective analog to the notion of an aﬃne variety. Let k be a ﬁeld; as in previous sections, we will often make the assumption that k is algebraically closed.

Deﬁnition. For n ∈ N, we deﬁne projective n-space over k to be the quotient space n n n+1 ∗ P = Pk = A r{0}/ ∼ where (a0, . . . , an) ∼ (b0, . . . , bn) if and only if there is some λ ∈ k n such that (b0, . . . , bn) = (λa0, . . . , λan). The coordinates of P are written [a0, . . . , an], called homogeneous coordinates.

Let S = k[t0, . . . , tn] be the polynomial ring in n + 1 indeterminates. Recall that S is a graded ring with graded pieces given by total degree:

∞ M S = Sd where Sd = {f ∈ S | deg f = d}. d=0

An arbitrary polynomial in S does not have a well-deﬁned vanishing set in Pn. However, homogeneous polynomials do have vanishing sets:

Deﬁnition. For f ∈ Sd, deﬁne the zero set of f to be

n Z(f) = {P ∈ P | f(P ) = 0}, where f(P ) = f(p0, . . . , pn) if P = [p0, . . . , pn]. This set is well-deﬁned, since f ∈ Sd implies d × f(λa0, . . . , λan) = λ f(a0, . . . , an) for all λ ∈ k .

h S∞ h Set S = Sd. For a collection of homogeneous polynomials F ⊆ S , deﬁne the zero d=0 T set of this collection by Z(F) = f∈F Z(f).

Deﬁnition. Let S = k[t0, . . . , tn] and suppose I ⊂ S is an ideal. Then S is homogeneous L∞ if I = d=0 Id where Id = I ∩ Sd for each d ∈ N0. Lemma 8.5.1. Let I ⊂ S be an ideal. Then P∞ (a) I is homogeneous if and only if for all f = d=0 fd ∈ I, each fd ∈ I. P T (b) For any family (Iα)α∈A of homogeneous ideals of S, α∈A Iα and α∈A Iα are also homogeneous ideals.

(c) For an element f ∈ Sd, fS is homogeneous. (d) I is a homogeneous ideal if and only if I is generated by ﬁnitely many homogeneous elements.

(e) If I is homogeneous then r(I) is homogeneous.

(f) A homogeneous ideal p ⊂ S is prime if and only if fg ∈ p implies f ∈ p or g ∈ p for all homogeneous elements f, g ∈ Sh.

150 8.5 Projective Varieties 8 Algebraic Geometry

Proof. (a) is routine and (b) follows from applying (a) and the deﬁnitions of sums and intersections of ideals. L∞ L∞ (c) Write S = j=d Sj−d. Then fS = j=d fSj−d and note that (fS)j = fSj−d makes fS into a homogeneous ideal. L∞ S∞ (d) If I is homogeneous, I = d=0(I ∩Sd) so I is generated by the set d=0(I ∩Sd) which consists of homogeneous elements. Finite generation follows from Hilbert’s basis theorem S∞ (3.3.2). Conversely, if I = (B) where B = Bd and Bd ⊆ Sd for each d ∈ N0, then for Pd=0 any a ∈ I, there exist rb ∈ S such that a = b∈B rbb and ﬁnitely many of the rb are nonzero. Then ∞ ∞ X X X M a = rbb = rbb ∈ (I ∩ Sd).

b∈B d=0 b∈Bd d=0 L∞ L∞ Therefore we have d=0(I ∩ Sd) ⊆ I ⊆ d=0(I ∩ Sd) and hence I is homogeneous. Pm (e) If f = d=0 fd ∈ r(I) then by induction on 0 ≤ d ≤ m, it follows that fd ∈ r(I) as well. For the base case, f ∈ r(I) implies there is some r ∈ N such that f r ∈ I. Then r r r ` (f )0 = f0 ∈ I0 so f0 ∈ r(I). Then f − f0 ∈ r(I) which implies (f − f0) ∈ I for some ` ∈ N, ` so f1 ∈ I`. Inductively, we get fd ∈ r(I). (f) One direction is trivial. For the other, assume the statement holds for homogeneous elements in S and suppose fg ∈ p and g 6∈ p for arbitrary elements f, g ∈ S. Write P` Pm f = d=0 fd and g = d=j gd with gj 6∈ p and fd, gd ∈ Sd. Then f0gj ∈ pj = p ∩ Sj and so f0 ∈ p. As in (e), we can continue inductively to get fd ∈ p for all d ≥ 0, so f ∈ p. Hence p is prime.

Deﬁnition. Let X ⊆ Pn be any subset. The (homogeneous) vanishing ideal of X is deﬁned to be J(X) = {f ∈ Sh | f(P ) = 0 for all P ∈ X}. X is called an algebraic subset if X = Z(I) for some homogeneous ideal I ⊂ S.

Lemma 8.5.2. Let I be an ideal of S and X ⊆ Pn a subset. Then Tm (a) If I = (f1, . . . , fm) then Z(I) = i=1 Z(fi). (b) J(X) is a homogeneous, radical ideal of S.

Proof. Similar to the proof of Lemma 8.1.1.

As in the aﬃne case, the sets Z(I) form the closed sets in the Zariski topology on Pn.

Deﬁnition. For an algebraic subset X ⊆ Pn, the coordinate ring of X is S[X] = S/J(X).

Lemma 8.5.3. An algebraic set X ⊆ Pn is irreducible if and only if J(X) is prime in S. Proof. See the proof of Lemma 8.1.9.

n Remark. By Hilbert’s basis theorem (3.3.2), S = k[t0, . . . , tn] is noetherian and so P is a noetherian space, as are its subspaces equipped with the Zariski topology.

We next deﬁne the projective version of an algebraic variety.

151 8.5 Projective Varieties 8 Algebraic Geometry

Deﬁnition. A projective variety is an irreducible closed subset of Pn.A quasi-projective variety is a nonempty, open subset of a projective variety.

