Algebraic Geometry Fall 2013 & Spring 2014

Eduard Looijenga Rings are always supposed to be commutative and to possess a unit 1. A ring homomorphism will always take unit to unit. We allow that 1 = 0, but in that case we get of course the zero ring {0}. If R is a ring, then we denote the multiplicative group of invertible elements (units) of R by R×. We say that R is a domain if R has no zero divisors, equivalently, if (0) is a prime ideal. An R-algebra is a ring A endowed with a ring homomorphism φ : R → A, but if φ is understood, then for every r ∈ R and a ∈ A, the product φ(r)a is often denoted by ra. We say that A is finitely generated as an R-algebra if we can find a1, . . . , an in A such that every element of A can be written as a polynomial in these elements with coefficients in R; in other words, the R-algebra homomorphism R[x1, . . . , xn] → A which sends the variable xi to ai is onto. This is not to be confused with the notion of finite generation of an R-module M resp. which merely means the existence of a surjective homomorphism of R-modules Rn → M. Similarly, a field K is said to be finitely generated as a field over a subfield k if we can find elements b1, . . . , bn in K such that every element of K can be written as a fraction of two polynomials in these elements with coefficients in k. Contents

Chapter 1. Affine varieties5 1. The Zariski topology5 2. Irreducibility and decomposition8 3. Finiteness properties and the Hilbert theorems 14 4. The affine category 18 5. The sheaf of regular functions 23 6. The product 25 7. Function fields and rational maps 27 8. Finite morphisms 31 9. Dimension 36 10. Nonsingular points 39 11. The notion of a variety 47

Chapter 2. Projective varieties 51 1. Projective spaces 51 2. The Zariski topology on a projective space 53 3. The Segre embeddings 56 4. Projections 57 5. Elimination theory and closed projections 58 6. Constructible sets 62 7. The Veronese embedding 63 8. Grassmannians 65 9. Multiplicities of modules 70 10. Hilbert functions and Hilbert polynomials 74

Chapter 3. Schemes 79 1. Presheaves and sheaves 79 2. The spectrum of a ring as a locally ringed space 84 3. The notion of a scheme 88 4. Formation of products 94 5. Elementary properties of schemes 95 6. Separated and proper morphisms 99 7. Quasi-coherent sheaves 109 8. When is a scheme afﬁne? 113 9. Divisors and invertible sheaves 116 10. The projective setting 124 11. Cohomology of coherent modules on a projective scheme 127 12. Ample and very ample sheaves 131 13. Blowing up 134 14. Flatness 138

3 4 CONTENTS

15. Serre duality 141

Chapter 4. Appendix 145 1. The basics of category theory 145 2. Abelian categories and derived functors 151 3. Sheaf cohomology 160

Bibliography 173 CHAPTER 1

Afﬁne varieties

Throughout these notes k stands for an algebraically closed field k. Recall that this means that every polynomial f ∈ k[x] of positive degree has a root x1 ∈ k: f(x1) = 0. This implies that we can split off the factor x − x1 from f with quotient a polynomial of degree one less than f. Continuing in this manner we then find × that f decomposes simply as f(x) = c(x − x1) ··· (x − xd) with c ∈ k = k − {0}, d = deg(f) and x1, . . . , xd ∈ k. Since an algebraic extension of k is obtained by adjoining to k roots of polynomials in k[x], this also shows that the property in question is equivalent to: every algebraic extension of k is equal to k. A first example you may think of is the field of complex numbers C, but as we proceed you should be increasingly be aware of the fact that there are many others: it is shown in a standard algebra course that for any field F an algebraic closure1 F¯ is obtained by adjoining to F the roots of every polynomial f ∈ F [x]. So we could take for k an algebraic closure of the field of rational numbers Q, of the finite field 2 Fq, where q is a prime power or even of the field of fractions of any domain such as C[x1, . . . , xr]. 1. The Zariski topology n Any f ∈ k[x1, . . . , xn] determines in an evident manner a function k → k. In such cases we prefer to think of kn not as vector space—its origin and vector addition will be irrelevant to us—but as a set with a weaker structure. We shall make this precise later, but it basically amounts to only remembering that elements of k[x1, . . . , xn] can be understood as k-valued functions on it. For that reason it n n is convenient to denote this set differently, namely as A (or as Ak , if we feel that we should not forget about the field k). We refer to An as the affine n-space over k.A k-valued function on An is then said to be regular if it is defined by some f ∈ k[x1, . . . , xn]. We denote the zero set of such a function by Z(f) and the n n complement (the nonzero set) by Af ⊂ A . If f is not a constant polynomial (that is, f∈ / k), then we call Z(f) a hypersurface of An.

EXERCISE 1. Prove that f ∈ k[x1, . . . , xn] is completely determined by the regular function it defines. (Hint: do first the case n = 1.) So the ring k[x1, . . . , xn] can be regarded as a ring of functions on An under pointwise addition and multiplication. (This would fail to be so had we not assumed that k is algebraically closed: q for instance the function on the finite field Fq defined by x − x is identically zero.)

1This can not be done in one step and involves an infinite process which involves in general many choices. This is reflected by the fact that the final result is not canonical, although it is unique up to a (in general nonunique) isomorphism; whence the use of the indefinite article in ‘an algebraic closure’. 2 (qn) Since the elements of any algebraic extension of Fq of degree n ≥ 2 are roots of x − x, we only need to adjoin roots of such polynomials.

5 6 1. AFFINE VARIETIES

EXERCISE 2. Prove that a hypersurface is nonempty. It is perhaps somewhat surprising that in this rather algebraic context, the language of topology proves to be quite effective: algebraic subsets of An shall appear as the closed sets of a topology, albeit a rather peculiar one. n LEMMA-DEFINITION 1.1. The collection {Af : f ∈ k[x1, . . . , xn]} is a basis of a 3 topology on An, called the Zariski topology . A subset of An is closed for this topology if and only if it is an intersection of zero sets of regular functions.

PROOF. We recall that a collection {Uα}α of subsets of a set X is a basis for a topology if and only if its union is all of X and any intersection Uα1 ∩ Uα2 is a union of members of {Uα}α. This is here certainly the case, for we have U(0) = X n and U(f1) ∩ U(f2) = U(f1f2). Since an open subset of A is by definition a union n of subsets of the form Af , a closed subset must be an intersection of subsets of the form Z(f). EXAMPLE 1.2. The Zariski topology on A1 is the cofinite topology: its closed subsets 6= A1 are the finite subsets. EXERCISE 3. Show that the diagonal in A2 is closed for the Zariski topology, but not for the product topology (where each factor A1 is equipped with the Zariski topology). So A2 does not have the product topology. We will explore the mutual relationship between the following two basic maps:

n I {subsets of A } −−−−→ {ideals of k[x1, . . . , xn]} ∪ ∩

n Z {closed subsets of A } ←−−−− {subsets of k[x1, . . . , xn]}. n where for a subset X ⊂ A , I(X) is the ideal of f ∈ k[x1, . . . , xn] with f|X = 0 and n for a subset J ⊂ k[x1, . . . , xn], Z(J) is the closed subset of A deﬁned by ∩f∈J Z(f). Observe that

I(X1 ∪ X2) = I(X1) ∩ I(X2) and Z(J1 ∪ J2) = Z(J1) ∩ Z(J2). In particular, both I and Z are inclusion reversing. Furthermore, the restriction of I to closed subsets deﬁnes a section of Z: if Y ⊂ An is closed, then Z(I(Y )) = Y . We also note that by Exercise1 I(An) = (0), and that any singleton {p = n (p1, . . . , pn)} ⊂ A is closed, as it is the common zero set of the degree one polynomials x1 − p1, . . . , xn − pn.

EXERCISE 4. Prove that I({p}) is equal to the ideal generated by these degree one polynomials and that this ideal is maximal.

EXERCISE 5. Prove that the (Zariski) closure of a subset Y of An is equal to Z(I(Y )). n Given Y ⊂ A , then f, g ∈ k[x1, . . . , xn] have the same restriction to Y if and only if f − g ∈ I(Y ). So the quotient ring k[x1, . . . , xn]/I(Y ) (a k-algebra) can be regarded as a ring of k-valued functions on Y . Notice that this k-algebra does not change if we replace Y by its Zariski closure.

3 n We shall later modify the deﬁnition of both A and the Zariski topology. 1. THE ZARISKI TOPOLOGY 7

n DEFINITION 1.3. Let Y ⊂ A be closed. The k-algebra k[x1, . . . , xn]/I(Y ) is called the coordinate ring of Y and we denote it by k[Y ].A k-valued function on Y is said to be regular if it lies in this ring. Given a closed subset Y ⊂ An, then for every subset X ⊂ An we have X ⊂ Y if and only if I(X) ⊃ I(Y ), and in that case IY (X) := I(X)/I(Y ) is an ideal of k[Y ]: it is the ideal of regular functions on Y that vanish on X. Conversely, an ideal of k[Y ] is of the form J/I(Y ), with J an ideal of k[x1, . . . , xn] that contains I(Y ), and such an ideal deﬁnes a closed subset Z(J) contained in Y . So the two basic maps above give rise to such a pair on Y : I {subsets of Y } −−−−→Y {ideals of k[Y ]} ∪ ∩ Z {closed subsets of Y } ←−−−−Y {subsets of k[Y ]}.

We ask: which ideals of k[x1, . . . , xn] are of the form I(Y ) for some Y ? Clearly, if f ∈ k[x1, . . . , xn] is such that some positive power vanishes on Y , then f vanishes on Y . In other words: if f m ∈ I(Y ) for some m > 0, then f ∈ I(Y ). This suggests: PROPOSITION-DEFINITION 1.4. Let R be a ring (as always commutative and with m 1) and let J ⊂ R be an ideal. Then the set of a ∈ R with the property√ that a ∈ J for some m > 0 is an ideal of R, called the√radical of J and denoted J. We say that J is a radical ideal if J = J. We say that the ring R is reduced if the zero ideal (0) is a radical ideal (in other words, R has no nonzero nilpotents: if a ∈ R is such that am = 0, then a = 0). √ √ m n PROOF. We show that J is an ideal. Let a, b ∈ J so that√ a , b ∈ J for m certain positive integers m,√ n. Then for every r ∈ R, ra ∈ J, since (ra) = rmam ∈ J. Similarly a−b ∈ J, for (a−b)m+n is a linear combination of monomials m n that are multiples of a or b and hence lie in J. EXERCISE 6. Show that a prime ideal is a radical ideal. Notice that J is a radical ideal if and only if R/J is reduced. The preceding shows that for every Y ⊂ An, I(Y ) is a radical ideal, so that k[Y ] is reduced. The dictionary between algebra and geometry begins in a more substantial manner with

THEOREM 1.5√ (Hilbert’s Nullstellensatz). For every ideal J ⊂ k[x1, . . . , xn] we have I(Z(J)) = J. We inclusion ⊃ is clear; the hard part is the opposite inclusion (which says that m if f ∈ k[x1, . . . , xn] vanishes on Z(J), then f ∈ J for some positive integer m). We postpone its proof, but discuss some of the consequences. n COROLLARY 1.6. Let Y ⊂ A be closed. Then the maps IY and ZY restrict to bijections: {closed subsets of Y } ↔ {radical ideals of k[Y ]}. These bijections are inclusion reversing and each others inverse. PROOF. We ﬁrst prove this for Y = An. We already observed that for every closed subset X of An we have Z(I(X)) = X . The Nullstellensatz says that for a radical ideal J ⊂ k[x1, . . . , xn], we have I(Z(J)) = J. An ideal of k[Y ] is of the form J/I(Y ). This is a radical ideal if and only if J is one. So the property also follows for an arbitrary closed Y . 8 1. AFFINE VARIETIES

Let m be a maximal ideal of k[x1, . . . , xn]. Such an ideal is certainly radical as it is a prime ideal and so it is also maximal among the radical ideals that are distinct from k[x1, . . . , xn]. Hence it is of the form I(Y ) for a closed subset Y . Since n the empty subset of A is deﬁned by the radical ideal k[x1, . . . , xn], Corollary 1.6 implies that Y will be nonempty and minimal for this property. In other words, Y is a singleton {y} (and if y = (a1, . . . , an), then m = (x1 − a1, . . . , xn − an), by Exercise4). Thus the above correspondence provides a bijection between the points of An and the maximal ideals of An. If we are given a closed subset Y ⊂ An and a point y ∈ An, then y ∈ Y if and only if the maximal ideal deﬁned by y contains I(Y ). But a maximal ideal containing I(Y ) is the preimage of a maximal ideal in k[Y ] = k[x1, . . . , xn]/I(Y ). We conclude that points of Y correspond to maximal ideals of k[Y ]. Via this (or a very similar) correspondence, algebraic geometry seeks to express geometric properties of Y in terms of algebraic properties of k[Y ] and vice versa. In the end we want to forget about the ambient An completely.

2. Irreducibility and decomposition We introduce a property which for most topological spaces is of little interest, but as we will see, is useful and natural for the Zariski topology.

DEFINITION 2.1. Let Y be a topological space. We say that Y is irreducible if it is nonempty and cannot be written as the union of two closed subsets 6= Y . An irreducible component of Y is a maximal irreducible subset of Y .

EXERCISE 7. Prove that an irreducible Hausdorff space must consist of a single point. Prove also that an inﬁnite set with the coﬁnite topology is irreducible.

EXERCISE 8. Let G1,...,Gs be closed subsets of a topological space Y . Prove that any irreducible subset of G1 ∪ · · · ∪ Gs is contained in some Gi.

LEMMA 2.2. Let Y be a topological space. (i) If Y is irreducible, then every nonempty open subset of Y is dense in Y and irreducible. (ii) Conversely, if C ⊂ Y is an irreducible subspace, then C¯ is also irreducible. In particular, an irreducible component of Y is always closed in Y .

PROOF. (i) Suppose Y is irreducible and let U ⊂ Y be open and nonempty. Then Y is the union of the two closed subspaces Y −U and U¯. Since Y is irreducible, and Y − U 6= Y , we must have U¯ = Y . So U is dense in Y . To see that U is irreducible, suppose that U is the union of two subsets that are both closed in U. These subsets will be of the form Gi ∩ U with Gi closed in Y . Then G1 ∪ G2 is a closed subset of Y which contains U. Since U¯ = Y , it follows that G1 ∪ G2 = Y . The irreducibility of Y implies that one of the Gi (say G1) equals Y and then U = G1 ∩ U. (ii) Let C ⊂ Y be irreducible (and hence nonempty). If C¯ is written as a union of two closed subsets G1,G2 of Y , then C is the union of the two subsets G1 ∩ C and G2 ∩ C that are both closed in C, and so one of these, say G1 ∩ C, equals C. ¯ ¯ This means that G1 ⊃ C and hence G1 ⊃ C. So C is irreducible. The following proposition tells us what irreducibility amounts to in the Zariski topology. 2. IRREDUCIBILITY AND DECOMPOSITION 9

PROPOSITION 2.3. A closed subset Y ⊂ An is irreducible if and only if I(Y ) is a prime ideal (which, we recall, is equivalent to: k[Y ] = k[x1, . . . , xn]/I(Y ) is a domain).

PROOF. Suppose Y is irreducible and f, g ∈ k[x1, . . . , xn] are such that fg ∈ I(Y ). Then Y ⊂ Z(fg) = Z(f) ∪ Z(g). Since Y is irreducible, Y is contained in Z(f) or in Z(g. So f ∈ I(Y ) or g ∈ I(Y ), proving that I(Y ) is a prime ideal. Suppose I(Y ) is a prime ideal, but that Y is the union of two closed subsets Y1 and Y2 that are both 6= Y . Since Yi 6= Y , there exists a fi ∈ I(Yi) − I(Y ). Then f1f2 vanishes on Y1 ∪ Y2 = Y , so that f1f2 ∈ I(Y ). The fact I(Y ) is a prime ideal implies that one of f1 and f2 is in I(Y ). Whence a contradiction.

One of our first aims is to prove that the irreducible components of any closed subset Y ⊂ An are finite in number and have Y as their union. This may not sound very surprising, but we will see that this reflects some nonobvious algebraic properties. Let us first consider the case of a hypersurface. Since we are going to use the fact that k[x1, . . . , xn] is a unique factorization domain, we begin with recalling that notion:

DEFINITION 2.4. We say that R is a unique factorization domain if R has no zero divisors and every principal ideal (a) := Ra in R which is neither the zero ideal nor all of R is in unique manner a product of principal prime ideals: (a) = (p1)(p2) ··· (ps) (so the ideals (p1),..., (ps) are unique up to order). Observe that the principal ideal generated by p ∈ R is prime if and only if for any factoring of p, p = p0p00, one of the factors must be a unit and that the identity (a) = (p1)(p2) ··· (ps) amounts to the statement that a is a unit times p1p2 ··· ps. The factors are then unique up to order and multiplication by a unit. For a field (which has no proper principal ideals distinct from (0)) the imposed condition is empty and hence a field is automatically unique factorization domain. A more substantial example (that motivated this notion in the first place) is Z: a principal prime ideal of Z is of the form (p), with p a prime number. Every integer n ≥ 2 has a unique prime decomposition and so Z is a unique factorization domain. A basic theorem in the theory of rings asserts that if R is a unique factorization domain, then so is its polynomial ring R[x]. This implies (with induction on n) that R[x1, . . . , xn] is one. This applies to the case when R is a field (think of our k): a nonzero principal ideal of this ring is prime precisely when it is generated by an irreducible polynomial of positive degree and every f ∈ R[x1, . . . , xn] of positive degree then can be written as a product of irreducible polynomials: f = f1f2 ··· fs, a factorization that is unique up to order and multiplication of each fi by a nonzero element of R. The following proposition connects two notions of irreducibility.

PROPOSITION 2.5. If f ∈ k[x1, . . . , xn] is irreducible, then the hypersurface Z(f) it deﬁnes is irreducible. If f ∈ k[x1, . . . , xn] is of positive degree and f = f1f2 ··· fs is a factoring into irreducible polynomials, then Z(f1),...,Z(fs) are the irreducible components of Z(f) and their union equals Z(f) (but we are not claiming that the Z(fi)’s are pairwise distinct). In particular, a hypersurface is the union of its irreducible components; these irreducible components are hypersurfaces and ﬁnite in number. 10 1. AFFINE VARIETIES

PROOF. If f ∈ k[x1, . . . , xn] is irreducible, then f generates a prime ideal and so Z(f) is an irreducible hypersurface by Proposition 2.3. If f ∈ k[x1, . . . , xn] is of positive degree and f = f1f2 ··· fs is as in the proposition, then it is clear that Z(f) = Z(f1) ∪ · · · ∪ Z(fs) with each Z(fi) irreducible. To see that Z(fi) is an irreducible component of Z(f), suppose that C ⊂ Z(f) is irreducible. By Exercise8, C is contained in some Z(fi). Since C is a maximal irreducible subset of Z(f), it follows that we then must have equality: C = Z(fi). It remains to observe that if any inclusion relation Z(fi) ⊂ Z(fj) is necessarily an identity (prove this yourself) so that each Z(fi) is already maximal and hence in irreducible component. The discussion of the general case begins with the rather formal

LEMMA 2.6. For a partially ordered set (A, ≤) the following are equivalent:

(i) (A, ≤) satisfies the ascending chain condition: every ascending chain a1 ≤ a2 ≤ a3 ≤ · · · becomes stationary: an = an+1 = ··· for n sufficiently large. (ii) Every nonempty subset B ⊂ A has a maximal element, that is, an element b0 ∈ B such that there is no b ∈ B with b > b0. PROOF. (i)⇒(ii). Suppose (A, ≤) satisfies the ascending chain condition and let B ⊂ A be nonempty. Choose b1 ∈ B. If b1 is maximal, we are done. If not, then there exists a b2 ∈ B with b2 > b1. We repeat the same argument for b2. We cannot indefinitely continue in this manner because of the ascending chain condition. (ii)⇒(i). If (A, ≤) satisfies (ii), then the set of members of any ascending chain has a maximal element, in other words, the chain becomes stationary. If we replace ≤ by ≥, then we obtain the notion of the descending chain condition and we find that this property is equivalent to: every nonempty subset B ⊂ A has a minimal element. These properties appear in the following pair of definitions.

DEFINITIONS 2.7. We say that a ring R is noetherian if its collection of ideals satisﬁes the ascending chain condition. We say that a topological space Y is noetherian if its collection of closed subsets satisﬁes the descending chain condition.

EXERCISE 9. Prove that a subspace of a noetherian space is noetherian. Prove also that a ring quotient of a noetherian ring is noetherian.

EXERCISE 10. Prove that a noetherian space is quasi-compact: every covering of such a space by open subsets contains a finite subcovering. The interest of the noetherian property is that it is one which is possessed by almost all the rings we encounter and that it implies many finiteness properties without which we are often unable to go very far. Let us give a nonexample first: the ring R of holomorphic functions on C is not noetherian: if Ik denotes the ideal of f ∈ R vanishing on all the integers ≥ k, then I1 ⊂ I2 ⊂ · · · is a strictly ascending chain of ideals in R. Obviously a field is noetherian. The ring Z is noetherian: if I1 ⊂ I2 ⊂ · · · is an ∞ ascending chain of ideals in Z, then ∪s=1Is is an ideal of Z, hence of the form (n) for some n ∈ Z. But if s is such that n ∈ Is, then clearly the chain is stationary as of index s. (This argument only used the fact that any ideal in Z is generated by a single element, i.e., that Z is a principal ideal domain.) That most rings we know are noetherian is a consequence of 2. IRREDUCIBILITY AND DECOMPOSITION 11

THEOREM 2.8 (Hilbert’s basis theorem). If R is a noetherian ring, then so is R[x]. As with the Nullstellensatz, we postpone the proof and discuss some of its consequences ﬁrst.

COROLLARY 2.9. If R is a noetherian ring (for example, a ﬁeld) then so is every ﬁnitely generated R-algebra. Also, the space An (and hence any closed subset of An) is noetherian.

PROOF. The Hilbert basis theorem implies (with induction on n) that the ring R[x1, . . . , xn] is noetherian. By Exercise9, every quotient ring R[x1, . . . , xn]/I is then also noetherian. But a finitely generated R-algebra is (by definition actually) isomorphic to some such quotient and so the first statement follows. n Suppose A ⊃ Y1 ⊃ Y2 ⊃ · · · is a descending chain of closed subsets. Then (0) ⊂ I(Y1) ⊂ I(Y2) ⊂ · · · is an ascending chain of ideals. As the latter becomes stationary, so will become the former. PROPOSITION 2.10. If Y is noetherian space, then its irreducible components are finite in number and their union equals Y .

PROOF. Suppose Y is a noetherian space. We first show that every closed subset can be written as a finite union of closed irreducible subsets. First note that the empty set has this property, for a union with empty index set is empty. Let B be the collection of closed subspaces of Y for which this is not possible, i.e., that cannot be written as a finite union of closed irreducible subsets. Suppose that B is nonempty. According to 2.6 this collection has a minimal element, Z, say. This Z must be nonempty and cannot be irreducible. So Z is the union of two proper closed subsets Z0 and Z00. The minimality of Z implies that neither Z0 nor Z00 is in B and so both Z0 and Z00 can be written as a finite union of closed irreducible subsets. But then so can Z and we get a contradiction. In particular, there exist closed irreducible subsets Y1,...,Yk of Y such that Y = Y1 ∪ · · · ∪ Yk (if Y = ∅, take k = 0). We may of course assume that no Yi is contained in some Yj with j 6= i (otherwise, omit Yi). It now remains to prove that the Yi’s are the irreducible components of Y , that is, we must show that every irreducible subset C of Y is contained in some Yi. But this follows from an application of Exercise8. If we apply this to An, then we find that every subset Y ⊂ An has a finite number of irreducible components, the union of which is all of Y . If Y is closed in An, then so is every irreducible component of Y and according to Proposition 2.3 any such irreducible component is defined by a prime ideal. This allows us to recover the irreducible components of a closed subset Y ⊂ An from its coordinate ring: COROLLARY 2.11. Let Y ⊂ An be a closed subset. If C is an irreducible component of Y , then the image IY (C) of I(C) in k[Y ] is a minimal prime ideal of k[Y ] and any minimal prime ideal of k[Y ] is so obtained: we thus get a bijective correspondence between the irreducible components of Y and the minimal prime ideals of k[Y ].

PROOF. Let C be a closed subset of Y and let IY (C) be the corresponding ideal of k[Y ]. Now C is irreducible if and only if I(C) is a prime ideal of k[x1, . . . , xn], or what amounts to the same, if and only if IY (C) is a prime ideal of k[Y ]. It is 12 1. AFFINE VARIETIES an irreducible component if C is maximal for this property, or what amounts to the same, if IY (C) is minimal for the property of being a prime ideal of k[Y ].

EXAMPLE 2.12. First consider the set C := {(t, t2, t3) ∈ A3 | t ∈ k}. This is a 3 closed subset of A : if we use (x, y, z) instead of (x1, x2, x3), then C is the common zero set of y − x2 and z − x3. Now the inclusion k[x] ⊂ k[x, y, z] composed with the ring quotient k[x, y, z] → k[x, y, z]/(y − x2, z − x3) is easily seen to be an isomorphism. Since k[x] has no zero divisors, (y − x2, z − x3) must be a prime ideal. So C is irreducible and I(C) = (y − x2, z − x3). We now turn to the closed subset Y ⊂ A3 deﬁned by xy−z = 0 and y3 −z2 = 0. Let p = (x, y, z) ∈ Y . If y 6= 0, then we put t := z/y; from y3 = x2, it follows that y = t2 and z = t3 and xy = z implies that x = t. In other words, p ∈ C in that case. If y = 0, then z = 0, in other words p lies on the x-axis. Conversely, any point on the x-axis lies in Y . So Y is the union of C and the x-axis and these are the irreducible components of Y . We brieﬂy discuss the corresponding issue in commutative algebra. We begin with recalling the notion of localization and we do this in the generality that is needed later.

2.13.L OCALIZATION. Let R be a ring and let S be a multiplicative subset of R: 1 ∈ S and S closed under multiplication. Then a ring S−1R, together with a ring homomorphism R → S−1R is defined as follows. An element of S−1R is by definition written as a formal fraction r/s, with r ∈ R and s ∈ S, with the understanding that r/s = r0/s0 if and only if s00(s0r − sr0) = 0 for some s00 ∈ S. This is a ring indeed: multiplication and subtraction is defined as for ordinary fractions: r/s.r0/s0 = (rr0)/(ss0) and r/s − r0/s0 = (s0r − sr0)/(ss0); it has 0/1 as zero and 1/1 as unit element and the ring homomorphism R → S−1R is simply r 7→ r/1. Observe that the definition shows that 0/1 = 1/1 if and only if 0 ∈ S, in which case S−1R is reduced to the zero ring. We also note that any s ∈ S maps to an invertible element of S−1R, the inverse of s/1 being 1/s. (This is also true when 0 ∈ S, for 0 is invertible in the zero ring.) In a sense (made precise in part (b) of Exercise 11 below) the ring homomorphism R → S−1R is universal for that property. This construction is called the localization away from S. It is clear that if S does not contain zero divisors, then r/s = r0/s0 if and only if s0r − sr0 = 0; in particular, r/1 = r0/1 if and only if r = r0, so that R → S−1R is then injective. If we take S maximal for this property, namely take it to be the set of nonzero divisors of R (which is indeed multiplicative), then S−1R is called the fraction ring Frac(R) of R. When R is a domain, S = R−{0} and so Frac(R) is a field, the fraction field of R. This gives the following corollary, which hints to the importance of prime ideals in the subject.

COROLLARY 2.14. An ideal p of a ring R is a prime ideal if and only if it is the kernel of a ring homomorphism from R to a ﬁeld.

PROOF. It is clear that the kernel of a ring homomorphism from R to a ﬁeld is always a prime ideal. Conversely, if p is a prime ideal, then it is the kernel of the composite R → R/p ,→ Frac(R/p).

Of special interest is when S = {sn | n ≥ 0} for some s ∈ R. We then usually write R[1/s] for S−1R. Notice that the image of s in R[1/s] is invertible. Note that when s is nilpotent, R[1/s] is the zero ring.

EXERCISE 11. Let R be a ring and let S be a multiplicative subset of R. (a) What is the the kernel of R → S−1R? 2. IRREDUCIBILITY AND DECOMPOSITION 13

(b) Prove that a ring homomorphism φ : R → R0 with the property that φ(s) is invertible for every s ∈ S factors in a unique manner through S−1R. (c) Consider the polynomial ring R[xs : s ∈ S] and the homomorphism of −1 R-algebras R[xs : s ∈ S] → S R that sends xs to 1/s. Prove that this homomorphism is surjective and that its kernel consists of the f ∈ R[xs : s ∈ S] which after multiplication by an element of S lie in the ideal generated the degree one polynomials sxs − 1, s ∈ S. EXERCISE 12. Let R be a ring and let p be a prime ideal of R. (a) Prove that the complement R − p is a multiplicative system. The resulting localization (R − p)−1R is called the localization at p and is usually denoted Rp. (b) Prove that pRp is a maximal ideal of Rp and that it is the only maximal ideal of Rp. (A ring with a unique maximal ideal is called a local ring.) (c) Prove that the localization map R → Rp drops to an isomorphism of ﬁelds Frac(R/p) → Rp/pRp. (d) Work this out for R = Z and p = (p), where p is a prime number. (e) Same for R = k[x, y] and p = (x).

LEMMA 2.15. Let R be a ring. Then the intersection of all the prime ideals of R is the ideal of nilpotents p(0) of R. Equivalently, for every nonnilpotent a ∈ R, there exists a ring homomorphism from R to a ﬁeld that is nonzero on a.

PROOF. It is easy to see that a nilpotent element lies in every prime ideal. Now for nonnilpotent a ∈ R consider the homomorphism R → R[1/a]. The ring R[1/a] has a maximal ideal4 and hence admits a ring homomorphism to some ﬁeld F : φ : R[1/a] → F . Then the kernel of the composite R → R[1/a] → F is a prime ideal and a is not in this kernel (for its image is invertible with inverse φ(1/a)). EXERCISE 13. Let R be a ring. Prove that the intersection of all the maximal ideals of a ring R consists of the a ∈ R for which 1 + aR ⊂ R× (i.e., 1 + ax is invertible for every x ∈ R). You may use the fact that every proper ideal of R is contained in a maximal ideal. We can do better if R is noetherian. The following proposition is the algebraic counterpart of Proposition 2.10. Note the similarity between the proofs.

PROPOSITION 2.16. Let R be a noetherian ring. Then any radical ideal of R is an intersection of finitely many prime ideals. Also, the minimal prime ideals of R are finite in number and their intersection is equal to the ideal of nilpotents p(0). PROOF. Let B be the collection of the radical ideals I ( R that can not be written as an intersection of finitely many prime ideals and suppose that B is nonempty. Since R is noetherian, B contains a maximal member I0. We derive a contradiction as follows. Since I0 cannot be a prime ideal,√ there exist a1, a2 ∈ R − I0 with a1a2 ∈ I0. Consider the radical ideal Ji := I0 + Rai. We claim that J1 ∩ J2 = I0. The inclusion ⊃ is obvious and ⊂ is seen as follows: if a ∈ J1 ∩J2, then for i = 1, 2, there ni n1+n2 exists an ni > 0 such that a ∈ I0 +Rai. Hence a ∈ (I0 +Ra1)(I0 +Ra2) = I0,

4Every ring has a maximal ideal. For noetherian rings, which are our main concern, this is obvious, but in general this follows with transfinite induction, the adoption of which is equivalent to the adoption of the axiom of choice. 14 1. AFFINE VARIETIES so that a ∈ I0. Since Ji strictly contains I0, it does not belong to B. In other words, Ji is an intersection of finitely many prime ideals. But then so is J1 ∩ J2 = I0 and we get a contradiction. p We thus find that (0) = p1 ∩ · · · ∩ ps for certain prime ideals pi. We may of course assume that no pi contains some pj with j 6= i (otherwise, omit pi). It now remains to prove that every prime ideal p of R contains some pi. If that is not the case, then for i = 1, . . . , s there exists a ai ∈ pi − p. But then a1a2 ··· as ∈ p p1 ∩ · · · ∩ ps = (0) ⊂ p and since p is a prime ideal, some factor ai lies in p. This is clearly a contradiction. √ EXERCISE 14. Let J be an ideal of the ring R. Show that J is the intersection of all the prime ideals that contain J. Prove that in case R is noetherian, the prime ideals that are minimal√ for this property are finite in number and that their common intersection is still J. What do we get for R = Z and J = Zn?

EXERCISE 15. Let R be a ring, S ⊂ R be a multiplicative system and denote by φ : R → S−1R the natural homomorphism. Prove that the map which assigns to every prime ideal of S−1R its preimage in R under φ deﬁnes a bijection between the prime ideals of S−1R and the prime ideals of R disjoint with S. Prove also that if 0 ∈/ S, then the preimage of the ideal of nilpotents of S−1R is the ideal of nilpotents of R.

3. Finiteness properties and the Hilbert theorems The noetherian property in commutative algebra is best discussed in the context of modules, even if one’s interest is only in rings. We ﬁx a ring R and ﬁrst recall the notion of an R-module.

The notion of an R-module is the natural generalization of a K-vector space (where K is some ﬁeld). Let us observe that if M is an (additively written) abelian group, then the set End(M) of group homomorphisms M → M is a ring for which subtraction is pointwise deﬁned and multiplication is composition (so if f, g ∈ End(M), then f − g : m ∈ M 7→ f(m) − g(m) and fg : m 7→ f(g(m))); clearly the zero element is the zero homomorphism and the unit element is the identity. It only fails to obey our convention in the sense that this ring is usually noncommutative. We only introduced it in order to be able state succinctly:

DEFINITION 3.1. An R-module is an abelian group M, equipped with a ring homomorphism R → End(M). So any r ∈ R defines a homomorphism M → M; we usually denote the image of m ∈ M simply by rm. If we write out the properties of an R-module structure in these terms, we get: r(m1 − m2) = rm1 − rm2, (r1 − r2)m = r1m − r2m, 1.m = m, r1(r2m) = (r1r2)m. If R happens to be field, then we see that an R-module is the same thing as an R-vector space. The notion of an R-module is quite ubiquitous if you think about it. A simple example is an ideal I ⊂ R. Any abelian group M is in a natural manner a Z-module. And a R[x]-module can be understood as an real linear space V (an R-module) endowed with an endomorphism (the image of x in End(V )). A more involved example is the following: if X is a manifold, f is a C∞-function on X and ω a C∞ differential p-form on X, then fω is also a C∞ differential p-form on X. Thus the linear space of C∞-differential forms on X of a fixed degree p is naturally a module over the ring of C∞-functions on X. Here are a few companion notions, followed by a brief discussion. 3. FINITENESS PROPERTIES AND THE HILBERT THEOREMS 15

3.2. In what follows is M an R-module. A map f : M → N from M to an R-module N is called a R-homomorphism if it is a group homomorphism with the property that f(rm) = rf(m) for all r ∈ R and m ∈ M. If f is also bijective, then we call it an R-isomorphism; in that case its inverse is also a homomorphism of R-modules. For instance, given a ring homomorphism f : R → R0, then R0 becomes an R-module by rr0 := f(r)r0 and this makes f a homomorphism of R-modules. A subset N ⊂ M is called an R-submodule of M if it is a subgroup and rn ∈ N for all r ∈ R and n ∈ N. Then the group quotient M/N is in a unique manner a R-module in such a way that the quotient map M → M/N is a R-homomorphism: we let r(m+N) := rm+N for r ∈ R and m ∈ M. Notice that a R-submodule of R (here we regard R as a R-module) is the same thing as an ideal of R. Given a subset S ⊂ M, then the set of elements m ∈ M that can be written as r1s1 + ··· + rksk with ri ∈ R and si ∈ S is a R-submodule of M. We call it the R-submodule of M generated by S and we shall denote it by RS. If there exists a ﬁnite set S ⊂ M such that M = RS, then we say that M is ﬁnitely generated as an R-module.

DEFINITION 3.3. We say that an R-module M is noetherian if the collection of R-submodules of M satisfies the ascending chain condition: any ascending chain of R-submodules N1 ⊂ N2 ⊂ · · · becomes stationary. It is clear that then every quotient module of a noetherian module is also noetherian. The noetherian property of R as a ring (as previously defined) coincides with this property of R as an R-module. The following two propositions provide the passage from the noetherian property to finite generation:

PROPOSITION 3.4. An R-module M is noetherian if and only if every R-submodule of M is ﬁnitely generated as an R-module.

PROOF. Suppose that M is a noetherian R-module and let N ⊂ M be a R- submodule. The collection of finitely generated R-submodules of M contained in N is nonempty. Hence it has a maximal element N0. If N0 = N, then N is finitely generated. If not, we run into a contradiction: just choose x ∈ N −N0 and consider N0 + Rx. This is a R-submodule of N. It is finitely generated (for N0 is), which contradicts the maximal character of N0. Suppose now that every R-submodule of M is finitely generated. If N1 ⊂ ∞ N2 ⊂ · · · is an ascending chain of R-modules, then the union N := ∪i=1Ni is a

R-submodule. Let {s1, . . . , sk} be a ﬁnite set of generators of N. If sκ ∈ Niκ , and j := max{i1, . . . , ik}, then it is clear that Nj = N. So the chain becomes stationary as of index j. PROPOSITION 3.5. Suppose that R is a noetherian ring. Then every ﬁnitely generated R-module M is noetherian.

PROOF. By assumption M = RS for a ﬁnite set S ⊂ M. We prove the proposition by induction on the number of elements of S. If S = ∅, then M = {0} and there is nothing to prove. Suppose now S 6= ∅ and choose s ∈ S, so that our induction hypothesis applies to M 0 := RS0 with S0 = S − {s}: M 0 is noetherian. But so is M/M 0, for it is a quotient of the noetherian ring R via the surjective R-module homomorphism R → M/M 0, r 7→ rs + M 0. Let now N1 ⊂ N2 ⊂ · · · be an ascending chain of R-submodules of M. Then 0 0 N1 ∩ M ⊂ N2 ∩ M ⊂ · · · becomes stationary, say as of index j1. Hence we only 0 need to be concerned with the stabilization of the submodules Nk/(Nk ∩ M ) of 16 1. AFFINE VARIETIES

0 0 M/M . These stabilize indeed (say as of index j2), since M/M is noetherian. So the original chain N1 ⊂ N2 ⊂ · · · stabilizes as of index max{j1, j2}. We are now sufficiently prepared for the proofs of the Hilbert theorems. They are gems of elegance and efficiency. We will use the notion of initial coefficient of a polynomial, which we re- d call. Given a ring R, then every nonzero f ∈ R[x] is uniquely written as rdx + d−1 rd−1x + ··· + r0 with rd 6= 0. We call rd ∈ R the initial coefficient of f and denote it by in(f). For the zero polynomial, we simply define this to be 0 ∈ R. Notice that in(f) in(g) equals in(fg) when nonzero.

PROOFOF THEOREM 2.8. The assumption is here that R is a noetherian ring. In view of Proposition 3.4 we must show that every ideal I of R[x] is finitely generated. Consider the subset in(I) := {in(f): f ∈ I} of R. We first show that this is an ideal of R. If r ∈ R, f ∈ I, then r in(f) equals in(rf) when nonzero and since rf ∈ I, it follows that r in(f) ∈ I. If f, g ∈ I with d := deg(f) − deg(g) ≥ 0, then in(f) − in(g) = in(f − tdg). So in(I) is an ideal as asserted. Since R is noetherian, in(I) is finitely generated: there exist f1, . . . , fk ∈ I such that in(I) = R in(f1) + ··· + R in(fk). Let di be the degree of fi, d0 := max{d1, . . . , dk} and R[x]

I = R[x]f1 + ··· + R[x]fk + (I ∩ R[x]

Our claim implies the theorem: R[x]

R-generators of I ∩ R[x]

PROPOSITION 3.6 (Artin-Tate). Let R be a noetherian ring, B an R-algebra and A ⊂ B an R-subalgebra. Assume that B is finitely generated as an A-module. Then A is finitely generated as an R-algebra if and only if B is so. Pm PROOF. By assumption there exist b1, . . . , bm ∈ B such that B = i=1 Abi. If there exist a1, . . . , an ∈ A which generate A as an R-algebra (which means that A = R[a1, . . . , an]), then a1, . . . , an, b1, . . . , bm generate B as an R-algebra. Suppose, conversely, that there exists a finite subset of B which generates B as a R-algebra. By adding this subset to b1, . . . , bm, we may assume that b1, . . . , bm also generate B as an R-algebra. Then every product bibj can be written as an A-linear combination of b1, . . . , bm: m X k k bibj = aijbk, aij ∈ A. k=1 3. FINITENESS PROPERTIES AND THE HILBERT THEOREMS 17

Let A0 ⊂ A be the R-subalgebra of A generated by all the (finitely many) coeffi- k P cients aij. This is a noetherian ring by Corollary 2.9. It is clear that bibj ∈ k A0bk P and with induction it then follows that B = R[b1, . . . , bm] ⊂ k A0bk. So B is finitely generated as an A0-module. Since A is an A0-submodule of B, A is also finitely generated as an A0-module by Proposition 3.4. It follows that A is a finitely generated R-algebra. This has a consequence for field extensions:

COROLLARY 3.7. A field extension L/K is finite if and only if L is finitely generated as a K-algebra.

PROOF. It is clear that if L is a finite dimensional K-vector space, then L is finitely generated as a K-algebra. Suppose now b1, . . . , bm ∈ L generate L as a K-algebra. We must show that every bi is algebraic over K. Suppose that this is not the case. After renumbering we can and will assume that (for some 1 ≤ r ≤ m) b1, . . . , br are algebraically independent over K and br+1, . . . , bm are algebraic over the quotient field K(b1, . . . , br) of K[b1, . . . , br]. So L is a finite extension of K(b1, . . . , br). We apply Proposition 3.6 to R := K, A := K(b1, . . . , br) and B := L and find that K(b1, . . . , br) is as a K-algebra generated by a finite subset S ⊂ K(b1, . . . , br). If g is a common denominator for the elements of S, then clearly K(b1, . . . , br) = K[b1, . . . , br][1/g]. Since K(b1, . . . , br) strictly contains K[b1, . . . , br], g must have positive degree. In N particular, 1 + g 6= 0, so that 1/(1 + g) ∈ K(b1, . . . , br) can be written as f/g , with f ∈ K[b1, . . . , br]. Here we may of course assume that f is not divisible by g in N K[b1, . . . , br]. From the identity f(1+g) = g we see that N ≥ 1 (for g has positive degree). But then f = g(−f + gN−1) shows that f is divisible by g. We thus get a contradiction.

EXERCISE 16. Prove that a field which is finite generated as a ring (that is, isomorphic to a quotient of Z[x1, . . . , xn] for some n) is finite. We deduce from the preceding corollary the Nullstellensatz.

PROOFOFTHE NULLSTELLENSATZ√ 1.5. Let J ⊂ k[x1, . . . , xn] be an ideal.√ We must show that I(Z(J)) ⊂ J. This amounts to: for every f ∈ k[x1, . . . , xn] − J there exists a p ∈ Z(J) for which f(p) 6= 0. Consider k[x1, . . . , xn]/J and denote by ¯ ¯ f ∈ k[x1, . . . , xn]/J the image of f. Since f is not nilpotent, we have defined ¯ A := (k[x1, . . . , xn]/J)[1/f]. ¯ Notice that A is as a k-algebra generated by the images of x1, . . . , xn and 1/f (hence is finitely generated). Choose a maximal ideal m ⊂ A. Then the field A/m is a finitely generated k-algebra and so by Corollary 3.7 a finite extension of k. But then it must be equal to k, since k is algebraically closed. Denote by φ : k[x1, . . . , xn] → A → A/m = k the corresponding surjection and put pi := φ(xi) n n and p := (p1, . . . , pn) ∈ A . So if we view xi as a function on A , then xi(p) = pi = φ(xi). The fact that φ is a homomorphism of k-algebras implies that it is given as ‘evaluation in p’: any g ∈ k[x1, . . . , xn] takes in p the value φ(g). Since the kernel of φ contains J, every g ∈ J will be zero in p, in other words, p ∈ Z(J). On the other hand, f(p) = φ(f) is invertible, for it has the image of 1/f¯ in A/m = k as its inverse. So f(p) 6= 0. 18 1. AFFINE VARIETIES

4. The afﬁne category We begin with specifying the maps between closed subsets of afﬁne spaces that we wish to consider. DEFINITION 4.1. Let X ⊂ Am and Y ⊂ An be closed subsets. We say that a map f : X → Y is regular if the components f1, . . . , fn of f are regular functions on X (i.e., are given by the restrictions of polynomial functions to X). Composition of a regular function on Y with f yields a regular function on X (for if we substitute in a polynomial of n variables g(y1, . . . , yn) for every variable yi a polynomial fi(x1, . . . , xm) of m variables, we get a polynomial of m variables). So f then induces a k-algebra homomorphism f ∗ : k[Y ] → k[X]. There is also a converse: PROPOSITION 4.2. Let be given closed subsets X ⊂ Am and Y ⊂ An and a k-algebra homomorphism φ : k[Y ] → k[X]. Then there is a unique regular map f : X → Y such that f ∗ = φ.

PROOF. Let fi := φ(yi|Y ) ∈ k[X] (i = 1, . . . , n) and define f = (f1, . . . , fn): n ∗ n X → A , so that f yi = fi = φ(yi|Y ). If j : Y ⊂ A denotes the inclusion, then we ∗ ∗ can also write this as f yi = φj yi. In other words, the k-algebra homomorphisms ∗ ∗ f , φj : k[y1, . . . , yn] → k[X] coincide on the generators yi. Hence they must be equal: f ∗ = φj∗. It follows that f ∗ is zero on I(Y ), which means that f takes its values in Z(I(Y )) = Y , and that the resulting map k[Y ] → k[X] equals φ. The same argument shows that if f : X → Y and g : Y → Z are regular maps, then so is their composite gf : X → Z. So we have a category (with objects the closed subsets and regular maps as defined above). In particular, we have a notion of isomorphism: a regular map f : X → Y is an isomorphism if is has a two-sided inverse g : Y → X which is also a regular map. This implies that f ∗ : k[Y ] → k[X] has a two-sided inverse g∗ : k[X] → k[Y ] which is also an isomorphism of k-algebras, and hence is an isomorphism of k-algebras. Proposition 4.2 implies that conversely, an isomorphism of k-algebras k[Y ] → k[X] comes from a unique isomorphism X → Y . We complete the picture by showing that any finitely generated reduced k- algebra A is isomorphic to some k[Y ]; the preceding then shows that Y is unique up to isomorphism. Since A is finitely generated as a k-algebra, there exists a surjective k-algebra homomorphism φ : k[x1, . . . , xn] → A. If we put I := Ker(φ), ∼ n then φ induces an isomorphism k[x1, . . . , xn]/I = A. Put Y := Z(I) ⊂ A . Since A is reduced, I is a radical ideal and hence equal to I(Y ) by the Nullstellensatz. It follows that φ factors through a k-algebra isomorphism k[Y ] =∼ A. We may sum up this discussion in categorical language as follows. PROPOSITION 4.3. The map which assigns to a closed subset of some An its coordinate ring defines an anti-equivalence between the category of closed subsets of affine spaces (and regular maps between them as defined above) and the category of reduced finitely generated k-algebras (and k-algebra homomorphisms). EXAMPLE 4.4. Consider the regular map f : A1 → A2, f(t) = (t2, t3). The image of this map is the hypersurface Z defined by x3 − y2 = 0. This is a homeomorphism on the image for it is a bijection of sets that have both the cofinite topology. The inverse sends (0, 0) to 0 and is on Z − {(0, 0)} given by (x, y) 7→ y/x. 4. THE AFFINE CATEGORY 19

In order to determine whether the inverse is regular, we consider f ∗. We have k[Z] = k[x, y]/(x3 − y2), A(A1) = k[t] and f ∗ : k[x, y]/(x3 − y2) → k[t] is given by x 7→ t2, y 7→ t3. This homomorphism is not surjective for its image misses t ∈ k[t]. So f is not an isomorphism.

EXAMPLE 4.5. An affine-linear transformation of kn is of the form x ∈ kn 7→ g(x) + a, where a ∈ kn and g ∈ GL(n, k) is a linear transformation. Its inverse is y 7→ g−1(y − a) = g−1(y) − g−1(a) and so of the same type. When we regard such an affine linear transformation as a map from An to itself, then it is regular: its coordinates (g1, . . . , gn) are polynomials of degree one. So an affine-linear transformation is also an isomorphism of An onto itself. When n ≥ 2, there exist automorphisms of An not of this form. For instance (x, y) 7→ (x, y + x2) defines an automorphism of A2 with inverse (x, y) 7→ (x, y − x2) (see also Exercise 18). 2 2 2 EXERCISE 17. Let C ⊂ A be the ‘circle’, defined by x + y = 1 and let p0 := (−1, 0) ∈ C. For every p = (x,√ y) ∈ C − {p0}, the line through p0 and p has slope f(p) = y/(x + 1). Denote by −1 ∈ k a root of the equation t2 + 1 = 0. 5 1 (a) Prove√ that when char(k) 6= 2, f defines an isomorphism onto A − {± −1}. √ (b) Consider the map g : C → A1, g(x, y) := x + −1y. Prove that when char(k) 6= 2, g defines an isomorphism of C onto A1 − {0}. (c) Prove that when char(k) = 2, the defining polynomial x2 + y2 − 1 for C is the square of a degree one polynomial so that C is a line.

EXERCISE 18. Let f1, . . . , fn ∈ k[x1, . . . , xn] be such that f1 = x1 and fi − xi ∈ n n k[x1, . . . , xi−1] for i = 2, . . . , xn. Prove that f defines an isomorphism A → A . 4.6.Q UADRATIC HYPERSURFACES IN CASE char(k) 6= 2. Let H ⊂ An be a hypersurface defined by a polynomial of degree two: n X X f(x1, . . . , xn) = aijxixj + aixi + a0. 1≤i≤j=n i=1 By means of a linear transformation (this involves splitting off squares, hence re- P quires the existence of 1/2 ∈ k), the quadratic form 1≤i≤j=n aijxixj can be brought in diagonal form. This means that we can make all the coefficients a √ ij with i 6= j vanish. Another diagonal transformation (which replaces xi by aiixi when aii 6= 0) takes every nonzero coefficient aii to 1 and then renumbering the coordinates (which is also a linear transformation) brings f into the form Pr 2 Pn f(x1, . . . , xn) = i=1 xi + i=1 aixi + a0 for some r ≥ 1. Splitting off squares Pr once more enables us to get rid of i=1 aixi so that we get r n X 2 X f(x1, . . . , xn) = xi + aixi + a0. i=1 i=r+1 We now have the following cases. If the nonsquare part is identically zero, then we end up with the equation Pr 2 i=1 xi = 0 for H.

5We have not really defined what is an isomorphism between two nonclosed subsets of an affine space. Interpret this here as: f ∗ maps k[x, y][1/(x + 1)]/(x2 + y2 − 1) (the algebra of regular functions 2 1 √ on C − {p0}) isomorphically onto k[t][1/(t + 1)] (the algebra of regular functions on A − {± −1}). This will be justified by Proposition 4.8. 20 1. AFFINE VARIETIES

Pn If the linear part i=r+1 aixi is nonzero (so that we must have r < n), then an Pn affine-linear transformation which does not affect x1, . . . , xr and takes i=r+1 aixi+ Pr 2 a0 to −xn yields the equation xn = i=1 xi . This is the graph of the function Pr 2 n−1 n−1 i=1 xi on A and so H is then isomorphic to A . If the linear part Pn a x is zero, but the constant term a is nonzero, then i=r+1 i i 0 √ we can make another diagonal transformation which replaces xi by −a0xi) and Pr 2 divide f by a0: then H gets the equation i=1 xi = 1. In particular, there are only a finite number of quadratic hypersurfaces up to isomorphism. (This is also true in characteristic two, but the discussion is a bit more delicate.) The previous discussion (and in particular Proposition 4.3) leads us to associate to any ring R a space with a topology so that for R = k[X] we get a space homeomorphic to X. Since the points of X correspond to maximal ideals of k[X], we choose the underlying set of this space (which we shall denote by Specm(R)) to be the collection of maximal ideals of R. In order to avoid confusion about whether a maximal ideal is to be viewed as a subset of R or as a point of Specm(R) we agree that if m is a maximal ideal of R, then the corresponding element of Specm(R) is denoted xm and if x ∈ Specm(R), then the corresponding maximal ideal of R is denoted mx. Any a ∈ R defines a ‘zero set’ Z(a) ⊂ Specm(R), being the set of x ∈ R with a ∈ mx. We denote the complement of Specm(R) − Z(a) by U(a). We 0 0 have U(aa ) = U(a) ∩ U(a ) and so the collection of {U(a)}a∈R is the basis of a topology on Specm(R). Note that a subset Specm(R) is closed precisely if it is an intersection of subsets of the form Z(a); this is equal to the common zero set of the set of functions defined by an ideal of R. The space Specm(R) is called the maximal ideal spectrum6 of R (but our notation for it is less standard). We observe for later reference:

LEMMA 4.7. The maximal ideal spectrum Specm(R) is quasi-compact: every open covering of Specm(R) admits a ﬁnite subcovering.

PROOF. It suffices to verify this for an open covering by basic open subsets. So let {gi ∈ R}i∈I be such that Specm(R) = ∪i∈I U(gi). This means that the ideal generated by the gi is not contained in any maximal ideal and hence must be all of R. In particular, this ideal contains 1 so that 1 = a1gi1 + ··· + angin for certain n i1, . . . , in ∈ I and a1, . . . , an ∈ R. It follows that Specm(R) = ∪ν=1U(riν ). Let us now take for our ring a finitely generated k-algebra A. Then for every x ∈ Specm(A), the residue field A/mx is finitely generated as a k-algebra and hence by Corollary 3.7 a finite extension of k. Since we assumed k to be algebraically closed it is equal to k. We write ρx : A → k for the reduction modulo mx. Now every a ∈ A defines a ‘regular function’ a¯ : Specm(A) → k which takes in x ∈ Specm(A) the value ρx(a) ∈ k. Its zero set is indeed Z(a) ⊂ Specm(A). Notice that if A = k[X], then the above discussion shows that Specm(A) can be identified with X as a topological space and that under this identification, k[X] becomes the ring of regular functions on Specm(A).

6I. Gelfand was presumably the ﬁrst to consider this, albeit in the context of functional analysis: he characterized the Banach algebras that appear as the algebras of continuous C-valued functions on compact Hausdorff spaces and so it might be appropriate to call this the Gelfand spectrum. 4. THE AFFINE CATEGORY 21

A homomorphism φ : A → B of ﬁnitely generated k-algebras gives rise to a map Specm(φ) : Specm(B) → Specm(A): if x ∈ Specm(B), then the composite −1 homomorphism ρxφ : A → k is the identity map when restricted to k so that φ mx is a maximal ideal of A with residue ﬁeld k. We thus get a map

Specm(φ) : Specm(B) → Specm(A), xm 7→ xφ−1m. For a ∈ A, the preimage of U(a) under Specm(φ) is U(φ(a)). This shows that Specm(φ) is continuous. We call the resulting pair (Specm(φ), φ) a morphism.

PROPOSITION 4.8. Let A be a ﬁnitely generated k-algebra and put X := Specm(A). Then for every s ∈ A, A[1/s] is a ﬁnitely generated k-algebra (which is reduced when A is) and the natural k-algebra homomorphism A → A[1/s] induces a homeomor- 0 phism of Specm(A[1/s]) onto Xs := X − Z(s). Moreover, for s, s ∈ A the following are equivalent:

(i) Xs ⊂ Xs0 , (ii) s0 divides some positive power of s, (iii) we have an A-homomorphism A[1/s0] → A[1/s].

PROOF. It is clear that A[1/s] is a k-algebra and is as such ﬁnitely generated (just add to a generating set for A the generator 1/s). If A is reduced, then so is A[1/s]. For if s is not nilpotent (so that A[1/s] 6= 0) then for any nilpotent a/sr ∈ A[1/s], the numerator a must be nilpotent in R. The preimage of a maximal ideal of A[1/s] in A is a maximal ideal of A which does not contain s. Conversely, if m is a maximal ideal of A, then mA[1/s] is a maximal ideal of A[1/s] unless s ∈ m. So the map A → A[1/s] induces an injection of Specm(A[1/s]) in Specm(A) with image Xs. This map is continuous. To see that it is also open, note that a basic open subset of Specm(A[1/s]) is given by some b ∈ A[1/s]. If we write b = a/sn, then this open subset is also given by a and its image in Specm(A) is U(as), hence open. 0 Assume that we have an inclusion Xs ⊂ Xs0 . Then Z(s ) ⊂ Z(s) and so by the Nullstellensatz, s ∈ p(s0). This implies that sn ∈ (s0) for some positive integer n. If we write sn = as0, then we have an A-homomorphism A[1/s0] → A[1/(as0)] = A[1/(sn] = A[1/s]. Conversely, the A-homomorphism A → A[1/s] factors through A[1/s0] if s0 becomes invertible in A[1/s] and its easily follows that s0 divides some positive power of s and Xs ⊂ Xs0 .