Theorem 8.5.4 (Hilbert’s Nullstellensatz, Projective Version). Let k be an algebraically closed ﬁeld and set S = k[t0, . . . , tn]. Then for any homogeneous ideal I ⊂ S,

(a) J(Z(I)) = r(I) if Z(I) 6= ∅.

(b) Z(I) = ∅ if and only if I = S or r(I) = (t0, . . . , tn). Proof. (a) By Lemma 8.1.9(b) and Corollary 1.1.3, r(I) is a homogeneous ideal and clearly r(I) ⊆ J(Z(I)). For the other direction, suppose f ∈ J(Z(I)) ∩ Sd and f 6= 0. Since Z(I) 6= ∅, d must be at least 1, so f ∈ (t0, . . . , tn). In particular, f(0) = 0, i.e. f vanishes at the origin in An+1. Set

n+1 Za(I) = {P ∈ A | f(P ) = 0 for all f ∈ I}.

n Then the above shows f ∈ J(Za(I)). Notice that [a0, . . . , an] ∈ Z(I) ⊆ P if and only if n+1 (a0, . . . , an) ∈ Za(I)r{0} ⊆ A . By the affine Nullstellensatz (Theorem 8.1.5), J(Za(I)) = r(I) so by the above, we must have f ∈ r(I). Therefore J(Z(I)) = r(I). (b) If r(I) ⊇ (t0, . . . , tn) then Z(r(I)) ⊆ Z(t0, . . . , tn) = ∅, implying Z(I) = ∅. Con- versely, Z(I) = ∅ implies Za(I) ⊆ {0}, where Za(I) is as defined in part (a). But Za(I) ⊆ {0} impies J(Za(I)) ⊇ J({0}) = (t0, . . . , tn) by Lemma 8.1.2(a), so we have r(I) ⊇ (t0, . . . , tn) by the affine Nullstellensatz (Theorem 8.1.5). This completes the proof.

Corollary 8.5.5. For an algebraically closed ﬁeld k, the algebraic subsets of Pn are in bijective correspondence with the homogeneous radical ideals I ⊂ S, I 6= (t0, . . . , tn). As in the aﬃne case, when k is algebraically closed we have correspondences

algebraic subsets homogeneous radical ideals n ←→ X ⊆ P I ⊆ S,I 6= (t0, . . . , tn) projective varieties homogeneous prime ideals n ←→ . X ⊆ P p ∈ Spec(S), p 6= (t0, . . . , tn)

n n Deﬁne the ith projective hyperplane by Hi = Z(ti) ⊂ P for 0 ≤ i ≤ n. Set Ui = P rHi, n n Sn an open set in P . Then P = i=0 Ui, that is, the complements of the coordinate hyperplanes are an open cover of Pn.

n n Proposition 8.5.6. Each Ui is homeomorphic to A . That is, P is locally aﬃne. n a0 ai−1 ai+1 an Proof. Deﬁne ϕi : Ui → by ϕi[a0, . . . , an] = ,..., , ,..., . This is Zariski- A ai ai ai ai continuous and has a continuous inverse given by ψi(b1, . . . , bn) = [b1, . . . , bi, 1, bi+1, . . . , bn]. Therefore ϕi is a Zariski-homeomorphism for each 0 ≤ i ≤ n.

n Sn Corollary 8.5.7. If Y ⊆ P is a projective variety, then Y = i=0(Y ∩ Ui). In particular, every projective variety may be covered by open sets which are homeomorphic to aﬃne varieties in An.

152 8.6 Nonsingular Varieties 8 Algebraic Geometry

8.6 Nonsingular Varieties

Assume k is an algebraically closed ﬁeld. Let X ⊆ An be a d-dimensional aﬃne variety.

Deﬁnition. If J(X) = (f1, . . . , fr) is the vanishing ideal of X in A = k[t1, . . . , tn] and x ∈ X, then the Jacobian of X at x is the r × n matrix ∂fi Jx = (x) , ∂tj

k k−1 where derivatives of polynomials are just the formal power rule tj 7→ ktj . We say X is nonsingular, or smooth, at x ∈ X if rank Jx = n−d. Otherwise, X is said to be singular at x.

Remark. X is nonsingular at a point x if and only if every open set U ⊆ X containing x is nonsingular at x, which is even equivalent to there existing some open U ⊆ X containing x such that U is nonsingular at x.

Lemma 8.6.1. Let A be a ring, M ⊂ A a maximal ideal and m ⊂ AM the corresponding maximal ideal in the localization at M. Then ∼ (a) A/M = AM /m.

2 ∼ 2 (b) A/M = AM /m . (c) M/M 2 ∼= m/m2 as A/M-vector spaces.

The following asserts that the geometric condition of smoothness is equivalent to the algebraic condition of a ring being regular local. This is a fundamental connection in algebraic geometry.

Theorem 8.6.2. An aﬃne variety X ⊆ An is nonsingular at a point x ∈ X if and only if the stalk OX,x is a regular local ring.

n Proof. Let x = (x1, . . . , xn) ∈ X ⊆ A and deﬁne the corresponding maximal ideal mx = (t1 − x1, . . . , tn − xn) ⊂ A = k[t1, . . . , tn]. Then there is a map

θ : A −→ kn ∂f ∂f f 7−→ (x),..., (x) . ∂t1 ∂tn

Notice that θ is linear, and θ(ti − xi) = (0,..., 1,..., 0) (with 1 in the ith coordinate), so θ 2 is surjective. In fact, this shows that θ|mx is surjective. We claim that ker θ|mx = mx. On one hand, for any 1 ≤ i, j, k ≤ n, we have

∂((ti − xi)(tj − xj)) = δik(tj − xj) + δjk(ti − xi) ∂tk

153 8.6 Nonsingular Varieties 8 Algebraic Geometry

2 which is zero on mx. Take

X 2 g = gij(ti − xi)(tj − xj) ∈ mx, i,j where gij ∈ A. Then for each 1 ≤ k ≤ n,

∂g X ∂((ti − xi)(tj − xj)) ∂gij = g + (t − x )(t − x ) ∂t ij ∂t ∂t i i j j k i,j k k

2 2 which is zero at x. Hence mx ⊆ ker θ|mx . On the other hand, suppose f ∈ ker θ|mx r mx. P 2 Then f = i gi(ti − xi) for some gi ∈ A. Since f 6∈ mx, we must have gj 6∈ mx for some j. Thus ∂f ∂ X ∂gi = (gj(tj − xj)) + (ti − xi) ∂tj ∂tj ∂tj i6=j which vanishes on x. Thus gj(x) = 0, implying gj ∈ J(x) = mx, a contradiction. Therefore 2 ker θ|mx = mx as claimed. 0 2 n We now get a k-linear isomorphism θ : mx/mx → k . Then J(X) = (f1, . . . , fr) ⊆ mx is a k-linear subspace, so θ(J(X)) is a k-subspace of kn. This implies (θ0)−11(θ(J(X))) = 2 2 J(X) + mx/mx and so the Jacobian at x is   θ(f1)  .  Jx =  .  . θ(fr)

2 2 Thus rank Jx = dimk θ(J(X)) = dimk J(X) + mx/mx. From Section 8.3, we know that

OX,x = k[X]mx and this is a local ring because mx is a maximal ideal. Let m be the unique maximal ideal of OX,x and let m be the corresponding ideal in k[X]. By Lemma 8.6.1, 2 ∼ 2 2 ∼ 2 m/m = mx/mx as k-vector spaces, so m/m = mx/mx + J(X). As a result, we get 2 dimk m/m + rank Jx = n. Since k is algebraically closed, dim X = dim k[X] by Propo-

sition 8.1.15, so dim OX,x = dim k[X]mx = ht(mx) by Lemma 2.3.6(a). Since X is aﬃne, k[X] is both a ﬁnitely generated k-algebra and an integral domain (Lemma 8.1.14), and ∼ k[X]/mx = k as k-algebras, so Lemma 2.6.5 and Theorem 2.6.7 imply that ht(mx) = d. Putting everything together, we get

2 OX,x is regular ⇐⇒ dimk m/m = d ⇐⇒ rank Jx = n − d ⇐⇒ X is smooth at x.

One can define analogs of the sheaf of regular functions on an arbitrary variety X (including affine, projective, quasi-affine and quasi-projective). The notion of smoothness generalize to these cases:

Deﬁnition. An arbitrary variety X is nonsingular at a point x ∈ X if the stalk OX,x is a regular ring. We say X is a nonsingular variety, or a smooth variety, if OX,x is regular for all x ∈ X.

154 8.6 Nonsingular Varieties 8 Algebraic Geometry

One of the most important types of varieties is a hypersurface. As in ordinary geometry, hypersurfaces are those objects with codimension 1.

Deﬁnition. A subset Y ⊆ An is a hypersurface if Y = Z(f) for some irreducible, non- constant polynomial f ∈ A.

n Notice that if Y = Z(f) is a hypersurface in A , with k[Y ] = k[t1, . . . , tn]/J(f), Propo- sition 2.6.6 implies that ht(f) = 1 and dim Y = dim k[Y ] = dim k[t1, . . . , tn] − 1 = n − 1. So indeed hypersurfaces are the varieties with codimension 1. The next proposition illustrates why hypersurfaces are ubiquitous in modern algebraic geometry. First, we need:

Lemma 8.6.3. If A is a UFD with field of fractions K and f ∈ A[t] is an irreducible polynomial, then K[t]/(f) ∼= Frac(A[t]/(f)). Proposition 8.6.4. Let X be a d-dimensional affine variety. Then X is birationally equivalent to a hypersurface in Ad+1. Proof. Let K = k(X) be the fractional field of X. Since k[X]/k is a finitely generated algebra, K/k is a finitely generated field extension. Moreover, assuming k is algebraically closed and therefore perfect, Proposition 1.4.9 says that K/k can be written as a tower K ⊃ k(t1, . . . , td) ⊃ k such that K/k(t1, . . . , td) is finite and separable (using that tr degk K = dim X = d from Corollary 8.1.16) and k(t1, . . . , td) is purely transcendental. By the primitive element theorem, we may write K = k(t1, . . . , td)(g) for some g ∈ K. In particular, there exists an irreducible polynomial f ∈ k(t1, . . . , td)[t] such that f(g) = 0 and ∼ K = k(t1, . . . , td)[t]/(f). Clearing denominators (e.g. using Gauss’s lemma), we get an irre- ˜ ˜ d+1 ducible polynomial f = λf ∈ k[t1, . . . , td][t] for some λ ∈ k[t1, . . . , td]. Let Y = Z(f) ⊆ A . ˜ Then k[Y ] = k[t1, . . . , td][t]/(f) so applying Lemma 8.6.3 with A = k[t1, . . . , td] and its field of fractions k(t1, . . . , td), we get ∼ ∼ ˜ k(X) = K = k(t1, . . . , td)[t]/(f) = Frac(k[t1, . . . , td][t]/(f)) = Frac(k[Y ]) = k(Y ).

Hence X and Y are birationally equivalent by deﬁnition.

For a d-dimensional aﬃne variety X ⊆ An, let Sing X denote the set of singular points of X, that is, Sing X = {x ∈ X | rank Jx < n − d}.

(We will prove in a moment that rank Jx ≤ n − d always holds.) The following result characterizes this set of singular points when X is aﬃne.

Theorem 8.6.5. Suppose X ⊆ An is a d-dimensional aﬃne variety. Then Sing X is a proper closed subset of X.