From now on we identify a basic open subset Xs with the maximal ideal spectrum Specm(A[1/s]).

EXERCISE 19. Give an example of ring homomorphism φ : S → R and a maximal ideal m ⊂ R, such that φ−1m is not a maximal ideal of S. (Hint: take a look at Exercise 12.)

EXERCISE 20. Prove that if A is a finitely generated k-algebra, then the map a ∈ A 7→ a¯ is a k-algebra homomorphism from A to the algebra of k-valued functions on Specm(A) with kernel p(0). Show that for every subset X ⊂ Specm(A), the set I(X) of r ∈ A with a¯|X = 0 is a radical ideal of A. 7 DEFINITION 4.9 (Preliminary definition). An affine k-variety is a topological space X endowed with a finitely generated k-algebra k[X] of regular functions on

7Since we ﬁxed k, we will often drop k and speak of an afﬁne variety. 22 1. AFFINE VARIETIES

X which identifies X with the maximal ideal spectrum of k[X]. A map f : X → Y between affine varieties is a morphism if composition with f takes regular functions on Y to regular functions on X, so that we have a k-algebra homomorphism f ∗ : k[Y ] → k[X]. Since k[X] appears here as a k-algebra of k-valued functions X → k, k[X] must be reduced. Indeed, there will exist a homeomorphism h of X onto a closed n ∗ subset of some A such that k[X] = h k[x1, . . . , xn]. We shall often denote an affine variety simply by the underlying topological space, e.g., by X, where it is then understood that k[X] is part of the data. It follows from Proposition 4.8 that any basic open subset U of X has the structure of an affine variety. With this terminology Proposition 4.3 takes now a rather trivial form: 4.10. The functor which assigns to an affine k-variety its k-algebra of regular functions identifies the category of affine k-varieties with the dual of the category of reduced finitely generated k-algebras, the inverse being given by the functor Specm. From our previous discussion it is clear that if X is an affine variety, then every closed subset X0 ⊂ X determines an affine variety X0 with k[X0] := k[X]/I(X0). Here I(X0) is of the course the ideal of regular functions on X that vanish on X0. Let f : X → Y be a morphism of affine varieties. Since f is continuous, a fiber f −1(y), or more generally, the preimage f −1Y 0 of a closed subset Y 0 ⊂ Y , will be closed in X. It is the zero set of the ideal in k[X] generated by f ∗I(Y 0). But this ideal need not be a radical ideal. Here is a simple example:

EXAMPLE 4.11. Let f : X = A1 → A1 = Y be defined by f(x) = x2. Then f ∗ : k[y] → k[x] is given by f ∗y = x2. If we assume k not to be of characteristic 2, and we take a ∈ Y − {0}, then the fiber f −1(a) consists of two distinct points and is define by the ideal generated by f ∗(y − a) = x2 − a. This a radical ideal and the fiber has coordinate ring k[x]/(x2 − a). However, the fiber over 0 ∈ Y = A1 is the singleton {0} ⊂ X = A1 and the ideal generated by f ∗y = x2 is not a radical ideal. This example indicates that there might good reason to accept nilpotent elements in the coordinate ring of f −1(0) by endowing f −1(0) with the ring of functions k[f −1(0)] := k[x]/(x2). Since this is a k-vector space of dimension 2, we thus retain the information that two points have come together and the fiber should be thought of as a point with multiplicity 2. Example 4.4 shows that a homeomorphism of affine varieties need not be an isomorphism. Here is another class of examples.

EXAMPLE 4.12 (THE FROBENIUSMORPHISM). Assume that k has positive characteristic p and consider the morphism Φ: A1 → A1, x 7→ xp. If we remember that A1 can be identified with k, then we observe that under this identification, Φ is a field automorphism: Φ(a − b) = (a − b)p = ap − bp = Φ(a) − Φ(b) and of course Φ(ab) = (ab)p = Φ(a)Φ(b) and Φ is surjective (every element of k has a pth root since k is algebraically closed) and injective. But the endomorphism Φ∗ of k[x] induced by Φ sends x to xp and has therefore image k[xp]. Clearly, Φ∗ is not surjective. The fixed point set of Φ (so the set of a ∈ A1 with ap = a) is via the identifi- 1 cation of A with k just the prime subfield Fp ⊂ k and we therefore denote it by 1 1 1 r A (Fp) ⊂ A . Likewise, the fixed point set A (Fpr ) of Φ is the subfield of k with r ¯ p elements. Since the algebraic closure Fp of Fp in k is the union of the finite 5. THE SHEAF OF REGULAR FUNCTIONS 23

1 subfields of k, the affine line over Fp equals ∪r≥1A (Fpr ). This generalizes in a straightforward manner to higher dimensions: by letting Φ act coordinatewise on An, we get a morphism An → An (which we still denote by Φ) which is also a r n n ¯ n bijection. The fixed point of Φ is A (Fpr ) and A (Fp) = ∪r≥1A (Fpr ). EXERCISE 21. Assume that k has positive characteristic p. Let q = pr be a power q of p with r > 0 and denote by Fq ⊂ k the subfield of a ∈ k satisfying a = a. We write F for Φr : An → An. Suppose Y ⊂ An is the common zero set of polynomials with coefficients in Fq ⊂ k. ∗ (a) Prove that f ∈ k[x1, . . . , xn] has its coefficients in Fq if and only if F f = f q. n (b) Prove that an affine-linear transformation of A with coefficients in Fq commutes with F . (c) Prove that F restricts to a bijection FY : Y → Y and that the fixed point m n set of FY is Y (Fqm ) := Y ∩ A (Fqm ). (d) Suppose that k is an algebraic closure of Fp. Prove that every closed subset of An is defined over a finite subfield of k. REMARK 4.13. After this exercise we cannot resist to mention the Weil zeta function. This function and its relatives—among them the Riemann zeta function— codify arithmetic properties of algebro-geometric objects in a very intricate manner. In the situation of Exercise 21, we can use the numbers |Y (Fqm )| (= the number of m P m fixed points of F in Y ) to define a generating series m≥1 |Y (Fqm )|t . It appears to be more convenient to work with the Weil zeta function: ∞ X tm Z (t) := exp |Y ( m )| , Y Fq m m=1 d which has the property that t dt log ZY yields the generating series above. This series has remarkable properties. For instance, a deep theorem due to Bernard Dwork (1960) asserts that it represents a rational function of t. Another deep theorem, due to Pierre Deligne (1974), states that the roots of the numerator and denominator have for absolute value a nonpositive half-integral power of q and that moreover, these powers have an interpretation in terms of an ‘algebraic topology for algebraic geometry’. All of this was predicted by André Weil in 1949. (This can be put in a broader context by making the change of variable t = q−s. Indeed, now numerator and denominator have their zeroes when the real part of s is a nonnegative half-integer and this makes Deligne’s result reminiscent of the famous conjectured property of the Riemann zeta function.) EXERCISE 22. Compute the Weil zeta function of affine n-space relative to the field of q elements.

5. The sheaf of regular functions In any topology or analysis course you learn that the notion of continuity is local: there exists a notion of continuity at a point so that a function is continuous if it is so at every point of its domain. We shall see that in algebraic geometry the property for a function to be regular is also local in nature.

Suppose X is an affine variety and x ∈ X. We write A for k[X] and m for mx. A basic neighborhood of x is of the form Xg with g∈ / m and a regular function 24 1. AFFINE VARIETIES f on that neighborhood is an element of A[1/g]. Let us say that two such pairs 0 (Xg, f) and (Xg0 , f ) have the same germ at x if there exists a neighborhood U of x 0 in Xg ∩ Xg0 such that f|U = f |U. This is an equivalence relation; an equivalence class is called a germ of a regular function on X at x. The germs of regular functions on X at x form an k-algebra, which we shall denote by OX,x. Evaluation in x defines a homomorphism ρx : OX,x → k whose kernel (denoted mX,x) is the unique maximal ideal of OX,x. So OX,x is a local ring. We claim that OX,x = Am. Clearly, any φ ∈ Am is given as a fraction a/g with a ∈ A and g ∈ A − m, hence comes from a regular function on the basic 0 0 0 neighborhood Xg of x. Moreover, if φ is also given as a /g , then we have (ag − a0g)g00 for some g00 ∈ A − m. This just means that a/g and a0/g0 define the same element of A[1/(gg0g00)], where we note that gg0g00 ∈/ m so that U(gg0g00) is a basic neighborhood of x in X. Now let φ be a k-valued function defined on an open subset U of X and suppose x ∈ U. We say that φ is regular at x if it defines an element of OX,x, in other words, if it defines a regular function on some basic neighborhood of x in X. So there should exist g ∈ A − m and some f ∈ k[X][1/g] which coincides with φ on some neighborhood of x (we then may take g in such a manner that φ is in fact equal to f on Xg). We denote by O(U) the set of k-valued functions U → k that are regular at every point of U. This is clearly a k-algebra.

PROPOSITION 5.1. A k-valued function φ on X which is regular at every point of X is regular, in other words, the natural k-algebra homomorphism k[X] → O(X) is an isomorphism. A map φ : X → Y between afﬁne varieties is a morphism if and only if φ is continuous and for any f ∈ O(V ) (with V ⊂ Y open) we have f ∗φ = φf ∈ O(f −1V ).

PROOF. It is clear that the map is injective: if f ∈ k[X] is in the kernel, then f must be identically zero as a function and hence f = 0. For surjectivity, let φ : X → k be regular in every point of X. We must show that φ is representable by some f ∈ k[X]. By assumption there exists for every x ∈ X, gx ∈ k[X] − mx and fx ∈ k[X] such that φ|U(gx) is representable as the fraction fx/gx. Since X is quasicompact (Lemma 4.7), the covering {U(gx)}x∈X of X pos- N sesses a ﬁnite subcovering {U(gxi )}i=1. Let us now write fi for fxi and gi for gxi . Then fi/gi and fj/gj deﬁne the same regular function on U(gi) ∩ U(gj) = U(gigj) mij and so gifj − gjfi is annihilated by (gigj) for some mij ≥ 0. Let m := max{mij} m+1 m m m+1 so that gi gj fj = gi gj fi for all i, j. Since φ|U(gi) is also represented by m m+1 (figi )/gi we may then without loss of generality assume that m = 0, so that gifj = gjfi for all i, j. Since the U(gi) cover X, the ideal generated by g1, . . . , gN must be all of k[X] PN and hence contain 1. So 1 = i=1 aigi for certain ai ∈ k[X]. Now consider PN f := i=1 aifi ∈ k[X]. We have for every j,

N N X X fgj = aifigj = aigifj = fj i=1 i=1 and so f represents φ on Uj. It follows that φ is represented by f. The last statement is left as an exercise. 6. THE PRODUCT 25

Let us denote by OX the collection of the k-algebras O(U), where U runs over all open subsets of X. The preceding proposition says that OX is a sheaf of k-valued functions on X, by which we mean the following:

DEFINITION 5.2. Let X be a topological space and K a ring. A sheaf O of K- valued functions8 on X assigns to every open subset U of X a K-subalgebra O(U) of the K-algebra of K-valued functions on U with the property that (i) for every inclusion U ⊂ U 0 of open subsets of X, ‘restriction to U’ maps O(U 0) in O(U) and (ii) given a collection (Ui)i∈I of open subsets of X and f : ∪i∈I Ui → K, then f ∈ O(∪iUi) if and only if f|Ui ∈ O(Ui) for all i.

If (X, OX ) and (Y, OY ) are topological spaces endowed with a sheaf of K- valued functions, then a continuous map f : X → Y is called a morphism if for −1 every open V ⊂ Y , composition with f takes OY (V ) to OX (f V ). The definition above is a way of expressing that we are dealing with a local property—just as we have a sheaf of continuous R-valued functions on a topological space, a sheaf of differentiable R-valued functions on a manifold and a sheaf of holomorphic C-valued functions on a complex manifold. In fact for the following definition, which includes our final definition of an affine variety, we take our cue from the definition of a manifold. With the notion of a morphism, we have a category of topological spaces endowed with a sheaf of K-valued functions. In particular, we have the notion of isomorphism: this is a homeomorphism f : X → Y which for every open V ⊂ Y −1 maps OY (V ) onto OX (f V ). Note that a sheaf O of K-valued functions on X restricts to a sheaf O|U for every open U ⊂ X. Here is our final definition of an affine variety.

DEFINITION 5.3. A topological space X endowed with a sheaf OX of k-valued functions is called an afﬁne k-variety resp. a quasi-afﬁne k-variety it is isomorphic to a pair as above.

Note that X can then be identiﬁed with the spectrum of OX (X) resp. an open subset thereof.

6. The product

Let m and n be nonnegative integers. If f ∈ k[x1, . . . , xm] and g ∈ k[y1, . . . , yn], then we have a regular function f ∗ g on Am+n deﬁned by

f ∗ g(x1, . . . , xm, y1, . . . , yn) := f(x1, . . . , xm)g(y1, . . . , yn). m+n m n It is clear that A (f ∗ g) = Af × Ag , which shows that the Zariski topology on Am+n refines the product topology on Am × An. Equivalently, if X ⊂ Am and Y ⊂ An are closed, then X × Y is closed in Am+n. We give X × Y the topology it inherits from Am+n (which is usually finer than the product topology). For the coordinate rings we have defined a map: k[X] × k[Y ] → k[X × Y ], (f, g) 7→ f ∗ g

8We give the general definition of a sheaf later. This will do for now. A defect of this definition is that a sheaf of K-valued functions on X need not restrict to one on an arbitrary subspace of X. 26 1. AFFINE VARIETIES which is evidently k-bilinear (i.e., k-linear in either variable). We want to prove that the ideal I(X × Y ) defining X × Y in Am+n is generated by I(X) and I(Y ) (viewed as subsets of k[x1, . . . , xm, y1, . . . yn]) and that X × Y is irreducible when X and Y are. This requires that we translate the formation of the product into algebra. This centers around the notion of the tensor product, the definition of which we recall. (Although we here only need tensor products over k, we shall define this notion for modules over a ring, as this is its natural habitat. This is the setting that is needed later anyhow.)

If R is a ring and M and N are R-modules, then we can form their tensor product over R, M ⊗R N: as an abelian group M ⊗R N is generated by the expressions a ⊗R b, a ∈ M, 0 0 b ∈ N and subject to the conditions (ra) ⊗R b = a ⊗R (rb), (a + a ) ⊗R b = a ⊗R b + a ⊗R b 0 0 and a ⊗R (b + b ) = a ⊗R b + a ⊗R b . So a general element of M ⊗R N can be written PN like this: i=1 ai ⊗R bi, with ai ∈ M and bi ∈ N. We make M ⊗R N an R-module if we stipulate that r(a ⊗R b) := (ra) ⊗R b (which is then also equal to r ⊗R (rb)). Notice that the map ⊗R : M × N → M ⊗R N, (a, b) 7→ a ⊗R b, is R-bilinear (if we ﬁx one of the variables, then it becomes an R-linear map in the other variable). In case R = k we shall often omit the sufﬁx k in ⊗k.

EXERCISE 23. Prove that ⊗R is universal for this property in the sense that every R- bilinear map M×N → P of R-modules is the composite of ⊗R and a unique R-homomorphism M ⊗R N → P . In other words, the map

HomR(M ⊗R N,P ) → BilR(M × N,P )), f 7→ f ◦ ⊗R is an isomorphism of R-modules.

EXERCISE 24. Let m and n be nonnegative integers. Prove that Z/(n) ⊗Z Z/(m) can be identiﬁed with Z/(m, n).

If A is an R-algebra and N is an R-module, then A ⊗R N acquires the structure of an A-module which is characterized by 0 0 a.(a ⊗R b) := (aa ) ⊗R b.

For instance, if N is an R-vector space, then C ⊗R N is a complex vector space, the complexi- ﬁcation of N. If A and B are R-algebras, then A⊗R B acquires the structure of an R-algebra characterized by 0 0 0 0 (a ⊗R b).(a ⊗R b ) := (aa ) ⊗R (bb ).

Notice that A → A ⊗R B, a 7→ a ⊗R 1 and B → A ⊗R B, b 7→ 1 ⊗R b are R-algebra homomorphisms. For example, A ⊗R R[x] = A[x] as A-algebras (and hence A ⊗R R[x1, . . . , xn] = A[x1, . . . , xn] with induction).

EXERCISE 25. Prove that C ⊗R C is as a C-algebra isomorphic to C ⊕ C with componentwise multiplication.

PROPOSITION 6.1. For closed subsets X ⊂ Am and Y ⊂ An the bilinear map k[X] × k[Y ] → k[X × Y ], (f, g) 7→ f ∗ g induces an isomorphism µ : k[X] ⊗ k[Y ] → k[X × Y ] of k-algebras (so that in particular k[X] ⊗ k[Y ] is reduced). If X and Y are irreducible, then so is X × Y , or equivalently, if k[X] and k[Y ] are domains, then so is k[X] ⊗ k[Y ].

PROOF. Since the obvious map

k[x1, . . . , xm] ⊗ k[y1, . . . , yn] → k[x1, . . . , xm, y1, . . . , yn] 7. FUNCTION FIELDS AND RATIONAL MAPS 27 is an isomorphism, it follows that µ is onto. In order to prove that µ is injective, PN let let ﬁrst observe that every u ∈ k[X] ⊗ k[Y ] can be written u = i=1 fi ⊗ gi with g1, . . . , gN are k-linearly independent. Given p ∈ X, then the restriction of PN ∼ PN µ(u) = i=1 fi∗gi to {p}×Y = Y is the regular function up := i=1 fi(p)gi ∈ k[Y ]. Since the gi’s are linearly independent, we have up = 0 if and only if fi(p) = 0 for all i. In particular, the subset X(u) ⊂ X of p ∈ X for which up = 0, is equal to N ∩i=1Z(fi) and hence closed. If µ(u) = 0, then up = 0 for all p ∈ X and hence fi = 0 for all i. So u = 0. This proves that µ is injective. Suppose now X and Y irreducible and suppose that u is zero divisor: uv = 0 for some v ∈ k[X × Y ] with u, v 6= 0. Since the restriction of uv = 0 to {p} × Y =∼ Y is upvp and k[Y ] is a domain, it follows that up = 0 or vp = 0. So X = X(u) ∪ X(v). Since X is irreducible we have X = X(u) or X(v). This means that u = 0 or v = 0, which contradicts our assumption. EXERCISE 26. Let X and Y be closed subsets of afﬁne spaces. Prove that each irreducible component of X × Y is the product of an irreducible component of X and one of Y .

It is clear that the projections πX : X × Y → X and πY : X × Y → Y are regular. We have observed that X × Y has not a topological product in general. Still it is the ‘right’ product in the sense of category theory: it has the following universal property, almost seems too obvious to mention: if Z is a closed subset of some affine space, then any pair of regular maps f : Z → X, g : Z → Y defines a regular map Z → X × Y characterized by the property that its composite with πX resp. πY yields f resp. g (this is of course (f, g)). 7. Function fields and rational maps In this section we interpret the fraction ring of an algebra of regular functions. Let Y be an affine variety. Recall that for f, g ∈ k[Y ], a fraction f/g ∈ Frac(k[Y ]) is defined if g not a zerodivisor. We need the following lemma.

LEMMA 7.1. Let Y be an afﬁne variety and let g ∈ k[Y ]. Then g is not a zero divisor if and only if g is nonzero on every irreducible component of Y . This is also equivalent to Yg being dense in Y .

PROOF. Suppose g is zero on some irreducible component C of Y and suppose C 6= Y . If Y 0 denotes the union of the irreducible components of Y distinct form C, then either Y 0 = ∅ (and so g = 0 in k[Y ]) or Y 0 6= ∅ and so there exists a nonzero 0 0 0 0 0 g ∈ IY (Y ). Then gg vanishes on C ∪ Y = Y and so gg = 0. It follows that g is then a zero divisor. Conversely, assume that g is a zerodivisor of k[Y ]. So there exists a nonzero g0 ∈ k[Y ] with gg0 = 0. Let C be an irreducible component of Y on which g0 is nonzero. Then we must have g|C = 0. The proof of the last clause is left to you.

This lemma tells us that this means that the affine variety Yg is open-dense in 0 0 Y . Clearly, f/g defines a regular function Yg → k. Another such fraction f /g yields a regular function on an open-dense U(g0) and their product f/g.f 0/g0 = ff 0/gg0 resp. difference f/g−f 0/g0 = (fg0 −f 0g)/gg0 in Frac(k[Y ]) defines a regular function on U(gg0) which is there just the product resp. difference of the associated 28 1. AFFINE VARIETIES regular functions. In particular, if f 0/g0 = f/g in k[Y ], then the two associated regular functions coincide on U(gg0). There is a priori no best way to represent a given element of Frac(k[Y ]) as a fraction (as there is in a UFD), and so we must be content with the observation that an element of Frac(k[Y ]) defines a regular function on some (unspecified) open dense subset of Y , with the understanding that two such functions are regarded as equal if they coincide on an open-dense subset contained in a common domain of definition. When an element of Frac(k[Y ]) is thus understood, then we call it a rational function on Y . This is the algebro- geometric analogue of a meromorphic function in complex function theory. When Y is irreducible, then Frac(k[Y ]) is a field, called the function field of Y , and will be denoted k(Y ). We shall now give a geometric interpretation of finitely generated field extensions of k and the k-linear field homomorphisms between them. We begin with a notion that generalizes that of a rational function.

DEFINITION 7.2. Let X and Y be affine varieties. A rational map from X to Y is given by a pair (U, F ), where U is an affine open-dense subset of X and F : U → Y is a morphism, with the understanding that a pair (V,G) defines the same rational map if F and G coincide on U ∩ V . We denote a rational map like this f : X 99K Y . We say that the rational map is dominant if for a representative pair (U, F ), F (U) is dense in Y . (This is then also so for any other representative pair. Why?)

PROPOSITION 7.3. Any finitely generated field extension of k is k-isomorphic to the function field of an irreducible affine variety. If X and Y are irreducible and affine, then a dominant rational map f : X 99K Y determines a k-linear field embedding f ∗ : k(Y ) ,→ k(X) and every k-linear field embedding k(Y ) → k(X) is of this form (for a unique f).

PROOF. Let K/k be a finitely generated field extension of k: there exist elements a1, . . . , an ∈ K such that every element of K can be written as a fraction of polynomials in a1, . . . , an. So if R denotes the k-subalgebra R of K generated by a1, . . . , an, (a domain since K is a field), then K is the field of fractions of R. Since R is the coordinate ring of a closed irreducible subset X ⊂ An (defined by the kernel of the obvious ring homomorphism k[x1, . . . , xn] → R), it follows that K can be identified with k(X). Suppose we are given an open-dense affine subvariety U ⊂ X and a morphism F : U → Y with F (U) dense in Y . Now F ∗ : k[Y ] → k[U] will be injective, for if g ∈ k[Y ] is in the kernel: F ∗(g) = 0, then F (U) ⊂ Z(g0). Since F (U) is dense in Y , this implies that Z(g) = Y , in other words, that g = 0. Hence the composite map k[Y ] → k[U] → k(U) = k(X) is an injective homomorphism from a domain to a field and therefore extends to a field embedding k(Y ) → k(X). It remains to show that every k-linear field homomorphism Φ: k(Y ) → k(X) is so obtained. For this, choose generators b1, . . . , bm of k[Y ]. Then Φ(b1),..., Φ(bm) are rational functions on X and so are regular on an open-dense affine subset U ⊂ X. Since b1, . . . , bm generate k[Y ] as a k-algebra, it follows that Φ maps k[Y ] to k[U] ⊂ k(X). This is a k-algebra homomorphism and so defines a morphism F : U → Y such that F ∗ = Φ|k[Y ]. The image of F will be dense by the argument above: if the image of F is contained in a closed subset of Y distinct from Y , then its maps to some Z(g) with g ∈ k[Y ] − {0}. But this implies that Φ(g) = F ∗(g) = 0, 7. FUNCTION FIELDS AND RATIONAL MAPS 29 which cannot happen, since Φ is injective. It is clear that Φ is the extension of F ∗ to the function fields. As to the uniqueness: if (V,G) is another solution, then choose a nonempty affine open-dense subset W ⊂ U ∩ V so that F and G both define morphisms W → Y . The maps k[Y ] → k[W ] they define are equal, for are given by Φ. Hence F and G coincide on W . Since W is dense in U ∩V , they also coincide on U ∩V .

EXERCISE 27. Let f ∈ k[x1, . . . , xn+1] be irreducible of positive degree. Its zero n+1 set X ⊂ A is then closed and irreducible. Denote by d the degree of f in xn+1. (a) Prove that the projection π : X → An induces an injective k-algebra ∗ homomorphism π : k[x1, . . . , xn] → k[X] = k[x1, . . . , xn]/(f) if and only if d > 0. (b) Prove that for d > 0, π is dominant and that the resulting ﬁeld homomorphism k(x1, . . . , xn) → k(X) is a ﬁnite extension of degree d.

COROLLARY 7.4. We have a category with the irreducible affine varieties as objects and the rational dominant maps as morphisms. Assigning to an irreducible affine variety its function field makes this category anti-equivalent to the category of finitely generated field extensions of the base field k.

PROPOSITION-DEFINITION 7.5. A rational map f : X 99K Y is an isomorphism in the above category (that is, induces a k-linear isomorphism of function fields) if and only if there exists a representative pair (U, F ) of f such that F maps U isomorphically onto an open subset of Y . If these two equivalent conditions are satisfied, then f is called a birational map. If a birational map X 99K Y merely exists (in other words, if there exists a k-linear field isomorphism between k(X) and k(Y )), then we say that X and Y are birationally equivalent.

PROOF. If f identifies a nonempty open subset of X with one of Y , then f ∗ : k(Y ) → k(X) is clearly a k-algebra isomorphism. Suppose now we have a k-linear isomorphism k(Y ) =∼ k(X). Represent this isomorphism and its inverse by (U, F ) and (V,G) respectively. Then F −1V is affine and open-dense in U and GF : F −1V → X is defined. Since GF induces the identity on k(X), GF is the identity on a nonempty open subset of F −1V and hence on all of F −1V . This implies that F maps F −1V injectively to G−1U. For the same reason, G maps G−1U injectively to F −1V . So F defines an isomorphism −1 −1 between the open subsets F V ⊂ X and G U ⊂ Y .

EXERCISE 28. We here assume k not of characteristic 2. Let for X ⊂ A2 be 2 2 1 defined by x1 + x2 = 1. Prove that X is birationally equivalent to the affine line A . (Hint: take a look at Exercise 17.) More generally, prove that the quadric in An+1 2 2 2 n defined by x1 + x2 + ··· xn+1 = 1 is birationally equivalent to A . What about the n+1 2 2 2 quadric cone in A defined by x1 + x2 + ··· xn+1 = 0 (n ≥ 2)? In case k(X)/k(Y ) is a finite extension, one may wonder what the geometric meaning is of the degree d of that extension, perhaps hoping that this is just the number of elements of a general fiber of the associated rational map X 99K Y . We will see that this is often true (namely when the characteristic of k is zero, or more generally, when this characteristic does not divide d), but not always, witness the following example. 30 1. AFFINE VARIETIES

EXAMPLE 7.6. Suppose k has characteristic p > 0. We take X = A1 = Y and let f be the Frobenius map: f(a) = ap. Then f is homeomorphism, but f ∗ : k[Y ] → k[X] is given by y 7→ xp and so induces the field extension k(y) = k(xp) ⊂ k(x), which is of degree p. From the perspective of Y , we have enlarged its algebra of regular functions by introducing a formal pth root of its coordinate (which yields another copy of A1, namely X). From the perspective of X, k[Y ] is just f ∗k[X]. This is in fact the basic example of a purely inseparable field extension, i.e., a field extension L/K with the property that every element of L has a minimal polynomial in K[T ] that has at precisely one root in L. Such a polynomial must be of r the form T p − c, with c ∈ K not a p-th power of an element of K when r > 0, where p is the characteristic of K (for p = 0 we necessarily have L = K). So if L 6= K, then the Frobenius map Φ: a ∈ K 7→ ap ∈ K is not surjective. Purely inseparable extensions are not detected by Galois theory, for they have trivial Galois group: after all, there is only one root to move around.

For any algebraic extension L/K, the elements of L that are separable over K make up an intermediate extension Lsep/K that is (of course) separable and is such that L/Lsep is purely inseparable. When L is an algebraic closure of K, then Lsep is called a separable algebraic closure of K: it is a separable algebraic extension of K which is maximal for that property. Then L is as an extension of Lsep obtained by successive adjunction of p-power roots of elements of Lsep: it is an extension minimal for the property of making the Frobenius map Φ: a ∈ L 7→ ap ∈ L surjective. In case L/K is a finite normal extension (i.e., a finite extension with the property that every f ∈ K[x] that is the minimal polynomial of some element of L has all roots in L), then the fixed point set of the Galois group of L/K is a subextension Linsep/L that is purely inseparable, whereas L/Linsep separable. In that case sep insep the natural map L ⊗K L → L is an isomorphism of K-algebras.

EXERCISE 29. Let f : X → Y be a dominant morphism of irreducible affine varieties which induces a purely inseparable field extension k(Y )/k(X). Prove that there is an open-dense subset U ⊂ Y such that f restricts to a homeomorphism f −1U → U. (Hint: show first that it suffices to treat the case when k(X) is obtained from k(Y ) by adjoining the pth root of an element f ∈ k(Y ). Then observe that if f is regular on the affine open-dense U ⊂ Y , then Y contains as an open dense subset a copy of the locus of (x, t) ∈ U × A1 satisfying tp = f.)

If K/k is a finitely generated field extension and generated by b1, . . . , bm, say, then we may after renumbering assume that for some r ≤ m, b1, . . . , br are algebraically independent (so that the k-linear field homomorphism k(x1, . . . , xr) → K which sends xi to bi is injective) and br+1, . . . , bm are algebraic over k(b1, . . . , br). This number r is invariant of K/k, called its transcendence degree. So if K = k(X), r then this defines a dominant rational map X 99K A . We now recall the theorem of the primitive element:

THEOREM 7.7. Every separable ﬁeld extension L/K of ﬁnite degree is generated by a single element (which we then call a primitive element for L/K).

It follows that if br+1, . . . , bm are separable over k(b1, . . . , br), then we can ﬁnd a b ∈ K that is algebraic over k(b1, . . . , br) such that K is obtained from k(b1, . . . , br) by adjoining b. This b is a root of an irreducible polynomial F ∈ k(b1, . . . , br)[T ]. If we clear denominators, we may assume that the coefﬁcients of 8. FINITE MORPHISMS 31

F lie in k[b1, . . . , br][T ] = k[b1, . . . , br,T ]. Then we can take for X the hypersurface r+1 in A defined by F , when viewed as an element of k[x1, . . . , xr, xr+1]. In particular, every irreducible affine variety Y is birationally equivalent to a hypersurface in Ar+1, where r is the transcendence degree of k(Y ). This suggests that this transcendence degree might be a good way to define the dimension of Y . We shall return to this.

Much of the algebraic geometry in the 19th century and early 20th century was of a birational nature: birationally equivalent varieties were regarded as not really different. This sounds rather drastic, but it turns out that many interesting properties of varieties are an invariant of their birational equivalence class.

Here is an observation which not only illustrates how affine varieties over algebraically nonclosed fields can arise when dealing with affine k-varieties, but one that is also a tell-tale sign that we ought to enlarge the maximal ideal spectrum. Let f : X → Y be a dominant morphism of irreducible affine varieties. This implies that f ∗ : k[Y ] → k[X] is injective and that f(X) contains an open-dense subset of Y . Then we may ask whether there exists something like a general fiber: is there an open-dense subset V ⊂ Y such that the fibers f −1(y), y ∈ V all “look the same”? The question is too vague for a clear answer and for most naive ways of making this precise, the answer will be no. But there is a cheap way out by simply refusing to specify one such V and allowing it to be arbitrarily small: we would like to take a limit over all open-dense subsets of Y . This we cannot do (yet) topologically, but we can do this algebraically by making all the nonzero elements of k[Y ] in k[X] invertible: we take −1 (k[Y ] − {0}) k[X] = k(Y ) ⊗k[Y ] k[X]

(this equality follows from the fact that k[X] = k[Y ] ⊗k[Y ] k[X]). This is a reduced finitely generated k(Y )-algebra and we may regard its maximal ideal spectrum Specm(k(Y )⊗k[Y ] k[X]) as an affine variety over the (algebraically nonclosed) field k(Y ): now every regular function on X which comes from Y is treated as a scalar (and will be invertible when nonzero). If we decide to add to Y =∼ Specm(k(Y )) a point η with residue field k(Y ), then we can simply say that the generic fiber of f is 9 the fiber Xη over η(Y ) . But once we decide to do this for Y , we should of course do this for every irreducible affine variety. In particular, we must then add for every irreducible subset Z of Y (or equivalently, for every prime ideal p ⊂ k[Y ]) a point to Specm(k(Y )) with residue field k(Z) (= Frac(k[Y ]/p)). We will later do this in the same kind of generality as the maximal ideal spectrum, namely for an arbitrary ring.

8. Finite morphisms Let A be a ring and B an A-algebra.

PROPOSITION-DEFINITION 8.1. We say that b ∈ B is integrally dependent on A if one the following equivalent properties is satisﬁed.

9It is often preferable to work over an algebraically closed field. We can remedy this simply by choosing an algebraic closure L of k(Y ). The maximal ideal spectrum of L ⊗k[Y ] k[X] is then an affine L-variety, and yields a notion of a general fiber that is even closer to our geometric intuition. 32 1. AFFINE VARIETIES

n n−1 (i) b is a root of a monic polynomial x + a1x + ··· + an ∈ A[x], (ii) A[b] is as an A-module finitely generated, (iii) A[b] is contained in a A-subalgebra C ⊂ B which is finitely generated as an A-module. In particular, if B is finite over A (which is short for: B is a finitely generated A-module), then B is integral over A (which is short for: every element of B is integral over A). n n−1 PROOF. (i) ⇒ (ii). If b is a root of x + a1x + ··· + an ∈ A[x], then clearly A[b] is generated as a A-module by 1, b, b2, . . . , bn−1. (ii) ⇒ (iii) is obvious. (iii) ⇒ (i). Suppose that C is as in (iii). Choose an epimorphism π : An → C n of A-modules and denote the standard basis of A by (e1, . . . , en). We may (and Pn will) assume that π(e1) = 1. By assumption, bπ(ei) = j=1 xijπ(ej) for certain n xij ∈ A. Put σ := (bδij − xij)i,j ∈ End(A[b] ). Then πσ = 0. 0 n 0 If σ ∈ End(A[b] ) is the matrix of cofactors of σ, then σσ = det(σ)1n by 0 Cramer’s rule. We find that det(σ) = det(σ)π(e1) = π(det(σ)e1) = π(σσ )(e1) = 0 (πσ)σ (e1) = 0. Clearly, det(σ) is a monic polynomial in b. The integral closure of A in B is the set elements of B that are integrally depen- B dent on A and will be denoted A . COROLLARY 8.2. The integral closure of A in B is an A-subalgebra of B. PROOF. It is enough to prove the following: if b, b0 ∈ B are integrally dependent over A, then every element of A[b, b0] is algebraically dependent over A. Since b0 is integrally dependent over A, it is so over A[b]. Hence A[b][b0] = A[b, b0] is a finitely generated A[b]-module. Since A[b] is a finitely generated A-module, it follows that A[b, b0] is a finitely generated A-module. Hence by (iii) every element of 0 A[b, b ] is integrally dependent over A. EXERCISE 30. Prove that ‘being integral over’ is transitive: if B is an A-algebra integral over A, then any B-algebra that is integral over B is as an A-algebra integral over A. We say that the ring homomorphism A → B is an integral extension if it is injective (so that A may be regarded as a subring of B) and B is integral over A. PROPOSITION 8.3. Let A ⊂ B be an integral extension and let p ⊂ A be a prime ideal of A. Then the going up property holds: p is of the form q ∩ A, where q is a prime ideal of B. If also is given is a prime ideal q0 of B with the property that q0 ∩ A ⊂ p, then we can take q ⊃ q0. Moreover the incomparability property holds: two distinct prime ideals of B having the same intersection with A cannot obey an inclusion relation (equivalently, if q0 ∩ A = p, then q = q0 is the only solution). If B is a domain, then Frac(B) is an algebraic field extension of Frac(A), which is finite whenever B is finite over A.

PROOF. Consider the localizations Ap resp. ApB obtained by making the elements of A − p invertible. Then Ap is a local ring with maximal ideal p˜ := pAp and ApB is an integral extension of Ap. If we ﬁnd a prime ideal ˜q of ApB with ˜q∩Ap = p˜, then q := ˜q ∩ B is a prime ideal of B the property that q ∩ A = ˜q ∩ A = p˜ ∩ A = p. Hence there is no loss in generality in assuming that A is a local ring and p is its unique maximal ideal mA. 8. FINITE MORPHISMS 33

We claim that mAB 6= B. Otherwise 1 ∈ B can be written as an mA-linear combination of a finite set elements of B. Denote by B0 the A-algebra generated by this finite set. Then B0 is an integral extension of A and finitely generated as an 0 0 A-algebra. This implies that B is a finitely generated A-module. Since 1 ∈ mAB 0 0 we have B = mAB , and it then follows from Nakayama’s lemma (recalled below) that B0 = 0. This is clearly a contradiction. Now that we know that mAB 6= B, we can take for q any maximal ideal of B which contains the ideal mAB: then qB ∩ A is a maximal ideal of A, hence equals mA. For the refinement and the incomparability property we can, simply by passing to the integral extension A/(q0 ∩ A) ⊂ B/q0, assume that q0 = 0. This reduces the refinement to the case already treated, while the incomparability property then amounts showing that q 6= (0) implies q ∩ A 6= (0). If b is a nonzero element of q, n n−1 then it is a root of a monic polynomial: b +a1b +···+an−1b+an = 0 (ai ∈ A), where we can of course assume that an 6= 0 (otherwise divide by b). We then find that 0 6= an ∈ Bb ∩ A ⊂ q ∩ A. Since B is a union of finitely generated A-modules, the last clause follows if we prove that Frac(B) = Frac(A)B. Let b ∈ B. It is a root of a monic polynomial n n−1 n−1 x + a1x + ··· + an = 0 (ai ∈ A) with an 6= 0 and so 1/b = −1/an.(b + n−2 a1b + ··· + an−1) ∈ Frac(A)B. REMARK 8.4. Note that the going-up property implies that for any strictly ascending sequence p0 ( p1 ( ··· ( pn of prime ideals in A is the intersection with A of a (necessarily strictly) ascending sequence of prime ideals q0 ⊂ q1 ⊂ · · · ⊂ qn in B. Conversely, the incomparability property implies that if we intersect a strictly ascending sequence of prime ideals q0 ( q1 ( ··· ( qn in B with A, we get a strictly ascending sequence of prime ideals of A. For completeness we state and prove:

LEMMA 8.5 (Nakayama’s lemma). Let R be a local ring with maximal ideal m and M a ﬁnitely generated R-module. Then a ﬁnite subset S ⊂ M generates M as a R-module if (and only if) the image of S in M/mM generates the latter as a R/m-vector space. In particular (take S = ∅), mM = M implies M = 0.

PROOF. In fact, we ﬁrst reduce to the case when S = ∅ by passing to M/RS. Our assumptions then say that M = mM. We must show that M = 0. Let π : Rn → M be an epimorphism of R-modules and denote the standard basis of Rn by Ps (e1, . . . , en). By assumption there exist xij ∈ m such that π(ei) = j=1 xijπ(ej). n So if σ := (δij − xij)i,j ∈ End(R ), then πσ = 0. Notice that det(σ) ∈ 1 + m. Since 1 + m consists of invertible elements, Cramer’s rule shows that σ is invertible. So −1 −1 π = π(σσ ) = (πσ)σ = 0 and hence M = 0. EXERCISE 31. Prove that an integral extension A ⊂ B deﬁnes surjective morphisms Spec(B) → Spec(A) and Specm(B) → Specm(A).

DEFINITION 8.6. We say that a morphism of affine varieties f : X → Y is finite if the k-algebra homomorphism f ∗ : k[Y ] → k[X] is finite, i.e., makes k[X] a finitely generated k[Y ]-module.

EXERCISE 32. Prove that if Y is an afﬁne variety, then the disjoint union of its irreducible components is ﬁnite over Y . 34 1. AFFINE VARIETIES

COROLLARY 8.7. Let f : X → Y be a finite morphism of affine varieties such that f ∗ : k[Y ] → k[X] is injective. Then f is surjective and every closed irreducible subset Y 0 ⊂ Y is the image of a closed irreducible subset X0 ⊂ X. If furthermore is given a closed irreducible subset X00 ⊂ X with f(X00) ⊃ Y 0, then we may choose X0 ⊂ X00. If in addition X is irreducible, then so is Y and f ∗ : k(Y ) → k(X) is a finite algebraic extension.

∗ PROOF. Apply Proposition 8.3 to f .

EXERCISE 33. Prove that a finite morphism f : X → Y of affine varieties is closed and that every fiber f −1(y) is finite (possibly empty).

THEOREM 8.8 (Noether normalization). Let K be a field. Every finitely generated K-algebra B is a finite extension of a polynomial algebra over K: there exist an integer r ≥ 0 and an injection K[x1, . . . , xr] ,→ B such that B is finite over K[x1, . . . , xr]. In fact, if we are given an epimorphism φ : K[x1, . . . , xm] → B of K-algebras, then φ is an isomorphism or we can take in this statement r < m. The proof will be with induction on m and the induction step is worth recording separately.

LEMMA 8.9. Let φ : K[x1, . . . , xm] → B be an epimorphism of K-algebras, which is not an isomorphism. Then there exists a K-algebra homomorphism φ0 : 0 K[x1, . . . , xm−1] → B such that φ(xm) is integral over the image of φ .

PROOF. We deﬁne for any positive integer s a K-algebra automorphism σs of sm−i K[x1, . . . , xm] by σs(xm) = xm and σs(xi) := xi + xm for i < m (its inverse sm−i m ﬁxes xm and sends xi to xi − xm for i < m). So if I = (i1, . . . , im) ∈ Z≥0, and I i1 im x := x1 ··· xm , then m−1 I s i1 s im−1 im σs(x ) = (x1 + xm ) ··· (xm−1 + xm) xm .

When viewed as an element of K[x1, . . . , xm−1][xm], this is a monic polynomial in m−1 m−2 m xm of degree pI (s) := i1s + i2s + ··· + im. Now give Z≥0 the lexicographic ordering. Then I > J implies pI (s) > pJ (s) for s large enough. Choose a nonzero −1 m f ∈ ker(φ). Then σs(f) ∈ ker(φσs ). If I ∈ Z≥0 is the largest multi-exponent of a monomial that appears in f with nonzero coefﬁcient, then for s large enough, σs(f) is a constant times a monic polynomial in xm of degree pI (s) with coefﬁ- −1 cients in K[x1, . . . , xm−1] and hence φσs (xm) = φ(xm) becomes integral over −1 0 −1 φσs K[x1, . . . , xm−1]. So φ := φσs |K[x1, . . . , xm−1] is then as desired.

PROOFOF NOETHER NORMALIZATION. Let φ : K[x1, . . . , xm] → B be an epimorphism of K-algebras which is not an isomorphism (otherwise we are done). We prove the reﬁned version of the theorem with induction on m. According to 0 Lemma 8.9, there exists a K-algebra homomorphism φ : K[x1, . . . , xm−1] → B 0 0 0 such that φ(xm) is integral over B := φ (K[x1, . . . , xm−1]). By induction, B is a ﬁnite extension of some polynomial algebra K[x1, . . . , xr] with r ≤ m − 1. Hence so is B.

REMARK 8.10. If A is a domain, then according to Proposition 8.3, Frac(A) will be a ﬁnite extension of K(x1, . . . , xr) and so r must be the transcendence degree of Frac(A)/K. In particular, r is an invariant of A. 8. FINITE MORPHISMS 35

COROLLARY 8.11. For every affine variety Y there exists an integer r ≥ 0 and a finite surjective morphism f : Y → Ar. This corollary gives us a better grasp on the geometry of Y , especially when Y is irreducible, for it shows that Y can be spread over Ar in a finite-to-one manner over affine r-space. The last assertion of Proposition 8.3 has a converse, also due to Noether, which we state without proof.

*THEOREM 8.12 (Emmy Noether). Let K be a field and A be a finitely generated L K-domain. Then for any finite field extension L/ Frac(A), the integral closure A of A in L is finite over A.

L If we take L = Frac(A), then A is called the normalization of A. If A equals its normalization, then we say that A is normal. We carry this terminology to the geometric setting by saying that an irreducible afﬁne variety X is normal when k[X] is. Any nonempty open subset of An is normal. More generally:

LEMMA 8.13. Any unique factorization domain A is normal.

d d PROOF. Suppose b ∈ Frac(A) is integral over A: b +a1b +···+ad with ai ∈ A. d d−1 d Write b = r/s with r, s relatively prime. Then r + a1rs + ··· + ads and we see d that r is divisible by s. This can only happen when s is a unit, so that b ∈ A. Proposition 8.12 has a remarkable geometric interpretation: let be given an irreducible affine variety Y and a finite field extension L/k(Y ). Then Proposi- L tion 8.12 asserts that A is a finitely generated k[Y ]-module. It is also an domain (because it is contained in a field) and so it defines an irreducible variety L L YL := Spec(A ). Since A ⊂ A is an integral extension, we have a finite surjective morphism YL → Y . Notice that this morphism induces the given field extension L/k(Y ). This shows that every finite field extension of k(Y ) is canonically realized by a finite morphism of irreducible affine varieties! If L is a separable algebraic closure of k(Y ), then this does not apply, for L/k(Y ) will not be finite, unless Y is a singleton. But L can be written as a mono- ∞ tone union of finite (separable) field extensions: L = ∪i=1Li with Li ⊂ Li+1 and Li+1/Li finite. This yields a sequence of finite surjective morphisms

Y ← YL1 ← YL2 ← YL3 ← · · · of which the projective limit can be understood as a “pro-affine variety” (a point of ∞ this limit is given by a sequence (yi ∈ YLi )i=1 such that yi+1 maps to yi for all i). Its ∞ Li L algebra of regular functions is ∪i=1k[Y ] = k[Y ] (which is usually not a finitely generated k-algebra) and its function field is L.

Of special interest is the case of a ﬁnite Galois extension. We begin with the corresponding result from commutative algebra which has also important applications in algebraic number theory:

PROPOSITION 8.14. Let A be a normal domain and let L/ Frac(A) be a ﬁnite normal extension with Galois group G. Then G leaves invariant the integral closure L A of A in L, and for every prime ideal p ⊂ A, G acts transitively on the set of prime L ideals q ⊂ A that lie over p (i.e., with q ∩ A = p). 36 1. AFFINE VARIETIES

L PROOF. That any g ∈ G leaves invariant A is clear, for g fixes the coefficients of an equation of integral dependence over A. 0 0 0 Let q and q lie over p. We first show that q ⊂ ∪g∈Ggq. If b ∈ q , then the norm Q of b over Frac(A), which is defined as the product g∈G gb, lies in Frac(A), and L L hence in A ∩Frac(A). Since A is assumed to be normal, we have A ∩Frac(A) = A. Q 0 Hence g∈G gb ∈ A ∩ q = p ⊂ q. So we must have gb ∈ q for some g ∈ G, or equivalently, b ∈ g−1q. 0 From q ⊂ ∪g∈Ggq and the fact that each gq is a prime ideal, it follows from the prime avoidance lemma below that q0 ⊂ gq for some g ∈ G. Since both q0 and gq 0 lie over p, the incomparability property implies that q = gq.

LEMMA 8.15 (The prime avoidance lemma). Let R be a ring, q1,..., qn prime n ideals in R and I ⊂ R an ideal contained in ∪i=1qi. Then I ⊂ qi for some i. PROOF. We prove this induction on n. The case n = 1 being trivial, we may assume that n > 1 and that for every i = 1, . . . , n, I is not contained in ∪j6=iqj. Choose ai ∈ I \ ∪j6=iqj. So then ai ∈ qi. Consider a := a1a2 ··· an−1 + an. Then a ∈ I and hence a ∈ qi for some i. If i < n, then an = a − a1a2 ··· an−1 ∈ qi and we get a contradiction. If i = n, then a1a2 ··· an−1 = a − an ∈ qn and hence aj ∈ qn for some j ≤ n − 1. This is also a contradiction. We transcribe this to algebraic geometry:

COROLLARY 8.16. Let Y be a normal variety and L/k(Y ) be a normal Galois extension with Galois group G. Then G acts on YL in such a manner that it commutes 0 with f : YL → Y : for all g ∈ G we have fg = f. If Y ⊂ Y is closed an irreducible, then G acts transitively on the irreducible components of f −1Y 0. This also leads for normal domains to a supplement of the going up property:

COROLLARY 8.17 (Going down property). Suppose A is a normal domain and B ⊃ A an integral extension without zero divisors. Given prime ideals p in A and q0 in B such that q0 ∩ A ⊃ p, then there exists a prime ideal q in B with q ∩ A ⊂ p.

PROOF. Choose an algebraic closure Frac(B) of Frac(B) and let L be the subfield of Frac(B) generated by all the Frac(A)-embeddings of Frac(B) in Frac(B) (so this is obtained by adjoining the roots of the irreducible polynomials in Frac(A)[x] having a root in Frac(B)). Galois theory tells us that this is a finite normal extension of Frac(B) (with Galois group G, say), which brings us in the situation of L Proposition 8.14 above. Since L contains Frac(B) , A contains B. Observe that L A is finite over B and (hence) over A. L Put p0 := q0 ∩ A so that p0 ⊃ p. According to Proposition 8.3 we can find in A nested prime ideals p˜0 ⊃ p˜ which meet A in p0 ⊃ p. Similarly, there exists a prime L ideal ˜q0 in A which meets B in q0. Since ˜q0 and p0 both meet A in p0, there exists according to Proposition 8.14a g ∈ G which takes p˜0 to ˜q0. Then upon replacing p˜0 by g0p˜0, we may assume that p˜0 = ˜q0. Then q := p˜ ∩ B is as desired: it meets A in p 0 0 and is contained in p˜ ∩ B = ˜q ∩ B = q. 9. Dimension One way to define the dimension of a topological space X is as follows: the empty set has dimension −1 and X has dimension ≤ n if it admits a basis of open 9. DIMENSION 37 subsets such that the boundary of every basis element has dimension ≤ n − 1. This is close in spirit to the definition that we shall use here (which is however adapted to the Zariski topology; as you will find in Exercise 34, it is useless for Hausdorff spaces).

DEFINITION 9.1. Let X be a nonempty topological space. We say that the Krull dimension of X is at least d if there exists an irreducible chain of length d in X, that is, a strictly descending chain of closed irreducible subsets Y 0 ) Y 1 ) ··· ) Y d of X. The Krull dimension of X is the supremum of the d for which an irreducible chain of length d exists and we then write dim X = d. We stipulate that the Krull dimension of the empty set is −1.

LEMMA 9.2. For a subset Z of a topological space X we have dim Z ≤ dim X.

PROOF. If Y ⊂ Z is irreducible, then so is its closure Y¯ in X. So if we have an irreducible chain of length d in Z, then the closures of the members of this chain yield an irreducible chain of length d in X. This proves that dim Z ≤ dim X.

EXERCISE 34. What is the Krull dimension of a nonempty Hausdorff space?

EXERCISE 35. Let U be an open subset of the space X. Prove that for an irreducible chain Y • in X of length d with U ∩ Y d 6= ∅, U ∩ Y • is an irreducible chain of length d in U. Conclude that if U is an open covering of X, then dim X = supU∈U dim U.

EXERCISE 36. Suppose that X is a noetherian space. Prove that the dimension of X is the maximum of the dimensions of its irreducible components. Prove also that if all the singletons (= one element subsets) in X are closed, then dim(X) = 0 if and only if X is ﬁnite. It is straightforward to transcribe this notion of dimension into algebra:

DEFINITION 9.3. Let R be a ring. We say that the Krull dimension of R is at least d if there exists an prime chain of length d in R, that is, a strictly ascending chain p• of prime ideals p0 ( p1 ( ··· ( pd in R. The Krull dimension of R is the supremum of the d for which this is the case and we then write dim R = d. We stipulate that the trivial ring R = {0} (which has no prime ideals) has Krull dimension −1.

The prime ideals of a ring R are the preimages of the prime ideals of Rred := R/p(o) and so these rings have the same dimension. It is clear that for a closed subset X ⊂ An, dim k[X] = dim X. Similarly, we have that for a ﬁnitely generated k-algebra A, dim A = dim Spec(A). Remark 8.4 shows immediately:

LEMMA 9.4. The Krull dimension is invariant under integral extension: if B is integral over A, then A and B have the same Krull dimension. The notion of Krull dimension was easily defined, but can be difficult to use in concrete cases. How can we be certain that given prime chain has maximal possible length? It is not even clear how we can tell whether the Krull dimension of a given ring is finite. We will settle this for a finitely generated k-domain (i.e., of the form k[Y ] with Y an irreducible affine variety), by showing that this is just the transcendence degree over k of its field of fractions. 38 1. AFFINE VARIETIES

THEOREM 9.5. Let K be a ﬁeld. For a ﬁnitely generated K-domain A, the Krull dimension of A equals the transcendence degree of Frac(A)/K. Moreover, every maximal prime chain in A (i.e., one that cannot be extended to a longer prime chain) has length dim A.

PROOF. We prove both assertions with induction on the transcendence degree of Frac(A). By Noether normalization there exists an integer r ≥ 0 and a K-linear embedding of K[x1, . . . , xr] in A such that A is finite over K[x1, . . . , xr]. Then Frac(A) is a finite extension of K(x1, . . . , xr) and so the transcendence degree of Frac(A)/K is r. Let (0) = q0 ( q1 ( ··· ( qm be a prime chain of A of length m. By the incomparability property, p• := q• ∩ Sr is then a prime chain in Sr, also of length m. Choose an irreducible f ∈ p1. After renumbering the coordinates, we may PN i assume that f does not lie in K[x1, . . . , xr−1] and so f = i=0 aN−ixr with ai ∈ K[x1, . . . , xr−1], a0 6= 0 and N > 1. Then the image of xr in Frac(Sr/(f)) is a N PN−1 i root of the monic polynomial xr + i=0 (aN−i/a0)xr ∈ K(x1, . . . , xr−1)[xr] so that Frac(Sr/(f)) is a finite extension of K(x1, . . . , xr−1). Hence Frac(Sr/(f) has transcendence degree r − 1 over K. By our induction induction hypothesis, the Krull dimension of Sr/(f) equals r − 1. The image of the subchain p1 ( ··· ( pm in Sd/(f) will still be strictly ascending (for it will be so in Sr/p1) and so m−1 ≤ r−1. Hence m ≤ r. On the other hand, r is attained as the length of a prime chain in A, for the prime chain of length r in Sr given by (0) ( (x1) ( (x1, x2) ( ··· ( (x1, x2, . . . , xr), lifts to one in A, by Remark 8.4.

The last assertion relies on the going down theorem. If we assume that q• is a maximal prime chain in A, then the going down property Corollary 8.17 (which applies here since Sr is normal) implies that p• is one in Sr. Since our irreducible f generates a prime ideal, it must then generate p1. The image of p1 ( ··· ( pm in Sr/p1 = Sr/(f) is then a maximal prime chain in Sr/(f) and hence of length r − 1 by induction hypothesis. This shows that m = r. If S ⊂ R is a multiplicative system, then the preimage of a prime ideal ˜q of S−1R under the ring homomorphism R → S−1R is a prime ideal q of R which does not meet S and we have ˜q = S−1q. This sets up a bijection between the prime ideals of S−1R and those of R not meeting S. If we take S = R − p, with p a prime ideal, then this implies that dim Rp is the supremum of the prime chains in R which end with p. Since the prime chains in R which start with p correspond to prime chains in R/p, it follows that dim Rp +dim R/p is the supremum of the prime chains in R which contain p. According to Theorem 9.5, the latter is for a domain R ﬁnitely generated over a ﬁeld always equal to the dimension of R.