Proof. We ﬁrst prove Sing X is closed. In the proof of Theorem 8.6.2, we saw that for any 2 x ∈ X, rank Jx = n − dimk m/m where m is the maximal ideal of the local ring OX,x. By 2 Example 6.2.7, dimk m/m ≥ dim OX,x = d and so rank Jx ≤ n − d (justifying the above description of Sing X). By linear algebra,

rank Jx = max{m ≥ 1 | there is some m × m minor C of Jx with det C 6= 0}.

155 8.7 Abstract Curves 8 Algebraic Geometry

∂fi Let J(X) = (f1, . . . , fr) and consider the operator J = . Then any determinant of a ∂tj square minor of J is an element of A = k[t1, . . . , td]. Thus the set

D = {det C | C is an (n − d) × (n − d) minor of J}

is an ideal of A, and Sing X = Z(D) ∩ X. This shows Sing X is closed in X. It remains to show Sing X 6= X. By Proposition 8.6.4, X is birationally equivalent to d+1 a hypersurface Y ⊆ A , say Y = Z(f) for an irreducible polynomial f ∈ k[t1, . . . , td+1]. Then ∂f Sing Y = y ∈ Y : (y) = 0 for all 1 ≤ i ≤ d + 1 . ∂ti For each i, we have that deg ∂f ≤ deg f − 1. If Sing Y = Y , we would have ∂f ∈ J(Y ) = (f) ∂ti ∂ti for all 1 ≤ i ≤ d + 1, and thus ∂f = 0 for all i. If char k = 0, this implies f is a constant, ∂ti ∂f which is a contradiction. In the case that char k = p > 0, ∂t = 0 for all i implies that p i f is a polynomial in the terms ti . Taking pth roots of the coeﬃcients, which is possible p since k is algebraically closed, we get that f = g for some g ∈ k[t1, . . . , td+1], contradicting the irreducibility of f. Therefore Sing Y is a proper subset of Y . Now since X and Y are birationally equivalent, there exist nonzero elements g ∈ k[X] and h ∈ k[Y ] such that ∼ Xg = Yh, by Proposition 8.3.3. Suppose Sing Yh = Yh. Then Sing Yh = Yh = Yh but this implies Y ⊆ Sing Y , a contradiction. Thus Sing Yh is a proper subset of Yh. Now by Proposition 8.3.3, we get ∼ ∼ ∼ ∼ ∼ Xg = Yh =⇒ k[Xg] = k[Yh] =⇒ OX,x = OXg,x = OYh,y = OY,y for all x ∈ Xg and y ∈ Yg. Thus Sing Yh ( Yh implies Sing Xg ( Xg, so it follows that Sing X ( X. Corollary 8.6.6. Let X be any variety. Then Sing X is a proper closed subset of X.

Proof. If X is quasi-affine, with X ⊆ Y ⊆ An, then X may be covered by finitely many principal open subsets of Y , and Sing X is the intersection with the sets of singularities of each of these. Apply Theorem 8.6.5 to get that Sing X is closed and Sing X ( X. If X is projective, Corollary 8.5.7 says that X has a covering by finitely many affine varieties, so Theorem 8.6.5 gives the result. Finally, if X is quasi-projective, then X has a finite covering by quasi-affine varieties, so apply the above.

8.7 Abstract Curves

Assume k is algebraically closed.

Definition. An affine curve is an affine variety X over k of dimension 1.

By Corollary 8.1.16(a), an affine curve is a variety whose function field k(X) has transcendence degree 1 over k. If X is a nonsingular affine curve, as in Section 8.6, for each P ∈ X, the ring OP := OX,P = k[X]mP ⊆ k(X) is a regular local ring (Theorem 8.6.2). In particular,

156 8.7 Abstract Curves 8 Algebraic Geometry

T OP is a DVR for each P ∈ X. Also, for any open subset U ⊆ X, let OX (U) = P ∈U OP be the corresponding ring of k-valued functions on U. We seek to reproduce this notion of an algebraic curve in a more general setting. Let K

be a finitely generated field extension of k with tr degk K = 1, that is, K is a function field of degree 1 over k. Suppose v : K× → Z is a discrete valuation, with discrete valuation ring R = O(v) = {x ∈ K | v(x) ≥ 0} and maximal ideal mv = {x ∈ K | v(x) > 0}. We say R is × a DVR over k if v(k ) = {0}. Let CK be the set of all DVRs over k contained in K. Lemma 8.7.1. Let Y be a quasi-projective variety and P,Q ∈ Y be two points. Then OP ⊆ OQ implies P = Q.

Proof. In general, OP only depends on open sets around P so we may replace Y with its Zariski-closure Y and assume Y is projective. Further, supposing P and Q don’t lie in a coordinate hyperplane, we may in fact assume Y is aﬃne. Then OP = k[Y ]mP and

OQ = k[Y ]mQ , so OP ⊆ OQ implies that mQ ⊆ mP . Since these are maximal ideals, we get mQ = mP , but then Corollary 2.4.3 gives us P = Q.

Lemma 8.7.2. If K is a function ﬁeld of degree 1 over k and x ∈ K, then {R ∈ CK | x 6∈ R} is a ﬁnite set.