COROLLARY 9.6. Let X be an irreducible afﬁne variety. Then every irreducible chain in X is a subchain of one of length dim X. Moreover, every nonempty open afﬁne U ⊂ X has the same dimension as X.

EXERCISE 37. Prove that a hypersurface in An has dimension n − 1.

EXERCISE 38. Let X be an irreducible afﬁne variety and Y ⊂ X an irreducible closed subvariety. Prove that dim X − dim Y is equal to the Krull dimension of k[X]I(Y ). 10. NONSINGULAR POINTS 39

EXERCISE 39. Prove that when X and Y are irreducible affine varieties, then dim(X ×Y ) = dim X +dim Y . (Hint: Embed each factor as a closed subset of some affine space. You may also want to use the fact that the equality to be proven holds in case X = Am and Y = An.) 10. Nonsingular points In this section we focus on the local properties of an affine variety X = Spec(A) (so A = k[X] is here a reduced finitely generated k-algebra) at a point o and so a −1 central role will be played by the local algebra OX,o = mo A whose maximal ideal −1 is mX,o = (A − mo) mo.

If k = C and X ⊂ Cn is a closed subset of dimension d, then we hope that there is a nonempty open subset of X where X is ‘smooth’, i.e., where X looks like a complex submanifold of complex dimension d. Our goal is to deﬁne smoothness in algebraic terms (so that it make sense for our ﬁeld k) and then to show that the set of smooth points of a variety is open and dense in that variety. Our point of departure is the implicit function theorem. One version states that n n if U ⊂ R is an open neighborhood of o ∈ R and fi : U → R, i = 1, . . . n − d are differentiable functions such that fi(o) = 0 and the total differentials at o, df1(o), . . . , dfn−d(o) are linearly independent in o (this is equivalent to: the Jaco- bian matrix of (f1, . . . , fn−d) at o is of rank n − d), then the common zero set of f1, . . . , fn−d is a submanifold of dimension d at o whose tangent space there is the common zero set of df1(o), . . . , dfn−d(o). In fact, one shows that this solution set is there the graph of a map: we can express n − d of the coordinates as differentiable functions in the d remaining ones. Conversely, any submanifold of Rn at o of dimension d is locally thus obtained.

We begin with the observation that for any ring R partial differentiation of a polynomial f ∈ R[x1, . . . , xn] is well-defined and produces another polynomial. −1 The same goes for a fraction φ = f/g in R[x1, . . . , xn][g ]: a partial derivative of φ is a rational function (in this case with denominator g2). We then define the total differential of a rational function φ ∈ R[x1, . . . , xn] as usual: n X ∂φ dφ := (x)dx , ∂x i i=1 i where for now, we do not worry about interpreting the symbols dxi: we think of dφ simply as a regular map from an open subset of An to a k-vector space of dimension n with basis dx1, . . . , dxn, leaving its intrinsic characterization for later. However, caution is called for when R is a field of positive characteristic:

EXERCISE 40. Prove that f ∈ k[x] has zero derivative, if and only if f is constant or (when char(k) = p > 0) a pth power of some g ∈ k[x]. Generalize this to: given f ∈ k[x1, . . . , xn], then df = 0 if and only if f is constant or (when char(k) = p > 0) a pth power of some g ∈ k[x1, . . . , xn]. We should also be aware of the failure of the inverse function theorem: EXAMPLE 10.1. Let C ⊂ A2 be the curve defined by y2 = x3 + x. By any reasonable definition of smoothness we should view the origin (0, 0) as a smooth point of C. Indeed, when k = C, the projection f : C → A1, (x, y) 7→ y, would 40 1. AFFINE VARIETIES be a local-analytic isomorphism. But the map is not locally invertible within our category: the inverse requires us to find a rational function x = u(y) which solves the equation y2 = x3 + x and it is easy to verify that none exists. (We can solve for x formally: x = u(y) = y2 − y6 + 3y10 + ··· , where it is important to note that the coefficients are all integers so that this works for every characteristic.) In fact, the situation is worse: no affine neighborhood U of (0, 0) in C is isomorphic to an open subset V of A1. The reason is that this would imply that k(C) = k(U) is isomorphic to k(V ) = k(x) and one can show that this is not so. Somewhat related to this is an issue illustrated by the following example.

EXAMPLE 10.2. Consider the curve C0 ⊂ A2 deﬁned by xy = x3 + y3 and let o = (0, 0). The polynomial x3 + y3 − xy is irreducible in k[x, y], so that k[C0] is 0 0 without zero divisors. Hence OC0,o ⊂ k(C ) is also without zero divisors. But C seems to have two branches at o which apparently can only be recognized formally: one such branch is given by y = u(x) = x2 + x5 + 3x8 + ··· and the other by interchanging the roles of x and y: x = v(y) = y2 + y5 + 3y8 + ··· . If we use ξ := x − v(y) and η := y − u(x) as new formal coordinates, then C0 is simply given at 0 by the reducible equation ξη = 0. These examples make it clear that for a local understanding of a variety X at o, the local ring OX,o still carries too much global information, and that one way to get rid of this overload might be by passing formal power series. This is accomplished by passage to what is known as formal completion.

10.3.F ORMALCOMPLETION. Let R be a ring and I ( R a proper ideal. We endow every R-module M with a topology, the I-adic topology of which a basis is the collection additive translates of the submodules InM, i.e., the collection of subsets a + InM, a ∈ M, n ≥ 0. This is a topology indeed: given two basic subsets 0 a + InM, a0 + In M, then for any element c in their intersection, the basic subset 0 c + Imax{n,n }M is also in their intersection. The fact that our basis is translation invariant implies that with this topology, M is a topological abelian group ((a, b) ∈ M × M 7→ a − b ∈ M is continuous). If we endow R also with the I-adic topology and R × M with the product topology, then the map (r, a) ∈ R × M → ra ∈ M which gives the action by R is also continuous. For the topology on M to be Hausdorff, it suffices that the intersection of all n neighborhoods of 0 is reduced to {0}: ∩n≥0I M = {0}. It is then even metriz- able: if φ : Z+ → (0, ∞) is any strictly monotonously decreasing function with −n limn→∞ φ(n) = 0 (one often takes u , where u > 1), then a metric ρ is defined by ρ(a, a0) := inf{φ(n): a − a0 ∈ InM}. This metric is nonarchimedean in the sense that ρ(a, a00) ≤ max{ρ(a, a0), ρ(a0, a00)}. ∞ and a sequence (an ∈ M)n=0 is then a Cauchy sequence if and only if for every integer n ≥ 0 all but finitely many terms lie in the same coset of InM in M; n in other words, there exists an index in ≥ 0 such that aj − ain ∈ I M for all j ≥ in. This makes it clear that the notion of Cauchy sequence is independent of the choice of φ. We also note that the Cauchy sequence defines a sequence of n cosets (αn ∈ M/I M)n≥0 with the property that αn is the reduction of αn+1 ∈ M/In+1. Recall that a metric space is said to be complete if every Cauchy sequence in that space converges. A standard construction produces a completion of every metric space M: its points are represented by Cauchy sequences in M, with the 10. NONSINGULAR POINTS 41 understanding that two such sequences represent the same point if the distance between the two nth terms goes to zero as n → ∞. In the present situation this yields what is called the I-adic completion, Mˆ I . From the above discussion we see n that an element of Mˆ I is uniquely given by a sequence (αn ∈ M/I M)n≥0 whose terms are compatible in the sense that αn is the reduction of αn+1 for all n. It is an R-module for componentwise addition and multiplication so that the obvious n map M → Mˆ I , a 7→ (a + I )n≥0 is a R-module homomorphism. (This makes Mˆ I a Q∞ n ˆ submodule of the R-module n=0 M/I M.) It is easy to verify that MI is complete for the Iˆ-adic topology and that the homomorphism M → Mˆ I is continuous and has dense image. ˆ Q∞ n ˆ Notice that RI is in fact a subring of n=0 R/I in a way that makes R → RI n n a ring homomorphism. Moreover, Mˆ I is a RÎ -module (use that M/I M is a R/I - module for every n) and an R-homomorphism φ : M → N of R-modules extends ˆ n n uniquely to an RÎ -homomorphism φI : Mˆ I → NÎ (use that φ sends I M to I N and hence induces a homomorphism M/In → N/In).

EXAMPLE 10.4. Take the ring k[x1, . . . , xn]. Its completion relative the maximal ideal (x1, . . . , xn) is just the ring of formal power series k[[x1, . . . , xn]]. We get the n same result if we do this for the localization OA ,o of k[x1, . . . , xn] at (x1, . . . , xn). EXAMPLE 10.5. Take the ring Z. Its completion with respect to the ideal (p), ˆ ˆ p prime, yields the ring of p-adic integers Z(p): an element of Z(p) is given by a i ∞ sequence (ai ∈ Z/(p ))i=1 with the property that ai is the image of ai+1 under the reduction Z/(pi+1) → Z/(pi). We get the same result if we do this for the localization Z(p) = {a/b : a ∈ Z, b ∈ Z − (p)}. If n is an integer ≥ 2, then it follows ˆ Q ˆ from the Chinese remainder theorem that Z(n) = p|n Z(p).

EXERCISE 41. Let I and J be ideals of a ring R and m a positive integer with J m ⊂ I. Prove that the J-adic topology is finer than the I-adic topology and that there is a natural continuous ring homomorphism RˆJ → RÎ . Conclude that when ˆ ˆ√ R is noetherian, RI can be identified with R I . *LEMMA 10.6 (Artin-Rees). Let R be a noetherian ring, I ⊂ R an ideal, M a finitely generated R-module and M 0 ⊂ M a R-submodule. Then there exists an integer n0 ≥ 0 such that for all n ≥ 0: M 0 ∩ In+n0 M = In(M 0 ∩ In0 M). The proof (which is ingeneous, but not difficult) can be found in any book on commutative algebra (e.g., [1]). The Artin-Rees lemma implies that for every n ≥ 0 0 0 n0 n 0 0 there exists a n ≥ 0 such that M ∩ I M ⊂ I M (we can take n = n + n0). This implies that the inclusion M 0 ⊂ M is not merely continuous, but that the I-adic topology on M 0 is induced by that of M. We will use the Artin-Rees lemma via this property only.

ˆ 0 ˆ COROLLARY 10.7. In the situation of Lemma 10.6, the homomorphism MI → MI induced by the inclusion M 0 ⊂ M is a closed embedding with the property that the 0 preimage of M is M : if the image of a ∈ M under M → Mˆ I lies in the image ˆ 0 ˆ 0 ˆ ˆ 0 of MI → MI , then a ∈ M . Moreover, MI /MI can be identiﬁed with the I-adic completion of M/M 0. 42 1. AFFINE VARIETIES

One might phrase this by saying that when R is noetherian, I-adic completion is an exact functor on the category of ﬁnitely generated R-modules.

PROOFOF COROLLARY 10.7. We ﬁrst part of the corollary amounts to the fol- 0 ∞ 0 lowing property: if a sequence (ai ∈ M )i=0 in M is a Cauchy sequence in M, then it is already one in M 0 and when its limit lies in the image of M, then this limit 0 ˆ 0 ˆ lies in fact in the image of M (an element of the kernel of MI → MI is given by 0 ∞ 0 a sequence (ai ∈ M )i=0 in M which converges in M to 0 and so this implies that ˆ 0 ˆ MI → MI is injective). 0 ∞ Suppose therefore given a sequence (ai ∈ M )i=0 which is a Cauchy sequence n in M. Then for every n there exists an in such that ai − ain ∈ I M for i ≥ in. So if 0 n+n0 n 0 n0 n 0 we take i ≥ in+n0 , then ai − ain ∈ M ∩ I M = I (M ∩ I M) ⊂ I M . Hence 0 ∞ 0 0 ∞ (ai ∈ M )i=0 is a Cauchy sequence in M . If (ai ∈ M )i=0 converges to the image 0 n 0 of a ∈ M, then for every n there exists an in such that ai − a ∈ I M for i ≥ in. i ≥ i0 a − a ∈ InM 0 So if we take n+n0 , then by the preceding argument, i . Hence n 0 0 0 ∞ 0 a ∈ ai + I M ⊂ M and (ai ∈ M )i=0 converges to the image of a in M . ˆ ˆ 0 As to the last statement, we observe that RI -homomorphism MI → M/M\ I 0 ˆ 0 induced by the surjection M → M/M is also surjective: any b ∈ M/M\ I is rep- 0 ∞ n 0 resentable by a Cauchy sequence (bi ∈ M/M )i=0 with bm − bn ∈ I (M/M ) for n 0 ∼ n n 0 m ≥ n. Since I (M/M ) = I M/(I M ∩ M ), we can represent bn+1 − bn by some n dn+1 ∈ I M. Let d0 ∈ M lift b0. Then the series d0 + d1 + ··· + dn + ··· converges in Mˆ and maps to ˆb. 0 0 The kernel of MI → M/M\ I clearly contains M and since a kernel is closed, ˆ 0 ˆ it will also contain its closure MI in MI . In order to see that it is not bigger, 0 let aˆ ∈ ker(MI → M/M\ I ) be arbitrary. If we represent aˆ by a Cauchy sequence ∞ 0 n (ai ∈ M)i=0, then for every n there exists an in such that for i ≥ in, ai ∈ M +I M. 0 0 0 n ∞ Write ai accordingly: ai = ai + di with ai ∈ M and di ∈ I M. Since (di ∈ M)i=0 0 0 ∞ converges to 0 ∈ M, (ai ∈ M )i=0 is also a Cauchy sequence converging to aˆ. As we saw above, this is then a Cauchy sequence in M 0 with the same limit and so ˆ 0 aˆ ∈ MI .

Let R be a noetherian local ring with maximal ideal m and residue field κ. We use simply b for completion with respect to m. Then m is a finitely generated R-module. Since the ring R acts on m/m2 via R/m = κ, m/m2 is a finite dimensional vector space over κ.

DEFINITION 10.8. The Zariski cotangent space T ∗(R) of R is the κ-vector space 2 2 m/m and its κ-dual, T (R) := Homκ(m/m , κ) (which is also equal to HomR(m, κ)), is called the Zariski tangent space T (R) of R. The embedding dimension embdim(R) is the dimension of m/m2 as a vector space over κ. ∗ If X is an affine variety and o ∈ X, then the Zariski cotangent space To X resp. the Zariski tangent space ToX resp. the embedding dimension embdimo X of X at o are by definition that of OX,o. For instance, the embedding dimension of An at any point o ∈ An is n. This n n follows from the fact that the map do : f ∈ mA ,o 7→ df(o) ∈ k defines an isomor- 2 ∼ n phism of k-vector spaces m n /m n = k . We note in passing that we here have A ,o A ,o a way of understanding the total differential at o in more intrinsic terms as the map 2 d : O n → m n /m n which assigns to f ∈ O n the image of f −f(o) ∈ m n o A ,o A ,o A ,o A ,o A ,o 10. NONSINGULAR POINTS 43

2 in m n /m n . Thus, a differential of f at o can be understood a k-linear function A ,o A ,o n n 2 df(o): T → k and (d (x ) = dx (o)) is a basis of m n /m n . oA o i i i=1 A ,o A ,o Notice that the embedding dimension and the Zariski tangent space of a local ring only depend on its formal completion.

EXERCISE 42. Let (R0, m) and (R, m0) be local rings with the same residue ﬁeld10 κ. Prove that a nonzero ring homomorphism φ : R0 → R induces an κ- linear map of Zariski tangent spaces T (φ): T (R) → T (R0) An application of Nakayama’s lemma to M = m yields:

COROLLARY 10.9. The embedding dimension of a noetherian local ring R is the smallest number of generators of its maximal ideal. The embedding dimension is zero if and only if R is a ﬁeld.

DEFINITION 10.10. A local ring R with maximal ideal m is said to be regular if it is noetherian and its Krull dimension equals its embedding dimension. A point o of an affine variety X is called regular if its local ring OX,o is and singular when it is not. (We write Xreg resp. Xsing for the subsets of X defined by this property.) An affine variety without singular points is said to be nonsingular. We will show that for a point o ∈ X this is indeed an intrinsic characterization of ‘being like a submanifold’. We begin with a formal version of the implicit function theorem. n n LEMMA 10.11. Let o ∈ A and let f1, . . . , fn−d ∈ mA ,o be such that their images 2 in m n /m n are linearly independent. Then f , . . . , f generate a prime ideal A ,o A ,o 1 n−d n n p in the local ring OA ,o and the formal completion of OA ,o/p with respect to its maximal ideal is isomorphic to k[[x1, . . . , xd]] as a complete local k-algebra. Moreover, n there exists an affine neighborhood U of o in A on which f1, . . . , fn−d are regular and generate a prime ideal in k[U] with the property that the corresponding irreducible affine variety X ⊂ U has o as a regular point.

n PROOF. Let us abbreviate OA ,o by O and its maximal ideal by m. Extend f1, . . . , fn−d to a system of regular functions f1, . . . , fn ∈ m such that their images in m/m2 are linearly independent. (After an afﬁne linear transformation, we may n 2 then just as well assume that o is the origin of A and that fi ≡ xi (mod m ).) N N+1 Hence the monomials of degree N in f1, ··· , fn map to a k-basis of m /m . This implies that the monomials of degree ≤ N in f1, ··· , fn make up a k-basis of O/mN+1. It follows that the map

yi ∈ k[[y1, . . . , yn]] 7→ fi ∈ Oˆ ∼ determines an isomorphism k[[y1, . . . , yn]] = Oˆ of complete local rings (a ring isomorphism that is also a homeomorphism). Its inverse deﬁnes a topological embedding of O in k[[y1, . . . , yn]] (sending fi to yi). The ideal generated by (y1, . . . , yn−d) in k[[y1, . . . , yn]] is the closure of the image of p. It is clearly a prime ideal. Accord- ing to Corollary 10.7 the preimage of that prime ideal in O is p (hence p is a prime ideal) and the embedding

O/p ,→ k[[y1, . . . , yn]]/(y1, . . . , yn−d) = k[[yn−d+1, . . . , yn]]

10By this we mean that φ induces an isomorphism R0/m0 =∼ R/m and that we identify the two residue fields via this isomorphism. 44 1. AFFINE VARIETIES realizes the m-adic completion of O/p. n Let U ⊂ A be an affine neighborhood of o on which the f1, . . . , fn−d are regular and denote by P ⊂ k[U] the preimage of p under the localization homor- morphism k[U] → O. This is a prime ideal which contains f1, . . . , fn−d and has the property that it generates the same ideal in O as (f1, . . . , fn−d). Choose a finite set Pn−d of generators φ1, . . . , φr of P and write φi = j=1 uijfj with uij ∈ O. By passing to a smaller U, we may then assume that each uij is regular on U so that P is in fact equal to the ideal (f1, . . . , fn−d) in k[U] generated by f1, . . . , fn−d. Hence (f1, . . . , fn−d) defines an irreducible affine variety X ⊂ U which has o is a regular point of dimension d. n THEOREM 10.12. Let X ⊂ A be closed and let o ∈ X. Then the local ring OX,o is regular of dimension d if and only there exist regular functions f1, . . . , fn−d on an n affine neighborhood U of o in A such that these functions generate IU (X ∩ U) and df1, . . . , dfn−d are linearly independent in every point of U. In that case, X ∩ U is regular and for every q ∈ X ∩ U the Zariski tangent space TqX is the kernel of the n n−d linear map (df1(q), . . . , dfn−d(q)) : TqA → k .

n PROOF. Suppose that OX,o is regular of dimension d. Let IX,o ⊂ OA ,o be the ideal of regular functions at o vanishing on a neighborhood of o in X, in other

n words, the kernel of OA ,o → OX,o. The latter is a surjective homomorphism of 2 n n local rings and so the preimage of mX,o resp. mX,o is IX,o + mA ,o = mA ,o resp. 2 2 ∼ 2 I + m n . It follows that the quotient m n /(I + m n ) = m /m has X,o A ,o A ,o X,o A ,o X,o X,o 2 2 dimension d. So the image (I +m n )/m n of I in the n-dimensional vector X,o A ,o A ,o X,o 2 space m n /m n must have dimension n − d. Choose f , . . . , f ∈ I such A ,o A ,o 1 n−d X,o that df1(o), . . . , dfn−d(o) are linearly independent. We show among other things that these functions generate IX,o (this will in fact be the key step). n According to Lemma 10.11, the ideal pi ⊂ OA ,o generated by f1, . . . , fi is prime and so we have a prime chain

(0) ⊆ p0 ( p1 ( ··· ( pn−d ⊆ IX,o.

As OX,o is of dimension d, there also exists a prime chain of length d containing IX,o:

n IX,o ⊆ q0 ( q1 ( ··· ( qd ⊆ OA ,o.

n Since OA ,o has dimension n, these two prime chains cannot make up a prime sequence of length n + 1 and so pn−d = IX,o = q0. 0 In particular, f1, . . . , fn−d generate IX,o. Let U be an afﬁne neighborhood n 0 of o in A on which the fi’s are regular and has the property that X ∩ U is the common zero set of f1, . . . , fn−d. Since the differentials df1(o), . . . , dfn−d(o) are linearly independent, there exist n − d indices 1 ≤ ν1 < ν2 < ··· < νn−d ≤ n such 0 0 that δ := det((∂fi/∂xνj )i,j) ∈ k[U ] is nonzero in o. Then U := U(δ) ⊂ U has all the asserted properties (with the last property following from Lemma 10.11). The converse says that if U, o and f1, . . . , fn−d are as in the theorem, then n n the functions f1, . . . , fn−d generate a prime ideal Io in OA ,o such that OA ,o/Io is regular. This follows Lemma 10.11. The last assertion is clear from the preceding.

PROPOSITION 10.13. The regular points of an afﬁne variety X form an open- dense subset Xreg of X. 10. NONSINGULAR POINTS 45

For the proof we will assume Proposition 10.14 below, which we will leave unproved for now. (Note that it tells us only something new in case k has positive characteristic.)

*PROPOSITION 10.14. Every finitely generated field extension L/k (so with k as in these notes, i.e., algebraically closed) is separably generated, by which we mean that there exists an intermediate extension k ⊂ K ⊂ L such that K/k is purely tran- scendental (i.e., of the form k(x1, . . . , xr)) and L/K is a finite separable extension. According to 7.7 and the subsequent discussion, this implies that every irreducible affine variety is birationally equivalent to a hypersurface.

PROOFOF PROPOSITION 10.13. Without loss of generality we may assume that X is irreducible. Since we already know that Xreg is open, it remains to see that it is nonempty. It thus becomes an issue which only depends on k(X). Hence it sufﬁces to treat the case of a hypersurface in An so that I(X) is generated by an irreducible polynomial f ∈ k[x1, . . . , n]. In view of Lemma Lemma 10.11 it then sufﬁces to show that df is not identically zero on X. Suppose otherwise, i.e., that each partial derivative ∂f/∂xi vanishes on X. Then each ∂f/∂xi must be multiple of f and since the degree of ∂f/∂xi is less than that of f, this implies that it is identically zero. But then we know from Exercise 40 that the characteristic p of k is positive p and that f is of the form g . This contradicts the fact that f is irreducible.

EXERCISE 43. Let X be a nonsingular variety. Prove that X is connected if and only if it is irreducible.

REMARK 10.15. This enables us to find for an affine variety X of dimension d a descending chain of closed subsets X = Xd ⊃ Xd−1 ⊃ · · · ⊃ X0 such that dim Xi ≤ i and all the (finitely many) connected components of Xi − Xi−1 are nonsingular subvarieties of dimension i (such a chain is called a stratification of X). d−1 We let X be the union of Xsing and the irreducible components of dimension < d and if (with downward induction) Xi has been defined, then we take for Xi−1 i the union of the singular locus Xreg and the irreducible components of dimension ≤ i − 1. Then dim Xi−1 ≤ i − 1 and every connected component of Xi − Xi−1 i is an nonempty open subset of some Xreg and hence a nonsingular subvariety of dimension i.

10.16.D IFFERENTIALS AND DERIVATIONS. The differential that we defined earlier has an intrinsic, coordinate free description that turns out to be quite useful. Let us begin with the observation that the formation of the total differential of a polynomial, φ ∈ k[x1, . . . , xn] 7→ Pn ∂φ dφ := (x)dxi is a k-linear map which satisfies the Leibniz rule: d(φψ) = φdψ+ψdφ. i=1 ∂xi This property is formalized with the following definition. Let us fix a ring R (the base ring) and an R-algebra A.

DEFINITION 10.17. Let M be a A-module. An R-derivation of A with values in M is a map D : A → M is an R-module homomorphism which satisﬁes the Leibniz rule: D(a1a2) = a1D(a2) + a2D(a2) for all a1, a2 ∈ A.

The last condition usually prevents D from being an A-module homomorphism. Let us note that (by taking a1 = a2 = 1) that we must have D(1) = 0. Then it also follows that for every r ∈ R, D(r) = rD(1) = 0. Note also that if a happens to be invertible in A, then 0 = D(1) = D(a/a) = 1/aD(a) + aD(1/a) so that D(1/a) = −D(a)/a2. 46 1. AFFINE VARIETIES

Given a1, . . . , an ∈ A, then the values of D on a1, . . . , an determine its values on 0 the subalgebra A by the ai’s, for if φ : R[x1, . . . , xn] → A denotes the corresponding R- homomorphism and f ∈ R[x1, . . . , xn], then n X ∂f Dφ(f) = φ Dai. ∂xi i=1 If we combine this with the formula for D(1/a), we see that this even determines D on 0 0 the largest subalgebra A of A generated by the ai’s and the invertible elements in A . In particular, if we are given a ﬁeld extension L/K, then a K-derivation of L with values in some L-vector space is determined by its values on a set of generators of L as a ﬁeld extension of K. Observe that the set of R-derivations of A in M form an R-module: if D1 and D2 are R-derivations of A with values in M, and a1, a2 ∈ A, then a1D1 + a2D2 is also one. We denote this module by DerR(A, M).

EXERCISE 44. Prove that if D1,D2 ∈ DerR(A, A), then [D1,D2] := D1D2 − D2D1 ∈ DerR(A, A). What do we get for R = k and A = k[x1, . . . , xn]? It is immediate from the deﬁnition that for every homomorphism of A-modules φ : M → N, the composition of a D as above with φ is an R-derivation of A with values in N. We can now construct a universal R-derivation of A, d : A → ΩA/R (where ΩA/R must of course be an R-module) with the property that every D as above is obtained by composing ¯ d with a unique homomorphism of A-modules D :ΩA/R → N. The construction that is forced upon us starts with the free A-module A(A) which has A itself as a generating set— let us denote the generator associated to a ∈ A by d˜(a)—which we then divide out by the (A) A-submodule of A generated by the expressions d˜(ra) − rd˜(a), d˜(a1 + a2) − d˜(a1) − d˜(a2) and d˜(a1a2) − a1d˜(a2) − a2d˜(a2), with r ∈ R and a, a1, a2 ∈ A. The quotient A-module ˜ is denoted ΩA/R and the composite of d with the quotient map by d : A → ΩA/R. The latter is an R-derivation of A by construction. Given an R-derivation D : A → M, then the map which assigns to d˜(a) the value Da extends (obviously) as a A-module homomorphism A(A) → M. It has the above submodule in its kernel and hence determines an A-module ¯ ¯ homomorphism of D :ΩA/R → M. This has clearly the property that D = Dd. We call ΩA/R the module of Kahler¨ differentials. We will see that the map d : A → ΩA/R can be thought of as an algebraic version of the formation of the (total) differential.

The universal derivation of a ﬁnitely generated R-algebra is obtained as follows. First we do the case of a polynomial algebra P := R[x1, . . . , xn]. Assigning to f ∈ P the row n vector (∂f/∂x1, . . . , ∂f/∂xn) is an R-derivation on P with values in P . By the universal n property of ΩP/R this must be given by a unique P -homomorphism ΩP/R → P . For any Pn R-derivation D : P → M we have Df = i=1(∂f/∂xi)Dxi and the Dxi can be prescribed n n arbitrarily. So if we take M = P as above, we see that ΩP/R → P is an isomorphism. Thus the universal R-derivation d : P → ΩP/R may be regarded as the intrinsic way of forming the total differential. Next consider a quotient A := P/I of P , where I ⊂ P is an ideal. If M is an A-module and D0 : A → M is an R-derivation, then its composite with the projection π : P → A, D = D0π : P → M, is an R-derivation of P with the property that Df = 0 for every f ∈ I. Conversely, every R-derivation D : P → M in an A-module M with this property factors through an an R-derivation D0 : A → M. We observe here that the map f ∈ I 7→ D(f) factors through I/I2, for if f, g ∈ I, then D(fg) = fDg + gDf ∈ IM = {0}. Since I/I2 is an A-module, we thus obtain a short exact sequence of A-modules 2 I/I → ΩP/R/IΩP/R → ΩA/R → 0.

If we are given a generating set f1, . . . , fm for I, then ΩA/R can be identiﬁed with the n quotient of A by the A-submodule generated by the row vectors (∂fi/∂x1, . . . , ∂fi/∂xn), 11. THE NOTION OF A VARIETY 47 i = 1, . . . , m. Note that if R is a noetherian ring, then, as we know, so is A and it also follows that ΩA/R is a noetherian R-module. This applies to the case when R = k and A = k[X] for some afﬁne variety X. We then write Ω(X) for Ωk[X]/k.

EXERCISE 45. Prove that ΩA/R behaves well with localization: if S ⊂ A is a multiplicative subset, then every R-derivation with values is some A-module M extends naturally to R-derivation of S−1A-with values in S−1M. Prove that we have a natural map −1 S ΩA/R → ΩS−1A/R and that this map is a A-homomorphism.

For an afﬁne variety X and x ∈ X, we write ΩX,x for ΩOX,x/k. The preceding exercise −1 implies that ΩX,x is the localization of Ω(X) at x: ΩX,x = (k[X] − px) Ω(X).

EXERCISE 46. Let X be an affine variety and let o ∈ X. Show that the Zariski tangent space of X at o can be understood (and indeed, be defined) as the space of k-derivations of OX,o with values in k, where we regard k as a OX,o-module via OX,o/mX,o =∼ k. Prove that this identifies its dual, the Zariski cotangent space, with ΩX,o/mX,oΩX,o.

n EXERCISE 47. Show (perhaps with the help of the preceding exercises) that if o ∈ A , n then ΩO n,o/k is the free OA ,o-module generated by dx1, . . . , dxn and that if X is an afﬁne A n variety in A that has o a regular point, then ΩX,o is a free OX,o-module of rank dim OX,x. We must be careful with this construction when dealing with formal power series rings. k O 1 For instance, as we have seen, the completion of the local -algebra A ,0 with respect to its k[[x]] O 1 ,→ k[[x]] maximal ideal is and the embedding of A ,0 is given by Taylor expansion. A k-derivation D of k[[x]] with values in some k[[x]]-module is not determined by Dx. All k O 1 it determines are the values on the -subalgebra A ,0. This issue disappears however if we require that D commutes with inﬁnite summation when the series converges relative to the P∞ r P∞ r−1 (x)-adic metric, for then D( r=0 crx ) will be equal to r=0 crrx Dx. ˆ EXERCISE 48. Prove that the composite d : A → ΩA/R → (ΩA/R)I factors through a ˆ ˆ ˆ derivation dI : AI → (ΩA/R)I and prove that it is universal among all the R-derivations of

A in R-modules that are complete for the I-adic topology (i.e., for which M = Mˆ I ).

So if o is a regular point of an afﬁne variety X, then Ωˆ X,o is a free OˆX,o-module of rank dim OX,x.

11. The notion of a variety We begin with a deﬁnition.

DEFINITION 11.1.A k-prevariety is a topological space X endowed with a sheaf OX of k-valued functions such that X can be covered by finitely many open subsets U such that (U, OX |U) is an affine k-variety. Given k-prevarieties (X, OX ) and (Y, OY ), then a morphism of prevarieties f :(X, OX ) → (Y, OY ) is simply a morphism in the category of spaces endowed with a sheaf OX of k-valued functions: f is continuous and for every open V ⊂ Y , composition with f takes OY (V ) to −1 OX (f V ). We often designate a prevariety and its underlying topological space by the same symbol, a habit which rarely leads to confusion. The composite of two morphisms is evidently a morphism so that we are dealing here with a category. The prefix ‘pre’ in prevariety refers to the fact that we have not imposed a separation requirement which takes the place of the Hausdorff property that one normally imposes on a manifold (see Example 11.4 below). 48 1. AFFINE VARIETIES

Let X be a prevariety. By assumption X is covered by finitely many affine open subvarieties U1,...,UN . Suppose κi is an isomorphism of Ui onto an affine variety Xi that sits as a closed subset in some affine space. Then Xi,j := κi(Ui ∩ Uj) is −1 an open subset of Xi and κi,j := κjκi is an isomorphism of Xi,j onto Xj,i ⊂ Xj. We can recover X from the disjoint union of the X1,...,XN by means of a gluing process: if we use κi,j to identify Xi,j with Xj,i for all i, j we get back X. The N collection {(Ui, κi)}i=1 is called an affine atlas for X.

EXERCISE 49. Let (X, OX ) be a prevariety. (i) Prove that X is a noetherian space. (ii) Prove that X contains an open-dense subset which is affine. (iii) Let Y ⊂ X be locally closed (i.e., the intersection of a closed subset with an open subset). Prove that Y is in natural manner a prevariety in such a manner that the inclusion Y ⊂ X is a morphism of prevarieties. Much of what we did for affine varieties extends in a straightforward manner to this more general context. For instance, a rational function f : X 99K k is defined as before: it is represented by a regular function on a subset of X that is open-dense in its set of closed points and two such represent the same rational function if they coincide on a nonempty open-dense subset in their common domain of definition. If X is irreducible, then the rational functions on X form a field k(X), the function field of X. If U ⊂ X is an open nonempty affine subset, then k(X) = k(U) = Frac(O(U)) (but we will see that it is not true in general that k(X) = Frac(O(X))). In particular, U1,...,UN all have the same dimension trdegk k(X). According to Exercise 35, this is then also the (Krull) dimension of X. Similarly, if X and Y are prevarieties, then a rational map f : X 99K Y is represented by morphism from a nonempty open-dense subset of X to Y with the understanding that two such defined the same map if they coincide on a nonempty open-dense subset. If some representative morphism has dense image in Y , then f is said to be dominant and then f induces a field extension f ∗ : k(Y ) ,→ k(X). Conversely, an embedding of fields k(Y ) ,→ k(X) determines a dominant rational map X 99K Y . If U ⊂ X is open and nonempty, then k(U) = k(X) and the inclusion is a birational equivalence. The notion of a regular point is local and so automatically carries over to this general setting. It is then also clear that the regular points of a prevariety X define an open-dense subset Xreg of X.

11.2.T HEPRODUCTOFTWOPREVARIETIES. Our discussion of the product of closed subsets of affine spaces dictates how we should define the product of two prevarieties X and Y : if (p, q) ∈ X × Y , then let p ∈ U ⊂ X and q ∈ V ⊂ Y be affine open neighborhoods of the components. We require that the topology on U × V be the Zariski topology so that a basis of neighborhoods of (p, q) consists of the loci (U × V )(h) where a h ∈ O(U) ⊗ O(V ) with h(p, q) 6= 0 is nonzero. We also require that ring of sections of OX×Y over such a basic neighborhood (U × V )(h) be O(U) ⊗ O(V )[1/h].

EXERCISE 50. Prove that this product has the usual categorical characterization: the two projections X × Y → X and X × Y → Y are morphisms and if Z is a prevariety, then a pair of maps (f : Z → X, g : Z → Y ) deﬁnes a morphism (f, g): Z → X × Y if and only both f and g are morphisms. 11. THE NOTION OF A VARIETY 49

The Hausdorff property is not of a local nature: a non-Hausdorff space can very well be locally Hausdorff. The standard example is the space X obtained from two copies of R by identifying the complement of {0} in either copy by means of the identity map. Then X is locally like R, but the images of the two origins cannot be separated. A topological space X is Hausdorff precisely when the diagonal of X×X is a closed subset relative to the product topology. As we know, the Zariski topology is almost never Hausdorff. But on the other hand, the selfproduct of the underlying space has not the product topology either and so requiring that the diagonal is closed is not totally unreasonable a priori. In fact, imposing this condition turns out to be the appropriate way of avoiding the pathologies that can result from an unfortunate choice of gluing data.

DEFINITION 11.3.A k-prevariety X is called a k-variety if the diagonal is closed in X × X (where the latter has the Zariski topology as deﬁned above).

The diagonal in An ⊂ An × An is closed, so An is a variety. This implies that the same is true for any quasi-afﬁne subset of An. Hence a quasi-afﬁne prevariety is in fact a variety.

EXAMPLE 11.4. The simplest example of a prevariety that is not a variety is the obvious generalization of the space described above: let X be obtained from two 1 1 1 1 1 copies A+ and A− of A by identifying A+ − {0} with A− − {0} by means of the 1 identity map. If o± ∈ X denotes the image of origin of A±, then (o+, o−) ∈ X × X lies in the closure of the diagonal, but is not contained in the diagonal. A subset of a variety X is called a subvariety if it is open in a closed subset of X. The proof of the following assertion is left as an exercise (see also Exercise 49).

PROPOSITION 11.5. A subvariety is in a natural manner a variety such that the inclusion becomes a morphism. The product of two varieties is a variety.

EXAMPLE 11.6. We can take this further: if U ⊂ Am is open and f : U → An is a morphism, then consider the graph of f, U˜ := {(x, y) ∈ Am+n : x ∈ U, y = f(x)}. It is easy to see that U˜ is a subvariety of Am+n. The map x ∈ U 7→ (x, f(x)) ∈ U˜ and the projection U˜ → U are regular and each others inverse. So they deﬁne an isomorphism U˜ → U. Notice that via this isomorphism f appears as a projection mapping: (x, y) ∈ U˜ 7→ y ∈ V .

EXERCISE 51. The goal of this exercise is to show that U := A2 − {(0, 0)} is not afﬁne. Let Ux ⊂ U resp. Uy ⊂ U be the complement of the x-axis resp. y-axis so that U = Ux ∪ Uy.

(a) Prove that Ux is afﬁne and that O(Ux) = k[x, y][1/x]. (b) Prove that every regular function on U extends to A2 so that O(U) = k[x, y]. (c) Show that U is not afﬁne.

CHAPTER 2

Projective varieties

1. Projective spaces Two distinct lines in the plane intersect in a single point or are parallel. In the last case one would like to say that the lines intersect at infinity so that the statement becomes simply: two distinct lines in a plane meet in a single point. There are many more examples of geometric configurations for which the special cases disappear by the simple remedy of adding points at infinity. A satisfactory approach to this which makes no a priori distinction between ordinary points and points at infinity involves the notion of a projective space. The following notion, perhaps a bit abstract, is quite useful. It will soon become more concrete.

DEFINITION 1.1.A projective space of dimension n over k is a set P endowed with an extra structure that can be given by a pair (V, `), where V is k-vector space of dimension n + 1 and ` is a bijection between P and the collection of 1- dimensional linear subspaces of V : p ∈ P 7→ `p ⊂ V . It is here understood that another such pair (V 0, `0) defines the same structure if and only if there exists a 0 0 k-linear isomorphism φ : V → V such that for every p ∈ P , φ sends `p to `p. (We are in fact saying that thus is defined an equivalence relation on the collection of such pairs and that a projective structure is given by an equivalence class.) In particular, if V is a finite dimensional k-vector space, then the collection of its 1-dimensional linear subspaces is in a natural manner a projective space; we n n n+1 denote it by P(V ). We often write P or Pk for P(k ) and call it simply projective n-space (over k).

EXERCISE 52. Prove that the linear isomorphism φ in Deﬁnition 1.1 is unique up to scalar multiplication.

DEFINITION 1.2. Given a projective space P of dimension n over k, then a subset L of P is said to be linear subspace of dimension d if, for some (or any) pair (V, `) as above, there exists a linear subspace VL ⊂ V of dimension d + 1 such that `(L) is the collection of 1-dimensional linear subspaces of VL. A map f : P → P 0 between two projective spaces over k is said to be linear morphism if for corresponding structural data (V, `) and (V 0, `0) for P resp. P 0 there 0 0 exists a linear injection F : V → V such that F sends `p to `f(p) for all p ∈ P . So a linear subspace has itself the structure of a projective space and its inclusion in the ambient projective space is a linear morphism. Conversely, the image of a linear morphism is linear subspace. A linear subspace of dimension one resp. two is often called a line resp. a plane. A linear subspace of codimension one (of dimension one less than the ambient projective space) is called a hyperplane. It is now clear that two distinct lines in a

51 52 2. PROJECTIVE VARIETIES plane intersect in a single point: this simply translates the fact that the intersection of two distinct linear subspaces of dimension two in a three dimensional vector space is of dimension one. Let P be a projective space of dimension n. We can of course describe its structure by a pair (V, `) for which V = kn+1. This gives rise to a ‘coordinate system n+1 on P ’ as follows: if we denote the coordinates of k by (T0,...,Tn), then every point p ∈ P(V ) is representable as a ratio [p0 : ··· : pn] of n + 1 elements of k that are not all zero: choose a generator p˜ ∈ `p and let pi = Ti(˜p). Any other generator is of the form λp˜ with λ ∈ k − {0} and indeed, [λp0 : ··· : λpn] = [p0 : ··· : pn]. This is why [T0 : ··· : Tn] is called a homogeneous coordinate system on P(V ) even though an individual Ti is not a function on P(V ) (but the ratios Ti/Tj are, albeit that for i 6= j they are not everywhere deﬁned).

1.3.L INEAR CHARTS AND STANDARD ATLAS. Let U ⊂ P be a hyperplane complement in P , i.e., the complement of a hyperplane H ⊂ P . We show that U is in a natural manner an affine space. To see this, let the structure on P be given by the pair (V, `). Then the hyperplane H corresponds to a hyperplane VH ⊂ V . A coset A of VH in V is an affine hyperplane in the classical sense of the word. Given a coset A of VH distinct from VH , then assigning to v ∈ A the 1-dimensional linear subspace spanned by v defines a bijection A =∼ P − H whose inverse were denote ∼ by κA : P − H = A. This puts on U a structure of an affine space. This is independent of the choice of A: any another choice A0 is obtained from A by multiplication by some nonzero scalar u ∈ k and hence κA0 is the composite of κA and the map ∼ 0 u· : A = A ; since the latter is an isomorphism of affine spaces, κA0 defines the same affine structure on P − H as κA. For a linear subspace L ⊂ P the bijection ∼ U = A restricts to one between L ∩ U and the affine-linear subspace VL ∩ A. So if L ∩ U 6= ∅, then L ∩ U is an affine subspace of U. Concretely, if we choose a homogeneous coordinate system (T0,...,Tn) such that H is given by T0 = 0, then n (x1 := T1/T0, . . . , xn := Tn/T0) maps A bijectively onto A and any other choice will differ from this one by an affine-linear transformation of An. We call such an isomorphism from a hyperplane complement in P onto An a linear chart for P . Thus the homogeneous coordinate system [T0,...,Tn] defines a chart for every i = 0, . . . , n: if Ui ⊂ P is the hyperplane complement defined by Ti 6= 0, then ∼ n κi : Ui = A , [T0 : ··· : Tn] 7→ (T0/Ti,..., T\i/Ti,...,Tn/Ti), is a chart with inverse (x1, . . . , xn) 7→ [x1 : ··· : xi : 1 : xi+1 : ··· : xn]. Notice n that the Ui’s cover P . We call a collection of linear charts (Ui, κi)i=0 thus obtained from a homogeneous coordinate system a standard atlas for P . Let us determine −1 the coordinate change for a pair of charts, say for κnκ0 . The image of U0 ∩ Un n under κ0 resp. κn is the open subset U(xn) resp. U(x1) of A and −1 κnκ0 : U(xn) → U(x1), (x1, x2, . . . , xn) 7→ (1/xn, x1/xn, . . . , xn−1/xn). −1 Notice that this defines an isomorphism of affine varieties (the inverse is κ0κn ). Thus P becomes a prevariety. In particular, P acquires a (Zariski) topology for which U ⊂ P is open if and only if κ(U ∩ Ui) is open in the set of closed points of n A . Then we have a prevariety (P, OP ) with the property that U ⊂ P is open if and n −1 only if κ(U ∩ Ui) is open in A and f ∈ OP (U) if and only if fκi ∈ Oκi(U) for all i. We shall see shortly that this structure is independent of the coordinate system (T0,...,Tn) and that P is in fact a k-variety. 2. THE ZARISKI TOPOLOGY ON A PROJECTIVE SPACE 53

We could also proceed in the opposite direction and start with an affine space and realize it as the hyperplane complement of a projective space. Changing our point of view accordingly, we might say that a projective space arises as a ‘completion with points at infinity’ of a given affine space. For instance, An embeds in Pn by (x1, . . . , xn) 7→ [1 : x1 : ··· : xn] with image the hyperplane complement defined by T0 6= 0.

2. The Zariski topology on a projective space We begin with giving a simpler, more classical characterization of the Zariski topology on a projective space. Let P be a projective space of dimension n over k and let [T0 : ··· : Tn] be a homogeneous coordinate system for P . Suppose F ∈ k[X0,...,Xn] is homogeneous d of degree d so that F (tT0, . . . , tTn) = t F (T0,...,Tn) for λ ∈ k. The property of this being zero only depends on [T0 : ··· : Tn] and hence the zero set of F deﬁnes a subset Z(F ) ⊂ P (even though F is not a function on P ). We denote its nonzero set by PF := P − Z(F ).

PROPOSITION 2.1. The collection UF , where F runs over the homogeneous polynomials in k[X0,...,Xn], is a basis for the Zariski topology on P . This topology is independent of the choice of our homogeneous coordinate system [T0,...,Tn] and every linear chart is a homeomorphism that identiﬁes the sheaf of regular functions on

n its domain with OA .

PROOF. That the collection {PF }F is a basis of a topology follows from the obvious equality PF ∩ PF 0 = PFF 0 . The independence of the coordinate choice results from the observation that under a linear substitution a homogeneous polynomial transforms into a homogeneous polynomial (of the same degree). Let now U = P − H be a hyperplane complement. Choose a homogeneous coordinate system [T0,...,Tn] such that U is deﬁned by T0 6= 0 and denote by n ∼ j : A = U, (y1, . . . , yn) 7→ [1 : y1 : ··· : yn] the inverse. If F ∈ k[T0,...,Tn] is −1 homogeneous, then j UF = Uf , where f(y1, . . . , yn) := F (1, y1, . . . , yn). So the inclusion j is continuous. Conversely, if f ∈ k[y1, . . . , yn] is nonzero of degree d, d then its ‘homogenization’ F (T0,...,Tn) := T0 f(T1/T0,...,Tn/T0) is homogeneous −1 of degree d and Uf = j PF . So j is also open. −1 By letting (U, j ) run over the charts (Uν , κν ), we see that have thus recovered the Zariski topology. The last statement follows from the fact that each κν j is an n isomorphism between two open subsets of A .

EXERCISE 53. Let 0 6= F ∈ k[X0,...,Xn] be homogeneous of degree d. r (a) Prove that every function PF → k of the form G/F , with r ≥ 0 and G homogeneous of the same degree as F r, is regular. (b) Prove that conversely every regular function on PF is of this form. We now discuss the projective analogue of the (afﬁne) I ↔ Z correspondence. Let us denote for an integer d ≥ 0 by k[X0,...,Xn]d the set of F ∈ k[X0,...,Xn] that are homogeneous of degree d. It is clear that k[X0,...,Xn] is the direct sum of the k[X0,...,Xn]d, d = 0, 1,... and that k[X0,...,Xn]d · k[X0,...,Xn]e = k[X0,...,Xn]d+e. A good way to understand the decomposition of k[X0,...,Xn] into its homogeneous summands is in terms of the action n+1 in A deﬁned by scalar multiplication: if F ∈ k[X0,...,Xn]d, then for every 54 2. PROJECTIVE VARIETIES

d t ∈ k, we have F (tT0, . . . , tTn) = t F (T0,...,Tn). So this decomposition is in fact the eigenspace decomposition for this action of the multiplicative group k× on k[X0,...,Xn]. (Indeed, this simple decomposition should be understood as the algebraic expression of that action.) This implies among other things that for any n+1 n+1 F ∈ k[X0,...,Xn]d the zero set Z(F ) ⊂ A = k is invariant under scalar multiplication. This is still true for an intersection of such zero sets, in other words, if I ⊂ k[X0,...,Xn] is an ideal generated by homogeneous polynomials of positive degree, then Z(I) ⊂ An+1 is invariant under scalar multiplication. Such a closed subset is called an affine cone; the origin is called the vertex of that cone. Observe that since we insisted that the degrees of the homogeneous generators of I be positive, we have I ⊂ (T0,...,Tn), and so Z(I) will always contain the vertex 0. n If Y ⊂ P is closed, Y can be written as an intersection ∩αZ(Fα), where each n+1 Fα is homogeneous. The common zero set of the Fα in A , ∩αZ(Fα), is the cone that as a set is just the union of the 1-dimensional linear subspaces of kn+1 parameterized by Y . We denote this affine cone by Cone(Y ). n Given a subset Y ⊂ P , then for d ≥ 1, we denote by IY,d the set of F ∈ k[X0,...,Xn]d with Y ⊂ Z(F ) (we put IY,0 = 0). This is a vector space and we have IY,d · k[X0,...,Xn]e ⊂ IY,d+e. So the direct sum of the IY,d, d = 1, 2,... , denoted by IY ⊂ k[X0,...,Xn], is an ideal of k[X0,...,Xn]. Notice that with this definition, I∅ is the ideal generated by T0,...,Tn. The k-algebra k[X0,...,Xn] is graded and the ideal IY is homogeneous in the sense below.

DEFINITION 2.2.A (nonnegatively) graded ring is a ring R whose underlying ∞ additive group comes with a direct sum decomposition R• = ⊕k=0Rd such that the P∞ d ring product maps Rd × Re in Rd+e, or equivalently, is such that d=0 Rdt is a subring of R[t]. An ideal I of such a ring is said to be homogeneous if it is the direct sum of its homogeneous parts Id := I ∩ Rd.

If R• is a graded ring, then clearly R0 is a subring of R. If I• is a homogeneous ∞ ideal, then R/I = ⊕d=0Rd/Id is again a graded ring. Of special interest is the ∞ homogeneous ideal R+ := ⊕k=1Rd, for which we have R/R+ = R0. We will be mostly concerned with the case when R0 = k, so that R+ is then a maximal ideal (and the only one that is homogeneous).

LEMMA 2.3.√ If I,J are homogeneous ideals of a graded ring R•, then so are I ∩J, IJ, I + J and I. Moreover, a minimal prime ideal of R• is a graded ideal.

PROOF. The proofs of the ﬁrst statement are not difﬁcult. We omit them.

As to the last, it sufﬁces to show that if p ⊂ R• is a prime ideal, then the direct sum of its homogeneous parts, p• := ⊕n(p ∩ Rn) is also a prime ideal. Indeed, suppose r, s ∈ R are such that rs ∈ p• and neither r nor s lies in p•. Write r and s as a sum of its homogenous parts and let rk ∈ Rk resp. sl ∈ Rl be the lowest degree part or r resp. s that is not in p. Then it is easily seen that rksl ∈ p and so rk ∈ p or sl ∈ p and we get a contradiction. LEMMA 2.4. If X ⊂ An+1 is an afﬁne cone (with vertex 0), then I(X) is a homogeneous ideal.

PROOF. Let F ∈ I(X). We must show that each homogeneous component of F lies in I(X). As X is invariant under scalar multiplication, the polynomial 2. THE ZARISKI TOPOLOGY ON A PROJECTIVE SPACE 55

P d F (tT0, . . . , tTn) = d≥1 t Fd(T0,...,Tn) (as an element of k[x1, . . . , xn][t]) vanishes on X ×A1 in An+1 ×A1. Clearly the zero set of I(X)[t] is X ×A1 ⊂ An+1 ×A1 and since the quotient is k[x1, . . . , xn][t]/I(X)[t] = k[X][t] is reduced, we have 1 I(X × A ) = I(X)[t]. It follows that Fd ∈ I(X) for all d. The ideal I(X) is of course a radical ideal and so we ﬁnd:

COROLLARY 2.5. The maps Y 7→ Cone(Y ) and Y 7→ IY set up bijections between (i) the collection of closed subsets of Pn, (ii) the collection of afﬁne cones in An+1 and (iii) the collection of homogeneous radical ideals of k[X0,...,Xn] contained in (T0,...,Tn). n+1 Note that here Y = ∅ ↔ {0} ⊂ A ↔ the homogeneous ideal (T0,...,Tn). The bijections above restrict to bijections between (i) the collection of irreducible subsets of Pn, (ii) the collection of irreducible afﬁne cones in An+1 strictly containing the vertex 0 and (iii) the collection of homogeneous prime ideals of k[X0,...,Xn] strictly contained in (T0,...,Tn). DEFINITION 2.6. The homogeneous coordinate ring of a closed subset Y of Pn is the graded ring S(Y )• := k[X0,...,Xn]/IY , S(Y )d = k[X0,...,Xn]d/IY,d.

Notice that if we ignore the grading of S(Y )•, then we just get the coordinate ring of the afﬁne cone over Y , k[Cone(Y )].

EXERCISE 54. Let R• be a graded ring. (b) Prove that if I is a prime ideal in the homogeneous sense: if rs ∈ I for some r ∈ Rk, s ∈ Rl implies r ∈ I or s ∈ I, then I is a prime ideal.

(c) Prove that the intersection of all homogeneous prime ideals of R• is its ideal of nilpotents. n EXERCISE 55. Prove that a closed subset Y ⊂ P is irreducible if and only if IY is a prime ideal.

Since k[X0,...,Xn] is a noetherian ring, any ascending chain of homogeneous ideals in this ring stabilizes. This implies that any projective space is noetherian (and according to Exercise 49 Pb, as any prevariety, is noetherian). In particular, every subset of a projective space has a ﬁnite number of irreducible components whose union is all of that subset.

EXERCISE 56. Let S• be a graded k-algebra that is reduced, ﬁnitely generated and has S0 = k.

(a) Prove that S• is as a graded k-algebra isomorphic to the homogeneous coordinate ring of a closed subset Y . (b) Prove that under such an isomorphism, the homogeneous radical ideals contained in the maximal ideal S+ := ⊕d≥1Sd correspond to closed subsets of Y under an inclusion reversing bijection: homogeneous ideals strictly contained in S+ and maximal for that property correspond to points of Y .

(c) Suppose S• a domain. Show that a fraction F/G that is homogeneous of degree zero (F ∈ Sd and 0 6= G ∈ Sd for some d) deﬁnes a function on Ug.

EXERCISE 57. Let Y be an afﬁne variety. 56 2. PROJECTIVE VARIETIES

(a) Show that a homogeneous element of the graded ring k[Y ][T0,...,Tn] defines a closed subset of Y × Pn as its zero set. (b) Prove that every closed subset of Y ×Pn is an intersection of finitely many zero set of homogeneous elements of k[Y ][T0,...,Tn]. (c) Prove that we have a bijective correspondence between closed subsets of n Y × P and homogeneous radical ideals in k[Y ][T0,...,Tn]. 3. The Segre embeddings First we show how a product of projective spaces can be realized as a closed subset of a projective space. This will imply among other things that a projective space is a variety. Consider the projective spaces Pm and Pn with their homogeneous coordinate systems [T0 : ··· : Tm] and [W0 : ··· : Wn]. We also consider a projective space whose homogeneous coordinate system is the set of matrix coeffi- mn+m+n cients of an (m + 1) × (n + 1)-matrix [Z00 : ··· Zij : ··· : Zmn]; this is just P with an unusual indexing of its homogeneous coordinates.