1 Proof. For any DVR R over k, x 6∈ R if and only if x ∈ mR so it suffices to show that for × × any nonzero y ∈ K , the set {R ∈ CK | y ∈ mR} is finite. If y ∈ k , then y 6∈ mR for any R ∈ CK since mR consists of the elements of K with positive valuation. Now assume y 6∈ k. Then k[y] ⊆ K, k[y] is isomorphic as a k-algebra to some polynomial ring over k, and by Lemma 1.4.3(b), K/k(y) is a finitely generated, algebraic extension of fields. Let B be the integral closure of k[y] in K. Then k[y] is a PID, and in particular a Dedekind domain, so by Theorem 4.5.3, B is also a Dedekind domain. In particular, B is a finitely generated k-algebra with field of fractions K. Suppose y ∈ R for some DVR R ∈ CK . Then k[y] ⊆ R implies B ⊆ R since R is integrally closed in K. Let n = mR ∩ B; we claim this is a maximal ideal in B. To see this, note that since n = mR ∩ B, this is a prime ideal, and since B is Dedekind, it suffices to prove n 6= 0. Let v be the valuation for R. If n = 0, then v(B r {0}) = {0} implying v(K×) = {0}, contradicting the fact that R is a discrete valuation ring. So n must in fact × be maximal in B. Now B ⊆ R and B r n ⊆ R so localizing at n gives us Bn ⊆ R. Since B is Dedekind and n is nonzero, Bn must in fact be a DVR by Proposition 4.2.2. 0 We proceed to show that Bn = R. To do this, let v be the valuation for Bn and let −1 0 S = B r n, i.e. so Bn = S B. If x ∈ R r Bn then we would have v (x) < 0, and 0 1 1 −1 1 thus v x > 0, so x ∈ S n ⊆ mR. Thus v x > 0 which is equivalent to v(x) < 0, −1 contradicting x ∈ R. Hence Bn = R. Now suppose y ∈ mR = S n. Since B is a finitely generated k-algebra, it is the coordinate ring for some affine variety Y over k. Moreover, dim B = 1 because it is a Dedekind domain, so dim Y = 1 and the localizations of B at all maximal ideals are all DVRs (Proposition 4.2.2), so it follows from Theorem 8.6.2 that Y is nonsingular. Then y ∈ k[Y ] = B so y may viewed as a regular function on Y . In particular, y −1 y = 1 ∈ S n = mR is equivalent to y vanishing at the point in affine space corresponding to mR by Corollary 2.4.3. However, Z(y) is a nonempty, proper subset of Y and dim Y = 1, so by Corollary 8.1.16(b), Z(y) must be a finite set of points. This says that y vanishes at

157 8.7 Abstract Curves 8 Algebraic Geometry

only finitely many points in affine k-space, so by the above, y ∈ mR for only finitely many R ∈ CK . Corollary 8.7.3. Any DVR of the field extension K/k is the local ring of a variety Y at some point P ∈ Y .

Proof. Let Y and y be as in the proof of Lemma 8.7.2. ∼ Corollary 8.7.4. If R ∈ CK , then R/mR = k.

Proof. By Corollary 8.7.3, R = OP = k[Y ]mP for some point P ∈ K. Further, Lemma 1.3.11 −1 ∼ gives us k[Y ]mP /S mP = k[Y ]/mP = k, where S = k[Y ] r mP .

Let K/k be a function field of degree 1 and as above, let CK be the set of all DVRs of K/k. For any affine variety X over k with k(X) = K, dim X = 1 and in particular X is infinite (as a set). By Corollary 8.7.3, CK contains the local ring OP for each point P ∈ X, but by Lemma 8.7.1, these OP are all distinct, so it follows that CK itself is an infinite set. Then the finite sets of CK along with CK itself define the closed sets of a topology on CK . A ‘point’ in CK is a discrete valuation ring RP ; we will conflate the notation P and RP . T For an open set U ⊆ CK , define its sheaf of regular functions O(U) = P ∈U RP , where an element f ∈ O(U) is represented by a regular function f : U → k such that f(P ) is the residue of f mod mP . Notice that if f, g ∈ O(U) define the same function, then f − g ∈ mR for infinitely many R ∈ CK . By Lemma 8.7.2, f = g so elements of O(U) are well-defined.

Definition. An abstract nonsingular curve over a field k is an open subset U ⊆ CK for some function field K/k of degree 1, with the induced topology and sheaf of regular functions as described above.

Deﬁnition. A morphism of nonsingular curves is a continuous map ϕ : X → Y , where X and Y are nonsingular curves, such that for all open subsets V ⊆ Y and all regular functions f : V → k, f ◦ ϕ : ϕ−1(V ) → k is regular.

The next result shows that abstract nonsingular curves properly generalize nonsingular quasi-projective curves as deﬁned in Section 8.6.

Theorem 8.7.5. If X is a nonsingular quasi-projective curve over k then X is isomorphic to an abstract nonsingular curve.

Proof. Let K = k(X) and recall that Corollary 8.1.16(a) implies tr degk K = 1. Define a map ϕ : X → CK that sends P 7→ OP . If we set U = ϕ(X), changing targets gives us a surjection ϕ : X → U. Moreover, Lemma 8.7.1 says that ϕ is injective. If we show U is an T ∼ T open set in CK , we will be done since O(U) = P ∈U OP = P ∈ϕ(U) RP = O(ϕ(U)) shows that ϕ is an isomorphism. Now to show U is an open set, it is enough to show U contains an open set of CK since open sets are by definition the complements of finite sets. By Corollary 8.5.7, X may be covered by affine varieties X1,...,X`. Then ϕ(X1), . . . , ϕ(X`) cover U = ϕ(X) so it suffices to show each ϕ(Xj) is open in CK . Thus we may assume X = Xj is affine. We claim that U = {R ∈ CK | k[X] ⊆ R}. Note that the statement k[X] ⊆ R is logical because K = k(X) and so k[X] ⊆ K. On one hand, if R ∈ U is a DVR over k, then

158 8.8 The Spectrum of a Ring 8 Algebraic Geometry

R = OP for some point P ∈ X, and in this case k[X] ⊆ R. On the other hand, suppose k[X] ⊆ R for R ∈ CK . Then k[X] is a Dedekind domain and R is a DVR, so as in the proof of Lemma 8.7.2, n = mR ∩ k[X] is a maximal ideal of k[X] and k[X]n = R. Thus R is the localization of k[X] at a maximal ideal corresponding to a point in X, i.e. R ∈ ϕ(X). Hence U has the desired description. Now use the fact that k[X] is a ﬁnitely generated k-algebra to write k[X] = k[x1, . . . , xn] for elements x1, . . . , xn ∈ k[X]. Then for any R ∈ CK , k[X] ⊆ R if and only if xi ∈ R for Tn each 1 ≤ i ≤ n. Thus U = i=1 Ui, where Ui = {R ∈ CK | xi ∈ R}. By Lemma 8.7.2, each Ui is the complement of a ﬁnite set in CK , so U is open as required. Theorem 8.7.6 (Extension). If X is an abstract nonsingular curve, P ∈ X is any point and ϕ : X r {P } → Y is a morphism into a projective variety Y , then there exists a unique morphism Φ: X → Y extending ϕ to all of X.