PROPOSITION 3.1 (The Segre embedding). The map f : Pm × Pn → Pmn+m+n defined by Zij = TiWj, i = 0, . . . , m; j = 0, . . . , n is an isomorphism onto a closed subset of Pmn+m+n. If m = n, then the diagonal of Pm × Pm is the preimage of the m2+2m m m linear subspace of P defined by Zij = Zji and hence is closed in P × P . PROOF. In order to prove that f defines an isomorphism onto a closed subset mn+m+n of P , it is enough to show that for every chart domain Uij of the standard mn+m+n −1 m n atlas of P , f Uij is open in P × P and is mapped by f isomorphically onto a closed subset of Uij. For this purpose we may (simply by renumbering) mn+m+n assume that i = j = 0. So then U00 ⊂ P is defined by Z00 6= 0 and is parametrized by the coordinates zij := Zij/Z00, (i, j) 6= (0, 0). It is clear that −1 f U00 is defined by T0W0 6= 0. This is just UT0 × UW0 and hence parametrized by x1 := T1/T0, . . . , xm := Tm/T0 and y1 := W1/W0, . . . , yn := Wn/W0. In terms of −1 these coordinates, f : f U00 → U00 is given by zij = xiyj, where (i, j) 6= (0, 0) and where we should read 1 for x0 and y0 (so that zi0 = xi and z0j = yj). It is now clear that the image is in fact the graph of the morphism Am × An → Amn with m n mn coordinates xiyj, 1 ≤ i ≤ m, 1 ≤ j ≤ n. This graph is closed in A × A × A and m n −1 isomorphic to A × A . So f indeed restricts to an isomorphism of f U00 onto a closed subset of U00. In case m = n, we must also show that the condition TiWj = TjWi for 0 ≤ i < j ≤ m implies that [T0 : ··· : Tm] = [W0 : ··· : Wm], assuming that not all Ti resp. × Wj are zero. Suppose Ti 6= 0 and put t := Wi/Ti ∈ k . Then TiWj = TjWi implies Wj = tTj for all j. So t 6= 0 and [W0 : ··· : Wm] = [T0 : ··· : Tm]. COROLLARY 3.2. A projective space over k is a variety. m m PROOF. Proposition 3.1 shows that the diagonal of P × P is closed. DEFINITION 3.3. A variety is said to be projective if it is isomorphic to a closed irreducible subset of some projective space. A variety is called quasi-projective if is isomorphic to an open subset of some projective variety.

COROLLARY 3.4. Every irreducible closed (resp. locally closed) subset of Pn is a projective (resp. quasi-projective) variety. The collection of projective (resp. quasi- projective) varieties is closed under a product. 4. PROJECTIONS 57

PROOF. The ﬁrst statement follows from 11.5 and the second from Proposition 3.1.

EXERCISE 58. (a) Prove that the image of the Segre embedding is the common zero set of the homogenenous polynomials ZijZkl − ZilZkj. (b) Show that for every (p, q) ∈ Pm × Pn the image of {p} × Pn and Pm × {q} in Pmn+m+n is a linear subspace. n (n2+3n)/2 (c) Prove that the map P → P defined by Zij = TiTj, 0 ≤ i ≤ j ≤ n is an isomorphism on a closed subset defined by quadratic equations. Find these equations for n = 2. (d) As a special case we find that the quadric hypersurface in P3 defined by 1 1 Z0Z1 − Z2Z3 = 0 is isomorphic to P × P . Identify in this case the two systems of lines on this quadric.

EXERCISE 59 (Intrinsic Segre embedding). Let V and W be ﬁnite dimensional k-vector spaces. Describe the Segre embedding for P(V ) × P(W ) intrinsically as a morphism P(V ) × P(W ) → P(V ⊗ W ).

4. Projections Let V be a finite dimensional k-vector space. If W ⊂ V is a linear subspace, then we can form the quotient vector space V/W and we have linear surjection π : V → V/W . We cannot projectivize this as a morphism from P(V ) to P(V/W ), but we shall see that it defines at least a morphism Pπ : P(V ) − P(W ) → P(V/W ) (which then determines a rational map P(V ) 99K P(V/W ) in case W 6= V ). As a map, Pπ is defined as follows: for p ∈ P(V ) − P(W ), `p is a one dimensional subspace of V not contained in W so that π(`p) is a one dimensional subspace of V/W ; we then put Pπ(p) := [π(`p)]. To see that this is morphism, let n := dim P(V ), m := dim P(V/W ) (so that dim V = n + 1 and dim W = n − m) and choose a coordinate system (T0,...,Tn) for V such that W is given by T0 = ··· = Tm = 0. Then (T0,...,Tm) serves as a coordinate system for V/W and π : V → V/W is simply given by (T0,...,Tn) → (T0,...,Tm). In projective coordinates this is of course given by [T0 : ··· : Tn] → [T0 : ··· : Tm] and so is indeed a morphism where it makes sense, namely where T0,...,Tm not all vanish, that is, on the complement of P(W ).

EFINITION EMMA D -L 4.1. The blowup of P(V ) along P(W ), denoted BlP(W ) P(V ), is the closure of the graph of Pπ in P(V ) × P(V/W ) (hence is a projective variety). It enjoys the following properties:

The projection on the ﬁrst factor, p1 : BlP(W ) P(V ) → P(V ) is an isomorphism over P(V ) − P(W ) whereas the preimage over P(W ) is P(W ) × P(V/W ). The projection to the second factor p2 : BlP(W ) P(V ) → P(V/W ) is a locally trivial bundle of projective spaces of dimension dim W , to be precise, if U ⊂ P(V/W ) is a hyperplane complement (hence an afﬁne space), then there exists an isomorphism −1 ∼ n−m p2 U = P × U whose second component is given by p2.

PROOF. We use a homogeneous coordinate system [T0 : ··· : Tn] for P(V ) as above (so that P(W ) is given by T0 = ··· = Tm = 0). If we denote the corresponding coordinate system for P(V/W ) by [S0 : ··· : Sm], then the graph of Pπ in P(V ) × P(V/W ) is given by the pairs ([T0 : ··· : Tn], [S0 : ··· : Sm]) with [T0 : ··· : Tm] proportional to [S0 : ··· : Sm] and (T0,...,Tm) 6= (0,..., 0). The 58 2. PROJECTIVE VARIETIES proportionality property is equivalent to: TiSj = TjSi for all 0 ≤ i < j ≤ m. Let Z be the subset of ⊂ P(V ) × P(V/W ) defined by these equations: TiSj = TjSi, 0 ≤ i < j ≤ m. In terms of the Segre embedding of Pn × Pm ,→ Pnm+n+m this amounts to Zij = Zji in this range and so Z is a closed subset of P(V ) × P(V/W ). The equations defining Z are trivially satisfied when (T0,...,Tm) = (0,..., 0) and so Z contains P(W ) × P(V/W ). We have just shown that the graph of Pπ equals the complement of P(W ) × P(V/W ) in Z. Let us now see what the projection Z → P(V/W ) is like over the open subset m US0 of P(V/W ) defined by S0 6= 0. We parametrize US0 by A via (y1, . . . , ym) ∈ m 7→ [1 : y : ··· : y ]. Then its preimage Z := Z ∩ ( (V ) × U ) in Z is A 1 m US0 P S0 n−m parametrized by P × US0 by means of the morphism n−m ([T0 : Tm+1 : ··· : Tn], [1 : y1 : ··· : yn]) ∈ P × US0 7→ ([T : T y : ··· : T y : T : ··· : T ], [1 : y : ··· : y ]) ∈ Z . 0 0 1 0 m m+1 n 1 n US0 This is an isomorphism (the inverse is obvious) which commutes with the projection on U . Notice that it makes Z ∩( (V )× (V/W )) correspond to the locus where S0 US0 P P n−m T0 = 0, which is the product of a hyperplane in P times US0 . In particular, Z ∩ ( (V ) × (V/W )) lies in the closure of the graph of π. This remains true, US0 P P P of course, if we replace US0 by USi , i = 1, . . . , m, or by any other hyperplane complement in P(V/W ). It follows that Z is the closure of the graph of Pπ and that Z enjoys the stated properties. It is worthwhile to describe this map in purely projective terms. This starts with the observation that the π-preimage of a one-dimensional linear subspace of V/W is a linear subspace of V which contains W as a hyperplane and that every such subspace so arises. So we can think of P(V/W ) as parameterizing the projective linear subspaces of P(V ) that contain P(W ) as a projective hyperplane. EXERCISE 60. Prove that the projective structure on this collection of linear subspaces is intrinsic: prove that if P is a projective space and Q ⊂ P is a projective linear subspace of codimension m > 0, then the collection of projective linear subspaces Q˜ ⊂ P which contain Q as a projective hyperplane is in a natural manner a projective space (denoted here by PQ) of dimension m−1. Show that the blowup BlQ P of P along Q can be identified with the set of pairs (p, [Q˜]) ∈ P × PQ with p ∈ Q˜.

5. Elimination theory and closed projections Within a category of reasonable topological spaces (say, the locally compact Hausdorff spaces), the compact ones can be characterized as follows: K is compact if and only if the projection K × X → X is closed for every space X in that category. In this sense the following theorem states a kind of compactness property for projective varieties. n THEOREM 5.1. If U is a variety, then the projection πU : P × U → U is closed. Moreover, if Z ⊂ Pn × U is closed and irreducible, then there exists an open-dense 0 0 subset Y ⊂ πU (Z) such that for every y ∈ Y we have dim Z = dim πU (Z) + dim Zy, n where Zy ⊂ P denotes the ﬁber πU |Z over y. We derive this theorem from the main theorem of elimination theory, which we will state and prove ﬁrst. 5. ELIMINATION THEORY AND CLOSED PROJECTIONS 59

Given an integer d ≥ 0, let us write Vd for k[T0,T1]d, the k-vector space of d−i i d homogeneous polynomials in k[T0,T1] of degree d. The monomials (T0 T1)i=0 form a basis, in particular, dim Vd = d + 1. Given F ∈ Vm and G ∈ Vn, then

uF,G : Vn−1 ⊕ Vm−1 → Vn+m−1, (A, B) 7→ AF + BG is a linear map between two k-vector spaces of the same dimension m + n. The resultant R(F,G) of F and G is deﬁned as the determinant of this linear map with respect to the monomial bases of the summands of Vn−1 ⊕ Vm−1 and of Vn+m−1. So R(F,G) = 0 if and only if uF,G fails to be injective. Notice that R(F,G) is a polynomial in the coefﬁcients of F and G: Pm m−i i Pn n−i i if F = i=0 aiT0 T1 and G = j=0 biT0 T1, then   a0 a1 ··· am 0 ······ 0  0 a0 a1 ··· am ······ 0     0 0 a0 ··· 0     ···     0 0 ··· 0 a0 a1 ··· am R(F,G) = det   b0 b1 ······ bn 0 ··· 0     0 b0 b1 ······ bn 0 ··· 0     0 0 b0 ··· 0     ···  0 ··· 0 b0 b1 ······ bn

So the resultant deﬁnes an element of k[Vm × Vn] = k[Vm] ⊗ k[Vn].

LEMMA 5.2. R(F,G) = 0 if and only if F and G have a common zero in P1.

PROOF. If R(F,G) = 0, then uF,G is not injective, so that there exist a nonzero (A, B) ∈ Vn−1 ⊕ Vm−1 with AF + BG = 0. Suppose that B 6= 0. It is clear that F divides BG. Since deg(B) = m − 1 < m = deg F , it follows that F and G must have a common factor. If conversely F and G have a common zero in P1, then they must have a common linear factor L, say: F = LF1, G = LG1 and we see that (G1, −F1) ∈ Vn−1 ⊕ Vm−1 is nonzero and in the kernel of uF,G. n PROOFOF THEOREM 5.1. Let Z ⊂ P × U. It is clear that πU (Z) is closed 0 0 0 in U if for every open affine subset U ⊂ U, πU (Z) ∩ U is closed in U . Since 0 n 0 πU (Z) ∩ U = πU 0 (Z ∩ (P × U )) we may (and will) assume that is U affine. We first treat the case n = 1 and then proceed with induction on n. 1 Let Z ⊂ P × U be closed. Denote by IZ,• the homogeneous ideal in the graded algebra k[U][T0,T1] of functions vanishing on Z. Then Z is the common zero set of the members of IZ,• (see Exercise 57). For every homogeneous pair F,G ∈ ∪mk[U][T0,T1]m, we can form the resultant R(F,G) ∈ k[U]. We claim that πU (Z) is the common zero set Z(R) ⊂ U of the set of resultants R(F,G) of pairs of homogeneous forms F,G taken in ∪mIZ,m, hence is closed in U. 1 Suppose that y ∈ πU (Z) so that (y, p) ∈ Z for some p ∈ P . Then p is a common zero of each pair Fy,Gy, where F,G ∈ ∪mIZ,m and the subscript y refers to substituting y for the first argument. So R(F,G) vanishes in y. It follows that y ∈ Z(R). 1 Next we show that if y∈ / πU (Z), then y∈ / Z(R). Since {y}×P is not contained in Z, there exists an integer m > 0 and a F ∈ IZ,m with Fy 6= 0. Denote by 60 2. PROJECTIVE VARIETIES

1 p1, . . . , pr ∈ P the distinct zeroes of Fy. We show that there exists a G ∈ IZ,n for some n such that Gy does not vanish in any pi. This suffices, for this means that R(Fy,Gy) 6= 0 and so y∈ / Z(R). To this end, we note that for any given 1 ≤ i ≤ r, S 1 (i) Z ∪j6=i{(y, pj)} is closed in U × P , so that there will exist a G ∈ ∪mIZ,m with (i) (i) Gy zero in all the pj with j 6= i, but nonzero in pi. Upon replacing each G by some positive power of it , we may assume that G(1),...,G(r) all have the same (1) (r) (i) degree n, say. Then G := G + ··· + G ∈ IZ,n and Gy(pi) = G (pi) 6= 0. In order to prove the dimension assertion, we assume Z irreducible so that 1 Y := πU (Z) is irreducible and closed. We show that either Z = P × Y (so that 0 dim Z = dim Y +1) or that there exists an open-dense Y ⊂ Y such that ZY 0 := Z ∩ 1 0 0 0 (P × Y ) → Y is a finite morphism (so that dim Y = dim Y = dim ZY 0 = dim Z). Consider the ideal in k[U] generated by the coefficients of all the members of IZ ⊂ k[U][T0,T1] and denote by Y1 ⊂ U its zero set. This is closed subset of Y 1 consisting of the y ∈ U with the property that P × {y} ⊂ Z. If Z1 = Y , then 1 Z = P × Y . Assume therefore that Z1 is a proper closed subset of Y . Choose p ∈ P1 such that {p} × Y is not contained in Z. By adapting the coordinates in P1, we may assume that p = ∞. Let Y2 := πU ({∞}×U ∩Z). This is also a proper closed 0 subset of Y and so Y contains an an affine open-dense subset Y ⊂ Y − Y1 − Y2 1 0 0 such that ZY 0 is contained in A × Y and has finite fibers over Y . Then ZY 0 can be given as a subset of A1 × Y 0 by a polynomial in k[Y 0][t]. After possible shrinking Y 0, we can assume that the top coefficient of this polynomial is a unit in k[Y 0]. Then 0 ZY 0 will be given by a monic polynomial and hence ZY 0 → Y is a finite morphism.

Now assume n ≥ 2. Let q = [0 : ··· : 0 : 1] and consider the blowup Pñ := n n ñ Bl{q} P → P . Recall that an element of P is the set of pairs in ([T0 : ··· : n n−1 Tn], [S0 : ··· : Sn−1]) in P × P with [T0 : ··· : Tn−1] = [S0 : ··· : Sn−1]. We n−1 have seen that over the open subset USi ⊂ P defined by Si 6= 0, the projection ñ n−1 1 P → P is isomorphic to the projection P × USi → USi . Hence the projection ñ n−1 1 π1 : P × U → P × U is over USi × U like P × USi × U → USi × U. So this projection is closed over USi × U. It follows that the projection π1 is closed. The preimage Z˜ of Z under the projection Pñ × U → Pn × U is closed and by what ˜ n−1 we just proved, π1(Z) is closed in P × U. By induction, the image of the latter n−1 under the projection π2 : P × U → U is closed. But this is just πU (Z). For the dimension assertion, we may without loss of generality assume that Z is not contained in {q} × U (otherwise choose a different q) and that πU (Z) = U (otherwise replace U by π(Z)). We then replace Z˜ by the closure of Z ∩ (Pn − {q}) × U in Pñ × U so that Z˜ is irreducible, is of the same dimension as Z and maps also onto U. It therefore suffices to prove the dimension assertion for the projection ˜ ˜ n−1 π˜ : Z → U. We put W := π1(Z) ⊂ P × U. ñ n−1 The case n = 1 treated above shows that for the projection π1 : P → P × U ˜ −1 ˜ 1 ˜ either Z = π1 W (so that Z is a P -bundle over W and dim Z = dim W + 1), or dim Z˜ = dim W and Z˜ is finite over W − V ⊂ W where V ⊂ W is a proper closed subset. In that last case, we note that for every closed irreducible subset C ⊂ W −1 ˜ which is not contained in V , we have dim(π1 C ∩ Z) = dim C: this is clear for −1 ˜ −1 an irreducible component of π1 C ∩ Z which meets π (W − V ) and any other irreducible component will be contained in a P1-bundle over C ∩ V and hence of dimension 1 + dim(C ∩ V ) ≤ dim C. 5. ELIMINATION THEORY AND CLOSED PROJECTIONS 61

n−1 If π2 : P × U → U is the projection, then (by induction) there exists an open-dense subset U 0 ⊂ U such that the ﬁbers of W → U resp. V → U all have dimension dim W − dim U resp. have dimension smaller than dim W − dim U. So ˜ 0 ˜ all the ﬁbers of Z → U over U have dimension dim Z − dim U. We mention a few corollaries.

COROLLARY 5.3. Let X be a projective variety. Then any morphism from X to a variety is closed (and hence has closed image).

PROOF. Assume that X is closed in Pn. A morphism f : X → Y to a variety Y can be factored as the obvious isomorphism of X onto the graph Γf of f, the inclusion of this graph in X × Y (which is evidently closed), the inclusion of X × Y in Pn ×Y (which is closed since X ⊂ Pn is closed) and the projection onto Y (which is closed by Theorem 5.1). So f is closed. It is an elementary result from complex function theory (based on Liouville’s thoerem) that a holomorphic function on the Riemann sphere is constant. This implies the corresponding assertion for holomorphic functions on complex projective n-space n (to see that a holomorphic function on n takes the same value on PC PC any two distinct points, simply apply the previous remark to its restriction to the complex projective line passing through them, viewed as a copy of the Riemann sphere). The following corollary is an algebraic version of this fact.

COROLLARY 5.4. Let X be a projective variety. Then any morphism from X to a quasi-afﬁne variety is constant. In particular, any regular function on X is constant.

PROOF. If f : X → Y is a morphism to a quasi-affine variety Y , then its composite with an embedding of Y in some affine space An is given by n regular functions on X. So it suffices to prove the special case when Y = A1. By the previous corollary this image is closed in A1. But if we think of f as taking its values in P1 (via the embedding y ∈ A1 7→ [1 : y] ∈ P1), then we see that f(X) is also closed in P1. So f(X) cannot be all of A1. Since X is irreducible, so is the image and it follows that f(X) is a singleton. In other words, f is constant. PROPOSITION 5.5. Let Z ⊂ Pn an irreducible and closed subset. If Q ⊂ Pn is a linear subspace with the property that Q ∩ Z = ∅, then dim Z + dim Q < n and the n n projection π : P → PQ maps Z onto an irreducible and closed subset π(Z) with the property that every fiber of Z → π(Z) is finite.

PROOF. Every fiber of Z → π(Z) is the intersection of Z with a linear subspace Q˜ ⊂ Pn which contains Q a hyperplane. It is closed in Q˜ and hence this fiber is projective (or strictly speaking, every irreducible component of that fiber is). On the other hand it is contained in the hyperplane complement Q˜ − Q, which is affine. So the fiber is also affine. Then Corollary 5.4 implies that each irreducible component of this fiber is a singleton. Hence the fiber is finite. n It follows from Theorem 5.1 that dim Z = dim π(Z) ≤ dim PQ − dim Q − 1.

EXERCISE 61. Let P be a projective space of dimension n. (a) The dual Pˇ of P is by deﬁnition the collection of hyperplanes in P . Prove that Pˇ has a natural structure of a projective space. (b) Identify the double dual of P with P itself. 62 2. PROJECTIVE VARIETIES

(c) The incidence locus I ⊂ P × Pˇ is the set of pairs (p, q) ∈ P × Pˇ with the property that p lies in the hyperplane Hq deﬁned by q. Prove that I is a nonsingular variety of dimension 2n − 1. (d) Show that we can ﬁnd homogeneous coordinates [Z0 : ··· : Zn] for P and ˇ Pn [W0 : ··· : Wn] for P such that I is given by i=0 ZiWi = 0.

EXERCISE 62. Let F ∈ k[X0,...,Xn]d deﬁne a nonsingular hypersurface H in Pn. Prove that the map H → Pˇn which assigns to p ∈ H the projectived tangent space of H at p is given by [ ∂F : ··· : ∂F ]. Prove that the image of this map is ∂Z0 ∂Zn closed in Pˇn (this image is called the dual of H). What can you say in case d = 2? 6. Constructible sets We now ask about the image of an arbitrary morphism of varieties. This need not be a variety as the following simple example shows. 2 2 EXAMPLE 6.1. Consider the morphism f : A → A , (x1, x2) 7→ (x1, x1x2).A 2 point (y1, y2) ∈ A is in the image of f if and only if the equation y2 = y1x2 has a solution in x2. This is the case precisely when y1 6= 0 or when y1 = y2 = 0. So the image of f is the union of the open subset y1 6= 0 and the singleton {(0, 0)}. This is not a locally closed subset, but the union of two such.

DEFINITION 6.2. A subset of variety is called constructible if it can be written as the union of ﬁnitely many (locally closed) subvarieties.

THEOREM 6.3. Let f : X → Y be a morphism of varieties. Then f takes constructible subsets of X to constructible subsets of Y . In particular, f(X) is constructible. We ﬁrst show that the theorem follows from

PROPOSITION 6.4. Let X be an irreducible variety and let f : X → Y be a morphism of varieties. Then f(X) contains a nonempty open subset of its closure (in other words, f(X) contains a locally closed subvariety of Y which is dense in f(X)).

PROOF THAT PROPOSITION 6.4 IMPLIES THEOREM 6.3. A constructible subset is a finite union of irreducible subvarieties and so it is clearly enough to prove that the image of each of these is constructible. In other words, it suffices to show that the image of a morphism f : X → Y of varieties with X irreducible is constructible. We prove this with induction on the dimension of X. According to Proposition 6.4 the closure of f(X) contains a nonempty open subset U such that f(X) ⊃ U. It is clear that f −1U is a nonempty open subset of X. If Z := X − f −1U, then f(X) = U ∪ f(Z). The irreducible components of Z have smaller dimension than X and so f(Z) is constructible by induction. PROOFOF PROPOSITION 6.4. Since X is covered by finitely many of its open affine subsets, we may without loss of generality assume that X is affine. So we can assume that X is closed in some An. Then the graph of f identifies X with a closed subset of An × Y and so we see that we only need to prove that for every n closed subset X ⊂ A × Y , πY (X) contains a nonempty open subset of πY (X). Since we can factor πY in an obvious manner into successive line projections (by forgetting the last coordinate) n n−1 1 A × Y → A × Y → · · · → A × Y → Y, 7. THE VERONESE EMBEDDING 63 it suffices to do the case n = 1. Upon replacing Y by πY (X), we are now left to 1 show that if X ⊂ A × Y is closed and such that πY (X) is dense in Y , then πY (X) contains a nonempty open subset of Y . Since X is irreducible, so is its image πY (X) and hence also Y . Upon replacing Y be a nonempty affine open subset and X by the preimage of that open subset, we may also assume that Y is affine. When X = A1 × Y , there is nothing to show. Otherwise X is contained in a 1 PN i hypersurface Z(f) ⊂ A × Y with equation f ∈ k[Y ][x], f(x, y) = i=0 ai(y)x , with ai ∈ k[Y ] and aN not identically zero. Next, by replacing Y by the nonzero set U(aN ) and X by its preimage, we may assume that aN is nowhere zero. We prove that then πY (X) = Y . Since is aN now invertible in k[Y ], we may assume that f is a monic polynomial. Then Z(f) → Y is a finite morphism and hence closed by Exercise 33. Since X is a closed subset of Z(f), it follows that πY (X) is closed in Y . Since πY (X) is also dense in Y , it follows that πY (X) ⊃ Y . 7. The Veronese embedding

Let be given a positive integer d. We index the monomials in Z0,...,Zn that are homogenous of degree d by their exponents: these are the sequences of non- n+d negative integers d = (d0, . . . , dn) with sum d. They are d in number. We use (n+d)−1 this index set to label the homogeneous coordinates Zd of P d . n (n+d)−1 PROPOSITION 7.1 (The Veronese embedding). The map fd : P → P d d0 dn deﬁned by Zd = T0 ··· Tn is an isomorphism onto a closed subset of the target projective space.

PROOF. In order to prove that fd is an isomorphism onto a closed subset, it is enough to show that for every chart domain Ud of the standard atlas of the target −1 n space, its preimage fd Ud is open in P and is mapped by fd isomorphically onto a d0 dn closed subset of Ud. This preimage is deﬁned by T0 ··· Tn 6= 0. Let us renumber the coordinates such that d0, . . . , dr are positive and dr+1 = ··· = dn = 0. Then −1 fd Ud = U0 ∩ · · · ∩ Ur ⊂ U0. So if we use the standard coordinates (x1, . . . , xn) on −1 U0, then fd Ud is deﬁned by x1 ··· xr 6= 0. The coordinates on Ud are the functions Ze/Zd with e 6= d. Let us write ze−d := ze0−d0,e1−d1,...,en−dn for this function. This notation is chosen as to make the expression fd in terms of these coordinates simply:

−1 e1−d1 en−dn fd : U0(x1 ··· xr) = fd Ud → Ud, ze−d = x1 ··· xn (e 6= d).

e1−d1 en−dn Since the Laurent monomial x1 ··· xn has degree d0 − e0, the components k1 kn of the above morphism are all the nonconstant Laurent monomials x1 ··· xn with ki ≥ −di and of total degree k1 + ··· + kn ≤ d0. In particular, each xi appears −d1 −dr as a component and so does the Laurent monomial x1 ··· xr . The graph of d1 dr the regular function x1 ··· xr on U0(x1 ··· xr) identiﬁes U0(x1 ··· xr) with the d1 dr n+1 closed hypersurface Z = Z(yx1 ··· xr − 1) in A . All other components of −d1 −dr fd|U0(x1 ··· xr): U0(x1 ··· xr) → Ud are products of x1, . . . , xn, x1 ··· xr and (n+d)−n−2 so its image is the graph of a morphism Z → A d . So we get a closed embedding. The following proposition is remarkable for its repercussions in intersection theory. 64 2. PROJECTIVE VARIETIES

PROPOSITION 7.2. Let H ⊂ Pn be a hypersurface. Then Pn − H is afﬁne and for every closed irreducible subset Z ⊂ Pn of positive dimension, Z ∩ H is nonempty and of dimension ≥ dim(Z) − 1, with equality holding if Z is not contained in H.

For the proof we need the following complement to Proposition 5.5.

PROPOSITION 7.3. For every proper closed subset Z (Pn there exists a linear subspace in P of dimension n − dim(Z) − 1 which misses Z.

PROOF. We may (and will) assume that Z is irreducible. Let i be a nonnegative integer < n−dim(Z). We prove with induction on i that Z misses a linear subspace of dimension i. If i = 0, then we must have dim(Z) < n and so any singleton in Pn − Z is as required. When i > 0, there exists by induction hypothesis a linear subspace Q ⊂ Pn of dimension (i − 1) which does not meet Z. Then projection n from Q deﬁnes a morphism π : Z → PQ from Z to a projective space of dimension n − i. Any ﬁber of this map is an intersection of Z with a linear subspace in Pn of dimension i passing through Q. We claim that at least one of them is empty. For if n not, then π is surjective and hence dominant. In particular, dim(Z) ≥ dim(PQ) = n − i, which evidently contradicts our assumption that i < n − dim(Z).

PROOFOF PROPOSITION 7.2. The hypersurface H is given by a homogeneous P d0 dn polynomial of degree d, say by d cdT0 ··· Tn . This determines a hyperplane n+d ˜ ( d )−1 P ˜ H ⊂ P defined by d cdZd. It is clear that H is the preimage of H under the Veronese morphism and hence the latter identifies Pn − H with a closed subset (n+d)−1 n of the affine space P d − H˜ . So P − H is affine. For the rest of the argument we may, by passing to the Veronese embedding, assume that H is a hyperplane. If dim(Z ∩ H) ≤ dim(Z) − 2, then by Proposition 7.3 there exists a linear subspace Q ⊂ H of dimension dim(H)−(dim(Z ∩H)−1 ≤ (n − 1) − (dim(Z) − 2) − 1 = n − dim(Z) which avoids Z ∩ H. Then Q is a linear subspace of Pn which avoids Z and we thus contradict Proposition 5.5. This proves that dim(Z ∩ H) ≥ dim(Z) − 1. If Z is irreducible and not contained in H, then we have also dim(Z ∩ H) ≤ dim(Z) − 1.

EXERCISE 63. Let d be a positive integer. The universal hypersurface of degree d n+d n ( d )−1 P d0 d1 dn is the hypersurface of P × P deﬁned by F (X,Z) := d ZdT0 T1 ··· Tn . (n+d)−1 We denote it by H and let π : H → P d be the projection. (a) Prove that H is nonsingular. (b) Prove that projection π is singular at (X,Z) (in the sense that the derivative of π at (X,Z) is not a surjection) if and only the partial derivatives of FZ ∈ k[X0,...,Xn] have X as a common zero. (c) Suppose from now on that d ≥ 2. Prove that the singular set of π is a n (n+d)−1 smooth subvariety of P × P d of codimension n + 1. (n+d)−1 (d) Prove that the set of Z ∈ P d over which π has a singular point is a hypersurface. This hypersurface is called the discriminant of π. (n+d)−1 (e) For d = 2 we denote the coordinates of P d simply by Zij (where it is understood that Zij = Zji). Prove that the discriminant of π is then the zero set of det(Zij). 8. GRASSMANNIANS 65

8. Grassmannians Let P be a projective space of dimension n and let d ∈ {0, . . . , n}. We want to show that the collection Grd(P ) of linear d-dimensional subspaces of P is a nonsingular projective variety. Let the projective structure on P be deﬁned by the pair (V, `) so that V is a (n + 1)-dimensional k-vector space and P has been identiﬁed with P(V ).

LEMMA 8.1. Let Q ⊂ P be a linear subspace of codimension d + 1. Then the collection of linear d-dimensional subspaces of P contained in P − Q has in a natural manner the structure of an afﬁne space AQ of dimension (n − d)(d − 1).

PROOF. Let Q correspond to the linear subspace W ⊂ V of dimension (n + 1) − (d + 1) = n − d. Then the elements of AQ correspond to linear subspaces L ⊂ V of dimension d + 1 with L ∩ W = {0}. This means that V = L ⊕ W . The affine structure is best seen by fixing some L0 with that property. Then every other such L is always the graph of a (unique) linear map L0 → W and vice versa. So this identifies AQ with Hom(L0,W ). It is easy to verify that this affine structure is independent of the choices (the translation vector space of AQ is canonically identified with Hom(V/W, W )).

It can now be shown without much difficulty that Grd(P ) admits a unique structure of a variety for which a subset as in this lemma is affine open and its identification with affine space an isomorphism. We will however proceed in a more direct manner to show that Grd(P ) admits the structure of a projective variety. V• ∞ Vp We recall that the exterior algebra V = ⊕p=0 V is the quotient of the ∞ ⊗p ⊗0 tensor algebra on V , ⊕p=0V (here V = k by convention), by the two-sided ideal generated by the ‘squares’ v ⊗ v, v ∈ V . It is customary to denote the product by the symbol ∧. So we can characterize V• V as (noncommutative) associative k-algebra with unit element by saying that is generated by the k-vector space V and is subject to the relations v ∧ v = 0 for all v ∈ V . It is a graded algebra (Vp V is the image of V ⊗p) and ‘graded-commutative’ in the following sense: if Vp Vq pq α ∈ V and β ∈ V , then β ∧ α = (−1) α ∧ β. If e0, . . . , en is a basis for V , then a basis of Vp V is indexed by the p-element subsets I ⊂ {0, . . . , n}:

I = {0 ≤ i1 < i2 < ··· < ip ≤ n} is associated to the basis element eI = ei1 ∧· · ·∧eip V0 Vp n+1 (where the convention is that e∅ = 1 ∈ k = V ). So dim V = p . Notice Vn+1 Vp that V is one-dimensional and spanned by e0 ∧ · · · ∧ en, whereas V = 0 for p > n + 1. Let us say that α ∈ Vp V is fully decomposable if the exist linearly independent v1, . . . , vp in V such that α = v1 ∧ · · · ∧ vp. This is equivalent to the existence of a p-dimensional subspace K ⊂ V such that α is a generator of ∧pK. p LEMMA 8.2. Let α ∈ V V be nonzero. Denote by K(α) the set of e ∈ V with e∧α = 0. Then dim K(α) ≤ p and equality holds if and only if α is fully decomposable and spans Vp K(α).

0 PROOF. Let e1, . . . , er be a basis of K(α) and let V ⊂ V be a subspace supplementary to K(α). Then we have a decomposition

• •−|I| ^ M ^ 0 V = eI ∧ V . I⊂{1,...,r} 66 2. PROJECTIVE VARIETIES

Wedging with ei kills all the summands in which ei appears and is injective on the Vp−r 0 sum of the others. So α ⊂ e1 ∧ · · · ∧ er ∧ V . In particular, r ≤ p with equality holding if and only if α is a multiple of e1 ∧ · · · ∧ ep. If L is a linear subspace of V of dimension d + 1, then Vd+1 L is of dimension 1 and will be thought of as a one dimensional subspace of Vd+1 V . We thus have Vd+1 Vd+1 deﬁned map δ : Grd(P ) → P( V ), [L] 7→ [ L]. It is called the Plucker¨ embedding because of: Vd+1 PROPOSITION 8.3. The map δ : Grd(P ) → P( V ) maps Grd(P ) bijectively Vd+1 onto a closed subset of P( V ). d+1 PROOF. Let α ∈ V V . According to Lemma 8.2, [α] is in the image of δ if and only if K(α) is of dimension d + 1 and if that is the case, then δ−1[α] has P(K(α)) as its unique element. In particular, δ is injective. Consider the linear map d+1 d+2 ^ ^ V → Hom(V, V ), α 7→ eα := α∧

The set of linear maps V → Vd+2 V with kernel of dimension ≥ d + 1 are those of rank < s := dim(Vd+2 V ) − d. If we choose a basis for V , then this locus is given by set a homogeneous equations in Hom(V, Vd+2 V ), namely the s × s-minors of the corresponding matrices. Hence the set of α ∈ Vd+1 V for which dim K(α) ≥ d + 1 is also given by a set of homogeneous equations and thus deﬁnes a closed subset of Vd+1 P( V ). This subset is the image of δ.

Proposition 8.3 gives Grd(P ) the structure of projective variety. In order to complete the construction, let Q ⊂ P be a linear subspace of codimension d. Let W ⊂ V correspond to Q and choose a generator β ∈ Vn−d W . Then we have a nonzero linear form d+1 n+1 ^ ^ ∼ eβ = β∧ : V → V = k. Vd+1 Its kernel deﬁnes a hyperplane in P( V ) and hence an afﬁne subspace Ueβ ⊂ Vd+1 P( V ).

LEMMA 8.4. The preimage of Ueβ under the Plucker¨ embedding is the afﬁne space AQ and δ maps AQ isomorphically onto its image. d+1 PROOF. If α is the generator of V L for some (d + 1)-dimensional subspace, then L ∩ W = {0} if and only if α ∧ β 6= 0: if L ∩ W contains a nonzero vector v then both α and β are divisible by v and so α ∧ β = 0; if L ∩ W = {0}, then we have decomposition V = L ⊕ V and it is then easily seen (by picking a compatible −1 basis of V for example) that α ∧ β 6= 0. This implies that δ Ueβ = AQ.

Let us now express the restriction δ : AQ → Ueβ in terms of coordinates. Choose a basis e0, . . . , en for V such that ed+1, . . . , en is a basis for W and β = Vd+1 ed+1 ∧ · · · ∧ en. Then eβ simply assigns to an element α of V the coefficient of ∼ e0 ∧ · · · ∧ ed in α. If L0 ⊂ V denotes the span of e0, . . . , ed, then AQ = Hom(L0,W ) is identified with the affine space A(d+1)×(n−d) of (d + 1) × (n − d)-matrices via n j X j d (ai )0≤i≤d

COROLLARY 8.5. The Plucker¨ embedding realizes Grd(P ) as a nonsingular sub- Vd+1 variety of P( V ) of dimension (n − d)(d + 1). This structure is compatible with the afﬁne structure that we have on each AQ.

PROOF. Every two open subsets of the form AQ a have nonempty intersection and so Grd(P ) is irreducible. The rest follows from the previous corollary.

REMARK 8.6. The image of Grd(P ) is a closed orbit of the natural SL(V )-action Vd+1 on P( V ). It lies in the closure of any other SL(V )-orbit. EXERCISE 64. Let P be a projective space. Given integers 0 ≤ d < e ≤ dim P , prove that the set of pairs of linear subspaces R ⊂ Q ⊂ P with dim R = d and dim Q = e is a closed subvariety of Grd(P ) × Gre(P ). The Grassmannian of hyperplanes in a projective space is itself a projective space (see Exercise 61). So the simplest example not of this type is the Grassman- nian of lines in a 3-dimensional projective space. Let V be vector space dimension 4. On the 6-dimensional space V2 V we have a homogeneous polynomial F : V2 V → k of degree two deﬁned by 4 ^ F (α) := α ∧ α ∈ V =∼ k (the last identiﬁcation is only given up to scalar and so the same is true for F ). In coordinates F is quite simple: if e1, . . . , e4 is a basis for V , then (ei ∧ ej)1≤i

F (T1,2,...,T3,4) = T1,2T3,4 − T1,3T2,4 + T1,4T2,3. Notice that F is irreducible. Its partial derivatives are the coordinates themselves (up to sign and order) and so F deﬁnes a smooth quadric hypersurface of dimension 4 in a 5-dimensional projective space. V2 PROPOSITION 8.7. The image of the Plucker¨ embedding of G1(P(V )) in P( V ) is the zero set of F .

PROOF. The image of the Plucker¨ embedding is of dimension 4 and so must be a hypersurface. Since the zero set of F is an irreducible hypersurface, it sufﬁces to show that the Plucker¨ embedding maps to the zero set of F . For this, let α be a V2 generator of L for some linear subspace L ⊂ V of dimension 2. If e1, . . . , e4 is a basis of V such that α = e1 ∧ e2, then it is clear that α ∧ α = 0. This proves that the Plucker¨ embedding maps to the zero set of F . 68 2. PROJECTIVE VARIETIES

d+1 REMARK 8.8. The image of the Plucker¨ embedding Grd(P ) ,→ P(∧ V ) is in fact always the common zero set of a collection of quadratic equations, called the Plucker¨ relations. To exhibit these, we first recall that every φ ∈ V ∗ defines a linear ‘inner contrac- • •−1 tion’ map ιφ : ∧ V → ∧ V of degree −1 characterized by the fact that for v ∈ V , 0 p • p ιφ(v) = φ(v) ∈ k = ∧ V and for α ∈ ∧ V, β ∈ ∧ V , ιφ(α∧β) = ιφ(α)∧β +(−1) α∧ιφ(β). d+1 d+1 d+2 d We then define a linear map B : ∧ V × ∧ V → ∧ V ⊗ ∧ V as follows: if (e0, . . . , en) ∗ ∗ ∗ is a basis of V and (e0, . . . , en) is the basis of V dual to (e0, . . . , en), then n X ∗ B(α, β) := (α ∧ er) ⊗ (ιer β). r=0 It is easy to check that this is independent of the choice of basis. In particular, if α is fully decomposable, then we can choose (e0, . . . , en) in such a manner that α = e0 ∧ · · · ∧ ed. We then find: X ∗ B(α, α) = (e0 ∧ · · · ∧ ed ∧ er) ⊗ (ιer e0 ∧ · · · ∧ ed ∧ er) = 0, r

∗ because e0 ∧ · · · ∧ ed ∧ er = 0 for r ≤ d and ιer e0 ∧ · · · ∧ ed ∧ er = 0 for r > d. This identity might be called the universal Plucker¨ relation. One can show that conversely, any α ∈ ∧d+1V for which B(α, α) = 0 is either zero or fully decomposable. Let us rephrase this in terms of algebraic geometry: every nonzero linear form ` on d+2 d d+1 ∧ V ⊗ ∧ V , determines a quadratic form Q` on ∧ V deﬁned by α 7→ `(B(α, α)) whose d+1 zero set is a quadratic hypersurface in P(∧ V ). This quadric contains the Plucker¨ locus and the latter is in fact the common zero set of the Q`, with ` running over the linear forms d+2 d on ∧ V ⊗ ∧ V . It can be shown that the Q` generate the full graded ideal deﬁned by the Plucker¨ locus.

PROPOSITION-DEFINITION 8.9. Let X be a closed subvariety of the projective space P . If d is an integer between 0 and dim P , then the set of d-linear subspaces of P which are contained in X deﬁnes a closed subvariety Fd(X) of Grd(P ), called the Fano variety (of d-planes) of X.

PROOF. An open affine chart of Grd(P ) is given by a decomposition V = L⊕W with dim L = d + 1 and dim W = n − d and is then parametrized by Hom(L, W ) by assigning to A ∈ Hom(L, W ) the graph of A. It suffices to prove that via this identification Fd(X) defines a closed subset of Hom(L, W ). Choose homogeneous coordinates [Z0 : ··· : Zn] such that L resp. W is given by Zd+1 = ··· = Zn = 0 resp. Z0 = ··· Zd = 0. A linear map A ∈ Hom(L, W ) ∗ P is then given by A Zd+i = j=0d aijZj, i = 1, . . . , n − d. It defines an element ∗ ∗ of Fd(X) if and only for all G ∈ ∪m≥0Im(X), G(Z0,...,Zd,A Zd+1,...A Zn) is m0 md identically zero. This means that the coefficient of every monomial Z0 ··· Zd in such an expression much vanish. Since this coefficient is a polynomial in the matrix coefficients aij of A, we find that the preimage of Fd(X) in Hom(L, W ) is the common zero set of a set of polynomials and hence is closed therein. EXAMPLE 8.10. Consider the case of a quadratic hypersurface X ⊂ P(V ) and assume for simplicity that char(k) 6= 2. So in terms of homogeneous coordinates 1 P [T0 : ··· : Tn], X can be given by an equation F (T0,...,Tn) = 2 0≤i,j≤n bijTiTj 1 2 with bij = bji (so 2 bii is the coefficient of Ti ). In more intrinsic terms: X is given by a symmetric bilinear form B : V × V → k (namely by B(v, v) = 0). Let us assume that X is nonsingular. This means that the partial derivatives of F have no common zero in P(V ). This translates into: B : V × V → k is nonsingular, that is, the linear map b : v ∈ V 7→ B( , v) ∈ V ∗ is an isomorphism of vector spaces. 8. GRASSMANNIANS 69

A subspace L ⊂ V determines an element of the Fano variety of X precisely when B(v, v) = 0 for all v ∈ L. This implies that B is identically zero on L × L. So b maps L to (V/L)∗ ⊂ V ∗. Since b is injective, this implies that dim L ≤ dim(V/L), 1 in other words that dim L ≤ 2 dim V . This condition is optimal. If for instance dim V = 2m + 2 is even, then it not difficult to show that we can find coordinates (T0, ··· ,T2m+1) such that F = Pm 0 i=0 TiTm+1+i. Let L resp. L be the linear subspace defined by Tm+1 = ··· = 0 T2m+1 = 0 resp. T0 = ··· = Tm = 0, so that V = L ⊕ L . Notice that both [L] and 0 0 [L ] are in Fm(X). The vector space Hom(L, L ) describes an affine open subset of the Grassmannian of m-planes in P(V ). An element A ∈ Hom(L, L0) is given by ∗ Pm A Tm+i = j=0 aijTj, j = 0, . . . , m. The corresponding m-plane is contained in Pm X precisely when i,j=0 aijTiTj is identically zero, i.e., if (aij) is antisymmetric. It follows that [L] ∈ Fm(X) has a neighborhood isomorphic to an affine space of 1 dimension 2 m(m + 1). EXERCISE 65. Let X be a quadratic hypersurface P(V ) of odd dimension 2m+1 and assume for simplicity that char(k) 6= 2. Prove that Fm+1(X) = ∅ and that Fm(X) 6= ∅ (Hint: use the fact that we can choose coordinates (T0,...,T2m+2) for 2 Pm+1 V such that F is given by T0 + i=1 TiTm+1+i.) Prove that Fm(X) is a smooth variety and determine its dimension. EXERCISE 66. Let X ⊂ Pn be a hypersurface of degree d and let 0 ≤ m ≤ n. n Prove that the intersection of Fm(X) with a standard affine subset of Grm(P ) is m+d given by d equations. EXERCISE 67. Consider the universal hypersurface of degree d in Pn, H ⊂ n (n+d)−1 P × P d . n (n+d)−1 (a) For every m-plane Q ⊂ P , let Yz denote the set of z ∈ P d for which the corresponding hypersurface Hz contains Q. Prove that Yz is a linear n+d ( d )−1 m+d subspace of P of codimension d . (n+d)−1 (n+d)−1 (b) Let Y ⊂ P d be the set of z ∈ P d for which Hz contains an (n+d)−1 m-plane. Prove that Y is a closed subset of P d of codimension at m+d least d − (m + 1)(n − m). (c) Prove that the family of m-planes contained in a generic hypersurface of n m+d degree d in P is of dimension (m + 1)(n − m) − d or empty. In m+d1 particular, this is a finite set when (m + 1)(n − m) = d . Let P be as before and let X be a nonsingular (not necessarily closed) subvariety of P of dimension d. For every p ∈ X, we have defined the tangent space TpX as a d-dimensional subspace of TpP . There is precisely one d-dimensional linear subspace Tˆ(X, p) of P which contains p and has the same tangent space at p as X. In terms of a standard affine chart: if U ⊂ P is the complement of a hyperplane which does not pass through p, then U has the structure of an affine space and Tˆ(X, p) ∩ U is then the affine subspace of U spanned by TpX.

PROPOSITION-DEFINITION 8.11. The map G : X → Grd(P ), p ∈ X 7→ Tˆ(X, p) is a morphism. This morphism is called the Gauss map.

1 3 For instance, every cubic surface in P (so here n = 3, d = 3 and m = 1) contains a line. If it is nonsingular, then it contains in fact exactly 27 lines. This famous result due to Cayley and Salmon published in 1849 is still subject of research. 70 2. PROJECTIVE VARIETIES

PROOF. Let p0 ∈ X. The tangent space Tˆ(X, p0) deﬁnes a linear subspace

L ⊂ V of dimension d + 1. This linear subspace contains the line `p0 defined by po. Choose homogeneous coordinates Z0,...,Zn on P such that p0 lies in the standard open subset U0 ⊂ P defined by Z0 6= 0. We use the standard coordinates zi := Zi/Z0 to parametrize U0 ⊂ P , but this time we find it convenient to include z0, which is of course constant 1 on U0. In this way U0 is identified with the affine hyperplane H1 ⊂ V defined by z0 = 1. The tangent space TpU0 (p ∈ U0) is then just p+H, where H ⊂ V is defined by z0 = 0. In particular, if p ∈ X ∩U0, then TpX can be regarded as a linear subspace K(p) ⊂ H0 of dimension d displaced over p. The span of p and K(p) is then a linear subspace of V which corresponds to Tˆ(X, p). According to Theorem 10.12 there exists a neighborhood U of p0 in U0, regular functions f1, . . . , fn−d on U such that X ∩ U is their common zero set and df1, . . . , dfn−d are linearly independent on all of U. Then for every p ∈ X ∩ U, K(p) is the common zero set of df1(p), . . . , dfn−d(p). We regard (df1, . . . , dfn−d) n−d as a matrix-valued function on U, denoted D : U → Hom(H0, k ), so that for p ∈ U ∩ X, K(p) is the kernel of D(p). We also assume our coordinates chosen in 0 00 such a manner that K(p0) is spanned by e1, . . . , ed and then write H = H ⊕ H 0 00 with H = K(p0) and H spanned by ed+1, . . . , en. We write D = (D0,D00), with D0 : U → Hom(H0, kn−d),D00 : U → Hom(H00, kn−d). 00 00 Since D (p0) is an isomorphism, the set of p ∈ U for which D (p) is an isomorphism defines an open neighborhood of p0 and so we may as well assume that that this is the case on all of U. Cramer’s rule shows that p ∈ U 7→ D00(p)−1 ∈ Hom(kn−d,H00) has regular entries: it is a morphism. For every p ∈ U, the kernel of D(p) is the graph of the linear map B(p) := −D00(p)−1D0(p) ∈ Hom(H0,H00). In particular, for p ∈ X ∩ U, TpX is parallel to the subspace spanned by e1 + B(p)(e1), . . . , ed + B(p)(ed). It follows that a linear subspace of V which corresponds to Tˆ(X, p) is spanned Pn by p = e0 + i=1 piei and the d vectors e1 + B(p)(e1), . . . , ed + B(p)(ed). This is 0 n−d the graph of the linear map A(p): ke0 ⊕ H → k defined by

d n X X A(p)(e0) := − zi(p)B(p)(ei) + pjej,A(p)(ei) := B(p)(ei)(i = 1, . . . d). i=1 j=d+1

Since A is a morphism, this deﬁnes a morphism X ∩ U → Grd(P ). For a closed irreducible subset X ⊂ P , the Gauss map is deﬁned on its nonsingular part: G : Xreg → Grd(P ). The closure of the graph of G in X × Grd(P ) is called the Nash blowup of X and denoted Xˆ. The projection morphism Xˆ → X is an isomorphism over the open-dense subset Xreg and hence birational. A remarkable property of Xˆ is that the Zariski tangent space of each point contains a distinguished d-dimensional subspace (prescribed by the second projection to Grd(P )) in such a manner that these subspaces extend the tangent bundle in a regular manner.

9. Multiplicities of modules Bézout’s theorem asserts that two distinct irreducible curves C,C0 in P2 of degrees d and d0 intersect in dd0 points. Strictly speaking this is only true if C and C0 intersect as nicely as possible, but the theorem is true as stated if we count each 9. MULTIPLICITIES OF MODULES 71 point of intersection with an appropriate multiplicity. There is in fact a generalization: the common intersection of n hypersurfaces in Pn has cardinality the product of the degrees of these hypersurfaces, provided that this intersection is as nice as possible and again, this is even true more generally when the locus of intersection is finite and each point of intersection is counted with an appropriate multiplicity. One of our aims is to define these multiplicities. The commutative algebra tools that we use for this have an interest in their own right.

DEFINITION 9.1. We say that an R-module has length ≥ d if there exist a d-step ﬁltration by submodules M = M 0 ) M 1 ) ··· ) M d = {0}. The length of M is the supremum of such d (and so may be ∞).

EXERCISE 68. Suppose R is a noetherian local ring with maximal ideal m and residue field K. Prove that the length of a finitely generated R-module M is finite d Pd−1 i i+1 precisely when m M = 0 for some d and is then equal to i=0 dimK (m M/m M). Prove that if R is a K-algebra, then this is also equal to dimK (M). We fix a noetherian ring R and a finitely generated R-module M. Recall that if p is a prime ideal of R, then Rp is a local ring with maximal ideal pRp whose residue field can be identified with the field of fractions of R/p. We define Mp := Rp ⊗R M. So this is a Rp-module.

−1 REMARK 9.2. We can describe Mp and more generally, any localization S R ⊗R M, as follows. Consider the set S−1M of expressions m/s with m ∈ M and s ∈ S with the understanding that m/s = m0/s0 if the identity s00s0m = s00sm holds in M for some s00 ∈ S (so we are considering the quotient of S×M by an equivalence relation). Then the following rules put on S−1M the structure of a R-module: m/s − m0/s0 := (s0m − sm0)/(ss0), r · m/s := rm/s. The map S−1R × M → S−1M, (r/s, m) → (rm)/s is R-bilinear and hence factors through −1 −1 −1 an R-homomorphism S R ⊗R M → S M. On the other hand, the map S M → −1 0 0 00 0 S R ⊗R M, m/s 7→ 1/s ⊗R m is also deﬁned: if m/s = m /s , then s (s m = sm) for some s00 ∈ S and so

0 00 0 00 0 00 00 0 1/s ⊗R m = 1/(ss s ) ⊗R s s m = 1/(ss s ) ⊗R ss m = 1/s ⊗R m. It is an R-homomorphism and it immediately veriﬁed that it is a two-sided inverse of the −1 −1 map above. So S R ⊗R M → S M is an isomorphism. This description shows in particular that if N ⊂ M is a submodule, then S−1N may be regarded as submodule of S−1M (this amounts to: S-localization is an exact functor).

DEFINITION 9.3. The multiplicity of M at a prime ideal p of R, denoted µp(M), is the length of Mp as an Rp-module. (For x ∈ Spec(R), we may also write µx(M) instead of µpx (M).)

REMARK 9.4.√ Let X be a variety, x ∈ X a closed point of X and I ⊂ OX,x r an ideal with I = mX,x. So I ⊃ mX,x for some positive integer r. Then r dimk(OX,x/I) is ﬁnite (since dimk(OX,x/mX,x) is) and according to Exercise 68 equal to the length of OX,x/I of as an OX,x-module. Hence it is the multiplicity of OX,x/I at the prime ideal mX,x: µx(OX,x/I) = dimk(OX,x/I). If X is afﬁne and we are given an ideal I ⊂ k[X] whose image in OX,x is I, then OX,x/I is the localization of k[X]/I at x and x is an isolated point of Z(I). Then still µx(k[U]/I) = dimk(OX,x/I). 72 2. PROJECTIVE VARIETIES

If X is nonsingular at x of dimension n and I has exactly n generators f1, . . . , fn, n then we will see that µp(OX,x/(f1, . . . , fn)) = dimk(OA ,p/(f1, . . . , fn)) can be in- terpreted as the multiplicity of p as a common zero of f1, . . . , fn. We wish to discuss the graded case parallel to the ungraded case. This means ∞ that if R is graded: R = ⊕i=0Ri, then we assume M to be graded as well, that is, M is endowed with a decomposition as an abelian group M = ⊕i∈ZMi such that Rj sends Mi to Mi+j (we do not assume that Mi = 0 for i < 0). For example, a homogeneous ideal in R is a graded R-module. In that case we have the notion of graded length of M, which is the same as the deﬁnition above, except that we only allow chains of graded submodules. The annihilator of an R-module M, Ann(M), is the set of r ∈ R with rM = 0. It is clearly an ideal of R. We denote by P(M) the set of prime ideals of R which contain Ann(M) and are minimal for that property. According to Exercise 14 these are ﬁnite in number and their common intersection equals pAnn(M). In the graded setting, Ann(M) is a graded ideal and then according to Lemma 2.3 the members of P(M) are all graded. We further note that the annihilator of any m ∈ M, Ann(m) := {r ∈ R : rm 6= 0} is an ideal of R, which is graded when m is homogeneous.

LEMMA 9.5. Let M be a ﬁnitely generated and nonzero (graded) R-module. Then the collection of annihilators of nonzero (homogeneous) elements of M contains a maximal element and any such maximal element is a (graded) prime ideal of R.

PROOF. We only do the graded case. The ﬁrst assertion follows from the noetherian property of R. Let now Ann(m) be a maximal element of the collection (so with m ∈ M homogeneous and nonzero). It sufﬁces to show that this is a prime ideal in the graded sense (see Exercise 54), i.e., to show that if a, b ∈ R are homogeneous and ab ∈ Ann(m), but b∈ / Ann(m), then a ∈ Ann(m). So bm 6= 0 and a ∈ Ann(bm). Since Ann(bm) ⊃ Ann(m), the maximality property of the latter implies that this must be an equality: Ann(bm) = Ann(m), and so a ∈ Ann(m). If l is an integer and M is graded, then M(l) denotes the same module M, but with its grading shifted over l, meaning that M(l)i := Ml+i. Let us call a (graded) R-module elementary if it is isomorphic to R/p ((R/p)(l)), where p is a (homogeneous) prime ideal (and l ∈ Z). So the above lemma says that every nonzero (graded) R-module contains an elementary submodule.

PROPOSITION 9.6. Every finitely generated (graded) R-module M can be obtained as a successive extension of elementary modules in the sense that there exists a finite filtration by (graded) R-submodules M = M 0 ) M 1 ) ··· ) M d = {0} such that each quotient M i/M i+1 is elementary.