Proof. (Sketch) Let Y ⊆ Pn for n ≥ 1. If the map ψ : X r {P } → Pn (obtained from ϕ by changing targets) can be extended to all of X, then ψ−1(Y ) is a closed set containing Xr{P }. In particular, X r {P } is nonempty and open, so this would imply ψ−1(Y ) = X. Thus n ϕ(X) = ψ(X) ⊆ Y so we may reduce to the case where Y = P . Let U0,...,Un be the aﬃne n Tn coordinate charts of P as in Corollary 8.5.7. Set U = i=0 Ui. If ϕ(X r {P }) ∩ U = ∅, we Sn n would have ϕ(Xr{P }) ⊆ i=0 Hi, where Hi = P rUi is the ith projective hyperplane. Note ∼ n−1 that X r{P } is irreducible, so ϕ(X r{P }) is irreducible and thus ϕ(X r{P }) ⊆ Hi = P for some 0 ≤ i ≤ n. Restricting the target of ϕ to projective n − 1 space, we may assume by induction that ϕ(X r {P }) ∩ U 6= ∅. xi xi −1 On U, each for 0 ≤ i, j ≤ n is a regular function. Let fij = ◦ ϕ : ϕ (U) → k for xj xj 0 ≤ i, j ≤ n; each fij is also a regular function, so view fij ∈ K. Then we can choose ` ≥ 1 n such that Φ(P ) = [f0`(P ), . . . , fn`(P )]. This deﬁnes an extension Φ : X → P ; one may now check that Φ is a morphism. One can prove:

Theorem 8.7.7. If K is a function ﬁeld over k of degree 1 then CK is isomorphic to a nonsingular projective curve over k.

Corollary 8.7.8. Every abstract nonsingular curve is isomorphic to some quasi-projective nonsingular curve over a ﬁeld k.

Thus the abstract theory of curves presented in this section captures the more concrete notions of aﬃne, projective and quasi-aﬃne/projective varieties detailed in earlier sections.

8.8 The Spectrum of a Ring

In this section we discuss another generalization of aﬃne varieties in the form of the prime spectrum Spec(A) for a ring A. This in fact forms the groundwork for scheme theory, which in many ways is the main language of modern algebraic geometry. Let Spec(A) denote the set of prime ideals of A and let MaxSpec(A) denote the set of maximal ideals. As we have seen with Corollary 2.4.3 (the algebraic Nullstellensatz), when A is the coordinate ring of some aﬃne

159 8.8 The Spectrum of a Ring 8 Algebraic Geometry

variety X ⊆ An, the points of An correspond bijectively with elements (ideals) of MaxSpec(A) via P ↔ mP . This correspondence transfers the Zariski topology of X to A. Further, if ϕ : X → Y is a morphism of varieties, with A = k[X] and B = k[Y ], then identiﬁcations X ↔ MaxSpec(A) and Y ↔ MaxSpec(B) induce a map ϕ∗ : MaxSpec(A) → MaxSpec(B), ∗ −1 ∗ given by ϕ∗(m) = (ϕ ) (m), where ϕ : B → A is the induced comorphism. We can mimic this in a general setting; however, it is not true in general that if ϕ : A → B is a ring homomorphism then a maximal ideal in B pulls back to a maximal ideal in A. To remedy this, we replace the spectrum of maximal ideals with the prime spectrum, Spec(A) = {p ⊂ A | p is a prime ideal}. For a subset E ⊂ A, let V (E) be the set of prime ideals of A containing E: V (E) = {p ∈ Spec(A) | p ⊇ E}.

Lemma 8.8.1. Let E be a subset of A and set a = (E), the ideal generated by E. Then

(i) If a is the ideal generated by E then V (E) = V (a) = V (r(a)).

(ii) V (0) = X,V (1) = ∅.

(iii) If (Ei)i∈I is any family of subsets of A, then ! [ \ V Ei = V (Ei). i∈I i∈I

(iv) V (a ∩ b) = V (ab) = V (a) ∪ V (b) for any ideals a, b of A.

Proof. As a preliminary remark, if E,F ⊆ Spec(A) such that N ⊆ M, then any prime ideal containing M also contains N. Thus V (M) ⊆ V (N). (i) E ⊆ a so by the preliminary remark, V (E) ⊇ V (a). On the other hand, if p is a prime ideal containing E then p ⊇ a since by deﬁnition a is the smallest ideal containing E. Thus p ∈ V (a) so we have V (E) = V (a). For the second inequality, Corollary 1.1.3 says that r(a) = T{P primes | P ⊇ a}, which directly implies V (a) = V (r(a)). (ii) Every prime ideal p ⊂ A contains 0, so V (0) = X. On the other hand, if p is a prime ideal containing 1 then p = A, which is impossible. Therefore V (1) = ∅. (iii) For each Ei, let ai = (Ei) be the ideal generated by Ei, and let a be the ideal generated S S by Ei. Then by (i), V Ei = V (a) and V (Ei) = V (ai) for each i. Clearly ai ⊆ a i∈I i∈I T for each i, so V (a) ⊆ V (ai), and since this holds for all i, V (a) ⊆ V (ai). On the other T S i∈I hand, if p ∈ V (ai) then p ⊇ ai ⊇ Ei for each i, so p ⊇ Ei. By deﬁnition of a, this i∈I T i∈I implies p ⊇ a, so p ∈ V (a). Hence i∈I V (ai) ⊆ V (a) so we have equality. (iv) By Lemma 1.1.4, r(a)r(b) ⊆ r(ab) so (i) gives us

V (ab) = V (r(ab)) ⊆ V (r(a)r(b)) ⊆ V (r(a)) ∪ V (r(b)) = V (a) ∪ V (b).