PROOF. We do the graded case only. Since M is noetherian, the collection of graded submodules of M that can be written as a successive extension of elementary modules has a maximal member, M 0, say. We claim that M 0 = M. If M/M 0 were nonzero, then according to Lemma 9.5, it contains an elementary submodule. But then the preimage N of this submodule in M is a successive extension of an extension of elementary modules which strictly contains M 0. This contradicts the maximality of M. 9. MULTIPLICITIES OF MODULES 73

PROPOSITION 9.7. In the situation of the preceding proposition, let pi be the prime ideal of R such that M i−1/M i =∼ R/pi. Then P(M) is the set of minimal members i d of the collection {p }i=1 and every p ∈ P(M) occurs precisely µp(M) times in the sequence (p1,..., pd). p PROOF. We first show that Ann(M) = p1 ∩ · · · ∩ pd. If r ∈ p1 ∩ · · · ∩ pd, then r maps M i−1 to M i and so rd ∈ Ann(M). This proves that p1 ∩· · ·∩pd ⊂ pAnn(M). Conversely, if r ∈ R and l ≥ 1 are such that rl ∈ Ann(M), then rl ∈ pi for all i. This implies that r ∈ pi for all i. i d So the collection of minimal members of {p }i=1 is just the set P(M) of minimal prime ideals containing Ann(M). In particular, no pi can be strictly contained in a member of P(M). 0 d Fix p ∈ P(M). We then have a filtration Mp = Mp ⊃ · · · ⊃ Mp = {0} (for an inclusion of R-modules induces an inclusion of Rp-modules). We have i−1 i ∼ i Mp /Mp = Rp/p Rp and the latter is the residue field Rp/pRp or trivial according to whether or not pi = p (if pi 6= p, then pi contains an element not in p; this gives i in Rp an invertible element, and hence p Rp = Rp). Following our definition the first case occurs precisely µp(M) times. We can of course pass from the graded case to the nongraded case by just forgetting the grading. But more interesting and useful is the following construction, where we pass from a projective setting to an affine one and vice versa. The example to keep in mind is when we associate to a homogeneous polynomial of degree d, F ∈ k[T0,...,Tn] the inhomogeneous polynomial f ∈ k[x1, . . . , xn] of degree ≤ d by f(x1, . . . , xn) := F (1, x1, . . . , xn) and go back via F (T0,...,Tn) := d T0 f(T1/T0,...,Tn/T0). Let p ⊂ R be a graded prime ideal. Then denote by Rp,0 the set of fractions r/s with r ∈ Ri and s ∈ Ri − pi for some i. This is clearly a subring of Rp; it is in fact a local ring whose maximal ideal mp is obtained by taking in the previous sentence r ∈ pi. For instance, if R = k[T0,...,Tn] and p = (T1,...,Tn), then p defines the n n point p = [1 : 0 : ··· : 0] ∈ P and Rp can be identified with OP ,p. This definition makes also sense for any graded R-module M: Mp,0 is the set of fractions m/s with m ∈ Mi and s ∈ Ri − pi for some i. Note that this is a Rp,0-module. −1 Suppose now that p1 6= R1 and choose s1 ∈ R1 − p1 so that s1 ∈ (Rp)−1. Then R(l)p,0 is the the set of fractions r/s with r ∈ Ri+l and s ∈ Ri − pi for some l l l l i. For such a fraction we have r/s = s1r/(s1s) ∈ s1Rp,0, so that R(l)p,0 = s1Rp,0, which as an Rp,0-module is of course isomorphic to Rp,0. So the graded iterated extension M = M 0 ) M 1 ) ··· ) M d = {0} of M by elementary R-modules yields an iterated extension of Mp,0 by trivial or by elementary Rp,0-modules: Mp,0 = 0 1 d Mp,0 ⊃ Mp,0 ⊃ · · · ⊃ Mp,0 = {0} and the number of times a nonzero successive quotient appears is µp(M). In other words, we have µp(M) = µmp (Mp,0). We use this observation mainly via the following example.

n EXAMPLE 9.8. Let J ⊂ k[T0,...,Tn] be a homogeneous ideal and let p ∈ P be an isolated point of ZJ . We take here M := k[T0,...,Tn]/J and take for p the graded ideal Ip ⊂ k[T0,...,Tn] defining p. Then k[T0,...,Tn]Ip,0 can be identified with the local k-algebra O n . Moreover, J defines an ideal J ⊂ O n and we P ,p √ P ,p n n can identify MIp,0 with OP ,p/J. Notice that I = mP ,p. According to the above n n discussion µI (M) = µp(OP ,p/I) and by Exercise 68 this is just dimk(OP ,p/J). 74 2. PROJECTIVE VARIETIES

10. Hilbert functions and Hilbert polynomials We shall be dealing with polynomials in Q[z] which take integral values on integers. Such polynomials are called numerical. An example is the binomial function of degree n ≥ 0: z z(z − 1)(z − 2) ··· (z − n + 1) := . n n! k It has the property that its value in any integer k is an integer, namely n if k ≥ n, n−k+n−1 0 if 0 ≤ k ≤ n − 1 and (−1) n if k ≤ −1. Let ∆ : Q[z] → Q[z] denote the difference operator: ∆(f)(z) := f(z +1)−f(z). It clearly sends numerical polynomials to numerical polynomials. It is also clear that z z the kernel of ∆ consists of the constants Q. Notice that ∆ n+1 = n . LEMMA 10.1. Every P ∈ Q[z] which takes integral values on sufﬁciently large integers is numerical and a Z-basis of the abelian group of numerical polynomials is provided by the binomial functions. If f : Z → Z is a function such that for sufﬁciently large integers ∆(f) is given by a polynomial, then so is f.

PROOF. The ﬁrst assertion is proved with induction on the degree d of P . If d = 0, then P is constant and the assertion is obvious. Suppose d > 0 and the Pd−1 z assertion known for lower values of d. So ∆(P )(z) = i=0 ci i for certain ci ∈ Z. Pd−1 z Then P (z) − i=0 ci i+1 is in the kernel of ∆ and hence is constant. As this expression takes integral values on large integers, this constant is an integer. This proves that P is an integral linear combination of binomial functions. For the second assertion, let P ∈ Q[z] be such that P (i) = ∆(f)(i) ∈ Z for P z large i. By the preceding, P (z) = i ai i for certain ai ∈ Z. So if we put P z Q(z) := i ai i+1 , then Q is a numerical polynomial with ∆(f − Q)(i) = 0 for large i. This implies that f − Q is constant for large i, say equal to a ∈ Z. So f(i) = Q(i) + a for large i and hence Q + a is as required. We shall see that examples of such functions are furnished by the Hilbert functions of graded noetherian modules.

REMARK 10.2. A function f : Z → Z which is zero for sufﬁciently negative integers de- P k k termines a Laurent series Lf := f(k)u ∈ ((u)). For the function k 7→ max{0, } k∈Z Z n this gives n n n n n X k(k − 1) ··· (k − n + 1) k u d X k u d 1 u u = u = = . n! n! dun n! dun 1 − u n!(1 − u)n k≥n k≥0 So if we also know that for sufﬁciently large integers f is the restriction of a polynomial function, then Lemma 10.1 implies that Lf is a Z-linear combination of powers of u and the un −1 −1 rational functions n!(1−u)n . In particular, Pf ∈ Q[u][u , (1 − u) ].

In the rest of this section R stands for the graded ring k[T0,...,Tn] where each Ti has degree one. An R-module is always assumed to be graded and ﬁnitely generated.

Let M be a graded noetherian R-module. We deﬁne the Hilbert function of M, φM : Z → Z, by φM (i) := dimk Mi. i+n EXAMPLE 10.3. The Hilbert function of R is given by φR(i) = n . 10. HILBERT FUNCTIONS AND HILBERT POLYNOMIALS 75

LEMMA 10.4. Let F ∈ Rd and denote by Ann(F,M) ⊂ M the set of m ∈ M with F m = 0 (this is a graded submodule of M). Then we have φM/F M (i) = φM (i) − φM (i − d) + φAnn(F,M)(i − d). PROOF. In degree i, multiplication by F deﬁnes a k-linear map between ﬁnite dimensional k-vector spaces Mi−d → Mi whose kernel is Ann(F,M)i−d and whose cokernel is (M/F M)i. The lemma easily follows from this. The zero set of the graded ideal Ann(M) in Pn will be called the support of M and denoted supp(M).

THEOREM 10.5 (Hilbert-Serre). Let M be a ﬁnitely generated graded module over the graded ring R = k[T0,...,Tn] (where each Ti has degree 1). Then there exists a unique numerical polynomial PM ∈ Q[z], the Hilbert polynomial of M, such that φM (i) = PM (i) for i sufﬁciently large. The degree of PM is equal to dim(supp(M)) if we agree that the zero polynomial has the same degree as the dimension of the empty set (namely −1).

PROOF. We proceed with induction on n ≥ −1. For n = −1, M is a finite dimensional graded k-vector space and hence Mi = 0 for i sufficiently large. So assume n ≥ 0 and the theorem verified for lower values of n. We first reduce to the case when M is of the form R/p, with p a graded prime ideal. If N is a graded submodule of M, then Ann(M) = Ann(N)∩Ann(M/N) and so supp(M) = supp(N) ∪ supp(M/N). Since dimk(Mi) = dimk Ni + dimk(Mi/Ni), we have φM = φN + φM/N . It follows that if the theorem holds for N and M/N, then it holds for M. As M is a successive extension of elementary modules, it suffices to do the case M = (R/p)(l) with p a graded prime ideal. Since φM (i) = φR/p(l + i), it is enough to do the case M = R/p. So let M = R/p. Then supp(M) = supp(R/p) is defined by the graded ideal p. In case p = (T0,...,Tn), the theorem holds trivially: we have dimk Mi = 0 for i > 0 (so that we may take PM to be identically zero) and supp(M) = ∅. Suppose therefore that p 6= (T0,...,Tn), say that Tn ∈/ p. Now multiplication by Tn sends M isomorphically onto TnM with quotient M¯ := M/TnM and so for ¯ i ≥ 0, φM¯ (i) = φM (i) − φM (i − 1) = ∆φM (i − 1). Since Ann(M) = Ann(M) + ¯ n−1 TnR, we have supp(M) = supp(M) ∩ P . According to Proposition 7.2 we then have dim supp(M¯ ) = dim supp(M) − 1. By regarding M¯ as a finitely generated module over R¯ := k[T0, ··· ,Tn−1], our induction hypothesis tells us that there ¯ exists a polynomial PM¯ of degree dim supp(M) such that φM¯ and PM¯ coincide on large integers. Since ∆φM (i − 1) = PM¯ (i) for large i, Lemma 10.1 implies that there exists a polynomial PM of degree one higher than that of PM¯ (so of degree dim(supp(M))) which coincides with φM for sufficiently large integers.

It follows from Lemma 10.1 that when PM is nonzero, then its leading term has d the form cdz /d!, where d is the dimension of supp(M) and cd is a positive integer.

REMARK 10.6. For M as in this theorem we may also form the Laurent series LM (u) := P i i dim(Mi)u (this is usually called the the Poincar´e series of M). It follows from Remark −1 −1 10.2 and Theorem 10.5 that LM (u) ∈ Q[u][u , (1 − u) ]. DEFINITION 10.7. If d = dim supp(M), then the degree deg(M) is d! times the leading coefﬁcient of its Hilbert polynomial (an integer, which we stipulate to be n zero in case supp(M) = ∅). For a closed subset Y ⊂ P , the Hilbert polynomial PY resp. the degree deg(Y ) of Y are those of R/IY as a R-module. 76 2. PROJECTIVE VARIETIES

Let M be a finitely generated graded R-module. Recall that P(M) denotes the set of minimal prime ideals containing Ann(M). The support of M is defined by the ideal Ann(M) and is graded. So it defines a closed subset in Pn which in our previous notation is just ZAnn(M). For every p ∈ P(M) not equal to (T0,...,Tn), n Zp ⊂ P is an irreducible component of ZAnn(M) whose degree is deg(R/p) and all irreducible components of ZAnn(M) are so obtained (for p = (T0,...,Tn) we get Zp = ∅, but deg(R/(T0,...,Tn) = 0).

PROPOSITION 10.8. Let M be a ﬁnitely generated graded R-module. Then X deg(M) = µp(M) deg(R/p), p∈P(M) where we note that the right hand side can also be written as a sum over the irreducible components of ZAnn(M) of maximal dimension, where we then may replace deg(R/p) by the degree of that component.

PROOF. Write M as an iterated extension by elementary modules: M = M 0 ) 1 d i−1 i ∼ Pd M ) ··· ) M = {0} with M /M = R/pi(li). Then φM (i) = i=1 φR/pi (i−li).

Now φR/pi is a polynomial of degree equal to the dimension of supp(R/pi) = Zpi ⊂ n P . This degree does not change if we replace the variable i by i − li. In view of Proposition 9.7 we only get a contribution to the leading coefﬁcient of φM when pi ∈ P(M) and for any given p ∈ P(M) this happens exactly µp(Mp) times. We only need to consider the cases p for which the degree of φR/p is maximal (i.e., equals the degree of φM ). The proposition follows.

EXERCISE 69. Compute the Hilbert polynomial and the degree of n (n+d)−1 (a) the image of the d-fold Veronese embedding of P in P n , (b) the image of the Segre embedding of Pm × Pn in Pmn+m+n. EXERCISE 70. Let Y ⊂ Pm and Z ⊂ Pn be closed and consider Y ×Z as a closed subset of Pmn+m+n via the Segre embedding. Prove that the Hilbert function resp. polynomial of Y ×Z is the product of the Hilbert functions resp. polynomials of the factors. One can show that there exists a nonempty open subset of (n − d)-planes Q ⊂ Pn which meet Y in exactly deg(Y ) points (see Exercise 71). This characterization is in fact the classical way of deﬁning the degree of Y . n Observe that if Y ⊂ P is nonempty, then deg(Y ) > 0. For then IY,d 6= Rd for every d ≥ 0 and so the Hilbert function of S(Y ) is positive on all nonnegative integers. This implies that PY is nonzero with positive leading coefﬁcient. We also note that since deg(Y ) only depends on the dimensions of the graded pieces S(Y )i of the homogenenous coordinate ring the degree of Y will not change if we regard Y as sitting in a higher dimensional projective space Y ⊂ Pn ⊂ Pm (with m ≥ n). We verify some other properties we expect from a notion of degree.

PROPOSITION 10.9. A hypersurface H ⊂ Pn deﬁned by an irreducible polynomial of degree d has degree d.

PROOF. Let F ∈ Rd be the polynomial in question and apply Lemma 10.4 to M = R. Since the Hilbert function of R takes at the positive integer i the value 10. HILBERT FUNCTIONS AND HILBERT POLYNOMIALS 77

n+i n+i n+i−d n , the one of H takes at i the value n − n . Its Hilbert polynomial is therefore z + n z − d + n d − = zn−1 + lower order terms, n n (n − 1)! and so deg(H) = d.

PROPOSITION 10.10. Let Y1,...,Yr be the distinct irreducible components of Y Pr of maximal dimension (= dim Y ). Then deg(Y ) = i=1 deg(Yi). 0 PROOF. Put Y := Y2 ∪ · · · ∪ Yr. By proceeding with induction on r, we see that 0 it is enough to show that deg(Y ) equals deg(Y1) + deg(Y ) or deg(Y1) depending on 0 whether r ≥ 2 or r = 1. Since IY = IY1 ∩ IY , we have an exact sequence of graded R-modules 0 0 → S(Y ) → S(Y1) ⊕ S(Y ) → M → 0,

0 0 where M := R/(IY1 + IY ). This implies that PY = PY1 + PY − PM . We have 0 supp(M) = Y1 ∩ Y and so dim supp(M) < dim Y . Hence deg(PM ) < dim(Y ). The dim Y assertion then follows from comparing the coefﬁcients of z . We can now state and prove a theorem of B´ezout type.

COROLLARY 10.11. Let M be a graded R-module. Let F ∈ Rd \ Ann(M). Then M/F M has degree d deg(M) as a R-module and if we put m := dim(ZAnn(M)), then X µZ (M/F M) deg(Z) = d deg(M).

Z irred. comp of ZAnn(M/F M) of dim. m − 1

PROOF. The exact sequence

0 → M[−d] −−−−→·F M → M/F M → 0 shows that PM/F M (z) = PM (z) − PM (z − d). Since PM is of degree m with leading coefficient deg(M)/m!, it follows that PM/F M is of degree m − 1 with leading coefficient d deg(M)/(m − 1)!. An irreducible component C of Ann(M) is not contained in ZF and so dim(C ∩ ZF ) = dim(C) − 1. This produces all the irreducible components of Ann(M/F M) = Ann(M) + FR. It remains to apply Proposition 10.8 to M/F M. COROLLARY 10.12. Let Y ⊂ Pn be closed with all irreducible components of the same dimension m. If Q ⊂ Pn is a linear subspace of codimension m with the property P that Q ∩ Y is finite, then the degree of Y is equal to p∈Y ∩Q µp(S•(Y )/IQS•(Y )). n PROOF. Let H1,...,Hm be hyperplanes in P such that Q = H1 ∩ · · · ∩ Hm and k m put Y := Y ∩ H1 ∩ · · · ∩ Hk (k = 0, . . . m). We are given that Y = Q ∩ Y is finite. Since intersecting a subvariety of Pn with with a hyperplane makes its dimension drop by at most one, we see that each irreducible component of Y k has dimension m − k. If we apply the first part of Corollary 10.11 to a defining linear form for Hk, k k−1 we find that Y has the same degree as Y .

n For p ∈ Q ∩ Y , let IY,p resp. IQ,p be the ideal in OP ,p deﬁning Y resp. Q at that point. Then we observed that

n µp(S•(Y )/IQ) = dimk OP ,p/(IY,p + IQ,p). 78 2. PROJECTIVE VARIETIES

EXERCISE 71. Prove that deg(Y ) is the cardinality of Q ∩ Y if and only if IY,p + n IQ,p = mP ,p for all p ∈ Q ∩ Y . (We then say that Q and Y are in general position with respect to each other.) n THEOREM 10.13 (Theorem of Bézout). Let or i = 1, . . . n, Hi ⊂ P be a hypersurface of degree di > 0 and assume that H1 ∩ · · · ∩ Hn is finite. Each Hi determines n n at p ∈ H1 ∩ · · · ∩ Hn a principal ideal in OP ,p; denote by Ip ⊂ OP ,p the sum of these ideals. Then X n d1d2 ··· dn = dimk(OP ,p/Ip). p∈H1∩···∩Hn k n PROOF. Put Z := H1∩· · ·∩Hk. Since Z is finite, all irreducible components of Zk are of dimension n − k (by the same argument as in Corollary 10.12). Repeated n application of Corollary10.11 then yields that d1d2 ··· dn is the degree of Z . With the help of Exercise 71 we see that this degree is computed as stated. EXAMPLE 10.14. Assume char(k) 6= 2. We compute the intersection multiplicities of the conics C and C0 in P2 whose affine equations are x2 + y2 − 2y = 0 and x2 − y = 0. There are three points of intersection: (0, 0), (−1, 1) and (1, 1) (so none at infinity). The intersection multiplicity at (0, 0) is the dimension of 2 2 2 2 2 2 2 2 OA ,(0,0)/(x + y − 2y, x − y) as a k-vector space. But OA ,(0,0)/(x + y − 2y, x − 4 2 2 2 1 1 y) = OA ,0/(x − x ) = k[x]/(x ) (for (x − 1) is invertible in OA ,0). Clearly 2 dimk(k[x]/(x )) = 2 and so this is also the intersection multiplicity at (0, 0). The intersection multiplicities at (−1, 1) and (1, 1) are easily calculated to be 1 and thus the identity 2 + 1 + 1 = 2 · 2 illustrates the Bézout theorem. n REMARK 10.15. If Y ⊂ P is closed, then PY (0) can be shown to be an invariant of Y in the sense that it is independent of the projective embedding. In many ways, it behaves like an Euler characteristic. (It is in fact the Euler characteristic of OY in a sense that will become clear once we know about sheaf cohomology.) For example, PY ×Z (0) = PY (0)PZ (0). n z+n We have seen that for a hypersurface Y ⊂ P of degree d > 0, PY (z) = n − z−d+n −d+n nd−1 n and so PY (0) = 1 − n = 1 − (−1) n . For n = 2 (so that Y is a 1 1 curve), we get PY (0) = 1− 2 (d−1)(d−2). The number 1−PY (0) = 2 (d−1)(d−2) is then called the arithmetic genus of the curve. If the curve is smooth and k = C, then we may regard it as a topological surface (a Riemann surface) and g is then just the genus of this surface (and so PY (0) is half its Euler characteristic). CHAPTER 3

Schemes

1. Presheaves and sheaves The notion of a prevariety as well as its generalizations are best understood in the general context of ringed spaces. This involves the even more basic notion of a sheaf.

DEFINITION 1.1. Let X be a topological space. An abelian presheaf F on X consists of giving for every open subset U ⊂ X an abelian group F(U) (whose elements are called sections of F over U) and for every inclusion of open subsets U ⊂ U 0 a homomorphism of groups (called the restriction map) F(U 0) → F(U) such that: (i) for the identity U = U we get the identity in F(U), (ii) if U ⊂ U 0 ⊂ U 00 are open sets, then the homomorphism F(U 00) → F(U) is equal to the composite F(U 00) → F(U 0) → F(U) and In categorical language, a presheaf is a contravariant functor F : O(X)◦ → Ab from the category O(X) of open subsets of X (whose morphisms are inclusions of open subsets of X) to the category of abelian groups Ab. The terminology ‘section over U’ and ‘restriction’ is suggestive, although it sometimes can be a bit misleading. This terminology is mirrored by the notation: if we are given an inclusion of open subsets U 0 ⊂ U, then the image of s ∈ F(U) under the restriction map F(U) → F(U 0) is often denoted s|U 0 (the obvious categorical notation for the restriction 0 map would be F(U ⊂ U), but this is rarely used; more common is resU,U 0 ). In case the groups F(U) come with the structure of a ring and the restriction maps are ring homomorphisms, then we say that F is a presheaf of rings (in other words, the contravariant functor takes values in the category Ring of rings). Like- wise, we have the notion of a presheaf of modules over a ﬁxed ring R. In some situations no structure on F(U) is imposed at all (the target category is then the one of sets), but we will mostly deal with presheaves that are at least abelian.

EXAMPLE 1.2 (The constant presheaf). Given an abelian group G and a topological space X, then we have an abelian presheaf deﬁned on X if we take for every nonempty open U ⊂ X the group G and for each inclusion between two such sets the identity map of G.

EXAMPLE 1.3. Given a topological space X and an abelian topological group G, then assigning to every nonempty open U ⊂ X the group of continuous maps U → G (with the obvious restriction maps) is an abelian presheaf CX,G. Of special interest are G = R and G = C, in which we get a presheaf of R-algebras (resp. of C- algebras). Another case of interest is when G has been given the discrete topology: then this assigns to U the space of locally constant maps U → G.

79 80 3. SCHEMES

EXAMPLE 1.4. For a smooth manifold M, we have deﬁned the presheaf EM of R-algebras which assigns to every nonempty open U ⊂ M the R-algebra of differentiable functions U → R. an EXAMPLE 1.5. On a complex manifold M, we have deﬁned the presheaf OM of C-algebras which assigns to every nonempty open U ⊂ M the C-algebra Oan(U) of holomorphic functions U → C.

EXAMPLE 1.6. On a k-prevariety X, we have defined the presheaf OX of regular functions. DEFINITION 1.7. Given a presheaf F on X and a point p ∈ X, then a germ of a section of F at p, is a section of F on an unspecified neighborhood of p, with the understanding that two such sections represent the same germ if they coincide on a neighborhood of p contained in their common domain of definition. The set of germs of sections of F at p is called the stalk of F at p and denoted by Fp. (In categorical language, this is expressed as an inductive limit F = lim F(U), p −→U3p where one observes that the collection of neighborhoods of p in X form a projective system: the intersection of two neighborhoods of p is another one.) For instance in the case of Example 1.5, when M is an open subset of C, then an the stalk OM,p can be identified with the ring of convergent power series in z − p. Let us observe that an s ∈ F(U) determines an element sp ∈ Fp for every + + Q p ∈ U. We will also write s (p) for sp, where we view s as a map U → p∈U Fp. The last three examples differ from the first in that they satisfy two additional properties: DEFINITION 1.8. An abelian presheaf F on X is called an abelian sheaf if for every open U ⊂ X and open covering {Ui}i∈I of U,

(iv) every system of sections {si ∈ F(Ui)}i∈I that is compatible in the sense that for each pair (i, j) in I, si|Ui ∩ Uj = sj|Ui ∩ Uj comes from some section s ∈ F(U) in the sense s|Ui = si for all i ∈ I, and 0 0 (v) two sections s, s ∈ F(U) coincide if s|Ui = s |Ui for all i ∈ I. In other words, the sequence Y Y 0 → F(U) → F(Ui) → F(Ui ∩ Uj) i∈I (i,j)∈I2 in which the ﬁrst map is given by s 7→ (s|Ui)i∈I and the second by (si)i∈I 7→ (si|Ui ∩ Uj − sj|Ui ∩ Uj)i,j is exact. This implies that F(∅) is the trivial group {0}. The argument relies on a categorical notion of product1, but for our purposes we may just as well stipulate that this is so.

A space X endowed with with a sheaf of rings is also called a ringed space. Examples are 1.3, 1.4, 1.5 and 1.6. A constant presheaf is usually not a sheaf: if U1,U2 are disjoint open subsets and we take gi ∈ F(Ui) = G distinct for i = 1, 2,

1 Q If we are given a product of abelian groups with index set I, i∈I Gi, then for every subset Q Q J ⊂ I, this product maps isomorphically to the product ( i∈J Gi) × ( i∈I−J Gi). By taking J = I, we see that a product with empty index set must be the trivial group. Similarly, every union of sets ∪i∈I Ti, can be written as (∪i∈J Ti) ∪ (∪i∈I−J Ti), from which deduce (by taking J = I), that a union with empty index set must be empty. If we then apply the above exact sequence to U = ∅ by taking an open covering with empty index set, we ﬁnd that F(∅) = {0}. 1. PRESHEAVES AND SHEAVES 81 then clearly these elements cannot be the restriction of a single g ∈ F(U1 ∪U2) = G. There is however a simple modiﬁcation which is a sheaf, namely the constant sheaf GX which assigns to U ⊂ X the space of continuous maps from to U to G, where we endow G with the discrete topology (so these are the maps which are locally constant). This turns it into a special case of Example 1.3.

1.9.S HEAFIFICATION. This modification is a special case of a general construction which produces a sheaf F + out of a presheaf F with the same stalks: a section s˜ of F + over an open subset U is by definition a map from U to the disjoint union + of the stalks Fp, p ∈ U, with the property that locally this map is of the form s : we may cover U by open subsets Ui for which there exist si ∈ F(Ui) such that s˜p = si,p for all p ∈ Ui. It is straightforward to verify that this is well-defined, defines a sheaf and that F + = F in case F happens to be a sheaf. The sheaf F + has a universal property. In order to explain this, we need the notion of a homomorphism of presheaves. Let C be one of our categories. If F and G are C-valued presheaves on X, then a homomorphism of presheaves φ : F → G is what is called in category theory a natural transformation of functors: for every open U ⊂ X, we are given C-homomorphism φ(U): F(U) → G(U) such that this collection is compatible with restriction: if U 0 ⊂ U and s ∈ F(U), then φ(U)(s)|U 0 = φ(U 0)(s|U 0). For example, the formation of F + defines a homomorphism of presheaves F → F +. It is universal in the sense that if φ : F → G is a homomorphism of presheaves and G is a sheaf, then there is unique sheaf homomorphism φ+ : F + → G such that φ is the composite of F → F + and φ+. In general a homomorphism of presheaves φ : F → G induces a sheaf homomorphism φ+ : F + → G+.

EXERCISE 72 (Local nature of a sheaf). Let X be a topological space and U a basis of open subsets of X. Prove that an abelian sheaf F on the space X is determined by its restriction to that basis, that is, by the collection F(U), and the restriction maps F(U 0) → F(U), for the pairs (U, U 0) ∈ U × U with U ⊂ U 0. Prove also a converse: suppose that the basis U is closed under ﬁnite intersec- tions and assume that for every U ∈ U is given an abelian group F(U) and for every inclusion U ⊂ U 0 of members of U a homomorphism F(U 0) → F(U) such that the properties (i) through (iii) are satisﬁed. Then F generates a sheaf F + on X.

EXERCISE 73 (Direct image of a sheaf). Suppose we are given a continuous map f : X → Y between topological spaces and an abelian presheaf F on X. Then −1 a presheaf f∗F on Y is defined by assigning to an open V ⊂ Y the group F(f V ). + + Prove that if F is a sheaf, then so is f∗F. Is it true that (f∗F) = f∗(F )? 1 1 2 EXAMPLE 1.10. Let f : S → S be defined by f(z) = z and let ZS1 be the 1 1 constant sheaf on S with stalks Z. Then f∗ZS1 is a sheaf on S which is locally 2 1 1 like the constant sheaf Z . But (f∗ZS1 )(S ) = ZS1 (S ) = Z and so is of rank less than 2. Other examples in the same spirit are gotten as the sheaf on C − {0} that is the direct image of the constant sheaf on C − {0} with stalk Z under the map z ∈ C − {0} 7→ zn ∈ C − {0} (n ∈ Z − {0}) or of the direct image of the constant sheaf on Z with stalk C under the map z ∈ C 7→ ez ∈ C − {0}. 1.11.K ERNELSANDCOKERNELS. We begin with some obvious definitions: given a presheaf F on X, then a subpresheaf of F is simply a presheaf F 0 on X 82 3. SCHEMES with the property that F 0(U) ⊂ F(U) for every open subset U of X. If F 0 happens to be a sheaf, then we call it a subsheaf of F. Suppose and F and G are abelian presheaves over X and let φ : F → G be a homomorphism of presheaves. So for every open U ⊂ X, we then have a homomorphism φ(U): F(U) → G(U). By assigning to U, Ker(φ)(U), Im φ(U), Coker(φ(U)) we have defined three presheaves, of first two being subpresheaves of F and G respectively. If Ker(φ) is the zero sheaf, then clearly F can be thought of a subpresheaf of G. Suppose now that F and G are in fact sheaves. Then Ker(φ)(U) is easily seen to be a sheaf (and indeed denoted by Ker(φ)). But the other two need not be (see Example 1.12 below) and so we define the image sheaf Im(φ) resp. the cokernel sheaf Coker(φ) as the sheaf associated to the corresponding presheaves. Since this process does not affect the stalks, we have that for every x ∈ X, Im(φ)x resp. Coker(φ)x is the image resp. the cokernel of φx : Fx → Gx. Moreover, Im(φ)x is a subsheaf of G. The possible failure of the image presheaf being a sheaf is expressed by the fact that the image of φ(U): F(U) → G(U) is contained in Im(φ)(U) but need not equal it and a similar statement can be made for the cokernel sheaf. We finally note that the homomorphisms from F to G form an abelian group; we denote that group by Hom(F, G). If F is a sheaf on X, then F(X) is called its group of global sections. One often denotes this group by Γ(X, F) or H0(X, F) instead. The reason for the latter notation will become clear later. An interesting twist to the notion of a constant sheaf is that of a local system (and called by some algebraic topologists a system of twisted coefficients): this is a locally constant sheaf. For instance, if G is an abelian group, then a local system of type G on X is an abelian sheaf F on X with the property that X can be covered by open subsets U such that F|U is isomorphic to the sheaf associated to the constant presheaf G on U. Examples were given in 1.10. Here is another one, which illustrates some other phenomena as well. EXAMPLE 1.12. Let X be C − {0} with its usual Hausdorff topology. On X we define a local system L of 2-dimensional complex vector spaces by letting L(U) be the space of C-valued functions that are of the form f + c with f : U → C continuous and satisfying ef(z) = z. In other words, f is a branch of the logarithm function. But recall that branch need not defined and that The constant functions define a constant subsheaf CX ⊂ L. The quotient sheaf L/CX is also isomorphic to CX , for it admits as a global generating section the function “log” (note that the multivaluedness of this function disappears if we work modulo constants). But the section “log” cannot be lifted to X: the only global sections of L(X) are the constants. So the exact sequence of abelian sheaves

0 → CX → LX → LX /CX → 0 evaluated on X yields a sequence of abelian groups

0 → Γ(X, CX ) → Γ(X, LX ) → Γ(X, LX /CX ) → 0. 0 We observed that the last sequence can be identified with 0 → C = C −→ C → 0. So it fails to be exact at Γ(X, CX ). This also shows that the image presheaf of L → LX /CX and the cokernel presheaf of CX → L are not sheaves. 1.13.T HEPREIMAGEOFASHEAF. Here is a relative version of the sheafification process. Suppose we are given a continuous map f : X → Y between topological 1. PRESHEAVES AND SHEAVES 83 spaces and an abelian sheaf G on Y . We want to define a sheaf f −1G on X with the −1 property that the stalk of f G at x ∈ X is Gf(x). If U ⊂ X is open and s assigns −1 to x ∈ X, sx ∈ Gf(x), then we stipulate that s defines a section of f F if for each x ∈ U there exists a neighborhood Vx of y in Y and a section t ∈ G(Vy) such that 0 −1 for x in a neighborhood of x in f Vx ∩ U, we have sx0 = tf(x). This is indeed a sheaf, called the sheaf pull-back of G. If Y is a single point {y}, so that G is given by a single group G = Gy, then this reproduces the sheaf of locally constant maps from open subsets of X to G. If Y ⊂ X is a subspace, then we often write Γ(Y, F) for the group of global sections of i−1F, where i : Y ⊂ X denotes the inclusion. Notice that when Y is open, this is just F(Y ).

EXERCISE 74 (Adjunction). Suppose that f : X → Y is a continuous map and let F resp. G be an abelian sheaf on X resp. Y . −1 (a) Prove that we may identify Hom(G, f∗F) with Hom(f G, F). (In cate- −1 gorical language: G 7→ f G is the left adjoint of F 7→ f∗F and the latter is the right adjoint of the former, the convention being that ‘left’ acts on the item before the comma in Hom and ‘right’ on what comes after it.) −1 (b) Deduce the existence of natural sheaf homomorphisms G → f∗f G and −1 f f∗F → F. (c) Give examples to show that neither of these needs to be an isomorphism. (Hint for the second case: look at Example 1.10.) (d) Suppose that f is a homeomorphism. Show that under the identiﬁcation −1 (a) a sheaf isomorphism G → f∗F yields a sheaf isomorphism f G → F and vice versa. (When such an isomorphism exists, we say that (X, F) and (Y, G) are isomorphic.)

A space X endowed with a sheaf of rings OX is called a ringed space; the sheaf in question is then often referred to as its structure sheaf. This notion becomes much more useful if we also have a notion of morphism of ringed spaces: if (X, OX ) and (Y, OY ) are ringed spaces, then by a morphism from (X, OX ) to (Y, OY ) we mean a pair consisting of a continuous map f : X → Y and a sheaf homomorphism of rings OY → f∗OX . According to Exercise 74 the latter is equivalent to giving a sheaf −1 homomorphism of rings f OY → OX . (We often abuse notation by denoting this pair by the single symbol f.)

REMARK 1.14. Via Exercise 72 we can (at least formally) characterize certain extra structures on spaces in a uniform manner. For example, given a topological m- manifold M, then a Ck-differentiable structure on M is simply given by a subsheaf of the sheaf R-valued continuous functions on M which has the property that locally it is isomorphic to the sheaf of Ck-differentiable functions on Rm. It is a sheaf of R-algebras. A Ck-map between two Ck-manifolds is then simply a morphism between the corresponding ringed spaces (or rather, spaces endowed with a sheaf of R-algebras). This is a more conceptual (and perhaps also more concise) than the definition based on an atlas. A holomorphic structure on M (which turns M into a complex manifold) and a holomorphic map between complex manifolds can be defined in a similar manner. An affine variety is naturally a ringed space and indeed, a k-prevariety, is a a ringed space locally isomorphic as a ringed space to an affine k-variety. This is why such sheaves are often referred to as structure sheaves. We shall define in the same spirit a structure sheaf on the spectrum of a ring. 84 3. SCHEMES

2. The spectrum of a ring as a locally ringed space Let R be a ring. The (prime ideal) spectrum Spec(R) is the set of its prime ideals. For a prime ideal p of R, the corresponding element of Spec(R) is denoted xp, and the prime ideal associated to x ∈ Spec(R) is denoted px ⊂ R. We write X for Spec(R). For s ∈ R, Z(s) ⊂ Spec(R) denotes the set of xp ∈ Spec(R) with s ∈ p and Xs := X − Z(s). The collection {Xs}s∈R, is basis for a topology on Spec(R), the Zariski topology. We therefore refer to any Xs as a basic open subset of Spec(R). So the closed subsets of this topology are of the form Z(S) = ∩s∈SZ(s), where S ⊂ R is a subset (but note that Z(S) does not change if we replace S by the radical of the ideal generated by S). The same argument as used in Lemma 4.7 shows that Spec(R) is quasicompact. The spectrum of a ring R is a singleton precisely when R has just one prime ideal. Since p(0) is the intersection of all the prime ideals of R, it follows that this prime ideal equals p(0) and must be a maximal ideal. (So if R is noetherian it will have finite length.) In particular, p(0) = (0) precisely when R is a field. The spectrum of a local ring has just one closed point (namely the one defined by its maximal ideal), but can have plenty of nonclosed points, of course. A ring homomorphism φ : R0 → R determines a map 0 Spec(φ): X = Spec(R) → Spec(R ), xp 7→ xp0 , 0 −1 −1 where p := φ p. It has the property that Spec(φ) (Xr0 ) = Xφ(r0) so that Spec(φ) 0 0 is continuous. The ring homomorphism R → R → Rp factors through Rp0 (by 0 Exercise 11) and so we have an associated homomorphism of local rings φp : Rp0 → −1 0 Rp. We have φp (pRp) = p Rp0 and so this homomorphism is local in the sense of:

DEFINITION 2.1. A homomorphism ψ : A0 → A of local rings is said to be a −1 local homomorphism if ψ mA = mA0 (so that it defines an embedding of residue 0 fields A /mA0 ,→ A/m). Hence a homomorphism of local rings is local precisely if the induced map on spectra sends the closed point to the closed point. For instance, the inclusion of the local ring k[[t]] in its quotient field k((t)) is not local, for the maximal ideal of k((t)) is the zero ideal, but the zero ideal is not equal to the maximal ideal tk[[t]] of k[[t]]. Thus the singleton Spec(k((t))) maps to the nonclosed point of Spec(k[[t]]) defined by the zero ideal. For every subset Y ⊂ X, we may consider the set I(Y ) of r ∈ R which ‘vanish’ on Y , that is for which Z(r) ⊃ Y . This is clearly an ideal in R. In this context, a Nullstellensatz comes easily as it is basically the content of Lemma 2.15: √ PROPOSITION 2.2. Given a ring R, then for every ideal J ⊂ R, I(Z(J)) = J.

PROOF. The property r ∈ I(Z(J)) means Z(r) ⊃ Z(J). This in turn means that every prime ideal which contains J must contain r. In other words, I(Z(J)) is the intersection of all the√ prime ideals that√ contain J. According to Exercise 14 this intersection is precisely J and so r ∈ J. PROPOSITION 2.3. Let R be a ring.

(i) The closure of the singleton {xp} consists of the xq ∈ Spec(R) with q ⊃ p. (ii) A singleton {xp} is closed if and only if p is a maximal ideal (so that Spec(R) contains Specm(R) as the set of its closed points). 2. THE SPECTRUM OF A RING AS A LOCALLY RINGED SPACE 85

(iii) A closed subset of Spec(R) is irreducible if and only if it is the closure of a singleton. (iv) A subset of Spec(R) is an irreducible component of Spec(R) if and only if it is of the form {xp} with p a minimal prime ideal of R.

PROOF. The proofs of (i) and (ii) are left to you as an exercise. As for (iii) and (iv), let C ⊂ Spec(R) be a closed irreducible subset. We ﬁrst prove that I(C) is prime. For suppose that r1r2 ∈ I(C) for certain r1, r2 ∈ R. Then Z(r1r2) = Z(r1) ∪ Z(r2) contains C and since C is irreducible it follows that Z(r1) ⊃ C or Z(r2) ⊃ C or equivalently, that r1 ∈ I(C) or r2 ∈ I(C). By deﬁnition, x ∈ C if and only if px ⊃ I(C) and so C is the closure of the singleton {xI(C)}. On the other hand, a singleton is irreducible and hence so is its closure. Now (iii) and (iv) follow.

EXERCISE 75. Let φ : R0 → R be a ring homomorphism. Prove that if I0 ⊂ R0 is a proper ideal, then the natural map Spec(R0/I0) → Spec(R0) is injective with image the closed subset of Spec(R0) deﬁned by I0. Show that the preimage of this closed subset under Spec(φ) : Spec(R) → Spec(R0) is the closed subset of Spec(R) deﬁned by the ideal of R generated by φ(I).

REMARK 2.4. The ring Z has as its prime ideals the zero ideal (0) and the ideal pZ, where p is a prime number. We denote the corresponding points of Spec(Z) by x0 resp. xp. The only nonclosed point of Spec(Z) is x0, which contains every xp in it closure. Given any ring R, then we have a natural ring homomorphism φ : Z → R and hence a continuous map f := Spec(φ) : Spec(R) → Spec(Z). Let us see what its fibers are. Let n ≥ 0 be such that ker(φ) = Zn. If p ⊂ R is a prime ideal, then φ−1p is a prime ideal containing Zn. When n = 0 (so that φ is injective), then this can be any prime ideal, otherwise this must be of the form Zp with p dividing n. In other words, in the first case f is onto and in the second case the image of f is a finite iset of closed points. A prime ideal p ⊂ R with the property that φ−1(p) = Zp −1 corresponds to a prime ideal of R/pR = Fp ⊗ R and so the closed fiber f (xp) can −1 be identified with Spec(Fp ⊗ R) (even as topological spaces). The fiber f (x0) is nonempty only when φ is injective. We can then localize with respect to the image −1 of Z − {0} and find that a point of f (x0) is given by a prime ideal of Q ⊗ R. Conversely, given a prime ideal of Q ⊗ R, then its inverse image in R under the localization homomorphism R → Q ⊗ R is a prime ideal of R whose preimage in Z −1 is trivial. Hence f (x0) = Spec(Q ⊗ R), even as topological spaces.

EXAMPLE 2.5 (The affine line). We define the affine line over a ring R as Spec(R[x]). If R is the algebraically closed field k, then its closed points are in bijection with the points of k (any prime ideal is of the form (x − a), with a ∈ k). When R = R, we get as closed point also the complex conjugate pairs (defined by a prime ideal of the form (x2 + bx + c) with b2 < 4c). In particular, x2 + 1 has a zero in 1 (namely at the point defined by the prime ideal (x2 + 1)). AR Next we turn to 1 = Spec( [x]). From the remark above we see that we have AZ Z natural map f : 1 → Spec( ) with f −1(x ) = Spec( [x]) = 1 and f −1(x ) = AZ Z 0 Q AQ p 1 1 Spec( p[x]) = . The latter is closed in . We treat these cases separately. F AFp AZ Since Q[x] is a principal ideal domain, every prime ideal of Q[x] is either the zero ideal or is generated by an irreducible polynomial P ∈ Q[x] of positive degree. 86 3. SCHEMES

In the first case, the prime ideal defines the (generic) point of Spec(Q[t]), whose closure is dense in Spec(Q[t]). In the second case we can make P unique by taking it monic. Galois theory tells us that if Q is an algebraic closure of Q, then the roots of P in Q make up an orbit under the Galois group Aut(Q) and that conversely, any Aut( )-orbit in is so obtained. It follows the closed points of 1 correspond Q Q AQ bijectively to the Aut( )-orbits in . The nonclosed point of 1 comes from the Q Q AQ zero ideal in [x], and so is, when viewed as a point of 1 , dense in 1 . Z AZ AZ 1 The case can be treated similarly. Fix an algebraic closure p of p. Since AFp F F Fp[x] is a principal ideal domain, every nonzero prime ideal of Fp[x] is generated by an irreducible monic polynomial f ∈ Fp[x] of positive degree r, say. Such a prime ideal is then maximal and so defines a finite extension of Fp of the same degree r. Since f has coefficients in Fp, its set of roots is invariant under the Frobenius p r−1 automorphism Φ: a ∈ Fp 7→ a ∈ Fp. If a is a root of f, then a, Φ(a),..., Φ (a) r Qr−1 i are pairwise distinct, whereas Φ (a) = a. So i=0 (x − Φ (a)) divides f, hence equals f. Conversely, for any a ∈ Fp which generates a field extension of Fp of degree r (equivalently, for which Φr(a) = a), the minimal polynomial for a is irreducible and so generates a maximal ideal in Fp[x]. It follows that the closed 1 points of are in bijective correspondence with the Φ-orbits in p. There is one AFp F other (nonclosed) point in 1 and its closure equals 1 . AFp AFp Let R be a ring and write X for Spec(R). When s ∈ R is nilpotent, then s is contained in every prime of R so that Xs = ∅. We then also have that R[1/s] is the zero ring and since a prime ideal must always be a proper ideal, it follows that Spec(R[1/s]) = ∅. The following proposition identifies basic open subsets with spectra of simple localizations and shows that inclusions between them come from obvious ring homomorphisms.

PROPOSITION 2.6. Given s ∈ R, denote by js : R → R[1/s] the natural map. Then Spec(js) : Spec R[1/s] → Spec(R) = X is a homeomorphism onto Xs. We have p 0 Xs ⊂ Xs0 if and only if s ∈ (s ) and the inclusion is then induced by the associated homomorphism R[1/s0] → R[1/s].

PROOF. When R[1/s] is the zero ring there is nothing to prove and so let us assume that this is not the case. Then s is not nilpotent. If p˜ is a prime ideal R[1/r], −1 then p = js p˜ is a prime ideal with s∈ / p and we have p˜ = p[1/s]. On the other hand, for a prime ideal p of R with s∈ / p, p[1/s] is a prime ideal of R[1/r] and −1 the preimage of js p[1/s] = p. In other words, Spec(js) is injective with image Xs. This map is continuous. It is also open: a nonempty basic open subset of Spec R[1/s] is deﬁned by a nonnilpotent s0 ∈ R[1/s]. If s00 ∈ R denotes the product of s and the numerator of s0, then R[1/s][1/s0] = R[1/s00] and so this basic open set is also a basic open subset of Spec(R). 0 We have Xs ⊂ Xs0 if and only if Z(s) ⊃ Z(s ) and by Proposition 2.2 this means that sn = s0r for some r ∈ R and some integer n > 0. Hence R[1/s0] maps 0 to R[1/(s r)] = R[1/s] and this map of course deﬁnes the inclusion Xs ⊂ Xs0 .

Henceforth we will use Spec(js) to identify Xs with Spec R[1/s]. The following theorem is of fundamental conceptual importance. It interprets in a geometric manner the localizations of R and the natural maps between them. 2. THE SPECTRUM OF A RING AS A LOCALLY RINGED SPACE 87

THEOREM-DEFINITION 2.1. Let R be a ring and put X := Spec(R). There is a sheaf of rings OX on X characterized by the property that for a basic open set Xs, OX (Xs) = R[1/s], and the restriction map deﬁned by an inclusion Xs ⊂ Xs0 is the 0 associated homomorphism R[1/s ] → R[1/s]. Its stalk OX,x at x ∈ X is the local ring

Rpx . We call OX the structure sheaf of X.

PROOF. For the existence and uniqueness of OX , it sufﬁces, in view of Exercise 72, to prove the following:

Let be given a collection of nonempty basic open subsets {Xsα }α∈A of X whose union is a basic open subset Xs and let {uα ∈ R[1/sα]}α∈A be a collection with the property that for every pair α, β ∈ A, uα and uβ become equal in R[1/(sαsβ)]. Then the members of this collection come from a unique element u ∈ R[1/s]. The verification of this property is modeled after the proof of Proposition 5.1. Since Xs = Spec R[1/s] is quasicompact, there is a finite subset K ⊂ A such that Nκ Xs = ∪α∈K Xsα . For κ ∈ K, we write uκ = rκ/sκ . We can take Nκ larger if we want to and so we may assume that they are all equal to some N 0. N 0 N 0 N 0 For κ, λ ∈ K, rκ/sκ and rλ/sλ become equal in R[1/(sκsλ)], and so rκsλ − 0 N Nκ,λ rλsκ is annihilated by (sκsλ) for some Nκ,λ ≥ 0. We can take Nκ,λ larger if we want to, and so we may assume that they are all equal to some N 00. Now N 00 0 00 N N N put r˜κ := rκsκ and N := N + N . Then uκ =r ˜κ/s and r˜κsλ =r ˜λsκ for all κ, λ ∈ K. Since Xs = ∪κ∈K U(sκ), we have N X N Z(s) = ∩κ∈K Z(sκ) = ∩κ∈K Z(sκ ) = Z( Rsκ ). κ∈K M P N Then s of s will lie in the ideal κ∈K Rsκ for some M > 0 by Proposition 2.2: M X N s = tκsκ κ∈K P for certain tκ ∈ R. We put r := κ∈K tκr˜κ ∈ R. Then for every λ ∈ K, N X N X N M rsλ = tκr˜κsλ = tκr˜λsκ =r ˜λs κ∈K κ∈K M This proves that u := r/s ∈ R[1/s] maps to uκ in R[1/sκ] for κ ∈ K. This does not prove yet that u has this property for an arbitrary α ∈ A. Still we could in the above argument replace K by K ∪ {α} and thus come up with a u0 ∈ R[1/s] which is such that it not only maps to uκ in R[1/sκ] for κ ∈ K, but also to uα 0 in R[1/sα]. It therefore remains to show u = u (which then also would establish 0 uniqueness). The fact that u − u ∈ R[1/s] maps to the zero element of R[1/sκ] nκ 0 n means that sκ (u − u ) = 0 for some nκ > 0. Since some power s of s lies in the P nκ n 0 0 ideal κ∈K Rs , it follows that s (u − u ) = 0. This means that u = u in R[1/s]. It remains to prove the assertion about the stalks. An element of this stalk at xp is represented by a fraction r/s ∈ R[1/s] with s∈ / p, with the understanding that two such fractions r/s ∈ R[1/s] and r0/s0 ∈ R[1/s0] are considered equal if they coincide in some R[1/s00] with s00 ∈/ p, where s and s0 divide s00. This means that rs0 − r0s is annihilated by some power of s00. This implies that they define the same element of Rp. In other words, the stalk at xp embeds in Rp. On the other hand, any element of Rp thus appears as it is represented by a fraction r/s with s∈ / p (and so nonnilpotent). 88 3. SCHEMES

REMARK 2.7. If in the previous proposition we take s = 1, we ﬁnd that R = Γ(X, OX ) and so no information is lost when we pass from R to the associated ringed space.

EXERCISE 76. Determine the points and the stalks of the spectrum of the local rings K[[t]], K[t](t) (with K a ﬁeld) and Z(p) (the fractions whose denominator is not divisible by p). Proposition 2.1 associated to a ring R a ringed space. We expect a ring homomorphism φ : R0 → R to induce a morphism of ringed spaces. This is indeed the case.

PROPOSITION 2.8. Let R and R0 be rings and put X := Spec(R), X0 := Spec(R0). Every ring homomorphism φ : R0 → R deﬁnes a continuous map f = Spec(φ): 0 ˜ X → X and gives rise to a natural sheaf homomorphism of rings φ : OX0 → f∗OX 0 ˜ 0 0 0 0 0 whose value on X , φ(X ): R = Γ(X , OX0 ) → Γ(X , f∗OX ) = Γ(X, OX ) = R, is ˜ −1 φ. If we think of φ as a sheaf homomorphism f OX0 → OX , then its stalk at x = xp 0 is given by the local sheaf homomorphism Rφ−1p → Rp induced by φ. ˜ 0 ˜ Conversely, any pair (f, φ) with f : X → X a continuous map and φ : OX0 → ˜ f∗OX a sheaf homomorphism φ : OX0 → f∗OX such that the induced map on stalks is local, comes from the ring homomorphism R0 → R deﬁned by φ˜(X0).

PROOF. For the first assertion, we explain how φ˜ is defined on a basic open 0 0 0 0 U(r ) ⊂ X , i.e., we give the homomorphism Γ(U(r ), OX0 ) → Γ(U(r ), f∗OX ). ˜ 0 0 0 This will determine φ as a sheaf homomorpism. We have Γ(U(r ), OX0 ) = R [1/r ]. Put r := φ(r0). We have already seen that f −1U(r0) = U(r). The ring homomorphism R0[1/r0] → R[1/r] induced by φ can be understood as a ring homomorphism 0 0 0 0 Γ(U(r ), OX0 ) = R [1/r ] → R[1/r] = Γ(U(r), OX ) = Γ(U(r ), f∗OX ) (when r is nilpotent, U(r) = ∅ and the right hand side is the zero ring). We thus ˜ have defined a homomorphism φ : OX0 → f∗OX of sheaves of rings. It is clear that this homomorphism is on the stalks as asserted. Suppose now given (f, φ˜) is as in the proposition and let φ : R0 → R be the ring homomorphism that is the value of φ˜ on X0. The sheaf property implies that for every x = xp ∈ X we have a commutative diagram

0 0 φ 0 R = Γ(X , OX0 ) −−−−→ Γ(X , f∗OX ) = R     y y ˜ 0 φp Rp0 = OX0,x0 −−−−→ OX,x = Rp where p0 is the prime ideal of R0 associated to x0 := f(x). The bottom map is local 0 0 and so the preimage of pRp is p Rp0 . The commutativity of the diagram implies that then φ−1p = p0. In other words, Spec(φ) and f take the same value on x and define ˜ the same map on the stalk at this point. Hence φ induces the pair (f, φ). 3. The notion of a scheme Let us say that a locally ringed space is a ringed space whose stalks are local rings and let us agree that a morphism of locally ringed spaces is a morphism of ringed spaces that are locally ringed and which has the additional property that it 3. THE NOTION OF A SCHEME 89 gives local homomorphisms on the stalks. This defines the category of locally ringed spaces. To any ring R we have associated a locally ringed space (which we denote Spec(R) by abuse of notation) and we found that ring homomorphisms R0 → R are in bijective correspondence with the morphisms Spec(R) → Spec(R0) as locally ringed spaces. Thus the spec construction identifies the dual of the category of rings as a full subcategory of the category of locally ringed spaces. We enlarge this category by taking the full subcategory whose objects are locally like the spectrum of a ring and thus arrive at the central notions of algebraic geometry:

DEFINITION 3.1. An affine scheme is a ringed space that is isomorphic to Spec(R) for some ring R.A quasi-affine scheme is a ringed space that is isomorphic to an open subset to Spec(R) for some ring R.A scheme is a ringed space that is locally an affine scheme (hence this is always a locally ringed space). A morphism of schemes is a morphism between such objects in the category of locally ringed spaces. We denote the category of schemes by Sch and its subcategory of affine schemes by Affsch. The following corollary essentially restates Proposition 2.8:

COROLLARY 3.2. The passage to spec defines a contravariant functor which identifies the opposite of the category of rings, Ring◦, with the category Affsch of affine schemes. We often denote a scheme by a single symbol, but often use that same symbol for its underlying topological space. But if the distinction must be made, we denote that space by the absolute value sign: a scheme X is a space |X| with structure sheaf OX .