Next, a ⊇ a ∩ b and b ⊇ a ∩ b so by the preliminary remark, V (a) ⊆ V (a ∩ b) and V (b) ⊆ V (a ∩ b). Thus V (a) ∪ V (b) ⊆ V (a ∩ b). Finally, ab ⊆ a ∩ b implies V (a ∩ b) ⊆ V (ab). So we have V (ab) ⊆ V (a ∪ V (a) ∪ V (b) ⊆ V (a ∩ b) ⊆ V (ab) and hence equality all the way through.

160 8.8 The Spectrum of a Ring 8 Algebraic Geometry

The above properties show that the sets V (E) for E ⊆ A form the closed sets of a topology on Spec(A), called the Zariski topology on Spec(A). Definition. For a ring A, the space Spec(A) equipped with the Zariski topology is called an affine scheme. Lemma 8.8.2. Suppose f : A → B is a ring homomorphism and q is a prime ideal of B. Then f −1(q) is a prime ideal of A. Proof. Suppose x, y ∈ A such that xy ∈ f −1(q) but x 6∈ f −1(q). Then f(x)f(y) = f(xy) ∈ f(f −1(q)) = q but f(x) 6∈ q, so since q is prime, f(y) ∈ q. This implies y ∈ f −1(q) so we have that f −1(q) is prime. Proposition 8.8.3. Let f : A → B be a ring homomorphism. Then (a) The induced map f ∗ : Spec(B) → Spec(A), q → f −1(q), is continuous with respect to the Zariski topology. (b) If f is surjective, f ∗ is a closed map. (c) More generally, if B/f(A) is an integral ring extension, f ∗ is a closed map. Proof. (a) By Lemma 8.8.1(i), it suffices to show that for any ideal I ⊂ A,(f ∗)−1(V (I)) = V (f(I)). Note that for a prime p ∈ Spec(B), p ∈ (f ∗)−1(V (I)) ⇐⇒ f ∗(p) = f −1(p) ∈ V (I) ⇐⇒ f −1(p) is a prime containing I, by (a) ⇐⇒ p ⊇ f(I) ⇐⇒ p ∈ V (f(I)). Therefore (f ∗)−1(V (I)) = V (f(I)) so closed sets pull back to closed sets. This implies f ∗ is continuous. (b) Let J ⊂ B be any ideal. We claim that f ∗(V (J)) = V (f −1(J)). By Lemma 8.8.1(i), this is enough to conclude that f ∗ is closed. Set I = f −1(J) ⊂ A. Since f is surjective, the map B/J → A/I is also surjective. Then V (J) = Spec(B/J) and V (I) = Spec(A/I) so replacing A with A/I and B with B/J, it suffices to show that f ∗(Spec(B)) = Spec(A). If p ∈ f ∗(Spec(B)) then p = f −1(Q) for some prime ideal q ⊂ B. By Lemma 8.8.2, p ∈ Spec(A). The other inclusion follows from surjectivity of f, so we have f ∗(Spec(B)) = Spec(A) and f ∗ is a closed map. (c) Viewing f as the composition f : A f(A) ,→ B, (b) shows that the surjection is always closed so it suffices to consider the case when A ⊆ B is a subring and f : A,→ B is the inclusion map. Let J ⊂ B be an ideal, so that V (J) is closed in Spec(B). Set I = J ∩ A. We claim that f ∗(V (J)) = V (I). Consider the inclusion A/I ,→ B/J. By Lemma 2.3.1(a), B/J is integral over A/I, so replacing B with B/J and A with A/I, it is enough to show that f ∗(Spec(B)) is closed. If p ∈ f ∗(Spec(B)) then p = f −1(Q) = Q ∩ A for some prime ideal Q ⊂ B, but Q ∩ A is prime in A, so this means p ∈ Spec(A). On the other hand, if p ∈ Spec(A) then by Lemma 2.3.4, there is a prime ideal Q ⊂ B such that Q ∩ A = p. In particular p = f −1(Q) so p ∈ f ∗(Spec(B)). This shows f ∗(Spec(B)) = Spec(A) so f is a closed map.

161 8.8 The Spectrum of a Ring 8 Algebraic Geometry

The operator V takes sets (and in particular, ideals) of A to closed sets in Spec(A) and reverses inclusions. We deﬁne an operator going in the opposite direction: for a subset T Y ⊆ Spec(A), let J(Y ) = p∈Y p. Then J(Y ) is an ideal of A and clearly the operator J is inclusion-reversing.

Proposition 8.8.4. Let A be a ring, I ⊂ A an ideal and Y ⊆ Spec(A) a subset. Then

(a) V (J(Y )) = Y , the Zariski closure of Y in Spec(A).

(b) J(V (I)) = r(I).

Proof. (a) On one hand,

V (J(Y )) = {p ∈ Spec(A) | p ⊇ J(Y )} ( ) \ = p ∈ Spec(A) | p ⊇ q q∈Y ⊇ {p ∈ Spec(A) | p ∈ Y } = Y so V (J(Y )) ⊇ Y . Since V (J(Y )) is a closed set in Spec(A), we have V (J(Y )) ⊇ Y . Con- versely, we may write Y = V (E) for some subset E ⊂ A. If E 6⊂ J(Y ) then Y = V (E) 6⊃ V (J(Y )) ⊇ Y , a contradiction. Hence E ⊆ J(Y ), implying Y = V (E) ⊇ V (J(Y )). This gives us the desired equality. T T (b) By deﬁnition, J(V (I)) = p∈V (I) p = p⊇I p where the latter intersection is over all prime ideals p containing I. Thus by Corollary 1.1.3, J(V (I)) = r(I). (c) ( =⇒ ) Suppose V (I) is irreducible and let f, g ∈ A with fg ∈ J(V (I)) = r(I). Then by (a), we have V (I) ⊆ V (J(V (I))) ⊆ V (fg) but by Lemma 8.8.1(iv), V (fg) = V ((f)(g)) = V ((f)) ∪ V ((g)) = V (f) ∪ V (g), so in particular we can write V (I) = (V (f) ∩ V (I)) ∪ (V (g) ∩ V (I)). Since V (I) is assumed to be irreducible, it follows that V (I) = V (f) ∩ V (I) or V (I) = V (g) ∩ V (I). In other words, V (I) ⊆ V (f) or V (I) ⊆ V (g), so f ∈ r(I) or g ∈ r(I). Hence r(I) is prime. ( ⇒ = ) Conversely, assume r(I) is prime and V (I) = Y1 ∪Y2 for closed sets Y1,Y2 ⊆ V (I). Then there exist closed sets X1,X2 ⊆ Spec(A) with Y1 = X1 ∩ V (I) and Y2 = X2 ∩ V (I). For some ideals a1, a2 ⊂ A, we have X1 = V (a1) and X2 = V (a2), so V (I) ⊆ X1 ∪ X2 = V (a1) ∪ V (a2) = V (a1a2) by Lemma 8.8.1(iv). Applying J, we get r(I) ⊇ J(V (a1a2)) = r(a1a2) ⊇ a1a2. Since r(I) is prime, a1 ⊆ r(I) or a2 ⊆ r(I); without loss of generality, assume a1 ⊆ r(I). Then by (a), V (I) ⊆ V (r(I)) ⊆ V (a1) = X1. This implies Y1 = X1∩V (I) = V (I), so V (I) is irreducible.

Proposition 8.8.5. Let A be a noetherian ring and X = Spec(A) its prime spectrum. Then

(a) X is noetherian.

(b) dim X = dim A.

162 8.8 The Spectrum of a Ring 8 Algebraic Geometry

Proof. (a) Suppose X1 ⊇ X2 ⊇ X3 ⊇ · · · is a descending chain of Zariski-closed subsets of X. We may assume each Xj = V (aj) for ideals aj ⊂ A. Applying J, we get an ascending chain of radical ideals in A:

r(a1) ⊆ r(a2) ⊆ r(a3) ⊆ · · ·

Since A is noetherian, there is an n ∈ N such that r(an) = r(an+1) = ··· . Then since each Xj was taken to be closed, Proposition 8.8.4(a) implies V (J(Xj)) = V (r(aj)) = Xj and so we get

V (r(a1)) ⊇ V (r(a2)) ⊇ V (r(a3)) ⊇ · · · ⊇ V (r(an)) = V (r(an+1)) = ···

=⇒ X1 ⊇ X2 ⊇ X3 ⊇ · · · ⊇ Xn = Xn+1 = ···

Thus every descending chain of closed subsets stabilizes, meaning X is noetherian. (b) Suppose Y0 ) Y1 ) ··· ) Y` is a chain of closed, irreducible subsets of X. Write Yk = V (Ik) for ideals Ik ⊂ A. By Proposition 8.8.4(c), each J(Yk) = r(Ik) is prime, so we get a proper chain r(I0) ( r(I1) ( ··· ( r(I`)

of prime ideals in A. (The inclusions must be strict, else r(Ij) = r(Ij+1) for some 0 ≤ j ≤ ` − 1, in which case we would have Yj = V (Ij) = V (r(Ij)) = V (r(Ij+1)) = V (Ij+1) = Yj+1 by Lemma 8.8.1(i).) By deﬁnition, this means ` ≤ dim A, so taking the supremum over all such chains of closed, irreducible subsets in X, we obtain dim X ≤ dim A. Conversely, if p0 ( p1 ( ··· ( pm is a strict chain of prime ideals in A, Xk = V (pk) is irreducible for each 0 ≤ k ≤ m, by Proposition 8.8.4(c). This determines a chain of closed, irreducible subsets in X: X0 ) X1 ) ··· ) Xm. (Again, if Xj = Xj+1 for any 0 ≤ j ≤ m − 1, then pj = r(pj) = J(V (pj)) = J(Xj) = J(Xj+1) = J(V (pj+1)) = r(pj+1) = pj+1 by Proposition 8.8.4(b), a contradiction.) Therefore m ≤ dim X, so taking the supremum over all such chains of primes in A gives us dim A ≤ dim X.

Example 8.8.6. The converse of Proposition 8.8.5(a) is not true: if Spec(A) is a noetherian, A need not be noetherian. For an example, consider the polynomial ring k[tn | n ∈ N] in 2 countably many variables. We have an ideal I = (tn | n ∈ N) in this ring, so one can take the quotient to get A = k[tn | n ∈ N]/I. The ideal J = (tn | n ∈ N) strictly contains I, so J = J/I is an ideal of A which is clearly not finitely generated. So A is not noetherian. ∼ However, k[tn | n ∈ N]/J = k is a field, so J is maximal and hence J is maximal in A. On the other hand, since every generator of J is nilpotent, every element of J is nilpotent, so J must also be a minimal prime ideal. This implies J is the unique prime ideal of A, so Spec(A) is finite and in particular a noetherian space.

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