REMARK 3.3. We have seen that the closed points of an affine scheme Spec(R) bijectively correspond to the maximal ideals of R. In particular, Spec(R) has a closed point. But an arbitrary scheme need not have a closed point (see [8] Exerc. 3.27 for an example). Base schemes. It is often useful to fix a ‘base scheme’ S and to consider the overcategory Sch/S as defined above. Its objects are S-schemes, that is, schemes X endowed with a morphism X → S. According to the definition, an S-morphism X/S → X0/S, is then a morphism X → X0 whose composite with the given morphism X0 → S yields the given morphism X → S. But when S = Spec(R), we usually say that X is an R-scheme (rather than a Spec(R)-scheme) and we write X/R accordingly. This simply means that OX is a sheaf of R-algebras and that the R-morphisms f : X/R → Y/R are those with the property that the induced homomorphisms on stalks are R-algebra homomorphisms. The category of R-schemes is of course denoted Sch/R. Since Z is an initial object for Ring, Spec(Z) is a final object for Affsch. It then also follows that it is a final object for Sch: every scheme X is one over Z: Sch = Sch/Z (to see this, cover X by affine open subsets Ui and observe that the morphisms Ui → Spec(Z) and Uj → Spec(Z) will have the same restriction to Ui ∩ Uj). We also observe that the category of k-prevarieties is a full subcategory of the category Sch/k of k-schemes. The proj construction. We describe a way of producing a scheme out of a graded ring, such that the relation between the algebra and the geometry generalizes the way a projective variety defines a homogeneous coordinate ring. We 90 3. SCHEMES

∞ start out with a graded ring A = ⊕k=0Ak (so AkAl ⊂ Ak+l; in particular A ∞ is a A0-algebra). This has a distinguished ideal, A+ := ⊕k=1Ak. We define a scheme Proj(A) whose underlying topological space | Proj(A)| has as its set of points the homogeneous prime ideals strictly contained in A+ (briefly: strict homogeneous prime ideals). In particular, if A is reduced to A0 (so that An = 0 for n > 0), then Proj(A) = ∅. Given a homogeneous element of positive degree d, s ∈ Ad, then the strict homogeneous prime ideals of A which contain s define a subset Zs ⊂ | Proj(A)|. We denote its complement by Proj(A)s. We have Proj(A)s ∩ Proj(A)s0 = Proj(A)ss0 as usual and so the subsets Proj(A)s are a basis for a topology on | Proj(A)|. We give | Proj(A)| this topology. When s is not nilpo- k tent, then put Fk := Adk/s ⊂ A[1/s]. We have F0 = A0, Fk ⊂ Fk+1, FkFl ⊂ Fk+l and so the monotone union A[1/s]0 := ∪k≥0Fk is a filtered ring (and we would then write Fk = Fk(A[1/s]0)); it is the subring of A[1/s] of degree zero. We claim that the topological space underlying Proj(A)s can be identified with | Spec(A[1/s]0)|. For if p ⊂ A is a strict homogeneous prime ideal which does k not contain s, then you easily check that ∪k≥0pdk/s is a (filtered) prime ideal of 0 A[1/s]0. Conversely, if p is a prime ideal of A[1/s]0, then denote by pk ⊂ Ak the dl 0 set of a ∈ Ak for which aAdl−k/s ⊂ p for l large enough (if this inclusion holds for some l, then it holds also for any larger l). This is a homogeneous ideal which 0 does contain s. It is also a prime ideal: if a ∈ Ak and a ∈ Ak0 are such that 0 dm 0 aa Adm−k−k0 /s ⊂ p for some m, then by enlarging m we may assume that we 0 0 0 dl 0 can write m = l + l such that k ≤ dl and k ≤ dl . If neither aAdl−k/s ⊂ p nor 0 dl0 0 0 dl a Adl−k0 /s ⊂ p , then choose r ∈ Adl−k and r ∈ Adl−k0 such that neither ar/s 0 nor a0r0/sdl is in p0. But since their product is in p0 we get a contradiction. You can check that the two constructions are each others inverse. 0 Given s ∈ Ad0 , then we have an inclusion of basic open sets Proj(A)s ⊂ 0 Proj(A)s0 if and only s divides some positive power of s, and then such an in- 0 clusion is covered by an obvious ring homomorphism A[1/s ]0 → A[1/s]0. So the map Proj(A)s 7→ A[1/s]0 defines a sheaf of rings on | Proj(A)| with the property that its restriction to Proj(A)s equals Spec(A[1/s]0) as a locally ringed space. This defines a scheme structure on | Proj(A)|; we denote that scheme by Proj(A) and usually refer to it as the ‘proj’ of A. The ring homorphism A0 → A[1/s]0 defines a morphism of schemes Proj(A)s → Spec(A0). These morphisms compatible (we get the same morphism if we restrict the morphism Proj(A)s0 → Spec(A0) to Proj(A)s and thus define together define a morphism Proj(A) → Spec(A0). Notice that if p is a strict homogeneous prime ideal of A, then the stalk of OProj(A) at the point x ∈ Proj(A) associated to p is the filtered subring Ap,0 = ∪k≥1Ak/(Ak − pk) of degree zero elements of Ap. The image of x ∈ Proj(A) in Spec(A0) is defined by the prime ideal p0 and so the local homomorphisms on stalks is given by the local homomorphism (A0)p0 → Ap,0.

PROPOSITION-DEFINITION 3.4. We have thus defined an A0-scheme Proj(A). The points of Proj(A) are in bijective correspondence with the strict homogeneous prime ∞ ideals of A. Every homogeneous element s ∈ ∪n≥1An defines a basic open affine subscheme Proj(A)s that we may identify with Spec(A[1/s]0). If p is a strict homogeneous prime ideal of A, then the stalk of OProj(A) at the point x associated to p is Ap,0 = ∪k≥1Ak/(Ak − pk) and the image of x in Spec(A0) is defined by p0 ⊂ A0. An A0-scheme X/A0 is said to be projective (over A0) if it is isomorphic to Proj(A) for some finitely generated A0-algebra. Such a scheme is quasi-compact. 3. THE NOTION OF A SCHEME 91

PROOF. We already established all the assertions, but the last one. If (si ∈ k Adi )i=1 generates A as an A•-algebra, then Proj(A•) is covered by the afﬁne open subsets Proj(A•)si and hence is quasi-compact.

In case R is a ring and we take for A the graded ring R[T0,...,Tn] (with deg(Ti) = 1 for all i), then the associated projective scheme Proj(R[T0,...,Tn]) n is called projective n-space over R and is denoted PR. In case R = k, its set of n closed points reproduces our Pk . We shall later discuss these projective schemes in more detail.

The functor of points. This is related to a general categorical notion. We can view a scheme X as defining a contravariant functor hX from the category of schemes Sch to the category Set of sets. This functor depends covariantly on X: a morphism f : X → Y induces a natural transformation h∗(f): hX ⇒ hY . General category theory (the Yoneda Lemma) tells us that nothing is lost this way. The restriction hX to affine schemes can be thought of as a covariant functor Ring → Set (for Affsch can be regarded as the dual of Ring). We call this the (o) functor of points defined by X. Mumford denotes it by hX , and so shall we, but we will often we abuse notation a bit by simply using X. So it assigns to a ring R the set X(R) of morphisms Spec(R) → X and if φ : R → R0 is a ring homomorphism, then composition with Spec(φ) : Spec(R0) → Spec(R) defines the map X(φ): X(R) → X(R0). A morphism f : X → Y of schemes induces a natural transformation from the functor associated to X to the functor associated to Y and it is now logical to denote this natural transformation also by f. This just expresses the fact that for every ring homomorphism the square

f(R) X(R) −−−−→ Y (R)     X(φ)y Y (φ)y f(R0) X(R0) −−−−→ Y (R0) commutes. In case X is affine, i.e., when X = Spec(A) for some ring A, then the associated functor is essentially hA, the (contravariant) functor which assigns to a ring R the set of ring homomorphisms Hom(R,A). According to the Yoneda lemma this embeds the category of affine schemes Affsch (which is the opposite of the category of rings) in the category of functors Ring → Set. (At first it looks like surprising fact until one realizes that X has a special A-valued point defined by the identity Spec(A) = X.) Again nothing is lost by this passage:

PROPOSITION 3.5. Let X and Y be schemes. Then the map which assigns to a (o) (o) (o) morphism f : X → Y the natural transformation h (f): hX ⇒ hY deﬁnes a (o) (o) 2 bijection between MorSch(X,Y ) and Nat(hX , hY ) .

PROOF. We checked this for preschemes. The result for general schemes follows from a rather straightforward glueing argument (use that both X and Y admit an open covering by afﬁne schemes).

2In other words, the functor of points deﬁnes an fully faithful functor from Sch to the functor category Func(Ring, Set). 92 3. SCHEMES

It is remarkable that the not so straightforward notion of a scheme (which involves the category of locally ringed spaces) amounts to a certain type functor from Ring to Set.

EXAMPLE 3.6. We first take for X the line over , 1 = Spec( [x]) (which we Z AZ Z now prefer to denote A1, thereby abandoning our convention that this stands for 1 1 Ak). Then R 7→ A (R) is just the forgetful functor Ring → Set: an element of A1(R) is by definition given by a ring homomorphism φ : Z[x] → R and this in turn is given by φ(x) ∈ R; conversely, any r ∈ R determines a ring homomorphism n φr : Z[x] → R that takes x to r. Likewise, if A stands for Spec Z[x1, . . . , xn] (to which we might refer as absolute affine n-space), then

n n φ ∈ A (R) = MorRing(Z[x1, . . . , xn],R) 7→ (φ(x1), . . . , φ(xn)) ∈ R n n identifies A (R) with R with the inverse assigning to r = (r1, . . . , rn) the ring homomorphism φr : Z[x1, . . . , xn] → R which sends xi to ri. Taking this one step further: if X is the spectrum of a quotient ring A of Z[x1, . . . , xn], so that X = Spec(Z[x1, . . . , xn]/(f1, . . . , fk)) for certain elements f1, . . . , fk of Z[x1, . . . , xn], then an element of X(A) is given by a ring homomorphism A → R, which amounts to giving a ring homomorphism φ : Z[x1, . . . , xn] → R with the property that fi(φ(x1), . . . , φ(xn)) = 0 for all i. In other words, X(R) can be identified with n the solution set in R of the system of equations fi(r1, . . . , rn) = 0, i = 1, . . . , k. For a ring homomorphism φ : R → R0 the induced map X(φ): X(R) → X(R0) is obtained by applying φ to each coordinate. So this just corresponds to the more intuitive notion idea of an algebraic set, but now repackaged in functorial form. When R = k, then X(k) this is the underlying set of an affine variety, but beware that we have not endowed X(k) with a topology, let alone with a sheaf of regular functions.

If we ﬁx a scheme with associated functor S : Ring → Set, then the category of schemes over S, Sch/S, can be described as schemes for which the associated functor X : Ring → Set is endowed with a natural transformation X → S. The functor of points approach is a natural one in the theory of algebraic groups since it affords the following simple deﬁnition:

DEFINITION 3.7. An algebraic group scheme is a scheme X for which we are given a factorization of the associated functor X : Ring → Set through the category of groups: X : Ring → Grp → Set.

This just means that for every ring R, the set of R-valued points X(R) has been endowed with a group structure in such a manner that for any ring homomorphism φ : R → R0 the associated map X(φ): X(R) → X(R0) is a group homomorphism.

EXERCISE 77. Prove that there is defined a group scheme Ga which assigns to every ring R the abelian group underlying R (so this really an additional structure on the affine line Spec Z[x]). Similarly, prove that there is defined a group scheme Gm (also denoted GL1) which assigns to every ring R the (multiplicative) group R× on invertible elements (so this really an additional structure on the punctured −1 affine line Z[x, x ]). More generally, show that we have a group scheme GLn which assigns to R the group GLn(R) of n × n-matrices with entries in R and determinant in R× (such matrices have an inverse with entries in R). 3. THE NOTION OF A SCHEME 93

Functors Ring → Set arise naturally in what are called moduli problems. It then becomes important to ﬁnd criteria for when such a functor is representable by a scheme.

Special points. Let R be a local ring. Then Spec(R) has just one closed point; if κ := R/mR denotes its residue field, then it is a κ-valued point. Given a scheme X, then an R-valued point of X defines a map Spec(R) → X and so its restriction to the closed point is going to be a κ-valued point of X: we find a p ∈ X and a local homomorphism OX,p → R. We will call this an R-valued point over p. Such a local homomorphism must have the maximal ideal mX,p as kernel so that κ appears as an extension of the residue field κ(p) of p. Conversely, any local homomorphism from a stalk of X to a local ring R defines an R-valued point of X. Of particular interest is the case when we take for R a field, so that R = κ. Then a local homomorphism OX,p → κ is just a field embedding ι : κ(p) ,→ κ. Observe that by our definition, composition of ι with an nontrivial automorphism σ of κ defines another point. When κ is algebraically closed, then the absolute Galois group Aut(κ) permutes the κ-valued points over p. Such points have a name:

DEFINITION 3.8. A point of a scheme in the above sense is called a geometric point if it takes values in an algebraically closed ﬁeld.

So when X is an affine k-variety, then a geometric point is given by an irreducible subset Y ⊂ X and an embedding of its field of rational functions k(Y ) in an algebraically closed field K. Note that K may be larger than an algebraic closure of k(Y ). For instance, if Y˜ → Y is a dominant morphism of irreducible k-varieties, then K could be an algebraic closure of k(Y˜ ).

EXERCISE 78. Let f ∈ Q[x] be nonzero and let X = Spec(k[x]/(f)). Fix an algebraic closure Q¯ of Q and describe X(Q¯ ).

EXERCISE 79. Prove that a noetherian ring is a DVR if and only if it has only two prime ideals, one of which is the zero ideal (so that it is without zero divisors) and the other principal.

EXERCISE 80. Let X be an irreducible variety over the algebraically closed field k of dimension n and let D ⊂ X be a subvariety of dimension n − 1 with the property that D is not contained in the singular part of X. Prove that this defines a Z-valuation vD on k(X) for which the associated DVR is the local ring OX,xD × of the (possibly nonclosed) point xD ∈ Spec(X) defined by D. (For f ∈ k(X) , vD(f) is to be understood as ‘the order of vanishing’ of f along D.) (This applies in particular to the the germ of a nonsingular curve: to be precise, the local ring of a k-variety X that is nonsingular of dimension 1 at the closed point x ∈ X is a DVR.)

EXERCISE 81. Prove that the completion of a DVR with respect to its maximal ideal is still a DVR. Describe this completion for κ[x](x) and Z(p). Since every prescheme determines and is determined by a functor Ring → Set, the question arises whether we can characterize all the functors Ring → Set thus obtained. This turns out to be a highly nontrivial (but important) issue. We may come back to it later. 94 3. SCHEMES

4. Formation of products Given a ring R, then the under category R/Ring has a coproduct: if we are given (commutative) R-algebras A and B (in other words, we are given ring homomorphisms φ : R → A and ψ : R → B), then we can form the R-algebra A⊗R B. It ﬁts in the commutative diagram

A −−−−→ A ⊗R B x x   φ  ψ R −−−−→ B which makes A ⊗R B in fact the coproduct: if a ring C has both the structure of a A-algebra and a B-algebra, but in such a manner that the resulting two R-algebra structures coincide, then C is in fact a A ⊗R B-algebra (this is trivial). Since the category Affsch of affine schemes is dual to the category Ring of rings, it has therefore arbitrary fibered products: if f : X → S and g : Y → S are Q morphisms of affine schemes, then the fibered product X ×S Y (= X S Y ) exists and is simply defined as Spec(A ⊗R B), where A, B resp. R is the ring of global sections of the structure sheaf of X, Y resp. S. The tensor product commutes with −1 −1 localization: first, if we localize in S, by taking h ∈ S, then A[h ]⊗R[h−1] B[h ] = −1 −1 −1 (A ⊗R B)[h ]. This proves that f Sh ×Sh g Sh is a basic open of X ×S Y .A similar argument shows that if X0 resp. Y 0 is an arbitrary basic open of X resp. 0 0 Y , then X ×S Y is a basic open of X ×S Y . This quickly leads to the existence of arbitrary fibered products for the full category of schemes:

THEOREM 4.1. The category of schemes, Sch, admits arbitrary ﬁbered products.

PROOF. Let f : X → S and g : Y → S are morphisms of schemes. First note that it suffices to treat the case when S is affine: S = Spec(R). Then cover X resp. Y by affine open subsets {Ui = Spec(Ai)}i∈I and {Vj = Spec(Bj)}j∈J . We can form Ui ×S Vj = Spec(Ai ⊗R Bj). Then (Ui ∩ Ui0 ) ×S (Vj ∩ Vj0 ) can be identified with open an affine open of both Ui ×S Vj and Ui0 ×S Vj0 . These identifications yield the scheme X ×S Y (see Hartshorne [7], Ch. II Thm. 3.3) for a detailed discussion). Notice that for S = Spec(k) and X and Y k-varieties gives gives us back the product defined in 6.1 of Chapter 1.

Base change and the fibers of a morphism. The formation of a fiber product is sometimes also called a base change. The use of this terminology is a only a psychological device, as is encourages us to think a bit differently about a fibered product, namely given a morphism S˜ → S, we get a functor Sch/S → Sch/S˜ simply by associating to every S-scheme X/S the S˜-scheme S˜ ×S X/S˜. If we take S˜ to be a point s of S, then we get the notion of the fiber of a morphism f : X → S: the fiber f −1(s) (also written as X ⊗ κ(s) or X(s) when f is understood) is Spec(κ(s)) ×S X. This naturally a scheme over the field κ(s). This observation leads one to think of X/S as a family schemes parametrized by S in an algebraic manner (which may be sometimes helpful). Note that in the generality we consider here this makes sense for any A-valued point of S (where A is a ring): the fiber is then the A-scheme Spec(A)×S X. If we take this to be a geometric point 5. ELEMENTARY PROPERTIES OF SCHEMES 95

(meaning that A is an algebraically closed ﬁeld), then we call this a geometric ﬁber of X/S.

Extension of scalars. If S is affine: S = Spec(R), so that A is a R-algebra, then the formation of Spec(A) ×Spec(R) X is sometimes called extension of scalars. (In the literature you will sometimes see this abbreviated as A ×R X or A ⊗R X.) If R is a domain, then (0) ⊂ R is a prime ideal and the corresponding point of Spec(R) (see also Section5 below), called the generic point of S, has the field of fractions, Frac(R), as its residue field. The fiber over this point is called the generic fiber of X/S.A general point of Spec(R) is by definition given by an embedding of R in algebraically closed field K. The corresponding fiber is then called a general fiber of X/S. Here is another important case. Let K ⊃ Ko be a field extension. The induced map Spec(K) → Spec(Ko) is a trivial one (the underlying spaces are singletons). But now treat this as a base change. Suppose Xo/Ko is an Ko-scheme. Then

X := K ×Ko Xo will be a scheme over K. For instance, we could take Xo = Spec(Ko[x]/(f)). If we assume f ∈ Ko[x] irreducible, then Ko[x]/(f) is a field and |Xo| is a singleton. But X = Spec(K[x]/(f)), where f is now viewed as an element of K[x] and its points correspond to the distinct irreducible factors of f in K[x]. We will later take closer look at this situation. Since every scheme X is one over Z, Remark 2.4 extends to schemes: a closed point of Spec(Z) is given by a prime p and has residue field Fp and its unique nonclosed point is defined by the prime ideal (0) and has residue field Q, and so the fibers X ⊗ Fp and X ⊗ Q over these point are schemes over Fp resp. Q (which may be empty, of course).

REMARK 4.2. Suppose X/S and Y/S are schemes over a scheme S and let R be a ring. Then an element of (X ×S Y )(R) is by definition given by a morphism Spec(R) → X ×S Y . The latter is by definition given by morphisms Spec(R) → X and Spec(R) → Y such that the composites Spec(R) → X → S and Spec(R) → Y → S coincide. This amounts to giving a point of X(R) ×S(R) Y (R). We thus see that the set (X ×S Y )(R) can be identified with the set X(R)×S(R) Y (R). The same argument works of course if we replace Spec(R) by an arbitrary scheme.

EXERCISE 82. Prove that if X/S and Y/S are schemes over a scheme S, then there is a natural map |X ×S Y | → |X| ×|S| |Y |. Show that this map is surjective, but need not be injective. (Hint and permission: you may use Exercise 84 of the 1 next section.) Work this out for the following two simple cases: for X = Y = Ak, S = Spec(k) and for the case where L/K is genuine ﬁnite separable ﬁeld extension and X = Y = Spec(L), S = Spec(K).

5. Elementary properties of schemes Integral schemes. We say that a scheme is connected resp. irreducible if the underlying topological space has that property. In particular, an irreducible scheme is connected. We say that a scheme X is reduced if its structure sheaf takes values in the category of reduced rings, that is, for every open U ⊂ X, OX (U) is reduced. If a ring R is reduced, then so are it localizations. Hence the spectrum of a reduced ring is reduced. 96 3. SCHEMES

PROPOSITION-DEFINITION 5.1. A scheme X is irreducible and reduced if and only if for every nonempty open U ⊂ X, OX (U) is an integral domain. We then say that X is an integral scheme.

PROOF. An integral domain is reduced. If X is reducible, then it contains nonempty pairwise disjoint open subsets U0,U1. Let U := U0 ∪ U1. If we choose fi ∈ OX (U) so that it takes the value 1 on Ui and zero on U1−i, then f0f1 = 0 and so OX (U) is not a domain. If on the other hand a nonempty open subset U is irreducible and OX (U) is reduced, then OX (U) must be a domain: if f0, f1 ∈ O(U) are such that f0f1 = 0, then U = Z(f0) ∪ Z(f1), and so U = Z(fi) for some i. Since OX (U) is reduced, this implies that fi = 0.

REMARK 5.2. An integral scheme X contains a special point whose local ring is a field. If U ⊂ X is affine open and nonempty, then U is dense in X and OX (U) is an integral domain so that (0) is a prime ideal. This prime ideal defines a point ηU ∈ U whose residue field is Frac(OX (U)). That point is independent of the choice of U: if U 0 is nonempty affine open subset of U, then O(U) → O(U 0) is a localization map and since O(U) is an integral domain, it is injective and induces 0 an isomorphism of fraction fields. In particular, the inclusion U ⊂ U maps ηU 0 to ηU . We call this the generic point of X and denote it by ηX . This argument also shows that for U as above, Frac(OX (U)) → OX,ηX is an isomorphism of fields. We call this field the function field of X and sometimes denote it K(X). It is the local ring of the generic point of X.

Subschemes and immersions. If (X, OX ) is a scheme, then for every open subset U ⊂ X, (U, OX |U) has of course the structure of a scheme. We say that a morphism f : Y → X of schemes is an open immersion if the image of f is an open subset U of X and f factors through an isomorphism of Y onto U. In other words, |f| : |Y | → |X| is an open embedding and for every y ∈ Y , fy : OX,f(x) → OY,y is an isomorphism. For a closed subset of |X| the situation is a bit different in the sense that a closed subset may have many distinct structures as a scheme. Before we give the relevant deﬁnition, we recall that a subspace Y of a space X is called locally closed if for every p ∈ Y there exists an open neighborhood Up of p in X such that Y ∩ Up is a closed subset of Up. This is equivalent to: there exists an open subset UY of X which contains Y and is such that Y is closed in UY (just take U = ∪p∈Y Up).

DEFINITION 5.3. A morphism of schemes f : Y → X is called an immersion if it maps |Y | homeomorphically onto a locally closed subset of |X| and the sheaf −1 homomorphism f OX → OY is surjective (which we recall means that is surjective on the stalks: for every y ∈ Y , fy : OX,f(y) → OY,y is onto). If |f| is an inclusion (so that |Y | is a locally closed subspace of |X|), then we say that Y is subscheme of X (we then may write Y ⊂ X). When |Y | is a closed subset of |X|, we say that f is a closed submersion (resp. that Y is a closed subscheme of X). For an afﬁne scheme this notion has a simple interpretation.

LEMMA 5.4. Let X is be an affine scheme: X = Spec(A). Then every ideal I ⊂ A defines a closed subscheme and every surjection of rings A → B defines a closed immersion. 5. ELEMENTARY PROPERTIES OF SCHEMES 97

PROOF. Clearly an ideal I ⊂ R defines a closed subset Z(I) of |X|. This closed subset underlies the scheme Y := Spec(R/I). More precisely, the ring homomorphism R → R/I induces a scheme morphism Y → X whose underlying map of topological spaces is the inclusion of Z(I) ⊂ |X|. Moreover, a point y ∈ |Y | ⊂ |X| is defined by a prime ideal p of R which contains I and the homomorphism on stalks ∼ OX,y → OY,y is just the natural ring homomorphism Rp → Rp/Ip = (R/I)p¯, where p¯ ⊂ R/I denotes the image of p. This ring homomorphism is clearly surjective. The second assertion is left to you. We will see (Corollary 7.5) that the converse also holds: any closed subscheme of an affine scheme Spec(R) is defined by some ideal of R.

EXAMPLE 5.5. Let k be an algebraically closed ﬁeld (as always). For t ∈ k, the ideal It ⊂ k[z] generated by z(z + t) (with t ∈ k) deﬁnes a closed subscheme of 1 Ak. It consists of two copies of Spec(k) (two points) when t 6= 0, but for t = 0 they coalesce: we get a single point with structure sheaf (stalk) given by k[z]/(z2).

PROPOSITION 5.6. Let f : X → Y be a morphism. Then the graph Γf : X → X × Y (which is associated to the identity 1X : X → X and f) is an immersion. In particular, f factors as as the composite of an immersion and a projection: Γ f : X −−→f X × Y → Y.

PROOF. Given x ∈ X, then choose first an affine open neighborhood V ⊂ Y of f(x) and then an affine open neighborhood U of x in f −1V so that U × V is an open affine neighborhood of Γf (x). Then Γf |U maps to U × V and the resulting morphism U → U × V is just the graph of the morphism fU,V : U → V that is the restriction of f. It now suffices to see that fU,V is a closed immersion. Write U = Spec(A) and V = Spec(B) so that fU,V is given by a ring homomophism φ : B → A.

Then ΓfU,V is associated to the ring homomorphism a ⊗ b ∈ A ⊗ B 7→ aφ(b) ∈ A, which is clearly surjective (take b = 1). So fU,V is a closed immersion. p EXERCISE 83. Let X be afﬁne: X = Spec(R). Show that Xred := Spec(R/ (0)) is a reduced afﬁne scheme that is a closed subscheme of X with |Xred| = |X|. Prove that this ‘reduction process’ generalizes to an arbitrary scheme: every scheme Z contains a closed subscheme Zred that is reduced and has the same underlying point set.

EXERCISE 84. Let X be a scheme. Every closed integral subscheme Y ⊂ X has a generic point ηY . Identify ηY with its image in X so that we have defined a map η from the collection of closed integral subschemes of X to the points of X. Prove 0 that this is bijection and that under this bijection we have Y ⊂ Y ⇔ ηY ∈ ηY 0 . Finiteness properties. We proved in Lemma 4.7 that the maximal ideal spectrum of a ring is quasi-compact. The same proof shows that this is true for the prime ideal spectrum. So every open covering of an affine scheme admits a finite subcovering.

PROPOSITION-DEFINITION 5.7. For a scheme X the following are equivalent: (i) X admits an open affine covering by the spectra of noetherian rings, (ii) every open affine subset of X is the spectrum of a noetherian ring. When these equivalent conditions are fulfilled, we say that X is locally noetherian. 98 3. SCHEMES

PROOF. The only nontrivial assertion is that (ii) implies (i). So assume (i) and let U ⊂ X be open affine: U = Spec(R). Let x ∈ U. We claim that there exists a s ∈ O(U), such that x ∈ Xs = Spec(R[1/s]) with R[1/s] is noetherian. Since the localization of a noetherian ring is noetherian, a basic open subset of the spectrum of a noetherian ring is one, too. Now by assumption x ∈ U is contained in an open subset U 0 of X with U 0 the spectrum some noetherian ring. By passing to a basic open neighborhood of x in U 0 contained in U, we may assume that U 0 ⊂ U. 0 0 Another such passage also ensures that U is basic in U: U = Xs = Spec(R[1/s]) for some s ∈ O(U) and so R[1/s] is noetherian. It follows that we can cover U by basic open subsets that are spectra of noetherian rings. Since U is affine, it is quasi-compact and so a finite number of these k will do: Spec(R) = ∪α=1 Spec(R[1/sα]) for certain sα ∈ R. Write φα : R → R[1/sα] for the natural map.

Claim: Given an ideal I ⊂ R, denote by Iα ⊂ R[1/sα] the ideal generated by −1 φα(I). Then I = ∩αφα Iα. −1 Proof. We check the nontrivial inclusion ⊃. If r ∈ ∩αφα Iα, then φα(r) = mα rα/sα for some rα ∈ I and mα ≥ 0. This means that there exists a nα ≥ 0 such nα mα mα+nα mα+nα that sα (rsα −rα) = 0. In particular, sα r ∈ I. Since the U(sα) = U(sα ) P mα+nα cover U, there exist tα ∈ R such that α tαsα = 1. It follows that

P mα+nα P r = αtαsα r = αtαsα ∈ I.

We can now complete the proof. If I• = (I1 ⊂ I2 ⊂ · · · ) is an monotone sequence of ideals in R, then each sequence (I•)α becomes stationary (for R[1/sα] −1 is noetherian). Hence the same is true for the sequence ∩αφα (I•)α = I•.

DEFINITION 5.8. We say that a scheme is noetherian if it is locally noetherian and quasi-compact (in other words, if it has an open covering by a ﬁnite number spectra of noetherian rings).

DEFINITION 5.9. We say that a morphism of schemes, f : X → S is locally of finite type if each point of S has an open affine neighborhood V = Spec(R) with the property that f −1V is covered by affine open subsets that are spectra of finitely generated R-algebras, quasi-compact if each point of S has an open affine neighborhood V with the property that f −1V is quasi-compact, of finite type if f is locally of finite type and quasi-compact (so that each point of S has an open affine neighborhood V = Spec(R) with the property that f −1V is covered by finitely many affine open subsets that are spectra of finitely generated R-algebras), affine if each point of S has an open affine neighborhood V such that f −1V is affine, finite if each point of S has an open affine neighborhood V = Spec(R) such that f −1V is affine and the spectrum of a finitely generated R-algebra. The notions ‘of finite type’ and ‘finite morphism’ clearly generalize their name- sakes for k-varieties. As the above definition illustrates, any property P for schemes admits a simple generalization to S-schemes: a morphism f : X → S is said to satisfy P if we can cover S by affine open subsets V such that P holds for f −1V (so that the original property is recovered by taking S = Spec(Z)). 6. SEPARATED AND PROPER MORPHISMS 99

EXAMPLE 5.10. If we take for X a reduced k-scheme X/k, then to say that X/k is of finite type means that X is covered by a finite number of open subsets that are the prime ideal spectra associated to affine k-varieties. This essentially means that X is a k-prevariety.

6. Separated and proper morphisms Separation property. Let X/S be an S-scheme. Its self-product in the category of S-schemes is the ﬁbered product X ×S X. The two identity maps from X to the factors determine a morphism ∆ : X → X ×S X, called for good reasons the diagonal morphism, for the composite with any of the two projections X ×S X ⇒ X is the identity. Were we working in the topological category (so that X → X ×S X would be a subspace of the product space X × X), then the requirement that ∆ is a closed is equivalent to each ﬁber of X/S being a Hausdorff space. Although the topological space underlying a scheme is almost never a Hausdorff space, this characterization renders in this setting almost the same service:

DEFINITION 6.1. We say that X/S is separated if the diagonal morphism is a closed immersion. We say that X is separated if it so as a scheme of Spec(Z) (i.e., when ∆ : X → X × X is a closed immersion). This property can in fact be tested on the underlying topological spaces. We ﬁrst observe

LEMMA 6.2. Suppose X/S is affine, i.e., suppose that every point of S has an open affine neighborhood whose preimage in X is affine. Then X/S is separated.

PROOF. It suffices to prove this locally over S and so we may as well assume that S is affine: S = Spec(R) so that also X is affine: X = Spec(A), where A is a R-algebra. Then ∆ comes from the ring homomorphism A ⊗R A → A, a1 ⊗R a2 7→ a1a2. This is a surjection and so ∆ is a closed immersion. COROLLARY 6.3. The S-scheme X/S is separated if and only if the image of the diagonal |∆| : |X| → |X ×S X| is closed.

PROOF. Suppose the map |∆| : |X| → |X ×S X| has a closed image. Then it is a homeomorphism onto its image, for a continuous inverse is given by restriction of a projection |X ×S X| → |X| to the image of |∆|. It remains to see that OX×S X → ∆∗OX is onto. If V is an affine open subset of S and U is one of X such that f(U) ⊂ V , then the restriction U → V defined by f determines the affine scheme U ×V U. We have a natural morphism U ×V U → X ×S X. This morphism is an open immersion and so we may identify U ×V U with an open affine subset of

X ×S X. According to Lemma 6.2, the morphism OU×V U → ∆∗OU is surjective. Since such open subsets cover the diagonal of X ×S X, the corollary follows. REMARKS 6.4 (Elementary properties of the separation condition). It is clear from the definition that the property of f : X → S being separated is local on S: if we are given an open covering {Vα}α of S, then f is separated if and only if every −1 f Vα → Vα is separated. It is also clear that if f : X → S is separated, then so is the restriction of f to any open subset of X. Since an open immersion is affine, it is separated by Lemma 6.2. The same is true for a closed immersion X,→ Y : just check that X ×Y X can be identified 100 3. SCHEMES with X and that the diagonal morphism is then the identity morphism. The latter is certainly a closed immersion. The following properties amount to an exercise in topology; for more details, you may consult [6] Prop. 1.3.26 (or alternatively, invoke the valuative criterion Theorem 6.17 discussed below). Closed under composition: if f : X → Y and g : Y → Z are separated, then so is gf : X → Z. Fullness: if X/S and Y/S are separated, then so is every S-morphism from X to Y . Closed under taking products: if we are given a base scheme S and two separated S-morphisms X → Y and X0 → Y 0, then so is the associated 0 0 product S-morphism X ×S X → Y ×S Y . Closed under base change: if X/S is separated and S˜ → S is a morphism, then S˜ ×S X is separated over S˜. So the first three properties say that if we fix a base scheme S, then the schemes that are separated over S make up a full subcategory of Sch/S that is closed under S-products. The last property implies that the fibers of separated morphism are separated.

PROPOSITION 6.5. If X is a separated scheme, and U, U 0 are afﬁne open subsets, 0 0 0 0 0 then U∩U is afﬁne and OX (U∩U ) is generated by OX (U)|U∩U and OX (U )|U∩U .

PROOF. First observe that U × U 0 is an afﬁne open subset of X × X and that its preimage under ∆ : X → X × X is U ∩ U 0. Since ∆ is a closed embedding, U ∩ U 0 0 0 0 must be afﬁne with OX (U ∩ U ) a quotient of OX×X (U × U ) = OX (U) ⊗ OX (U ). 0 0 0 0 This means that OX (U ∩U ) is generated by OX (U)|U ∩U and OX (U )|U ∩U .

REMARK 6.6 (The graph of a morphism). Let S be a scheme and let f : X/S → Y/S be a morphism of S-schemes. The identity map of X and f combine to deﬁne an S-morphism Γf : X → X ×S Y , the graph of f. The two S-morphisms X ×S Y → X → Y and the identity of Y combine to deﬁne an S-morphism f ×S 1Y : X×S Y → Y ×S Y . We thus obtain a commutative diagram

Γf X −−−−→ X ×S Y     fy yf×S 1Y

∆Y Y −−−−→ Y ×S Y.

This is in fact a ﬁber product: if we are given S-morphisms Z → X×S Y and Z → Y such the composites Z → X ×S Y → Y ×S Y and Z → Y → Y ×S Y coincide, then there is unique morphism Z → X which makes everything commute, namely the composite Z → X ×S Y → X. In particular, if Y/S is separated, so that ∆Y is a closed immersion, then Γf is a closed immersion (for this is a property preserved under base change, as you can easily check) so that we may identify X with a closed subscheme of X ×S Y .

REMARK 6.7. Following up on Example 5.10, we can now identify the category of k-varieties with the full subcategory of Sch/k of k-schemes that are of ﬁnite type and are separated over k. 6. SEPARATED AND PROPER MORPHISMS 101

Valuation rings. We now want to discuss the valuative criterion of separated- ness. This necessitates a short excursion into commutative algebra, for a central role is played here by the notion of a valuation ring. This is a generalization of that of a discrete valuation ring. For more on this, see [1], Ch. 5.

DEFINITION 6.8.A valuation ring is an integral domain R with the property that every element of Frac(R) − R has its inverse in R. We say that a valuation ring is discrete (abbreviated as a DVR) if it is noetherian, but is not a ﬁeld. We will write K for Frac(R). So such a ring R contains a large part (‘more than half’) of K.

LEMMA 6.9. A a valuation ring R is a local ring. Moreover Almost PID: Any ﬁnitely generated ideal in R is principal. Ideal total ordering: The ideals of R are totally ordered by inclusion: if I and J are ideals of R, then I ⊂ J or J ⊂ I. Normality: R is integrally closed in its ﬁeld of fractions

PROOF. Let a, b ∈ R. Then one of ab−1 or a−1b is in R. In other words, we have a ∈ (b) or b ∈ (a). It follows (with induction) that any ﬁnitely generated ideal in R is generated by one element. Let I and J be ideals of R such that I 6⊂ J. Choose a ∈ I such that a∈ / J. Then for every b ∈ J, we cannot have a ∈ (b) and so b ∈ (a). In other words, J ⊂ (a) ⊂ I. In particular, R has only one maximal ideal, so is a local ring. n n−1 Let x ∈ K be integral over R. Then x + a1x + ··· + an = 0 for certain −1 n and ai ∈ R. If x ∈ R we are done. Otherwise x ∈ R, and so we have −1 −2 −(n−1) −1 x = −(a1x + a2x + ··· + anx ) ∈ R[x ] = R. Our deﬁnition of a DVR is equivalent to a more familiar one:

LEMMA 6.10. A DVR has the property that every nonzero ideal is a power of its maximal ideal. Conversely, any integral domain that is not a ﬁeld and of which every nonzero ideal is a power of a the maximal ideal is a DVR.

PROOF. Let R be a DVR. The assumption that R is not a field implies that its maximal ideal m is nonzero. Since R in noetherian, m is finitely generated and hence admits a generator π ∈ m. If r ∈ R is arbitrary, then r ∈ (π) = m or π ∈ (r) (and hence m ⊂ (r)). Since m is maximal, the last possibility can only occur when we have (r) = m or (r) = R. We conclude that either r is a unit or r is divisible by π: r = r1π for some r1 ∈ R. In the last case we continue with r1 and find that either r1 is a unit or that r1 = r2π for some r2 ∈ R. This produces a strictly ascending sequence (r) ( (r1) ( (r2) ( ··· and so this process will terminate, because R is noetherian. It follows that r = uπn with u ∈ R× and n ≥ 0. In particular, every ideal of R is of the form mn with n a nonnegative integer. The proof of the second assertion is left to you. Let R be a DVR and let π be a generator of its maximal ideal (this is often called a uniformizer). Then its field of fractions K := Frac(R) is simply obtained by inverting π so that every a ∈ K× = K − {0} is uniquely written as uπn with u ∈ R× and n ∈ Z. We denote this exponent by v(a). We now have obtained a Z- valued valuation, that is, a surjective homomorphism v : K× → Z with the property that the nonarchimedean inequality holds: v(a + b) ≥ min{v(a), v(b)} (assuming a + b 6= 0). Notice that v has kernel R× and so defines an isomorphism of groups 102 3. SCHEMES

K×/R× =∼ Z. A field endowed with such a valuation is called a local field. Given a local field (K, v), then the union of {0} and the subset of K where this valuation is ≥ 0 is then automatically a DVR whose field of fractions is K.

REMARK 6.11. For a valuation ring R we can also define a valuation, but one that takes values in a totally ordered abelian group, the group in question being Γ := K×/R× (for which we shall write the group law additively) and the total order being defined by aR× ≤ bR× ⇔ (a) ⊃ (b). This makes Γ a totally ordered group in the sense that γ ≤ γ0 implies γ + γ00 ≤ γ0 + γ00 for all γ00 ∈ Γ. The resulting group homomorphism v : K× → Γ satisfies the nonarchimedean inequality, for if a, b, a + b ∈ K× and for instance v(a) ≤ v(b) so that b ∈ (a), then a + b ∈ (a) and so v(a + b) ≥ v(a) = min{v(a), v(b)}.

EXAMPLES 6.12 (DVR’s). A standard example of a DVR is the local ring of a smooth k-variety C of dimension one (a curve) at a closed point x: the associated field is Frac(OC,x) and if f ∈ Frac(OC,x), then the valuation v(f) is the order of vanishing of a rational function at x (this is negative when it has a pole in x). Another is the C-algebra of convergent power series C{z}. But the residue field and the field of fractions may have different characteristic, witness the example Z(p), with p a prime, where they are Fp and Q respectively. More generally, suppose that R is a noetherian domain and p ⊂ R is a nonzero prime ideal that is minimal for this property. Then Rp has no other prime ideals than (0) and pRp and so by Exercise 79 to say that this is a DVR is to say that pRp has a single generator. Exercise 80 shows why DVR’s are also important in classical algebraic geometry.

EXAMPLE 6.13 (A valuation ring that is not a DVR). The ring of Laurent series × Ro := k[[t]] is a DVR with fraction field Ko := k((t)). The valuation νo : Ko → Z assigns to any nonzero element its order: if a = P a tj with a 6= 0, then n≥no j no ν(a) = no so that Ro is the union of {0} and the part where νo ≥ 0. The maximal ideal mo = (t) is the union of {0} and the part where νo ≥ 0. × Now suppose k is of characteristic zero. Then the multiplicative group Ro is × × divisible: for every positive integer n and every u ∈ Ro there exist a unit un ∈ Ro n such that (un) = u (first extract an nth root from the constant term, which is possible since k is algebraically closed. This allows you to assume that the constant 2 1/n term of u is 1 so that the Taylor series of (1 + a1t + a2t + ··· ) will produce 1/n the desired root). So if we adjoin to Ro for every n > 1 an element t satisfying (t1/n)n = t, then we obtain a ring R such that every element has an nth root. (We 0 must do this in way that if n = mm0 then (t1/n)m = t1/m .) Then R is a valuation ring having K := R[t−1] as its field of fractions. In fact, any element of K can be written as a Laurent series in some t1/n. We thus see that × νo can be extended to a map ν : K → Q and is such that R is the union of {0} and the part where ν ≥ 0. This implies that R is a valuation ring. Note also that the maximal ideal m of R is the union of {0} and the part where ν > 0. But any f ∈ m has a square root in K, which is then of course also in m. In other words, m = m2. Were R noetherian, then it would follow that m = 0, which is clearly not the case. So R is not noetherian and hence not a DVR. One can show that K is an algebraic closure of Ko and that R is the integral closure of Ro in K.

2 EXAMPLE 6.14 (Not a valuation ring). Let C ⊂ Ak be the cuspidal curve defined 3 2 3 2 by x = y and let o := (0, 0) ∈ C. The the local ring OC,o = k[x, y]/(x − y ) (x,y) 6. SEPARATED AND PROPER MORPHISMS 103 is not integrally closed in its field of fractions and hence is not a valuation ring. To see this, observe that t := y/x satisfies the equation t2 − x = 0, hence is integral over OC,o. Observe that the spectrum of a DVR R has just two points: the closed point o defined by the ideal m and with local ring R and residue field κ = R/m and the generic point η defined by the zero ideal (0) with local ring equal its residue field K = Frac(R). So if X is a scheme, then any x ∈ X(R) determines two points with values in a field: one in X(κ) and one in X(K). If we think of x as a morphism Spec(R) → X, then these are x(o) and x(η) respectively. Let K be a field. We consider the collection of subrings of K that are local. On this collection we define a partial ordering as follows: given local subrings A, B of K, we say that B dominates A (and write A ≤ B) if A ⊂ B and the inclusion is 1/n local: mB ∩A = mA. For instance in Example 6.13, for the local rings Rn := Ro[t ] in K we have Rn ≤ Rm if and only of n divides m. PROPOSITION 6.15. A local subring of K which is maximal relative to the partial order ≤ is a valuation ring having K as its field of fractions. In particular, (by Zorn’s lemma) any local subring of K is contained in a valuation ring in K having K as its field of fractions. PROOF. Let R ⊂ K be a local subring that is maximal for ≤ and let x ∈ K − R. We must show that x−1 ∈ R. We first observe that R must be integrally closed (otherwise its integral closure yields a local ring which strictly dominates R). We next claim that mR[x] = R[x]. For otherwise mR[x] is contained in a maximal ideal m˜ of R[x]. Then the localization R˜ := R[x]m˜ is a local ring contained in K which contains R. The homomorphism R → R˜ local: since the maximal ideal m˜ R˜ does not contain 1, its intersection with R is a proper ideal in R and since this ideal contains m, it must be equal to m. But then the maximality of R implies that R˜ = R. This contradicts our assumption that x∈ / R. Since mR[x] = R[x], there exist n ≥ 0 and r0, . . . , rn ∈ m such that n 1 = r0 + r1x + ··· rnx .

We can arrange that r0 = 0, simply by subtracting from both members r0 and then dividing by the unit 1 − r0. Assuming this to be the case, then we get an equation −1 −n −(n−1) of integral dependence for x : x = r1x + ··· + rn. Since R is integrally −1 closed, it follows that x ∈ R. In algebraic geometry valuation rings often come up via: COROLLARY 6.16. Let X be a scheme and let p, q ∈ X be such that p lies in the closure of {q}. Then there exists a valuation ring R and a morphism Spec(R) → X which sends the closed point of Spec(R) to p and the generic point of Spec(R) to q. PROOF. Let Q ⊂ X be the closed integral subscheme that has q as its generic point. So κ(q) is the function field of Q. Since p ∈ Q, OQ,p is a local ring inside κ(q). According to Proposition 6.15, this ring is dominated by a valuation ring R ⊂ κ(q) having κ(q) as its field of fractions. The resulting morphism Spec(R) → Q ⊂ X is then as desired: it sends the generic point to q (and defines the identity on their residue fields) and the closed point defined to p (this is the local homomorphism OX,p → OQ,p ⊂ R on the stalks). In the situation of the corollary above one says that p is a specialization of q. 104 3. SCHEMES

Valuative approach to the separation property. If R is a valuation ring with ﬁeld of fractions K, then for every scheme X we have a natural map X(R) → X(K). This need not be injective or surjective. If we are given some x ∈ X(K), then we call any x˜ ∈ X(R) mapping to x a lift of x. The main result of this subsection is:

THEOREM 6.17 (Valuative characterization of being separated). Let f : X → S be a morphism of schemes. If f is separated if and only if ∆ is quasi-compact and for every valuation ring R with ﬁeld of fractions K, the natural map X(R) → S(R)×S(K) X(K) is injective. In other words, if we are given a K-valued point x ∈ X(K) whose image f(x) ∈ Y (K) has been lifted to an R-valued point y˜ ∈ Y (R), then there is at most one lift x˜ ∈ X(R) of x which maps to y. Exercise 85 shows that the property of ∆ being quasi-compact (in which case one also says that f is quasi-separated) should be regarded as a ﬁniteness condition and Exercise 86 should help to make the valuative criterion more useful.

EXERCISE 85. Let f : X → S be a morphism of schemes. Prove that if ∆ : X → 0 X ×S X → X is quasi-compact, then for any two open afﬁne subschemes U, U of X, the restriction f|U ∩ U 0 : U ∩ U 0 → S is quasi-compact.

EXERCISE 86. Let f : X → S be a morphism of schemes. (a) Prove that if f is of ﬁnite type and S is locally noetherian, then S can be covered by open subsets V such that f −1V is noetherian. (b) Prove that if S can be covered by open subsets V with f −1V is noetherian, then ∆ : X → X ×S X is quasi-compact. (We may thus conclude that if f is of ﬁnite type and S is locally noetherian, then f is quasi-separated.) We shall need:

LEMMA 6.18. Let f : X → Y be a quasi-compact morphism of schemes. Then every po ∈ Y in the closure of f(X) lies in the closure of f(q) for some q ∈ X. In fact, there exist a valuation ring R with ﬁeld of fractions K, a morphism φ : Spec(K) → X and an extension Spec(R) → Y of fφ which maps the closed point to po (so po then lies in the closure of the fφ-image of generic point of Spec(R)).

PROOF. Every point of Y lies in an affine open subset and so without loss of generality we may assume that Y is affine. Then X is quasi-compact and hence a union of a finitely many open affines U1,...,Un. We denote by Yi the closure of f(Ui) in Y , endowed with its reduced structure. If we put Xi := (Ui)red, then f maps Xi to Yi so that f restricts to a dominant morphism fi : Xi → Yi between reduced affine schemes. The closure of f(X) in Y is the union of the closures of the (finitely many) fi(Xi) (for the closure of a finite union of subsets of a subspace is the union of their closures). So without loss of generality we assume that both X and Y are affine and reduced: X = Spec(A) and Y = Spec(B) with A and B reduced, and that f : X → Y is dominant: the associated ring homomorphism φ : B → A is injective (otherwise ker(φ) would define a proper closed subscheme of Y to which f maps). Let now po ∈ Y be defined by a prime ideal po ⊂ B. By Zorn’s lemma here exists a minimal prime ideal p in B contained in po. This defines a point p ∈ Y 6. SEPARATED AND PROPER MORPHISMS 105 which contains po in its closure. So this closure is a closed integral subscheme P of Y that contains po and is maximal for this property. Notice that the residue field −1 κ(p) of p is equal to the localization Bp. The scheme theoretic fiber f p is equal to Spec(κ(p) ⊗B A) by definition. It is nonempty: since localization is exact, the ring homomorphism φp : κ(p) → κ(p) ⊗B A is injective. In particular, κ(p) ⊗B A is not the zero ring. Choose q ∈ f −1p and put K := κ(q). This contains κ(p) as a subfield and κ(p) contains OP,po as a local ring. According to Corollary 6.16 there exists a valuation ring R of K which dominates OP,po . Then this defines a morphism φ : Spec(R) → X which sends the generic point to q and for which fφ : Spec(R) → X sends the closed point to po (for mP,po ⊂ mR).

PROOFOF THEOREM 6.17. Suppose first that f is separated. Then ∆ is a closed immersion and hence quasi-compact. Let now R be a valuation ring with field of fractions K and let φ1, φ2 ∈ X(R) have the same image S(R) ×S(K) X(K). That is, we have morphisms φ1, φ2 : Spec(R) → X which become equal after composing with f : X → S or after restriction to Spec(K). This means that we have a morphism (φ1, φ2) : Spec(R) → X ×S X which maps the generic point to the diagonal. Since the diagonal is closed, (φ1, φ2) maps to the diagonal also and hence φ1 = φ2. Suppose now that f possesses the valuative property and is such that ∆ : X → X ×S X is quasi-compact. We must show that ∆ : X → X ×S X has closed image. Since ∆ is quasi-compact, it suffices by (the first part of) Lemma 6.18 to show that when p ∈ ∆(X) and po ∈ X ×S X lies in the closure of p, then po ∈ ∆(X). According to Corollary 6.16 there exists a valuation ring R and a morphism φ : Spec(R) → X ×S X which sends the generic point η to p and the closed point o to po. Since φ(η) lands in the image of ∆, we must have π1φ(η) = π2φ(η) (where πi : X ×S X → X denotes projection on the ith factor). If we denote this common image by q, then ∆(q) = p. Since fπ1 = fπ2, π1φ and π2φ define two elements of X(R) with the same image in S(R) ×S(K) X(K). Our assumption tells us that they ˜ must then coincide: π1φ = π2φ. If we denote this last morphism by φ : Spec(R) → X, then we must have ∆φ˜ = φ (by the universal property of the fiber product) and ˜ so po = φ(o) = ∆(φ(o)) ∈ ∆(X).

Proper morphisms. A continuous map f : X → S between topological spaces is said to be proper if for every compact subset K ⊂ S, its pre-image f −1K is also compact. In particular, such a map has compact fibers. This notion is the most useful for the category of locally compact Hausdorff spaces (which contains all spaces homeomorphic to a closed subset of some Rn) and if we restrict to that category, then there is an alternate definition: f : X → S is proper if and only if it is universally closed, which means that for every locally compact Hausdorff space S˜ and continous map S˜ → S, the natural map S˜ ×S X → S˜ is closed: the image of every closed subset of S˜ ×S X in S˜ is closed. We take this as our cue for the definition of proper in the category of schemes:

DEFINITION 6.19. A morphism of schemes f : X → S is proper when it is separated, of ﬁnite type and universally closed in the category of schemes. We say that a scheme X is proper, if it is so over Z. 106 3. SCHEMES

REMARK 6.20. Some authors use an apparently weaker condition for universally closed and only ask that for any scheme T , the morphism 1T × f : T × X → T × S is closed. The difference is indeed only apparent. For if g : T → S is a morphism, then according to Proposition 5.6 its graph identiﬁes T with a locally closed subspace of T × S. Via this identiﬁcation, the restriction of 1T × f to this subspace gives us back the base change morphism T ×S X → T . But if 1T ×f : T ×X → T ×S is closed, then its restriction over any locally closed subspace of T × S is closed as well.

REMARKS 6.21 (Elementary properties of the properness condition). Most of the remarks we made for the separation condition extend easily to properness. The property of f : X → S being proper is local on S: if we are given an open covering −1 {Vα}α of S, then f is proper if and only if every f Vα → Vα is proper. A closed immersion X,→ Y is proper. Closed under composition: if f : X → Y and g : Y → Z are proper, then so is gf : X → Z. Fullness: if X/S and Y/S are proper, then so is every S-morphism from X → Y . Closed under taking products: if we are given a base scheme S and two proper S-morphisms X → Y and X0 → Y 0, then the associated product 0 0 S-morphism X ×S X → Y ×S Y is also proper. Closed under base change: if X/S is proper and S˜ → S is a morphism, then S˜ ×S X is proper over S˜. So the first three properties say that if we fix a base scheme S, then the schemes that are proper over S make up a full subcategory of Sch/S that is closed under S-products. The last property implies that any scheme theoretic fiber of a proper morphism is proper over its base ring. Note that the first property implies that when f : X → S is proper and Y ⊂ X is a closed subscheme of X, then f|Y : Y → S is also proper. Integral affine schemes are rarely proper, witness:

PROPOSITION 6.22. Let X be an integral scheme proper over the algebraically closed ﬁeld k. Then any f ∈ Γ(X, OX ) must be constant. In particular, X cannot be afﬁne unless it is a copy of Spec(k).

PROOF. Since X is integral, the first assertion follows if we show that f is constant on any open affine subscheme U ⊂ X. Define a closed subscheme Cf 1 of Ak ×k X as the subset defined by the ideal tf − 1. So if U = Spec(A), with 1 A a reduced k-algebra, then Ak ×k U = Spec R[t], f|U ∈ A, and the restriction 1 of Cf to Ak ×k U is Spec(A[t]/(tf|U − 1). The image of Cf under the projection 1 1 1 π : Ak ×k X → Ak will not contain the origin of Ak (the closed point defined by the maximal ideal (t)). Since X is proper over k, this projection is closed, and so ∗ π(Cf ) cannot contain the generic point. In particular, (f|U) : k[t] → A is not injective: its kernel contains a nonzero polynomial. Then f|U will have its image in the (finite) zero set of this polynomial. But A is an integral domain, and so f|U is then constant. If X is affine, then we can take X = U. It then follows the finitely generated k-algebra A consists of constants. Since A is a finitely generated field extension of k and k is algebraically closed, it follows that A = k. 6. SEPARATED AND PROPER MORPHISMS 107

THEOREM 6.23 (Valuative characterization of being proper). Let f : X → S be a morphism of schemes that is of ﬁnite type and for which the diagonal ∆ : X → X×S X is quasi-compact. Then f is separated if and only if for every valuation ring R with ﬁeld of fractions K, the natural map X(R) → S(R) ×S(K) X(K) is bijective.

PROOF. We first assume that f is proper and show that for R a valuation ring, X(R) → S(R) ×S(K) X(K) is bijection. Since f is separated, we know this map is injective and so we only need to show it is onto. So let be given morphism φ : Spec(K) → X and a lift ψ : Spec(R) → Spec(S) of fφ. We make a base change over ψ: let Xψ := Spec(R) ×S X and denote by fψ : Xψ → Spec(R) the projection. We can now think of φ as defining a section of fψ over the generic point η ∈ Spec(R). Put q := φ(η) and denote by Q the integral closed subscheme of Xψ which has q as its generic point. From fψ(q) = η we get that κ(q) ⊃ K and from φ(η) = q we get the opposite inclusion: K ⊃ κ(q). So κ(q) = K. Since f is universally closed, fψ(Q) is closed. Hence it must contain the closed point o ∈ Spec(R): there exists a qo ∈ Q such that fψ(qo) = o. In other words, K ⊃ OQ,qo ⊃ R. In view of the maximality of R, this implies that OQ,qo = R and hence we thus have defined a morphism Spec(R) → Q whose composite with Q ⊂ Xψ → X equals φ and whose composite with Q ⊂ Xψ → X → S (which is the composite Q → Spec(R) → S) equals ψ. We now assume that f has the valuative property and prove that f is universally ˜ closed. Let S˜ → S be a morphism and let f : X˜ := S˜ ×S X → S˜ be the associated morphism. We must show that for any closed subset Z ⊂ X˜, f˜(Z) is closed. Since ˜ ˜ f is quasi-compact, so is f|Z. Let po ∈ S˜ be in the closure of f(Z). According to Lemma 6.18 we can find a valuation ring R with field of fractions K and a morphism φ : Spec(K) → X˜ such that fφ : Spec(K) → S˜ lifts to a morphism ψ : Spec(R) → S˜. Then the composites Spec(K) → X˜ → X and Spec(R) → S˜ → S define an element of S(R)×S(K)X(K). By assumption that element originates from a unique morphism ψˆ : Spec(R) → X. Then the pair (ψ, ψˆ) defines a morphism ˜ ˜ φ : Spec(R) → S˜ ×S X = X˜. Since φ maps the generic point η to Z and Z is closed, ˜ ˜ ˜ ˜ φ(o) is also a point of Z. We have f(φ(o)) = po and so po ∈ f(Z).

Our main example of a proper scheme will be absolute projective n-space, n n n n P := P and the schemes PS := S ×Spec(Z) P obtained by base change. We recall n Z that P is obtained as the proj construction applied to the graded ring Z[T0,...,Tn].

It is covered by n + 1 affine open pieces U0,...,Un, where Ui = UTi is defined by Ti 6= 0 and is identified with the spectrum of the ring

[ T0 ,..., Ti−1 , Ti+1 ,..., Tn ]. Z Ti Ti Ti Ti

In particular, Pn is of ﬁnite type. The following lemma links the schematic to the conventional deﬁnition of projective space. Its proof is modeled after the proof of following more classical fact: any meromorphic map φ from the punctured disk 0 < |z| < 1 to Pn(C) extends holomorphically across 0. This is shown as follows: represent φ as [φ0(z): ··· : φn(z)] with each φi meromorphic. If k ∈ Z is the smallest integer k such that each z φi is holomorphic at 0 (so that not all vanish there), then it is clear k k that [φ0(z): ··· : φn(z)] = [z φ0(z): ··· : z φn(z)] is holomorphic at 0. 108 3. SCHEMES

LEMMA 6.24. Let R be a valuation ring with field of fractions K. Then the natural map Pn(R) → Pn(K) is a bijection and Pn(K) is naturally identified with the classical n n+1 projective space Pclss(K) of nonzero vectors in K given up to a nonzero scalar. n n PROOF. We first define a map P (R) → Pclss(K) and prove it is a bijection. We n n then observe that this also shows that a similarly defined map P (K) → Pclss(K) is a bijection and that the resulting bijection Pn(R) =∼ Pn(K) is the obvious map. An element of Ui(R) is given by a ring homomorphism φ : [ T0 ,..., Ti−1 , Ti+1 ,..., Tn ] → R. Z Ti Ti Ti Ti

By assigning to φ its values on T0/Ti,...,Ti−1/Ti,Ti+1/Ti,...,Tn/Ti, we get an n identification of Ui(R) with R , but for reasons that, if not clear now, will be clear in a moment, we prefer to assign to φ the element [φ( T0 ): ··· : φ( Ti−1 ) : 1 : φ( Ti+1 ): ··· : φ( Tn )] ∈ n (K). Ti Ti Ti Ti Pclss n This defines a map ei : Ui(R) ,→ Pclss(K) which is easily seen to be injective with image the set of [a0 : ··· : an] for which ai 6= 0 and aj/ai ∈ R for all j. The embeddings ei and ej coincide on Ui(R) ∩ Uj(R) and so the ei’s assemble to n n produce a map e : P (R) → Pclss(K). This map is still injective. It is also surjective: n n+1 given a p ∈ Pclss(K), then represent p by some (a0, . . . , an) ∈ K − {0}. After multiplication by a common denominator, we can arrange that each ai is in R. The ideal in R generated by the a0, . . . , an is generated by one of these elements, say by ai, so that ai divides each aj. This implies that p ∈ e(Ui). We conclude that e is a bijection. The field K is also an example of a valuation ring (one which equals its field of fractions) and so we also have a bijection n ∼ n n ∼ n P (K) = Pclss(K). The resulting bijection P (R) = P (K) is indeed induced by the obvious map. EXERCISE 87. Let C be a curve over Spec(k) and let p ∈ C be a smooth point of n C. Prove that every morphism f : C − {p} → Pk extends uniquely to a morphism n C → Pk . (Hint: observe that OC,p is a valuation ring.) COROLLARY 6.25. The absolute projective space n = n is proper (over ) and P PZ Z hence for every scheme S, n := n × S is proper over S. PS PZ PROOF. We may conclude from either Exercise 85 or Exercise 86( Z is noe- n n n therian) that the diagonal P → P ×Spec(Z) P is quasi-compact. By the valuative criterion, it then suffices to show that for every valuation ring R with field of fractions K, the natural map n n P (R) → P (K) ×Spec(Z)(K) Spec(Z)(R) is a bijection. But for any ring A, Spec(Z)(A) is a singleton (Z is an initial object of Ring) and so the right hand side is simply Pn(K). The desired property therefore amounts to the assertion that Pn(R) → Pn(K) is a bijection, and this is part of Lemma 6.24. DEFINITION 6.26. A morphism X → S is called projective (resp. quasi-projective) n if it factors through a closed S-immersion (resp. a S-immersion) in PS for some n. THEOREM 6.27. A projective morphism is proper. n PROOF. Suppose f factors through a closed S embedding i : X,→ PS. Both i n and the projection PS → S are proper, hence so is f. 7. QUASI-COHERENT SHEAVES 109

7. Quasi-coherent sheaves

For a scheme X we introduce here an important class of OX -modules. (We refer to the appendix for a discussion of modules over a sheaf.) If M is an OX - module, and x ∈ X, we write Mx for its stalk at x (an OX,x-module). The κ(x)- vector space κ(x) ⊗OX,x Mx := Mx/mX,xMx is called the ﬁber of M at x and is often denoted M(x).

PROPOSITION-DEFINITION 7.1. Let R be a ring and write X for its spectrum. Then for every R-module M there is a sheaf of OX -modules M on X (the sheafification of M) characterized by the property that for a basic open set Xs, M(Xs) = M[1/s], and the restriction map defined by an inclusion Xs ⊂ Xs0 is the associated homomorphism 0 M[1/s ] → M[1/s]. Its stalk at x ∈ X is the localization Mpx and M(X) = M. The R-module M and its sheafification M have the same support. Sheafification is compatible with the tensor product: if M and N are R-modules, then M ⊗OX N is sheafification of M ⊗R N and with arbitrary direct sums: {Mi}i is a collection of R-modules, then ⊕iMi is the sheafification of ⊕iMi. It defines a functor ModR → ModOX which is exact and fully faithful.

PROOF. We omit the proof of the first part as it is similar to the one of 2.1. The proofs of the second and third statements then follow from the fact that the natural R[1/s]-homomorphisms (M ⊗R N)[1/s] → M[1/s] ⊗R[1/s] N[1/s] resp. (⊕iMi)[1/s] → ⊕iMi[1/s] are isomorphisms. Sheafification is exact: if 0 → M → M 0 → M 00 → 0 is an exact sequence of 0 R-modules, then for every prime ideal p of R, the sequence 0 → Mp → Mp → 00 Mp → 0 is still exact (localization is exact) and this is just the sequence of stalks of sheafified sequence at x = xp. Sheafification is fully faithful: we must show that any sheaf homomorphism φ : M → N comes from a unique R-homomorphism Φ: M → N. There is no uniquess issue, for we must Φ = φ(X): M = M(X) → N (X) = N. But then existence is also clear: on a basic open subset Xs, φ(Xs): M[1/s] → N[1/s] must be induced by Φ. The following supplements Proposition 2.8. The proof is straightforward and therefore omitted.

PROPOSITION 7.2. Let φ : R0 → R be ring homomorphism and denote by f : X → X0 the associated morphism of affine schemes. Sheafification is compatible with the restriction functor ModR → ModR0 : for an R-module M, f∗M is the sheafification of M regarded as a R0-module. It is also compatible with its left adjoint, the induction 0 0 ∗ 0 0 functor: for an R -module M , f M is the sheafification of R ⊗R0 M . We will be mostly concerned with sheaves that arise locally from sheafifying modules.

PROPOSITION-DEFINITION 7.3. Let X be scheme. For an OX -module F the following are equivalent: (i) X admits a covering by affine open subsets U such that F|U is the sheafification of the OX (U)-module F(U). (ii) for every affine open affine subset U of X, F|U is the sheafification of the OX (U)-module F(U). 110 3. SCHEMES

When these equivalent conditions are fulfilled, we say that F is quasi-coherent. In particular, for an affine scheme Spec(R), the sheafification functor defines an equivalence between the category of R-modules and the category of quasi-coherent OSpec(R)-modules and hence the global section functor sends a short exact sequence of quasi-coherent OSpec(R)-modules to a short exact sequence of R-modules.

PROOF. We assume (i) and prove (ii). We do this in steps. Let U = Spec(R) ⊂ X be open and affine and put M := F(U). We must show that F|U is the sheafification of the R-module M. r Step 1: There exist a finite set {si ∈ R}i=1 such that the Usi cover U and F|Usi is the sheafification of the O(Usi )-module Mi := F(Usi ). Given a point x ∈ U, let U(x) = Spec(Rx) by an affine open subset containing x such that F|U(x) is the sheafification of the Rx-module M(x) := F(U(x)). Choose a basic open neighborhood U 0(x) of x contained in U ∩ U(x), so that there exists 0 0 0 s ∈ R s ∈ R U (x) = U(x) 0 = U F|U (x) an x x and an x such that sx sx . Then is the 0 0 sheafification of F(U (x)) = M(x)[1/sx]. Since U is quasi-compact, finitely many of the U 0(x)’s suffice to cover U: U = ∪r U . Then {s := s }r is as desired. i=1 sxi i xi i=1 We have a natural sheaf homomorphism M → F|U. To prove that this is an isomorphism, it suffices to show that this induces an isomorphism M(U 0) → F(U 0) for U 0 running over a basis of open subsets of U. So it suffices to show that for an arbitrary s ∈ O(U), the ring homomorphism M[1/s] → F(Xs). We prove injectivity and surjectivity separately.

Step 2: The homomorphism M[1/s] → F(Us) is injective. Since sheaﬁﬁcation is compatible with localization, we have that F(U(ssi)) = Mi[1/s]. Now let a ∈ M = F(U) be such that its restriction to F(Us) is zero. N We must show that s a = 0 for some N ≥ 0. The image of a in F(Us ∩ Usi ) =

F(Ussi ) = Mi[1/s] is of course also zero. So if ai ∈ Mi = F(Usi ) is the image of a, Ni then s ai = 0 for some integer Ni ≥ 0. Then N = maxi Ni has the property that N N s ai = 0 in Mi = F(Usi ) for all i. This means that s a = 0 in F(U) = M.

Step 3: The homomorphism M[1/s] → F(Us) is surjective. Ni Let b ∈ F(Us). Then its image in F(Us ∩ Usi) = Mi[1/s] is of the form bi/s for some bi ∈ Mi and some Ni ≥ 0. We multiply numerator and denominator by a positive power of s so that we can take Ni independent of i: Ni = N. Now bi and bj have the same image in F(Us ∩ Usi ∩ Usj ) = F(Usisj )[1/s] and so there exists Nij an integer Nij ≥ 0 such that bi|Usi ∩ Usj − bj|Usi ∩ Usj is killed by s . So if we 0 N 0 N 0 let N := maxi,j Nij, then s bi and s bj have the same restriction to Usi ∩ Usj for all i, j. This implies that there exists a ˜b ∈ F(U) = M having the property that ˜ N 0 ˜ N+N 0 b|U(si) = s bi for all i. Then b/s ∈ M[1/s] restricts to b|F(Us ∩ Usi ) for all i and hence restricts to b ∈ F(Us).

COROLLARY 7.4. Let f : X → Y be a morphism of schemes. Then (a) the kernel, cokernel and image of a homomorphism between quasi-coherent OX -modules is also quasi-coherent, in other words, the category Qcoh(X) of

quasi-coherent OX -modules is abelian subcategory of the category ModOX of OX -modules, (b) the tensor product over OX of two quasi-coherent OX -modules is quasi- coherent, ∗ (c) f takes QcohY to QcohX , 7. QUASI-COHERENT SHEAVES 111

(d) if f is quasi-compact and quasi-separated (note that this is the case when X is noetherian), then f∗ takes QcohX to QcohY . PROOF. The first two assertions are of a local nature, hence follow from Propo- sition 7.3 and the fact that ModR is abelian. The same is true for (c) (this is local on X). For (d) we may assume that Y is affine: Y = Spec(S). If X is also affine: X = Spec(R), then we are done of course, for any quasi-coherent sheaf M on X is the sheafification of an R-module M and f∗M is then the sheafification of M viewed as a S-module via the ring homomorphism S → R. We now turn to the general case. Since f is quasi-compact, this implies that N X is covered by finitely many affine open subsets {Ui)}i=1. Since f is also quasi- Nij separated, Ui ∩ Uj is covered by finitely many affine open subsets {Uijk}k=1. Let now F be quasi-coherent on X. The sheaf property of F implies that we have an exact sequence of OY -modules

0 → f∗F → ⊕if∗(F|Ui) → ⊕i,j,kf∗(F|Uijk).

By the case treated above, f∗(F|Ui) and f∗(F|Uijk) are quasi-coherent OY -modules. Hence, f∗F, being the kernel of a OY -homomorphism between two such modules, is also quasi-coherent. COROLLARY 7.5. Let f : Y → X be a morphism of schemes that is quasi-compact and quasi-separated. Then ker(OX → f∗OY ) is a quasi-coherent ideal sheaf in OX and hence defines a closed subscheme of X whose underlying subset is the closure of the image of f. We call this the scheme theoretic image of f. In particular any closed subscheme of a scheme X is defined by a quasi-coherent ideal sheaf in OX and conversely any quasi-coherent ideal sheaf in OX defines a closed subscheme of X.

PROOF. Since f is quasi-compact and quasi-separated, part (d) of Corollary 7.4 implies that f∗OY is quasi-coherent. By part (a) of that corollary, the ideal sheaf ker(OX → f∗OY ) is then quasi-coherent as well. It is clear that a quasi-coherent ideal sheaf in OX deﬁnes a closed subscheme. Since f∗OY is zero on any open subset of X which does not meet the image of Y , the underlying subset is the closure of the image of f. The second statement follows from the fact that the inclusion of a closed subscheme of X in X is separated (and hence quasi-compact and quasi-separated). We use this corollary to prove an ‘almost converse’ to Theorem 6.27 in case the base is noetherian. It is known an Chow’s Lemma and helps us to extend properties of projective morphisms to proper morphisms.

PROPOSITION 7.6 (Chow’s Lemma). Let f : X → S be a proper morphism with S noetherian. Then there exists a projective morphism π : X0 → X that is an isomorphism over an open-dense subset of X (so that π will be surjective)3 and has the property that fπ : X0 → S is also projective.

PROOF. It is not hard to see that it suffices to prove this for each irreducible component of X separately and so we will assume that X is irreducible. Since f is of finite type, we can cover X by a finite number of nonempty affine r open subsets {Ui}i=1 such that f(Ui) is contained in an affine open subset Vi of S

3Such a morphism is often called a projective modification of X. 112 3. SCHEMES and OX (Ui) a finitely generated OS(Vi)-algebra. Since each Ui is nonempty (hence dense in X), so is U := ∩iUi. Choose a finite set of OS(Vi)-generators of OX (Ui) so that we have a closed immersion U ,→ ki and hence a locally closed immersion i AVi U ,→ ki ⊂ ki . We write P for ki and P for P × · · · × P . Note that P is i AVi PS i PS 1 S S r projective over S. The diagonal morphism j : U → X ×S P is an immersion so that 0 we can regard U as a locally closed subscheme of X ×S P . We then take to be X the closure of U as a subscheme (here we invoke Corollary 7.5) and π : X → X0 the projection on the first factor. 0 Clearly π is projective. Let πi : X → Pi denote the projection. We claim that −1 −1 0 πi Ui = π Ui and that the projection of this open subset of X onto Ui is an 0 open immersion. For this we note that the projection (π, πi): X → X ×S Pi is projective, hence closed. Its restriction to the open-dense U ⊂ X0 is the diagonal 0 U → U ×S U ⊂ X ×S Pi and so the image of X lies in the closure of U ×S U. This is also the closure of Ui ×S Ui. Since X/S is separated, this implies our claim. −1 We show that this implies the proposition. Since the Ui cover X, the π Ui will 0 0 cover X . It then follows from our claim that the projection (π1, . . . , πn): X → P is an immersion. Since X is proper over S, this projection is also closed and hence X0 is thus realized as a closed subscheme of P . In particular, the projection fπ : X0 → S is projective. It also follows that π−1U is the diagonally embedded U in X ×S P so that π is an isomorphism over U.

PROPOSITION-DEFINITION 7.7. Let X be a noetherian scheme. For an OX - module F the following are equivalent: (i) X admits a covering by affine open subsets U such that F|U is the sheafification of a finitely generated OX (U)-module F(U). (ii) for every affine open affine subset U of X, F|U is the sheafification of the finitely generated OX (U)-module F(U). When these equivalent conditions are fulfilled, we say that F is coherent4. In particular, for a noetherian ring R the sheafification functor defines an equivalence between the category of finitely generated R-modules and the category of coherent OSpec(R)-modules. Moreover, any closed subscheme of Spec(R) is defined by a unique ideal of R. Furthermore: (a) the category Coh(X) of coherent sheaves is abelian subcategory of the cate-

gory ModOX of OX -modules, (b) the tensor product of two coherent OX -modules is a coherent OX -module, (c) if f : X → Y is a morphism between noetherian schemes, then f ∗ takes CohY to CohX , (d) any closed subscheme of X is defined by a coherent sheaf of ideals. PROOF. We first prove that (i) implies (ii), the other assertions follow as in the quasi-coherent case. Let U ⊂ X be open and affine and put R := OX (U). Since X is noetherian, then by Proposition 5.7, so is R. According to the previous proposition, F|U is the sheafification of the R-module M := F(U) and (in the situation of the proof of that proposition), we can assume that each M[1/si] is a

4This definition is fine when the scheme X is noetherian (and we will only consider coherent modules on such schemes), but for a general ringed space (X, O) it takes the following form: a O- module F is coherent if we can cover X by open subsets U for which F|U and the kernel of any O|U- homomorphism Or|U → F|U are finitely generated O|U-modules. This generalization is for instance the one that is customary in the setting of complex-analytic spaces. 8. WHEN IS A SCHEME AFFINE? 113 noetherian R[1/si]-module. Then the argument given in Proposition 5.7 shows that M must be noetherian as well. The proofs of the other assertions straightforward. REMARK 7.8. A direct image of a coherent module on a projective scheme is in general not coherent. The simplest example is perhaps the direct image of O 1 Ak 1 1 under the projection map → Spec(k): this is the k-vector space Γ( , O 1 ) = Ak Ak Ak k[t], which is not a finite dimensional. We will see however that this is the case for a projective morphism. The following results show among other things that on a noetherian scheme quasi-coherent modules are well approximated by coherent ones. PROPOSITION 7.9. Let X be a noetherian scheme, U ⊂ X an open subset and G a quasi-coherent OX -module whose restriction to U is coherent. Then there exists a coherent submodule F ⊂ G with F|U = G|U. PROOF. First consider the case when X = Spec(R) is affine. Then G is the sheafification of an R-module N. Cover U by finitely many basic open affines

Xf1 ,...,Xfr . Since G|U is coherent, N[1/fi] is a finitely generated R[1/fi]-module. Let Si ⊂ N be a finite subset whose image in N[1/fi] generates the latter as a R[1/fi]-module. Then the R-submodule M ⊂ N generated by the finite set ∪iSi yields after sheafification a coherent subsheaf F ⊂ G with the desired properties. In general we proceed as follows. Cover X with finitely many affine open subsets U1,...,Un and put U0 := U, Vk := U0 ∪ U1 ∪ · · · ∪ Uk and Gk := G|Vk. Let 1 ≤ k ≤ n and assume with induction that we have constructed a coherent submodule Fk−1 of Gk−1 which on U coincides with G (this is an empty statement for k = 1). The difference Zk := Vk − Vk−1 is a closed subset of Vk that is contained in the affine open Uk and is disjoint with U. Denoting the inclusion Vk−1 ⊂ Vk by j, then we have natural homomorphisms of quasi-coherent OVk -modules Gk → 0 j∗Gk−1 and j∗Fk−1 → j∗Gk−1. Let Fk ⊂ Gk be the preimage of the image of j∗Fk−1 → j∗Gk−1 in Gk. This is a quasi-coherent OVk -submodule of Gk whose restriction to Vk−1 = Vk − Zk equals Fk−1 (and hence is coherent). By the affine 00 0 case we already treated, there exists a coherent submodule Fk ⊂ Fk|Uk which 0 extends Fk|Uk − Zk. We then complete the induction step by letting Fk ⊂ G|Vk be 00 the coherent submodule characterized by Fk|Uk = Fk and Fk|Vk−1 = Fk−1. COROLLARY 7.10. Let X be a noetherian scheme. Then

(i) if U ⊂ X is open, then every coherent OU -module extends to a coherent OX -module and (ii) every quasi-coherent OX -module is the union of its coherent submodules.

PROOF. If F is a coherent OU -module, then assertion (i) follows by applying the above proposition to its direct image G := j∗F under j : U ⊂ X. For (ii) we must show that if G is a quasi-coherent OX -module, then any local section s ∈ Γ(U, G) is contained in a coherent OX -submodule. This follows by applying (i) to the OU -submodule OU s ⊂ G|U, which is indeed coherent. 8. When is a scheme afﬁne? The main theorem of this short section is a characterization for a noetherian scheme to be afﬁne in terms of cohomology (for sheaf cohomology we refer to the appendix). 114 3. SCHEMES

THEOREM 8.1 (Serre). For a noetherian scheme X the following are equivalent: (i) X is afﬁne, (ii) every quasi-coherent sheaf on X is Γ-acyclic: Hk(X, F) = 0 when k 6= 0, 1 (iii) for every quasi-coherent ideal sheaf I ⊂ OX , H (X, I) = 0. The implication (ii) ⇒ (iii) is trivial and so it sufﬁces to establish (i) ⇒ (ii) and (iii) ⇒ (i). For the former we shall need the following general result on sheaf cohomology.

LEMMA 8.2 (G. Kempf). Let F be an abelian sheaf on a space X and let U be a basis of open subsets. Suppose that for some positive integer n we have Hk(X, F) = 0 for 0 < k < n (so this condition is empty for n = 1). Then for every a ∈ Hn(X, F) n −1 we can cover X by U ∈ U with the property that a dies in H (X, jU∗jU F) (where −1 jU : U ⊂ X so that jU∗jU F(V ) = F(U ∩ V )). PROOF. Embed F in a flasque sheaf I (for instance the sheaf defined by I(U) = Q x∈U Fx) and put G := I/F. Then the long exact sequence associated to this quotient yields an exact sequence 0 → Γ(X, F) → Γ(X, I) → Γ(X, G) → H1(X, F) → 0 k ∼ k+1 and isomorphisms H (X, G) = H (X, F) for k ≥ 1. For U ∈ U, put GU := −1 −1 −1 −1 jU∗jU I/jU∗jU F. This embeds in jU∗jU G (jU∗ is left exact). Since jU∗jU I is also flasque, we have likewise an exact sequence −1 −1 1 −1 0 → Γ(X, jU∗jU F) → Γ(X, jU∗jU I) → Γ(X, GU ) → H (X, jU∗jU F) → 0 k ∼ k+1 −1 and H (X, GU ) = H (X, jU∗jU F) for k ≥ 1. We prove the assertion with induction on n and so we first treat the case n = 1. The first exact sequence shows that we can lift a ∈ H1(X, F) to an element a0 ∈ Γ(X, G). We can cover X by basic open subsets U ∈ U such that a0|U lifts to a sec- 00 −1 1 −1 tion a ∈ Γ(U, I) = Γ(X, jU∗jU I). Then the image of a in H (X, jU∗j F) is the 00 −1 1 −1 image of a under the composite Γ(X, jU∗jU I) → Γ(X, GU ) → H (X, jU∗j F) and hence zero. Now assume n > 1. We have Hk−1(X, G) =∼ Hk(X, F) = 0 for k > 1, so that G satisfies our hypothesis with n replaced by n − 1. By our induction assumption, we n−1 −1 can then cover X by U ∈ U such that H (X, jU∗jU G) = 0. Obviously GU |U = n −1 ∼ n−1 −1 n−1 −1 G|U and so H (X, jU∗jU F) = H (X, jU∗jU GU ) = H (X, jU∗jU G) = 0.

PROOFOF (i) ⇒ (ii). We prove (ii) with induction: assume that we have established that any quasi-coherent sheaf on an affine scheme has zero cohomology in degrees 1, . . . , n − 1 (this assertion being empty for n = 1). Then the hypotheses of Lemma 8.2 hold for the quasi-coherent sheaf F on X. Since X is noetherian, we find that for any a ∈ Hn(X, F) we can cover X by a finite number of affine open r n subsets {Ui}i=1 such that a dies in H (Ui, F), i = 1, . . . , r. Since the Ui’s cover X, the sheaf homomorphism F → ⊕r j j−1F is a monomorphism of quasi- i=1 Ui∗ Ui coherent sheaves. So its cokernel, denoted G, is also quasi-coherent. Since X is affine, say equal to Spec(R), this comes from an exact sequence of R-modules 0 → Γ(X, F) → ⊕r Γ(X, j j−1F) → Γ(X, G) → 0. i=1 Ui∗ Ui Feeding this in the long exact sequence for cohomology shows that then H1(X, F) → ⊕r H1(X, j j−1F) is injective. So if n = 1, then a = 0. For n > 1 we find that i=1 Ui∗ Ui 8. WHEN IS A SCHEME AFFINE? 115 a lies in the image of Hn−1(X, G) → Hn(X, F). But then it must be zero, since by n−1 our induction assumption H (X, G) = 0. For the proof that (iii) ⇒ (i) we need the following criterion for affineness:

LEMMA 8.3. A scheme X is afﬁne if and only if there exist f1, . . . , fr ∈ Γ(X, OX ) such the corresponding open subschemes Xfi ⊂ X are afﬁne and cover X.

PROOF. Let X be a scheme and put A := Γ(X, OX ). Then there is an obvious morphism X → Spec(A), which of course is an isomorphism precisely when X is affine. We only prove the nontrivial direction and assume that A contains f1, . . . , fr as in the lemma. Then each fi defines a basic open subset Spec A[1/fi]) ⊂ Spec(A) whose preimage in X is Xfi . So it will be enough to show that for every f ∈ A, the φ restriction Xf −→ Spec A[1/f]) is an isomorphism. In other words, we must show that the natural ring homomorphism A[f] → OX (Xf ) is an isomorphism. For this we will use the observation that for every i, Xf ∩ Xfi is a basic open subset of the ¯ ¯ affine Xfi so that OX (Xf ∩ Xfi ) = O(Xfi )[1/fi], where fi := f|Xfi . To prove injectivity, suppose g ∈ A is such that g|Xf = 0. Then for every i, ni g|Xf ∩ Xfi = 0 and so exists an integer ni ≥ 0 such that f g|Xfi = 0. So if we n n let n := max ni, then f g|Xfi for all i and hence f g is zero in Γ(X, OX ) = A. In other words, g has zero image in A[1/f]. ¯ To prove surjectivity, let g ∈ OX (Xf ). Then g|Xfi ∩ Xf ∈ O(Xfi )[1/fi] has the ni property for some an integer ni ≥ 0, f g extends to a section of OX |Xfi . So if we n let n := max ni, then f g extends to a section h ∈ A of OX and g is the image of n h/f ∈ A[1/f]. PROOFOF (iii) ⇒ (i). We will show that the hypotheses of Lemma 8.3 are satisfied. Let x ∈ X be a closed point and let Ux be an open affine neighborhood of x x. Let I ⊂ OX denote the ideal sheaf that defines the closed subscheme X − Ux x x x and let J ⊂ I be the kernel of the natural surjection I → ix∗κ(x). Since H1(X, J x) = 0, the evaluation map Γ(X, Ix) → κ(x) is surjective and so there x exists a fx ∈ Γ(X, I ) whose image in κ(x) is 1. Notice that Xfx can be understood as a basic open subset of Ux and is therefore an affine neighborhood of x in X. Since X is noetherian a finite number of such subsets will cover X. Now apply Lemma 8.3. We turn our attention to higher direct images of quasi-coherent sheaves. In the appendix we show that a for morphism (X, O) → (X0, O0) of ringed spaces, the right derived functors of ‘ringed direct image functor’ ModO → ModO0 composed the forgetful functor ModO0 → AbX0 produce the right derived functors of the ‘ordinary direct image functor AbX → AbX0 . PROPOSITION 8.4. Let f : X → Y be a morphism of schemes with X noetherian. k Then R f∗ sends Qcoh(X) to Qcoh(Y ). In fact, if V ⊂ Y is open affine, then for every k quasi-coherent sheaf F on X, R f∗(F)|V is the sheafification of the OX (V )-module Hk(f −1V, F).

PROOF. It suffices to prove the last assertion and so we assume that Y = Spec(R) is affine. For a quasi-coherent sheaf F on X, we denote by T k(F) the k 0 sheafification of the R-module H (X, F). It is clear that then T (F) = f∗F. It is also clear that the T k’s are part of a δ-package and so it remains to see that k n it each T is effaceable. For this let {jν : Uν ⊂ X}ν=1 be a finite affine open 116 3. SCHEMES

k ∗ covering of X. For every affine open subset V ⊂ Y , we have T (jν∗jν F)(V ) = k −1 ∗ k −1 −1 H (f V, jν∗jν F) = H (f V ∩ Uν , F) and since f V ∩ Uν is affine, Serre’s criterion for affineness implies that this is zero when k > 0. So if we put F 0 := ∗ k 0 ⊕ν jν∗jν F), then T (F ) = 0 for k > 0. We have a natural OY -homomorphism ∗ 0 F → jν∗jν F of quasi-coherent sheaves which combine to a monomorphism F → 0 k F . So T is effacable. EXERCISE 88. Let f : X → Y be a quasi-compact morphism of schemes with X locally noetherian. Prove that f is affine if and only if for every quasi-coherent 1 k sheaf, R f∗(F) = 0. Show that in either case, R f∗(F) = 0 for all k > 0.

9. Divisors and invertible sheaves

We ﬁx a scheme X. An OX -module L is called invertible if we can cover X ∼ by open subsets U such that L|U = OU . So for every x ∈ X, the ﬁber L(x) =

κ(x)⊗OX,x Lx = Lx/mX,xLx is then a κ(x)-vector space of dimension one. We note ∨ that the OX -dual, defined as L := HomOX (L, OX ) is also invertible and so is its tensor product L ⊗ L0 with another invertible sheaf L0. We also write Ln for resp. ⊗n ∨ ⊗(−n) L , OX , (L ) when resp. n > 0, n = 0, n < 0. The tensor product turns the set isomorphism classes of invertible sheaves on X into a group (with the inverse represented by the dual), the Picard group Pic(X) of X. The group structure is written addively so that [Ln] = n[L]. It turns out that the Picard group often has the structure of a group scheme. Observe that for any morphism of schemes f : X → Y , f ∗ takes an invertible sheaf L on Y to one on X. This is verified on affine open subsets U = Spec(R) ⊂ X that map to an affine open subset V = Spec(S) of Y on which L is trivial and this it is a simple consequence of the fact that for the associated ring homomorphism φ : S → R, the map r ∈ R 7→ r ⊗ 1 ∈ R ⊗S S is an isomorphism with inverse r ⊗ s ∈ R ⊗S S 7→ rφ(s) ∈ R. This pull-back is compatible with ⊗ and takes isomorphic invertible sheaves to isomorphic invertible sheaves. Hence f induces a group homomorphism: Pic(f) : Pic(Y ) → Pic(X). We thus have defined a contravariant functor Pic : Sch◦ → Ab. This notion of a Pi- card group makes sense for any ringed space and in the appendix we prove that in 1 × that generality Pic(X) can be identified with H (X, OX ). Here we describe Pic(X) (at least part of it) in terms of Cartier divisors.

Cartier divisors. For a ring R, we have defined its fraction ring Frac(R) (2.13) as the localization with respect to the multiplicative set S(R) of all nonzero divisors. It has the property that R → Frac(R) is injective. We note that the property of being (not) a zero divisor can be tested ‘locally’: For every prime ideal p of R, we have the localization map R → Rp. Clearly, a zero divisor of R will map to a zero divisor of Rp or to zero. On the other hand, for every prime ideal p of R, S(R) maps to S(Rp): if s ∈ S(R) kills a/b ∈ Rp, then sac = 0 in R for some c ∈ R − p. Since s is not a zero divisor, we then have ac = 0, which means that a/b = 0. We phrase this in scheme language: if U ⊂ X is affine, then s ∈ Γ(U, OX ) is not a zero divisor if and only if its image in each stalk OX,x with x ∈ U is not a zero divisor. It is now clear that we have defined on X a presheaf of rings which assigns to an arbitrary open subset U ⊂ X, the localization of Γ(U, OX ) with respect to the 9. DIVISORS AND INVERTIBLE SHEAVES 117 multiplicative system of s ∈ Γ(U, OX ) which map to a nonzero divisor in OX,x for 5 all x ∈ U. We denote the associated sheaf by KX . It is characterized by the property that for an affine open subset U ⊂ X, KX (U) = Frac(OX (U)). The natural sheaf homomorphism OX → KX is a monomorphism and we will therefore think × of OX as a subsheaf of KX . We denote by KX ⊂ KX the subsheaf of invertible elements. Notice that when X is an integral scheme, then KX is the constant sheaf that assigns to any nonempty open subset of X the function field K(X) of X.

A quasi-coherent OX -submodule I ⊂ KX is called a fractional ideal if we can cover X by affine open subsets U with the property that fI ⊂ OX for some nonzero divisor f ∈ Γ(U, OX ). Note that a fractional ideal is automatically coherent when X is locally noetherian. Invertible fractional ideals arise as follows. Let L be an invertible OX -module. Any section s ∈ Γ(X, L) defines a subscheme Z(s) ⊂ X whose defining ideal sheaf IZ(s) is locally principal: it is characterized by the prop- ∼ erty that if U ⊂ X is an open affine subset and φ : L|U = OX |U is an isomorphism, then IZ(s)|U = OU φ(s). This is indeed independent of the choice of φ, for any two such differ by a factor in O(U)×, hence generate the same principal ideal sheaf. We can also view s as defining a sheaf homomorphism OX → L which maps 1 to s. This is an embedding precisely when for every (U, φ) as above, φ(1) is not a zero divisor in O(U). In that case, we can reconstruct L from Z := Z(s) as a fractional −1 ideal IZ ⊂ KX . This submodule is characterized by the property that for every −1 −1 local generator f ∈ IZ (U) of IZ , IZ |U := OU f . Then the sheaf homomorphism −1 ∼ OX → L defined by s extends to an isomorphism IZ = L. This observation is the point of departure for the discussion that follows. DEFINITION 9.1.A Cartier divisor on X is an element of the (additively written) × × abelian group Γ(X, KX /OX ). We say that a Cartier divisor is principal if it lies in × × × the subgroup that is the image of Γ(X, KX ) → Γ(X, KX /OX ). The quotient group is denoted CaCl(X). Two Cartier divisors are said to be linearly equivalent if they have the same image in CaCl(X), i.e., if their difference (really their quotient, but we agreed to write the group law additively) is principal. We will see that often a Cartier divisor has the simple geometric meaning as formal linear combinations of integral subschemes of codimension one that can locally be given by a single equation. This explains why we write the group law additively. In any case, we note that a Cartier divisor H on X can always be given by a collection of pairs (Uα, fα)α, where U = (Uα)α is an affine covering of X and × fα ∈ K(U) is such that the restrictions of fα and fβ in Uα ∩Uβ have a quotient that × lies in O(Uα ∩ Uβ) (in a sense only the zeroes and the poles of the fα’s matter). But beware that we are not claiming that any open affine covering of X will do for this purpose: indeed, quite often an affine variety has Cartier divisors that are not principal. EXAMPLE 9.2. Consider the quadric Q = Spec(k[x, y, z]/(xy − z2 + z)). The line ` ⊂ Q defined by the ideal (y, z) (the ‘x-axis’) defines a Cartier divisor: we cover Q by the open subsets U0 defined by z 6= 0 and U1 defined by z 6= 1 (so 2 that U0 ∩ U1 is defined by z − z 6= 0, or equivalently, xy 6= 0) and we take on U0

5Because of this roundabout deﬁnition there are a number of subtle issues here (see S.L. Kleiman: Misconceptions about KX ). For instance, for x ∈ X, the natural map KX,x → Frac(OX,x) is always injective, but need not be surjective. In particular, KX need not be quasi-coherent. 118 3. SCHEMES the regular function y|U0 and on U1 the regular function constant 1. Then their quotient y/1 = y will be indeed nonzero on U0 ∩ U1. But this Cartier divisor is not principal: there is no rational function on f on Q with the property that both f/y|U0 and f|U1 are invertible. For this would imply that f is a regular function on Q which ˜ generates the ideal sheaf I` in OQ deﬁning `. But then a representative f ∈ k[x, y, z] of f must generate the ideal (y, z) in the coordinate ring k[x, y, z]/(xy − z2 + z). In other words, (f,˜ xy−z2 +z) = (y, z) in k[x, y, z]. This is easily seen to be impossible.

EXAMPLE 9.3. This example shows that not every curve on a surface defines a Cartier divisor. Consider the quadric cone C = Spec(k[x, y, z]/(xy − z2)) and let ` ⊂ C be the line defined by the ideal (y, z) (the ‘x-axis’). Then the zero sets x, y, z restricted to C are resp. the y-axis with multiplicity 2, the x-axis with multiplicity 2, the union of the x-axis and the y-axis. It follows that no regular function on C can vanish on ` with multiplicity 1. This remains true after localization at the vertex o ∈ C: if f, g ∈ k[x, y, z] are such that g(o) 6= 0, then the image of f/g in OC,o can not vanish of order one on the x-axis and not anywhere else near o (for then f would have the same property). So ` can not be defined by a Cartier divisor. On the other hand, the principal divisor defined by y|Q yields twice `.

PROPOSITION 9.4. A Cartier divisor H on X determines an invertible fractional ideal L(H) ⊂ KX characterized by the property that if H is represented on an open × −1 subset U ⊂ X by f ∈ K(U) , then L(H)|U := OU f . Conversely, any invertible fractional ideal L ⊂ KX deﬁnes a Cartier divisor D(L) characterized by the property ∼ −1 that given a local isomorphism φ : OU = L|U, then φ(1) ∈ KX (U) represents D(L)|U. This sets up a bijection between the group of Cartier divisors on X and the group of invertible fractional ideals of KX , where the group structure on the latter corresponds 0 to the (tensor) product: given invertible subsheaves L ⊂ KX and L ⊂ KX , then we 0 can form L⊗OX L ⊂ KX . Under this correspondence the principal divisors correspond to the invertible fractional ideals of KX that are trivial (i.e., isomorphic to OX ).

PROOF. If H is represented on an affine open subset U 0 ⊂ X by f 0 ∈ K(U)×, then the images of f and f 0 in K(U ∩ U 0)× have a quotient in O(U ∩ U 0)× and so they generate the same subsheaf of KU∩U 0 . This proves that L(H) is well-defined. It is clearly invertible. Of the other assertions we only verify the last one (the others are staightfor- ward). If we are given a OX -monomorphism φ : OX ,→ KX , then φ(1) defines × a global section of KX and the associated divisor is then the principal divisor defined by φ(1). Conversely, if the divisor for L ⊂ KX is principal, say given by × −1 f ∈ Γ(X, KX ), then L = OX f ⊂ KX and so multiplication by f defines an ∼ isomorphism L = OX .

REMARK 9.5. The following may help to explain the inversion convention in Proposition 9.4. Let L be an invertible sheaf on X and s ∈ Γ(X, L) a section. We observed that s deﬁnes a principal ideal sheaf IZ ⊂ OZ and that under some −1 assumption, L can be identiﬁed with the invertible fractional ideal IZ ⊂ KX . ∼ If U ⊂ X and φ : OU = L|U is an isomorphism, then s|U = fφ(1) for some −1 −1 f ∈ OX (U) and so f generates IZ |U. Hence f is a generator of IZ |U and by −1 our convention, the Cartier divisor associated to IZ is on U represented by the pair (U, f). 9. DIVISORS AND INVERTIBLE SHEAVES 119

Proposition 9.4 immediately implies:

COROLLARY 9.6. The group of Cartier divisors modulo its subgroup of principal divisors (denoted CaCl(X)) can be identified with the subgroup of Pic(X) of isomorphism classes of invertible sheaves that are embeddable in KX . We can obtain this embedding also in terms of the cohomological description of the Picard group: the short exact sequence × × × × 0 → OX → KX → KX /OX → 0 gives rise to the long exact sequence 0 × 0 × 0 × × 1 × 1 × 0 → H (X, OX ) → H (X, KX ) → H (X, KX /OX ) → H (X, OX ) → H (X, KX ) 0 × 0 × × and we thus see that CaCl(X) = Coker(H (X, KX ) → H (X, KX /OX )) maps in- 1 × ∼ 1 × jectively to H (X, OX ) = Pic(X) (with the cokernel contained in H (X, KX )). COROLLARY 9.7. If X is an integral scheme, then the map CaCl(X) → Pic(X) is an isomorphism. × PROOF. Then KX is a constant sheaf (with value the units of the function field 1 × of X) on an irreducible space and hence flasque. So H (X, KX ) = 0. Now feed this into the exact sequence above. Weil divisors. As some of the examples discussed below will show, the Picard group of a scheme can be quite interesting, but is in general hard to determine. For computations it is helpful to think of a Cartier divisor as a formal linear combination of hypersurfaces in the sense that if the Cartier divisor is given by the system P (Uα, fα), then we would want to define a linear combination i riPi with Pi an integral subscheme of codimension one and ri ∈ Z such that fα has along Pi ∩ Uα a zero of order ri (when this ri < 0 is to be understood as a pole of order −ri) and fα has no other zeros or poles. It is then reasonable to assume that each Pi which may possibly occur is at least generically given by a single equation and that we can speak of order of vanishing. This just amounts to insisting that for such a P = Pi, the generic point ηP defined by P , OX,ηP is a DVR (whose valuation

Frac(OX,ηP ) → Z we denote by vP ): when P ∩ Uα 6= ∅, then fα ∈ Frac(O(Uα) × has a nonzero image in Frac(OX,ηP ) and so the valuation on Frac(OX,ηP ) has a well-deﬁned value vP (fα) on fα. This value only depends on the Cartier divisor, for if we also have P ∩ Uβ 6= ∅, then P ∩ Uα ∩ Uβ 6= ∅ (for P is irreducible) and × since fβ/fα is a unit in O(Uα ∩ Uβ), it is also a unit of Frac(OX,ηP ) and hence vP (fβ) = vP (fα). This motivates the following deﬁnition.

DEFINITION 9.8. A closed integral subscheme P of X (or its associated point ηP ∈ X) whose local ring OX,η is a DVR is called a prime divisor of X. A Weil P P divisor on X is a finite formal Z-linear combination of prime divisors D = P nP P (the sum is over prime divisors, but for only finitely many we have nP 6= 0) and we then denote by OX (D) ⊂ KX the subsheaf of f ∈ KX with vP (f) ≥ −nP for all P . We say that the Weil divisor is effective if all the coefficients are ≥ 0 (so that OX (D) ⊃ OX ).

EXAMPLE 9.9. Here are a few simple examples. Let K be a ﬁeld. (a) Any nonconstant irreducible polynomial f ∈ K[x1, . . . , xn] deﬁnes a prime n n divisor in Ak and every prime divisor on Ak is so obtained. 120 3. SCHEMES

(b) Any irreducible homogeneous polynomial F ∈ K[T0,...,Tn] of positive n n degree defines a prime divisor on PK and every prime divisor on PK is so obtained. (c) Any closed point on a nonsingular curve C defined over K defines a prime divisor and every prime divisor on C is so obtained. (d) Suppose K/Q is a number field and denote by OK its ring of integers. Then the prime divisors of Spec(OK ) are defined by the prime ideals of OK . The above program works best if we impose a few extra conditions on X. Con- sider for instance the affine cuspidal curve C := Spec(k[x, y]/(y2 − x3)). As we have seen, the local ring at the origin OC,o is not a DVR, yet x|C and y|C have a zero at o. So if we focus on local rings that are DVR’s such a zero goes undetected. The simplest way out is to assume that there are no such points. This we can do by stipulating every closed integral subscheme of X that is not an irreducible component of X is contained in a prime divisor. This is equivalent to: the closure of every x ∈ X is an irreducible component of X or is contained in a prime divisor D. This implies that OX,x is a field or contained in a DVR and this implies that OX,x has no zero divisors (hence is reduced). So X is reduced and every point of X is contained in exactly one irreducible component. In other words, each connected component of X is integral. It is then hardly a restriction to assume that X itself is integral, for we are then just asking that X be connected. Another natural condition to impose is that for every f ∈ K(X)× only a finite number valuations are nonzero on it. We will see that this is automatic when X is noetherian. So in the rest of this section we assume the DVR domination property: X is noetherian and every closed integral subscheme of X distinct from X is contained in a prime divisor (so that X is integral). All the schemes described in the examples in 9.9 have this property. Another interesting case is that of (the spectrum of) a Dedekind domain: this is a noetherian domain with the property that every maximal ideal is a a DVR. These occur in algebraic geometry as affine normal curves over a field F and in algebraic number theory as rings of integers in number fields. More generally, any integral noetherian scheme all of whose closed points are prime divisors satisfies these hypotheses. This includes the case of a nonsingular curve over a field.

EXERCISE 89. Prove that an irreducible variety over k of dimension n whose singular locus is of dimension ≤ n − 2 satisﬁes the DVR domination property.

PROPOSITION-DEFINITION 9.10. Let X be a scheme having the DVR domination property. × P (i) Then for every f ∈ K(X) , the sum (f) := P vP (f)P (with D running over the prime divisors) is finite and hence defines a Weil divisor. This defines a group homomorphism from K(X)× to the group of Weil divisors. The image of this homomorphism is called the group of principal Weil divisors. Its cokernel is called the divisor class group of X and will be denoted Cl(X). (ii) Any Cartier divisor H on X determines a Weil divisor (H) on X characterized by the property that if H is on an open affine U represented by f ∈ KX (U), then (H)|U = (f). This defines a homomorphism from the group of Cartier divisors to the group of Weil divisors, and this induces a monomorphism Pic(X) =∼ CaCl(X) ,→ Cl(X). 9. DIVISORS AND INVERTIBLE SHEAVES 121

PROOF. Since X is noetherian, it is for (i) enough to show that for any affine open U ⊂ X, (f)|U is finite. In other words, we may assume that X = Spec(R). We first assume that f is regular: f ∈ R. A prime divisor for U is given a by a prime ideal p with the property that Rp is a DVR. The only a prime ideal contained in such a p is the zero ideal. We have vp(f) > 0 if and only f ∈ p. In that case, p appears in the primary decomposition of p(f) and since this involves only a finite P number of primes, the sum (f) := D vD(f)D is finite. For an arbitrary nonzero f = f1/f0 ∈ Frac(R), (f) = (f1) − (f0) and so this sum is also finite. At the same it is clear that (fg) = (f) + (g). If the Cartier divisor H on X is on the affine open U represented by f and g, then f/g ∈ O(U)× and so (f) = (g). This implies that (H) is well-defined. This also implies that H 7→ (H) is a homomorphism of groups. If (H) is principal (as a Weil divisor), then H is principal (as a Cartier divisor), and so the induced ∼ homomorphism Pic(X) = CaCl(X) ,→ Cl(X) will be injective. REMARK 9.11. In order that Pic(X) =∼ CaCl(X) ,→ Cl(X) be also surjective, we want that every prime divisor P represents a Cartier divisor. This amounts to: the ideal sheaf IP defining P is locally principal. Then for every Weil divisor D, OX (D) is an invertible subsheaf of KX and this defines the inverse Cl(X) → Pic(X). This is so when the local rings of X are all UFD’s. An important example in algebraic number theory is the spectrum of the ring of integers of a number field K/Q (by abuse of language this is then often called the divisor class group of K). A theorem of Auslander-Buchsbaum asserts that every regular local ring is a UFD. This implies that for instance that for a nonsingular irreducible variety X, we have Pic(X) =∼ Cl(X) (but this is not hard to prove independently of the Auslander-Buchsbaum theorem).

REMARK 9.12. We elaborate on Remark 9.5 a bit. Suppose X satisfies the DVR domination property. Let L be an invertible sheaf on X and s ∈ Γ(X, L) a nonzero section. Then s defines a principal ideal sheaf IZ(s) ⊂ OX and an isomorphism of −1 ∼ L onto the invertible fractional ideal IZ(s) ⊂ KX . If U ⊂ X is open, φ : OU = L|U an isomorphism and f ∈ OX (U) such that s|U = fφ(1), then the Cartier divisor −1 associated to IZ(s) is on U represented by (U, f). The Weil divisor attached to this P Cartier divisor is then on U represented by P vP (f)P . Clearly, vP (f) ≥ 0 and can be understood as the order of vanishing of s along P . So the Weil divisor attached this situation and has then all its coefficients ≥ 0 and can be understood as the ‘zero divisor’ of the section s. We write (s) for that divisor. Since IZ(s) is the sheaf P of g ∈ OX with P vP (g)P ≥ (s), (s) and Z(s) determine each other. This generalizes to the case of a ‘rational section’ of L, that is, a nonzero element s of the generic fiber Lη: then s determines an embedding L ,→ KX such that s ∈ Lη maps to 1 ∈ KX,η = K(X). The Weil divisor of the image (an invertible fractional ideal) is the divisor (s) of s.

EXAMPLES 9.13. We revisit some of the Examples 9.9 (so K is a field). n n (a) Since K[x1, . . . , xn] is a UFD, we have Pic(AK ) = Cl(AK ) = 0. n (b) A prime divisor P on Pk is given by a nonzero homogeneous irreducible × polynomial F ∈ K[T0,...,Tn]. This polynomial is unique up a scalar in K and so has has a well-defined degree, n, say. One such prime divisor is the hyperplane n n n H in PK defined by the ideal T0. Now F/T0 is a rational function on PK and its n n divisor is P − nH. It follows that Pic(PK ) = Cl(PK ) is generated by the class of 122 3. SCHEMES

H. On the other hand, for n > 0, nH is not principal (otherwise we would get a n rational function on PK without poles and we know that such a function must be n ∼ constant). It follows that the ‘degree’ defines an isomorphism Pic(PK ) = Z. (c) We now treat of the cuspidal curve C := Spec(K[x, y]/(x3 − x2). This curve ˆ 1 ∗ 2 ∗ 3 is parametrized by the morphism f : C := AK → C, f x = t , f y = t and thus 2 3 identifies Γ(C, OC ) with the subalgebra K[t , t ] of K[t]. This is the algebra of all the polynomials in t without linear term. Similarly, if we regard OC,o as a subring OC,oˆ , then the linear terms ct form a supplementary one-dimensional subspace and any uˆ ∈ O× is uniquely written u(1 + λt) with u ∈ O× and λ ∈ K. In fact, the C,oˆ C,o map uˆ 7→ λ defines a group homomorphism O× → (K) with kernel O× . C,oˆ Ga C,o Now let L be an invertible sheaf on C. Then f ∗L is trivial and so admits a nowhere zero section sˆ. Note that sˆ is unique up to scalar and that Γ(C,ˆ f ∗L) = K[t]ˆs. The latter contains Γ(C, L) as a K[t2, t3]-submodule. A generator s for the ∗ free rank one OC,o-module Lo will also be a generator for f L as a O 1 -module AK ,0 × and so will be of the form uˆsˆ with uˆ ∈ O 1 . Since s is unique up to an element of AK ,0 × × OC,o, the image of uˆ in OC,oˆ /OC,o only depends on L. We identified this quotient × with Ga(K ). Notice that (1 + λt)ˆs can then be understood as a section of L which has a zero only in f(−λ−1) (assuming λ 6= 0).

This gives us a complete invariant of L. If λ ∈ K −{0}, then we let Lλ ⊂ OCˆ be −1 the OC -submodule which over Cˆ − {−λ } is generated by 1 + λt and over C − {o} is generated by 1. This is clearly an invertible OC -module. We have thus obtained ∼ a group isomorphism Ga(K) = Pic(C). (d) We compute the Picard variety of the nodal curve C over a field K defined by Spec(K[x, y]/((x − y)3 + xy)) (the node is at o = (0, 0)). Put Cˆ := P1 − {1}. Then f : Cˆ → C, f(t) := (t(t − 1)−3, t2(t − 1)−3) parametrizes C. Notice that ∗ f(0) = f(∞) = o. Thus f identifies Γ(C, OC ) with the subalgebra of K[t][1/(t−1)] of functions which take the same value in t = 0 and t = ∞. Now let L be an ˆ ∼ 1 ∗ invertible sheaf on C. Since C = AK has trivial Picard group, f L is trivial and so admits a nowhere zero section sˆ. Because sˆ(0) and sˆ(∞) are both generators of the × one-dimensional K-vector space L(o) = κ(o) ⊗OC,o Lo there exists a λ ∈ K with sˆ(∞) = λsˆ(0). Since s := (λt − 1)/(t − 1)ˆs has the property that s(0) =s ˆ(0) and s(∞) =s ˆ(∞) we find that s defines a section of L. It has a zero in t = 1/λ. This ˆ identifies L with the sub OC -module of OCˆ which over C − {1/λ} is generated by (λt − 1)/(t − 1) and over Cˆ − {0, ∞} is generated by 1. This identifies Pic(C) with the multiplicative group Gm(K).

Divisors on a nonsingular curve. Let C be a connected nonsingular curve over k. Then a Weil divisor D on C can be given as a formal linear combination P of closed points of C: D = p∈C(k) np(p) with np nonzero for only ﬁnitely many P p ∈ C(k). We then deﬁne the degree deg(D) of D as the sum p np. In order to better understand this notion we need the following proposition.

PROPOSITION-DEFINITION 9.14. Let π : C → C0 a nonconstant proper morphism of connected nonsingular curves over k. Then π∗OC is locally free of ﬁnite rank; this rank is called the degree of π and denoted deg(π). If we deﬁne the local degree 9. DIVISORS AND INVERTIBLE SHEAVES 123 degp(π) of π at p ∈ C as the k-dimension of OC,p/mC0,π(p)OC,p, then X deg(π) = degp(π). p∈π−1(p0)

PROOF. Since π is proper, π(C) is closed in C0. Since π(C) is connected, but not a singleton, it follows that π is onto and that any fiber over a closed point of C0 is finite. In particular, π is dominant: π∗ embeds k(C0) in k(C). For p ∈ C we 0 ∗ ∗ have OC,p ⊂ k(C), OC0,f(p) ⊂ k(C ) and π maps OC0,f(p) to OC,p. So π restricts ∗ to a monomorphism πp : OC0,f(p) ,→ OC,p. Since OC,p is an integral domain, this implies that OC,p is a torsion free OC0,f(p)-module. The fact that π is proper further implies that π∗OC is a coherent OC0 -module (we will prove a general result of this kind in 11.5). Let p0 ∈ C0 and choose a neigh- 0 0 0 borhood U of p and f1, . . . , fd ∈ Γ(U , π∗OC ) such that their images in the fiber 0 (π∗OC )(p ) = ⊕p∈π−1(p0)OC,p/mC0,p0 OC,p make up a k-basis. Then their images in the stalk (π∗OC )p0 = ⊕p∈π−1(p0)OC,p generate that stalk as a OC0,p0 -module by Nakayama’s Lemma. As we have seen, each summand of this stalk is torsion free. Since a torsion free finitely generated module over a DVR is free, (f1, . . . , fd) will map to a OC0,p0 -basis of (π∗OC )p0 . So the homomorphism of coherent OU 0 -modules d 0 0 OU 0 → π∗OC |U defined by the f1, . . . , fd has its kernel and cokernel zero in p and 0 hence is an isomorphism on a neighborhood of p . This proves that π∗OC is locally free. The identity is obtained by taking dimensions. Now suppose that in the above situation we are given a Weil divisor D0 = P 0 0 0 0 p0 np0 (p ) on C . Associated to D is the invertible sheaf OC (D ) of f ∈ KC0 0 ∗ 0 with vp(f) ≥ np0 for all p . Now π OC (D ) is also an invertible sheaf and as a straightforward local computation shows, it comes with a divisor denoted π∗D0 P 0 that it is equal to p∈C(k) degp(π)nf(p). So if we define the degree of the divisor D 0 P ∗ 0 0 by deg(D ) = p0∈C0(k) np0 , then the degree of π D is equal to deg(π) deg(D ).

PROPOSITION 9.15. If C is a connected nonsingular curve proper over k, then the degree of a Weil divisor on C only depends on its equivalence class; in fact the degree factors through a surjective group homomorphism deg : Pic(C) → Z. ∼ 1 Moreover the map p ∈ C(k) 7→ (p) ∈ Pic(C) is injective, unless C = Pk. PROOF. For the first assertion we must show that a principal divisor D on C has zero degree. By definition D is given by a nonzero f ∈ K(C). We can think of ˜ 1 such an element as defining a morphism f : C → Pk. To be precise, let U0 resp. U1 be the open subset where f resp. 1/f is regular (a closed point p ∈ C(k) is in U0 ˜ ˜ resp. U1 if and only if vp(f) is ≥ 0 resp. ≤ 0). Then f|U0 resp. f|U1 is defined by ˜∗ ˜∗ ˜ f (T1/T0) = f resp. f (T0/T1) = 1/f. The morphism f is proper (for C is proper over k and nonconstant so that Proposition 9.14 applies. Since D is obtained as f˜∗ applied tot the degree zero divisor (0) − (∞)) is has degree zero. For the second assertion we must show that if p, q ∈ C(k) are distinct and ∼ 1 (p) − (q) is principal, then C = Pk. For such a divisor (p) − (q) the preceding ˜ 1 ˜ yields a morphism f : C → P := Pk of degree one. This means that f is an isomorphism. REMARK 9.16. It is clear that the image of C(k) in Pic(C) will generate Pic(C) as a group. If we let Pic0(C) denote the kernel of deg : Pic(C) → Z, then it is also easy to see that the image of (p, q) ∈ C(k) × C(k) 7→ (p) − (q) ∈ Pic0(C) generates 124 3. SCHEMES the latter as a group. When k = C, C determines a compact connected Riemann surface, of genus g say, and then Pic0(C) can be identified with the Jacobian of C. This is an abelian variety of dimension g: as a complex manifold it is a copy of the real torus of dimension 2g, endowed with a translation invariant complex structure, i.e., of the form Cg/L, where L ⊂ Cg is the Z-span of some R-basis of Cg and admits an embedding a complex projective space. One can show that something like this is true in general: the abelian group Pic0(C) is naturally the set of k-valued points of a connected projective group scheme over k that is characterized by the property that C(k) × C(k) → Pic0(C) comes from a morphism of schemes.

10. The projective setting ∞ Projective sheafification. Let A = ⊕k=0Ak be grading ring and write R for its subring A0. We recall that the proj construction associates to A a scheme Proj(A) (which we shall here often abbreviate by P ) over Spec(R). The points of P are in + bijection with the homogeneous prime ideals p strictly contained in A . For s ∈ Ad (d > 0), Z(s) ⊂ P denotes the set of points associated to homogeneous prime ideals containing s. Then the points of Ps = P − Z(s) are also associated to the k homogeneous prime ideals of A[1/s]0 = ∪k≥0Adk/s (this is an increasing union, k k+1 k+1 for Adk/s = sAdk/s ⊂ Ad(k+1)/s ) and a scheme structure on P is characterized by the property that any such Ps is open and identified with Spec(A[1/s]0). The stalk of OP at the point associated to p is then (Ap)0.

REMARKS 10.1. This notion leads us to make a number of simple observations.

First, if {si ∈ Adi }i is a set of homogeneous elements of positive degree (di > 0) which generates A as a R-algebra, then the associated open subsets {Psi }i cover P : if p ( A+ is a strict homogeneous prime ideal, then there exists a s ∈ A+ such that s∈ / p and since s can be written as a polynomial in the si’s, there must exist a si such that si ∈/ p and so the point of P deﬁned by p lies in Psi . k km Let us also note that for every positive integer m, Adk/s ⊂ Adkm/s , and so ∞ Proj(A) does not change if we replace A by its subalgebra A(m) := ⊕k=0Amk. Finally, a surjective homomorphism φ : A → B of graded algebras gives rise to a closed immersion Proj(B) → Proj(A). To see this, let s ∈ Ad, d > 0 be as above. Then φ induces a graded ring epimorphism A[1/s] → B[1/φ(s)]. This restricts to a ring epimorphism A[1/s]0 → B[1/φ(s)]0, which amounts to a closed immersion of afﬁne schemes Proj(B)φ(s) → Proj(A)s. Since the Proj(B)φ(s) cover Proj(B), our claim follows. A graded A-module gives rise to a quasi-coherent sheaf on P :

PROPOSITION-DEFINITION 10.2. A graded A-module M defines a quasi-coherent sheaf MP on P characterized by the property that for any d ≥ 0 and s ∈ Ad, k Γ(Ps, MP ) = M[1/s]0 = ∪k∈ZMdk/s . It has the property that the stalk of MP + at the point associated to a strict homogeneous prime ideal p ( A equals (Mp)0. We call this the projective sheafification of M. The sheaf is coherent when A is noetherian and M is finitely generated.

PROOF. The proof is straightforward.

Notice that any element of M0 gives rise to a section of MP : we have an obvious R-homomorphism M0 → Γ(P, MP ). 10. THE PROJECTIVE SETTING 125

In general this is neither injective nor surjective. For instance, if we take M = + A/A , then M = M0. For every s ∈ Ad, d > 0, we have sM = 0 and so M[1/s] = 0. Hence MP = 0. If we take A = R[T ] (with T of degree 1 and R in degree 0) and M = (T ), then M0 = 0, but it is not hard to check that Γ(P, MP ) = Γ(P, OP ) = R. The reason is that the passage from Spec(A) to Proj(A) goes via the union of the Spec(A[1/s]) taken over all homogeneous s ∈ A+. This union is equal to Spec(A) − {o}, where o is the ‘vertex point’ defined by A+, and hence ignores any possible uggliness of Spec(A) at o. Similarly, MP only depends on the sheafification of M restricted to Spec(A) − {o}. Given a graded A-module M and an integer n ∈ Z, then we have defined a new graded A-module M(n) by shifting degrees: M(n)k := Mk+n. The associated OP -module M(n)P is characterized by the property that for any d ≥ 0 and s ∈ Ad, it assigns to Ps the A[1/s]0-module k M[1/s]n = ∪k∈ZMdk+n/s so that MP (n)xp = (Mp)n.

We have of course a natural R-homomorphism Mn → Γ(P, M(n)P ). So if we form the quasi-coherent OP -module M(•)P := ⊕n∈ZM(n)P , then we have natural homomorphism M → ⊕n∈ZΓ(P, M(n)P ) = Γ(P, M(•)P ) of graded R-modules. If we apply this to A we find an associated quasi-coherent sheaf O(n)P and a natural graded homomorphism A → Γ(P, O(•)P ). But there is more structure present, for it is immediate from the definitions that multiplication defines for every pair n, r ∈ Z a sheaf homomorphism

O(n)P ⊗OP M(r)P → M(n + r)P . so that O(•)P is in fact a sheaf of graded OP -algebras, and M(•)P is a graded module over this sheaf. In particular, Γ(P, M(•)P ) = ⊕n∈ZΓ(P, M(n)P ) is a graded ⊕n∈ZΓ(P, O(n)P )-module. This is compatible with the natural graded maps

A → Γ(P, O(•)P ),M → Γ(P, M(•)P ) in the sense that the former is a ring homomorphism via which the latter becomes an A-module homomorphism.

PROPOSITION 10.3. Suppose A is generated as a R-algebra by A1. Then for every n and r, the OP -module O(n)P is invertible and the OP -module homomorphism

O(n)P ⊗OP M(r)P → M(n + r)P is an isomorphism.

PROOF. Our assumption implies that P is covered by the open subsets Ps with s ∈ A1. So it suffices to show that O(n)P ⊗OP M(r)P → M(n + r)P is an isomorphism on an open subset Ps with s ∈ A1. For r = 0 this amounts to the assertion n that multiplication by s defines an isomorphism of Γ(Ps, MP ) = M[1/s]0 onto Γ(Ps, M(n)P ) = M[1/s]n, which is obvious. If we apply this to MP = OP we find n ∼ that multiplication by s defines an isomorphism OP |Ps = O(n)P |Ps. The general case then follows immediately. Under the assumption of this proposition (i.e., A is generated as a R-algebra by A1), O(1)P is called the Serre twisting sheaf. Proposition 10.3 then allows us to define for any quasi-coherent OP -module F, its Serre twist F(n) := O(n)P ⊗P F (so that we can write MP (n) for M(n)P and in particular OP (n) for O(n)P ).

We can now also go in the opposite direction and form F(•) := ⊕n∈ZF(n) and F := Γ(P, F(•)) (so that Fn := Γ(P, F(n))). This is a graded module over the graded algebra Γ(P, OP (•)) and hence, via the graded ring homomorphism A → 126 3. SCHEMES

Γ(P, OP (•)), also one over the graded algebra A. This has an associated quasi- coherent sheaf FP . A natural sheaf homomorphism FP → F is then defined as m follows: for s ∈ A1, an element of Γ(Ps, FP ) can be given by a fraction a/s with n a ∈ Fm = Γ(P, F(m)) for some m with the understanding that b/s defines the m n same element if s b − s a is annihilated by some power of s. Now a|Ps is a section −m of F(m)|Ps and s |Ps is a section of OP (−m)|Ps and so their product defines a ∼ section of OP (−m) ⊗OP F(m) |Ps = F|Ps (by 10.3). This defines a natural map Γ(Ps, FP ) → Γ(Ps, F). Its sheafification defines sheaf homomorphism FP → F on Ps and it is easily checked that the sheaf homomorphisms thus defined agree on overlaps and thus produce a sheaf homomorphism FP → F.

PROPOSITION 10.4. Suppose A is generated as a R-algebra by a ﬁnite subset of A1. Then the OP -homomorphism FP → F is an isomorphism. In particular, every quasi-coherent sheaf on P comes from a graded A-module.

PROOF. Let s1, . . . , sr ∈ A1 generate A as a R-algebra. Then P = ∪Psi and so P is quasi-compact. Since both FP and F are quasi-coherent, it is enough to verify that for every s ∈ A1, the map Γ(Ps, FP ) = F [1/s]0 → Γ(Ps, F) is injective and surjective. The proof proceeds exactly as that of Step 2 resp. Step 3 of Proposition 7.3. We do this this only for injectivity and leave the surjectivity to you. An element of the kernel is represented by a fraction a/sn with a ∈ Γ(P, F(n)) and a|Ps = 0. We must show that then a is killed by some positive power of s. The image of a in Γ(Ps ∩ Psi , F(n)) = Γ(Psi , F(n))[1/s] is zero and hence the image of mi a in Γ(Psi , F(n)) is killed by s for some mi ≥ 0. So if we take m = maxi mi, then m m s a|Psi = 0 for all i. This means that s a = 0.

PROPOSITION 10.5. Let A be the graded R-algebra R[T0,...,Tr] (with deg(Ti) = r 1) so that P = PR. Then the R-algebra homomorphism A → Γ(P, OP (•)) is an ∼ r isomorphism. (So for n ≥ 0, An = Γ(PR, OP (n)) is a free R-module with basis the r monomials in T0,...,Tr of degree n, whereas Γ(PR, OP (n)) = 0 for n < 0.)

PROOF. For r = 0 there is nothing so show and so we assume r > 0. Let n−ni F ∈ Γ(P, OP (n)). Then F |PTi can be written as FiTi ∈ A[1/Ti]n for some n−ni n−nj Fi ∈ Ani (so ni ≥ 0). Since FiTi and FjTj have the same restriction to

PTi ∩ PTj = PTiTj we ﬁnd that they are equal in A[1/TiTj]. Now assume i 6= j. The left hand side lies in A[1/Ti] and the right hand side in A[1/Tj]. Since n−ni n−nj A[1/Ti] ∩ A[1/Tj] = A it follows that FiTi = FjTj ∈ An. So n ≥ 0 and n−ni n−n0 FiTi is independent of i. Hence F is uniquely represented by F0T0 .

r Let us take R = k so that P = Pk. We observed that the isomorphism class of the invertible sheaf OP (1) generates the inﬁnite cyclic group Pic(P ). The other generator is then of course OP (−1) and can be distinguished from the former by the property that it has no nonzero sections. In particular, for any automorphism ∗ ∼ σ of P there exists an isomorphism σ OP (1) = OP (1) (which will be unique up to × × ∼ an element of Γ(P, OP ) = k ). The obvious k-linear isomorphism Γ(P, OP (1)) = ∗ Γ(P, σ OP (1)) can then be composed with an isomorphism as above to produce ∗ a k-linear automorphism σ˜ of Γ(P, OP (1)) that is unique up multiplication by a scalar. Proposition 10.5 shows that T0,...,Tr is a basis of Γ(P OP (1)). If we ∗ P then write σ˜ (Ti) = i,j σ˜i,jTj, then we recover σ as the projectivization of the 11. COHOMOLOGY OF COHERENT MODULES ON A PROJECTIVE SCHEME 127 adjoint of (˜σi,j) ∈ GLr+1(k). In particular, we ﬁnd that σ is given by an element of PGLr+1(k). We conclude:

r COROLLARY 10.6. The natural group homomorphism PGLr+1(k) → Autk(Pk) is an isomorphism. The preceding discussion also enables us to describe the closed subschemes of r P = PR in terms of A = R[T0,...,Tr] as projective schemes: a closed subscheme i : Y ⊂ P is by definition given by a quasi-coherent ideal sheaf I ⊂ OP so that OP /I = i∗OY . Since OP (n) is invertible, I(n) → OP (n) is an injection and so we may then form I := ⊕n≥1In with In := Γ(P, I(n)). Clearly In ⊂ Γ(P, OP (n)) and by Proposition 10.5 the latter is just An. It follows that I is a strict homogeneous ideal of A. According to Proposition 10.4 we recover I from I by means of projective sheafification. Let us now form the graded R-algebra B := A/I (so that we 0 have B0 = R) and put Y := Proj(B). If T¯j denotes the image of Tj in B1, then 0 −1 Γ(Y , OY 0 ) is by definition B[T¯ ]0. Since localization is exact, we can also ob- T¯j j −1 −1 tain this by taking A[Tj ]0 and reduce it modulo the ideal I[Tj ]0. But the former is Γ(PTj , OP ) and the latter is Γ(PTj , I) and so the quotient is Γ(PTj , i∗OY ) (for

PTj is affine). Since the PTj cover P , we conclude that Y may be identified with Proj(B) and that via this isomorphism the sheaf OP (n)/I(n) can be understood as the Serre twist OY (n). But we are not claiming that Γ(Y, OY (n)) can be identified with Bn = An/In: the exact sequence 1 0 → Γ(P, I(n)) → Γ(P, OP (n)) → Γ(Y, OY (n)) → H (P, I(n)) → ... merely tells us that we have an injection Bn ,→ Γ(Y, OY (n)). We record part of this discussion as

r PROPOSITION 10.7. Every closed subscheme Y of PR is given by a homogeneous + ideal in I ⊂ R[T0,...,Tr] (so that Y can be identiﬁed with Proj(R[T0,...,Tr]/I)). In particular, Y is projective over R.

11. Cohomology of coherent modules on a projective scheme r Let R be a noetherian ring, r an integer ≥ 1 and and put P := PR. We have 0 seen that H (P, OP (n) is zero for n < 0 and equal to R[T0,...,Tr]n for n > 0. The following proposition gives all the cohomology of each OP (n). r PROPOSITION 11.1. The R-module H (P, OP (−r −1)) is free of rank one and the natural homomorphism of R-modules r r HomOP (OP (n), OP (−r − 1)) → HomR(H (P, OP (n), H (P, OP (−r − 1)) r which assigns to φ ∈ Hom(OP (n), OP (−r − 1)) the induced R-homomorphism H (φ) r is an isomorphism (hence the choice of a generator of H (P, OP (−r − 1)) identi- r ∼ ﬁes HomR(H (P, OP (n)),R) with the R-module HomOP (OP (n), OP (−r − 1)) = 0 ∼ H (P, OP (−n − r − 1) = R[T0,...,T − r]−n−r−1). The cohomology of OP (n) is zero in degrees 6= 0, r. We will see that the proof will at the same time show that

k r+1 k H ( − {0}, O r+1 ) = ⊕n∈ H (P, OP (n)). AR AR −{0} Z 128 3. SCHEMES

PROOF. Note that OP (n) is the degree n-part of the graded OP -algebra A := ˇ ⊕n∈ZOP (n). It is clear that F is quasi-coherent. We make explicit the Cech complex r for F relative to the standard open covering U := {PTi }i=0: Γ(PTi , A) can as a graded R-module be identiﬁed with the localization R[T0,...,Tr][1/Ti]. Likewise, if 0 ≤ i < i < ··· < i ≤ r, then Γ(P ∩ · · · ∩ P , A) = Γ(P , A) = 0 1 k Ti0 Tik Ti0 ···Tik

R[Ti0 ,...,Tr][1/(T0 ··· Tik )] and so k C (U, A) = ⊕0≤i0

Tr · 0 → R[T0,...,Tr] −−→ R[T0,...,Tr] → R[T0,...,Tr−1] and ends with Tr · Dr −−→ Dr → Dr−1 → 0. It is clear from the definitions that the last map of the former is onto and the first map of the latter is into. So if we take 0 < k < r, then it follows that k ∼ k multiplication by Tr induces an isomorphism H (P, A(−1)) = H (P, A). If we ignore the grading, then A(−1) = A. So this implies that Hk(P, A) is in fact a k k R[T0,...,Tr][1/Tr]-module so that H (P, A) = H (P, A)[1/Tr]. But the latter is k just H (PTr , A) (localization is exact) and hence zero, since is PTr is affine. 11. COHOMOLOGY OF COHERENT MODULES ON A PROJECTIVE SCHEME 129

r THEOREM 11.2 (Serre). Let R be a noetherian ring, X ⊂ PR a closed subscheme and F a coherent OX -module. Then

(i) for n sufficiently large, F(n) is generated as a OX -module by its global sections, k (ii) for every k and n ∈ Z, H (X, F(n)) is a finitely generated R-module. It is zero when k 6= 0 and n is sufficiently large. We first make the simple observation that there is no loss in assuming that X r coincides with P := PR. The reason is that since i : X ⊂ P is closed immersion, i∗F will be a coherent OP -module and (i∗F)(n) = i∗(F(n)). Furthermore, i i H (X, F(n)) = H (P, i∗F(n)) for all i (this is because i∗ applied to flasque sheaf on X is a flasque sheaf on P ). We first prove part (ii), derive a Corollary and then continue with the proof of the other assertions.

PROOF OF PART (I). Since F is coherent and PTi is afﬁne, F|PTi is the sheaﬁ-

fication of finitely generated Γ(PTi , OP )-module. In particular, F|PTi is generated as a Γ(PTi , OP )-module by a finite number of its sections, {sij}j, say. The usual nij argument shows that there exists an integer nij ≥ 0 such that Ti sij extends to an element s˜ij ∈ Γ(P, F(nij)). Let n0 := max nij. For n ≥ n0, the collection n−nij {Ti s˜ij ∈ Γ(P, F(n))}i,j has the property that for i = 0, . . . , r, its restriction to

PTi generates F(n)|PTi . Hence this collection generates F(n).

r COROLLARY 11.3. For every coherent sheaf F on P = PR with R noetherian there ` exists a surjective OP -homomorphism OP (−n) F for certain n ≥ 0 and ` ≥ 0. In particular, F admits a left resolution P• → F by such OP -modules.

PROOF. According to Assertions (i) (case i = 0) and (ii), there exists a positive integer n such that F(n) is generated as a OP -module by a ﬁnite number of global sections, say by s1, . . . , s`. This means that the the OP -homomorphism ` P (f1, . . . , f`) ∈ OP 7→ i fisi ∈ F(n) is onto. If we twist this with OP (−n) we get a ` OP -homomorphism OP (−n) → F that will still be surjective.

COROLLARY 11.4. Let X be a projective scheme over a noetherian ring R. Then every coherent OX -module F admits a left resolution P• → F by locally free OX - modules.

r PROOF. Let i : X → P = PR be closed submersion. Then there exists an ∗ epimorphism P˜0 → i∗F with P˜0 locally free. This induces an epimorphism i P˜0 → ∗ ∗ F and i P˜0 is locally free as a OX -module. Take P0 := i P˜0 and continue with ker(P0 → F).

PROOF OF PART (II) OF THEOREM 11.2. By Theorem 11.1, the assertions hold for any F of the form OP (m) and hence also feo any F that is a ﬁnite direct sum k of such modules. The fact that H (P, F(n)) = 0 for all n ∈ Z and all k > r, allows us to prove the corollary with downward induction on k. According to Corollary 11.3 there exists an epimorphism φ : P F, with P a ﬁnite direct sum of copies of O(m)’s. Then K := ker(φ) is coherent OP -module. The kernel of P(n) F(n) is K(n) and we have an associated long exact sequence ··· Hk(P, K(n)) → Hk(P, P(n)) → Hk(P, F(n)) → Hk+1(P, K(n)) → · · · 130 3. SCHEMES

Our induction hypothesis says that for l > k, Hl(P, K(n)) is a finitely generated R- module that is zero for n sufficiently large. The exact sequence above then shows k that H (P, P(n)) has the same property. Here is an important consequence of part (ii) (case i = 0) of Theorem 11.2 and Chow’s lemma: COROLLARY 11.5. Let f : X → S be a proper morphism of noetherian schemes. Then f∗ maps Coh(X) to Coh(S). PROOF. According to Chow’s lemma 7.6 there exists a projective modification 0 ˜ π : X → X such that f := fπ is projective. Let F be a coherent OX -module. We ∗ have a natural in adjoint homomorphism F → π∗π F. Since π is surjective, this ∗ ∗ homomorphism is injective and so f∗F → f∗π∗π F = (πf)∗π F is also injective. ∗ But π F is a coherent OX0 -module and Theorem 11.2-(i) tells us that that for every affine open subset V = Spec(R) ⊂ S, Γ(f˜−1V, π∗F) is a finitely generated R-module. It contains Γ(f −1V, F) as a R-submodule and since R is noetherian, this submodule must also be finitely generated. So f∗F is a coherent OS-module. Euler characteristics. Serre’s theorem allows us to define the Euler characteristic for a coherent sheaf F on a projective scheme X over k as X k χ(F) := dimk H (X, F). i r This is indeed a finite sum. If i : X,→ Pk is a closed immersion, then i∗F is a coherent O r -module with the same cohomology as F and so χ(F) = χ(i F). It is Pk ∗ therefore here no restriction to assume that X is a projective space.

EXAMPLE 11.6. Proposition 11.1 enables to compute χ(O r (n)). According to Pk this proposition, O r (n) has for n ≥ 0 only cohomology in degree zero and that Pk r+n cohomology is then k[T ,...,T ] . So χ(O r (n)) = when n > 0. When 0 r n Pk r n ≤ −r − 1, O r (n)) has only cohomology in degree r and this is k-dual to Pk r−n−1 n+r k[T ,...,T ] . Hence χ(O r (n)) = (−1) = when n ≤ −r − 1. 0 r −n−r−1 Pk r r For −r ≤ n ≤ −1, O r (n) has no cohomology and so χ(O r (n)) = 0, which also Pk Pk n+r n+r happens to be . We thus ﬁnd that χ(O r (n)) = for all n ∈ . r Pk r Z A short exact sequence 0 → F 0 → F → F 00 → 0 of coherent sheaves gives a long exact sequence for cohomology and from this we see that χ is additive in the sense that χ(F) = χ(F 0) + χ(F 00). The same argument shows that when 0 → Pn → Pn−1 → · · · → P0 → F is an exact sequence of coherent sheaves, then P k χ(F) = k(−1) χ(Pk). If X is a projective space over k, then by Corollary 11.3 we can always take the Pk to be locally free (we will later see that we then may take Pk = 0 for k > dim X).

These Euler characteristics enter in the famous Hirzebruch-Riemann-Roch theorem. They also lead to a somewhat different approach to the Hilbert-Serre polynomials introduced earlier. The support of a quasi-coherent sheaf F, supp(F), is the closure of the set of x ∈ P with Fx 6= 0. PROPOSITION 11.7. Let P be a projective space over k. Then for every coherent OP -module F, the function PF : n ∈ Z 7→ χ(F(n)) is given by a numerical polynomial in Q[T ] of degree ≤ dim(supp(F)), called the Hilbert polynomial of F. This is also the polynomial we associated in Theorem 10.5 to the graded module Γ(P, F(•)). 12. AMPLE AND VERY AMPLE SHEAVES 131

PROOF. We prove this with induction on dim(supp(F)). When F = 0, the support is empty and we get the zero polynomial. Let therefore F have nonempty support of dimension r and assume the proposition proved for coherent OP -modules whose support has dimension < r. Choose a hyperplane i : Q ⊂ P that is not contained in supp(F) unless supp(F) = P . This ensures that dim(supp(i∗F)) < r. The defining ideal IQ ⊂ IP can be identified with OP (−1), so that we a short exact sequence 0 → OP (−1) → OP → i∗OQ → 0. By the right exactness of the tensor ∗ product. Tensoring with F yields the exact sequence F(−1) → F → i∗i F → 0. The first map need not be injective, but its kernel K will have support of dimension < r. From the exact sequences

∗ 0 → K(n) → F(n − 1) → F(n) → i∗i F(n) → 0 we read off the formula PF (n) − PF (n − 1) = Pi∗F (n) − PK(n). Our induction hypotheses tell us that Pi∗F −PK is given by a numerical polynomial of degree < r. This implies that PF is given by a numerical polynomial of degree ≤ r. Part (ii) of Theorem 11.2 tells us that for n sufﬁciently large, we have χ(F(n)) = 0 dimk(H (P, P(n))). This implies the last assertion.

12. Ample and very ample sheaves

Let S be a scheme. We have seen that the homogeneous coordinates T0,...,Tr r of P = PS can be understood as sections of OP (1) and that they in fact generate OP (1). So if X is a scheme and f : X → P is morphism, then Ti gives rise to ∗ ∗ a section f Ti of invertible sheaf f OP (1). Since T0,...,Tr generate OP (1) as ∗ ∗ ∗ a OP -module, f T0, . . . , f Tr generate f OP (1) as a OX -module. The following proposition shows that these sections completely determine f once X is given as an S-scheme:

PROPOSITION 12.1. Let X/S be a S-scheme, L an invertible OX -module that comes with r + 1 generating sections s0, . . . , sr. Then we have a morphism f : X → r ∗ ∼ ∗ PS =: P and an isomorphism F : f OP (1) = L which takes f Ti to si. The pair (f, F ) is unique.

An S-morphism Ui → PTi is given by a homomorphism R[T0/Ti,...,Tn/Ti] →

OX (Ui) of R-algebras. We define fi : Ui → PTi to be the morphism that sends i ∗ Tj/Ti to fj . Since si|Ui is a generating section of L|Ui and f Ti|Ui is a generating ∗ section of f O(1) we have defined over Ui a unique isomorphism of trivialized ∗ ∼ ∗ invertible sheaves Fi : f OP (1)|Ui = L|Ui that takes f Ti|Ui to si|Ui and hence ∗ i ∗ i f Tj|Ui = fj f Ti to fj si|Ui = sj|Ui. The pairs (fi,Fi) and (fj,Fj) coincide on Ui ∩ Uj and hence such pairs define a pair (f, F ) as asserted. The uniqueness assertion is left to you.

The following proposition addresses the question of when the f found in Propo- sition 12.1 is a closed immersion. We state and prove this only in case X is proper over S = Spec(k), although the proposition below holds in a setting where X and S are noetherian and f is proper. 132 3. SCHEMES

PROPOSITION 12.2. In the situation of the previous proposition, assume that R = k and that X is proper over k. Then f is a closed immersion (so that X is in fact projective over k) if and only if: Two-point separation: For any two distinct closed points x, x0 ∈ X the im- 0 6 ages of the si in L(x) ⊕ L(x ) span the latter as a k-vector space. Infinitesimal separation: For every closed point x ∈ X, the images of the si 2 in L(2x) := LX,x/mX,xLX,x span the latter as a k-vector space. PROOF. We prove that these properties imply that f is a closed immersion, the converse is left as an exercise. The separation property implies that f is injective on closed points and hence is injective. Since f is a morphism between proper k- schemes, f is also proper and hence is closed. It follows that f is a homeomorphism ∗ onto its image. So it remains to see that fx : OP,f(x) → f∗OX,x is onto for all x ∈ X. Again its suffices to check this for closed points only (why?). By corollary 11.5, f∗OX is a coherent OP -module so that for every x ∈ X, ∗ fx : OP,f(x) → OX,x is a homomorphism of finitely generated OP,f(x)-modules. By ∗ ∗ means of a local generator of Lx we can identify fx with Fx : OP (1)f(x) → Lx. We ∗ 0 0 have si,x = Fx (Ti) and so if si ∈ OX,x corresponds to si,x, then si is in the image of ∗ fx . Our assumption regarding infinitesimal separation amounts to the projections 0 2 of the si in OX,x/mX,x spanning the latter over k. Lemma 12.3 below then implies ∗ that fx : OP,f(x) → OX,x is onto. LEMMA 12.3. Let φ : A → B a local homomorphism of noetherian rings that is finite (i.e., is such that B is a finitely generated A-module). If the composite of φ with 2 B → B/mB is onto, then so is φ.

PROOF. By assumption the ideal in B generated by φ(mA) = φ(A) ∩ mB maps 2 surjectively to mB/mB. Nakayama’s lemma applied to the B-module mB then tells us that this ideal is all of mB. It follows that B = φ(A) + φ(mA)B, in other words, the A-module Coker(φ) is annihilated by mA. Since Coker(φ) is a ﬁnite generated A-module, another application of Nakayama’s lemma implies that it is trivial. The preceding discussion leads us to make the following deﬁnition.

DEFINITION 12.4. Let f : X → S be a morphism of schemes. An invertible sheaf L on X is said to be very ample relative to f if we can cover S by afﬁne open −1 r subsets V such that there exists an r ≥ 0 and a V -immersion i : f V,→ PV and ∗ −1 an isomorphism i O r (1) ∼ L|f V . PV = We have the following companion notion.

DEFINITION 12.5. We say that an invertible sheaf L on a noetherian scheme X is ample if for every coherent OX -module F, there exists an integer n0 such that n F ⊗ L is generated by its global sections for all n ≥ n0. REMARKS 12.6. This notion is empty when X is afﬁne, for then even every quasi-coherent OX -module is generated by its global sections. We also note that if Lm is ample for some positive integer m, then so is L: for given a quasi-coherent mn OX -module, then there exists an n0 such that F ⊗ L is generated by its global −k sections for n ≥ n0. Next we apply then this property to F ⊗L for k = 0, . . . , n0 − mn−k 1 and ﬁnd that there exist an nk such that F ⊗ L is generated by its global

6 L(x) stands for the ﬁber of L over x, i.e., the 1-dimensional κ(x)-vector space LX,x/mX,xLX,x. 12. AMPLE AND VERY AMPLE SHEAVES 133

n sections for n ≥ nk. So for n ≥ m maxk nk, F ⊗ L is generated by its global sections.

PROPOSITION 12.7. Let X be a scheme of ﬁnite type over a noetherian ring R. Then an invertible OX -module L is ample if and only if some positive tensor power Lm is very ample over Spec(R).

PROOF. Suppose L is is ample. We first show that there exists a positive integer n n and a finitely many sections s1, . . . , sr of L such that {X − Z(si)}i is a finite covering of X by affine open subsets. ∼ Given x ∈ X, let φ : L|U = OU be a trivialization of L on an open affine neighborhood of x in X. Put Y := X −U (a closed subscheme). By our assumption, nx there exists an nx > 0 such that IY ⊗ L is generated by its sections and so there nx nx exists an s˜ ∈ Γ(X, IY ⊗ L ) with x∈ / Z(˜s). The image s of s˜ in Γ(X, L ) has then the property that x ∈ X − Z(s) ⊂ U. Since Z(s|U) is defined by the section of φ(s) of OU , its complement in the affine open U (which is X − Z(s)) is a basic affine open neighborhood of x in X. Since X is quasi-compact a finite number of these basic open subsets will cover X. The corresponding finitely many sections ni si ∈ Γ(X, L ) are in possibly different powers of L, but we make these equal by n/ni taking for n a common multiple of these and replacing si by si . Put Ui := X − Z(si). Since X is of finite type over R, OX (Ui) is a finitely generated R-algebra. Choose generators fi0, . . . , firi of OX (Ui) with fi0 = 1. Upon replacing n by a positive multiple if necessary, we then may assume that each fijsi n extends to a section sij of Γ(X, L ). Then ∩i,jZ(sij) = ∅ so that by Proposition 12.1 we have defined an R-morphism

f : X → P := Proj R[(Tij)ij] ∗ n ∼ ∗ and an isomorphism F : L = f OP (1) which takes sij to Tij. The preimage of ∗ the standard open set PTi0 is the affine open Ui and (f|Ui) takes Tij/Ti0 to fij. f Since the (fij)j generate O(Ui) as a R-algebra, it follows that Ui −→ PTi0 is a closed embedding. It follows that f maps X onto a closed subscheme of ∪iPTi0 (this union is open in P ). Now assume Lm very ample for some m > 0. By Remark 12.6 it suffices to show m r that L is ample. By definition there exists an R-immersion f : X → P := PR for m ∼ ∗ some r > 0 and an isomorphism L = f OP (1) and so we may as well assume that X is locally closed in P . If F is a coherent sheaf on X, then according to Corollary 7.10 it admits a coherent extension to the closure X. This in turn can be regarded as a coherent OP -module G. By part (ii) of Serre’s theorem, there exists an integer n0 such that G(n) is generated by its global sections for n ≥ n0. The same is is then ∗ ∼ m true for f G = L .

REMARK 12.8. Let C be a connected proper nonsingular curve. With the help of the Riemann-Roch theorem on can show that an invertible sheaf on C of positive degree is ample. In particular, for any p ∈ C(k), the invertible sheaf O(p) has ∼ 1 degree one and so is ample. But if O(p) is very ample, then we must have C = Pk. To see this, choose q ∈ C(k) − {p}. The separation property implies that there is a section of O(p) which is zero in q. In other words, its divisor has all its coefﬁcients ≥ 0 with the one in q positive. Since its degree is 1, this divisor must be just (q). So ∼ 1 (p) − (q) is principal. But according to Proposition 9.15 this implies that C = Pk. 134 3. SCHEMES

For an invertible module on a proper scheme over a noetherian base we have the following ampleness criterion.

PROPOSITION 12.9. Let X a proper scheme over a noetherian ring R and L an n invertible OX -module. If L is ample then for every coherent OX -module F, F ⊗L has no cohomology for n sufﬁciently large. Conversely, if for every coherent OX -module F, there exist an n > 0 such that H1(X, F ⊗ Ln) = 0, then L is ample.

PROOF. Suppose L is ample. Then Lm is very ample for some m > 0. As a r consequence, we can assume that X is a closed subscheme of PR for some r and m mn that L = OX (1). Then F(n) = F ⊗ L has no cohomology in nonzero degree for n ≥ n0, say. Similarly, we find for k = 1, . . . , mn0 − 1 an integer nk such that −k mn−k F ⊗ L (n) = F ⊗ L has no cohomology in nonzero degree for n ≥ nk. It n follows that F ⊗ L has no cohomology in nonzero degree for n ≥ mn0 maxk nk. 1 n Now suppose that for every coherent OX -module F, H (X, F ⊗ L ) = 0 for n sufficiently large. We must show that F ⊗ Ln is generated by its sections when n is sufficiently large. Let x ∈ X be a closed point and denote by i : {x} ⊂ X ∗ the inclusion. Then we have an epimorphism from F to the skyscraper sheaf i∗i F (its stalk at x is the finite dimensional κ(x)-vector space F(x) = κ(x) ⊗ F) whose kernel is IxF, where Ix ⊂ OX denotes the ideal defining x. We have for every integer n ≥ 0 a short exact sequence n n ∗ n 0 → IxF ⊗ L → F ⊗ L → i∗i F ⊗ L → 0.

By assumption, there exists a positive integer nx(F) such that this gives an exact sequence of sections on X. Since Fx is a ﬁnitely generated OX,x-module, Nakayama’s nx(F) nx(F) lemma then implies that Γ(X, F ⊗ L ) → Fx ⊗ Lx is onto. The coherence nx(F) of F implies that the global sections of F ⊗ L will generate this OX -module on a neighborhood Ux(F) of x. Put

nx(F)−1 \ −k U˜x(F) :=Ux(OX ) ∩ Ux(F ⊗ L ), k=0 nx(F)−1 −k n˜x(F) := max nx(OX ), max nx(F ⊗ L ) k=0 n and then observe that for n ≥ n˜x(F), the global sections of F ⊗ L generate that sheaf over U˜x(F). Since X is quasi-compact, the open covering {U˜x(F)}x∈X admits ˜ ` ` n a finite subcovering {Uxi (F)}i=1. Then for n ≥ maxi=1 n˜xi (F), F ⊗L is generated by its global sections. 13. Blowing up In this section we fix a noetherian scheme X. Suppose we are given a graded ∞ sheaf of OX -algebras A = ⊕d=0Ad with the property that A0 = OX , A1 is a coherent OX -module which locally generates A by as OX -algebra. In other words, X can be covered by a finite number open affines of the form U = Spec(R) such ∞ that R is noetherian and the R-algebra A(U) = ⊕d=0Γ(U, Ad) is as in Proposition 10.4: it is a graded R-algebra with R in degree zero, and generated as a graded R-algebra by a finite subset of the degree one summand. We then have defined the projective morphism Proj A(U) → U and on Proj A(U) an invertible Serre twisting 0 sheaf OProj A(U)(1) that is ample relative the projection. If U ⊂ U is a basic open subset, then the data we get for U 0 are obtained by the restricting the ones we 13. BLOWING UP 135 just defined and so these morphisms and invertible sheaves glue together to form a projective morphism that we naturally denote as Proj(A) → X plus an invertible Serre twisting sheaf OProj(A)(1). There is one particular case of interest and that is when E := A1 is locally free of rank r. Then we may form the graded symmetric OX -algebra Sym• E on E. The latter means that if U ⊂ X is open and T1,...,Tr is a basis of sections of E(U), then Sym• E(U) = E(U)[T1,...,Tr]. In more invariant terms, Symd E is the biggest quotient of the d-fold tensor power E ⊗OX E ⊗OX · · · ⊗OX E invariant under the r−1 permutation group Sd. Then Proj(Sym• E) → X is locally like PU → U: it is a projective space bundle that we denote by Pˇ(E) → X. So for r = 0, Pˇ(E) = ∅ and when r = 1, i.e., E invertible, Pˇ(E) → X is an isomorphism. Notice that Proj(A) ˇ can now be regarded as a closed subscheme of P(A1). To explain the notation: suppose X = Spec(k) so that E is just an (n + 1)- dimensional k-vector space. Then Sym• E is the algebra of regular functions on the k-dual E∨ of E and so Pˇ(E) is the projective space associated to E∨. This is also the projective space of hyperplanes in E. We will here focus on a special case, namely the ‘blowup’ of closed subscheme. Let us first do an example which in a way revisits the discussion of Chapter 2, section 4.

n n EXAMPLE 13.1. Let X := Ak and 1 ≤ r ≤ n−1. Denote by Y ⊂ Ak be the closed n−r subscheme defined by x1 = ··· = xr = 0 (so this a copy of Ak ). Consider the r−1 ˜ r−1 morphism X −Y → Pk defined by (x1, . . . , xn) 7→ [x1 : ··· : xr] and let X ⊂ PX r−1 be the closure of its graph. If we identify PX with Proj(Γ(X, OX )[T1, ··· ,Tr]), then X˜ is given by the homogeneous ideal in k[x1, . . . , xn][T1, ··· ,Tr] generated ˜ by xiTj − xjTi, 1 ≤ i < j ≤ r. Notice also that on XTr we have r − 1 regular r−1 ˜ functions (ξj := Tj/Tr)j=1 so that xj = ξjxr for j < r. We may thus identify XTr with the affine space Spec k[ξ1, . . . , ξr−1, xr, . . . , xn]. The projection X˜ → X is an r−1 isomorphism over X − Y , whereas a fiber over a closed point p ∈ Y is all of Pk . Pr ˜ The ideal sheaf I ⊂ OX defining Y is j=1 OX xj. But on XTr we have xj = ξjxr, ∗ ˜ and so π I|XTr it simply the principal ideal generated by xr. Notice that this ideal ˜ ˜ ˜ defines the preimage Y of Y in XTr . A similar property holds for XTi , i = 1, . . . , r, ˜ ∗ ∗ and as these cover X, it follows that π I ⊂ OX˜ is a principal ideal sheaf: π I is invertible and the subscheme Y˜ ⊂ X˜ is a prime divisor. The case that concerns us here is a far reaching generalization of this situation, where X is a noetherian scheme and Y is a closed subscheme of X. Then Y defines d (and is defined by) a coherent ideal sheaf I = IY and we take Ad = I ⊂ OX , the image of the d-fold tensor power I ⊗OX · · · ⊗OX I → OX . The preceding construc- ∞ d ∞ d tion applies to ⊕d=0I and produces a projective morphism Proj(⊕d=0I ) → X. Since Id ⊂ Id+1, it is often helpful to keep track of the degree by means of an extra ∞ d variable T so that ⊕d=0I can be regarded as a subalgebra of OX [T ]: ∞ d ∼ P∞ d d ⊕d=0 I = k=0 I T ⊂ OX [T ]. P∞ d d We then write BlY (X) for Proj( d=0 I T ) and call the morphism π : BlY (X) → d X the blowup of X along (or with center) Y . Notice that I |X − Y = OX−Y and P∞ d d so d=0 I T |X − Y = OX−Y [T ]. The projection Proj(OX−Y [T ]) → X − Y is an isomorphism and so π is an isomorphism over X−Y . This blowup has a remarkable 136 3. SCHEMES property with respect to the ideal sheaf I: recall that the OProj(A)-module OProj(A) P d d is associated to the graded A-module A(1). Since here A = d≥0 I T and A(1) = P d+1 d d≥0 I T , it is clear that A(1) = IA. This identity shows that OBlY (X)(1) may −1 be identified with the ideal pull-back OBlY (X)π I ⊂ OBlY (X) of I. In particular, this ideal pull-back is invertible. This property characterizes the blowup:

PROPOSITION 13.2. Let f : X0 → X be a morphism with X0 noetherian and let Y 0 := f −1Y be the closed subscheme deﬁned by the the ideal pull-back I0 := −1 ˜ OX0 f I ⊂ OX0 . Then f is uniquely covered by a morphism f : BlY 0 (X) → BlY (X). 0 0 Moreover, I is invertible if and only if f lifts to a morphism X → BlY (X). 0 PROOF. The homomorphism of OX -modules OX → f∗OX0 takes I to f∗I and so we have a homomorphism of OX -algebras d d 0d d 0d ⊕d≥0I T → ⊕d≥0f∗I T = f∗(⊕d≥0I ). ˜ 0 0 This induces a morphism between proj schemes: f : BlY 0 (X ) → BlY (X). If I 0 0 is invertible, then BlY 0 (X ) → X is an isomorphism, and so we then ﬁnd a lift 0 0 0 X → BlY (X) of f. Conversely, if such a lift X → BlY (X) exists, then I is ∗ invertible, as it is the pull-back of the invertible sheaf π I. The morphism f˜ obtained above is sometimes called the strict transform of f.

REMARK 13.3. If we replace I by Im for some positive integer m, then we get a closed subscheme Y (m) ⊂ X which contains Y , but according to Remark 10.1, BlY (m) (X) can be identified with BlY (X). 0 The same is true for a closed subscheme Y ⊂ Y ⊂ X for which IY 0 = LI ∼ for some invertible ideal L ⊂ OX : BlY 0 (X) = BlY (X). For if U = Spec(R) is an affine open subset of X on which L has a generating section s, and I|U is given by the ideal I ⊂ R, then I0 is given by I0 = sI. For every f ∈ I, we have n −n 0n −n ∪n≥0I f = ∪n≥0I (sf) and hence an isomorphism between the two blow- ups over U; this identification is evidently independent of the choice of s. This also leads to the observation that the obvious generalization of a blowup ∞ k k 0 to one of a fractional ideal I ⊂ KX as ProjX (⊕k=0I T ) → X (where I := OX ) hardly yields anything new: by definition we may cover X by affine open subsets U for which there exist a nonzero divisor f ∈ Γ(U, OX ) such that fI is an ideal in OU and if YU ⊂ U is the corresponding closed subscheme, then BlI (X)|U can be identified with BlYU U. In fact, if L ⊂ KX is an invertible submodule such that LI ⊂ OX and Y ⊂ X is the closed subscheme defined by LI, then we may identify BlI (X) with BlY (X). Such an L always exists in case X is quasi-projective and integral: then X supports an ample invertible OX -module O(1). Since I is a coherent OX -module, so is Hom(I, OX ). We have Hom(I, OX )x 6= 0 for all x ∈ X, because there exists a nonzero fx ∈ OX,x with fIx ⊂ OX,x. Following Part (ii) of Theorem 11.2 there exists an n such that Hom(I, OX )(n) has a global section which does not vanish on a nonempty open subset. Such a section amounts to a nonzero OX -homomorphism I(−n) → OX . Since KX has no zero divisors, this homomorphism must be injective and so L := OX (−n) will do.

EXAMPLE 13.4 (Tangent cone). Let X be a scheme and x ∈ X a closed point. We consider the blow-up of {x} in X. Since this does not affect X − {x}, we may just as well assume that X is the spectrum of a local ring (R, m) (so that R = OX,x). 13. BLOWING UP 137

P k k We denote by κ := R/m the residue field. Then Bl{x} X = Proj( k≥0 m T ) → Spec(R) (read R for m0). The fiber over x ∈ X is the Proj of the graded κ- P k k+1 k algebra k≥0(m /m )T . This defines a closed subscheme of the projective space Pˇ(m/m2). If X is (the germ) of a variety over k, then we recognize Pˇ(m/m2) as the projectivized Zariski tangent space of X at x. The closed subscheme defined P k k+1 k by Proj( k≥0(m /m )T ) is the called the projectivized tangent cone of X at x. It is the full projectivized Zariski tangent space precisely when X is nonsingular at P k k+1 k x. (The tangent cone of X at x is Spec( k≥0(m /m )T ); it is a homogeneous 2 subscheme of the Zariski tangent space Spec(Sym• m/m ).)

Here is a somewhat more concrete model for the blowup π : BlY (X) → X which also makes the connection with Example 13.1. Assume that X is affine, so X = Spec(R) with R noetherian. Then I is defined by a finitely generated ideal I ⊂ R. If f0, . . . , fr is a set of generators of I, then [f0 : ··· : fr] defines a morphism r r X − Y → Pk, but we prefer to view this as defining a section f of PX /X over r X − Y . The scheme theoretic image of f : X − Y → PX as defined in Corollary r 7.5 is a closed subscheme f(X − Y ) of PX whose ideal sheaf is the kernel of the O r -homomorphism O r → f O . PX PX ∗ X−Y ∼ PROPOSITION 13.5. There is a natural X-isomorphism φ : f(X − Y ) = BlY (X). In particular, if X − Y is irreducible, then so is BlY (X).

r PROOF. We deﬁne a closed immersion of BlY (X) in PX with image f(X − Y ).

Since the fi generate I, BlY (X)fi is covered by the affine open subsets BlY (X)fi and we will specify this immersion on BlY (X)fi for i = 0, . . . , r as a closed embed- r r ding in (PX )Ti with image f(X − Y ) ∩ (PX )Ti . r r Let us do this for i = 0: if (t1, . . . , tr) ∈ AX 7→ [1 : t1, . . . , tr] ∈ (PX )T0 is r the standard parametrization of a standard open subset of PX , then we have a ∗ homomorphism of R-algebras φ0 : R[t1, . . . , tn] → R[1/f0] which sends ti to fi/f0. ∗ n −n The image of φ0 is the subring ∪n≥0I f0 whose spec is just BlY (X)f0 , and so φ r defines a closed embedding of BlY (X)f0 in (PX )T0 . It is also clear that the ideal ∗ ker(φ0) is generated by fi − tif0 (i = 1, . . . , r). This ideal defines the subscheme r r f(X − Y ) ∩ (PX )T0 in (PX )T0 and hence this subscheme must be the image of φ0. Likewise for i = 1, . . . , r, and one checks that these morphisms agree on overlaps. To complete the proof that we get a closed immersion we still need to check r that the preimage of (PX )Ti is BlY (X)fi . This can be done directly, or by observing that the ideal pull-back of I to f(X − Y ) is invertible (so that by Proposition 13.2 the inverse is defined).

REMARK 13.6. The same argument works for the blowup of a fractional ideal I ⊂ KX : a set of OX -generators f0, . . . , fr ∈ Γ(X, I) determines a closed embed- r ding of BlI (X) in PX . Blowing up on a variety. Here we confine ourselves to the case when X is a k-variety in the sense of this chapter: i.e., a separated integral scheme of finite type over k. We let Y be a subscheme 6= X so that I 6= 0. Since X is irreducible, so is X − Y and hence is BlY (X) by Proposition 13.5. If U ⊂ X is affine and f ∈ Γ(U, I) k −k nonzero, then ∪k≥0I f is subring of the integral domain Γ(U, OX )[1/f] and hence an integral domain. It follows that BlY (X) is integral as well and that the 138 3. SCHEMES projection BlY (X) → X is birational. The following theorem shows that very often every birational projective morphism of varieties is thus obtained.

THEOREM 13.7. A variety X has the property that every birational, projective morphism f : X˜ → X is X-isomorphic to the blowup of a fractional ideal on X. When X is quasi-projective, we can assume this in fact to be an ideal sheaf (so that X˜ is the blowup of a closed subscheme of X). ˜ r PROOF. By assumption f factors through a closed X-embedding i : X → PX . We thus have defined a very ample invertible sheaf OX˜ (1) that comes with r + 1 sections T0,...,Tr defining the embedding. Moreover, f∗OX˜ (1) is a coherent OX - ˜ module. Since X is integral there exists a OX˜ -monomorphism OX˜ (1) ,→ KX˜ . This produces a OX -monomorphism f∗OX˜ (1) ,→ f∗KX˜ . Since f is birational, f∗KX˜ = KX . Thus, f∗OX˜ (1) is realized as a fractional ideal I ⊂ KX . Let U = Spec(R) be an affine open subset of X and let (f0, . . . , fk) be set of generating sections of I|U, with −1 k fi = Ti for i ≤ r. These define a morphism [f0, . . . , fk]: f U → PU whose image r r r r r lies in ∪i=0(PU )Ti and whose composite with the projection ∪i=0(PU )Ti → PX is −1 k just i. So f U → PU is a closed U-immersion as well. According to Proposition 13.5 and its subsequent Remark 13.6 this also models the blowup of I over U. If X is quasi-projective, then following Remark 13.3 there exists an invertible subsheaf L ⊂ OX such that LI ⊂ OX . Now use that BlLI (X) = BlI (X).

14. Flatness We first recall the definition of a flat module. Given a ring R , then for any R-module M, the functor ⊗RM : ModR → ModR is a left exact, but not in general exact. If it is, then we say that M is a flat R-module. The jth derived functor of R ⊗RM is denoted by Torj (M, −) and so this can be expressed by saying that these functors are zero for j > 0. The flatness property can be made more concrete: it amounts to saying that any Pn R-linear relation among elements m1, . . . mn of M: i=1 rimi = 0 is necessarily of the type that we would have if M were a free R-module: we then can write P 0 P mi = j rijmj and we have i ririj for all j. In particular, a free R-module is flat. But a flat R-module need not be free. For instance, if R is a domain, then its field of fractions Frac(R) is rarely free over R, but is a flat R-module. Here are some basic properties:

Flatness is local: M is a flat R module if and only if every localization Mp at a prime ideal p is a flat Rp-module. Transitivity of flatness: Let R → R0 be a ring homomorphism which makes R0 a flat R-module. Then any flat R0 module is also flat as a R0-module. Base change: Let R → R0 be a ring homomorphism. If M is a flat R mod- 0 0 ule, then R ⊗R M is a flat R -module. Flatness over a noetherian ring: Let M be a finitely generated module over a noetherian ring R. Then: M is flat ⇔ M is locally free ⇔ the function x ∈ Spec(R) 7→ dimκ(x) κ(x) ⊗ M is locally constant. Flatness preserved by kernels: The kernel of an R-homomophism between flat R-modules is flat. Flatness preserved by extension: Let M be an R -module and N ⊂ M a submodule. If N and M/N are exact, then so is M. 14. FLATNESS 139

Completion is flat: If R is a local noetherian ring, then its m-adic completion is a flat R-module. Fiber dimension: Let φ : R → R0 be a local homomorphism of noether- 0 0 0 ian local rings. Then dim(R ) − dim(R) ≤ dim(R /mRR ) and we have equality if R0 is flat over R. We refer to [1] (and for the last property to [2]) for proofs. We will explore the geometric meaning of this notion via the following definition. DEFINITION 14.1. Let f : X → Y be a morphism of schemes and let F be a OX -module. We say that F is flat over Y if for every x ∈ X, Fx is flat as a OY,f(y)- module. If this is the case for F = OX , then we say that f is a flat morphism. It is clear that flatness is local property. The properties listed above imply among other things: Sheafification of flat modules: Let M be an R-module. Then M is a flat R-module if and only if its sheafification M is a flat over Spec(R). f g Transitivity of flatness: Let X −→ Y and Y −→ Z be morphisms of schemes with g flat. Then an OX module F that is is flat over Y is also flat over Z. Base change: Let f : X → Y a morphism of schemes and F an OX -module 0 0 that is flat over Y and g : Y → Y a morphism. Denoting by g˜ : Y ×Y X → X the associated morphism, then g˜∗F is flat over Y 0. Flat coherent modules: Let X be a locally noetherian scheme and F a coherent OX -module. Then F is flat over X ⇔ F is locally free ⇔ x ∈ X 7→ dimκ(x) F(x) is locally constant. Flatness preserved by kernels: The kernel of an homomophism between flat quasi-coherent OX -modules is flat. Flatness preserved by extension: Let M be quasi-coherent OX -module and N ⊂ M a submodule. If N and M/N are exact, then so is M. Flat implies constant fiber dimension: Let f : X → Y a morphism of lo- −1 cally noetherian schemes and x ∈ X. Then dimx X−dimy Y ≤ dimx f (y) with equality when f is flat. Here y := f(x). We further mention that a morphism of between noetherian schemes is open. REMARK 14.2. Let f : X → Y be a morphism of noetherian schemes and let F be a quasi-coherent OX -module. We may then resolve F by f∗-acyclic quasi- coherent OX -modules in a Cechlikeˇ manner as follows. Cover X by a finite number n ∗ of affine open subsets {jν : Uν ⊂ X}ν=1. Then jν∗jν F is a quasi-coherent OX - ∗ module. In case F is flat over Y , then f∗jν∗jν F = (f|Uν )∗F|Uν is a flat quasi- coherent OY -module. k ∗ k Since Uν is affine we have R f∗(jν∗jν F) = R (f|Uν )∗F|Uν = 0 when k > 0. 0 ∗ 0 So if we put C := ⊕ν jν∗jν F, then we have an monomorphism F ⊂ C of quasi- 0 coherent OX -modules, with C f∗-acyclic and f∗C0 a flat OY -module, whenever F is flat over Y . With induction we thus obtain a resolution F → C• by such OX -modules. Let now be given a commutative diagram of morphisms of schemes g˜ X0 −−−−→ X   0  f y fy g Y 0 −−−−→ Y 140 3. SCHEMES

We found in Proposition 3.14 that for any OX -module F on X and any j = 0, 1,... , there exists by a natural homomorphism j ∗ j j 0 ∗ φ : g R f∗F → R f∗g˜ F. 0 0 PROPOSITION 14.3. Suppose that the above diagram is cartesian: X = Y ×Y X 0 with X,Y,Y are noetherian and g is ﬂat. When F is a quasi-coherent OX -module, then each φj is an isomorphism.

PROOF. Since this is local issue on Y 0 we may assume that both Y 0 and Y are affine: Y 0 = Spec(R0) and Y = Spec(R) with R noetherian and R0 a flat R-module. 0 k k 0 ∗ We must then show that the natural map R ⊗R H (X, F) → H (X , g˜ F) is an isomorphism. Notice that g˜ will be affine and flat by base change. • Let F → C be a resolution by quasi-coherent Γ-acyclic OX -modules as in Remark 14.2 above. Since g˜ is flat, g˜∗F → g˜∗C• is still a resolution and since g˜ is affine, this resolution is Γ-acyclic. Therefore we have Hk(X, F) = hk(C•(X)) and k 0 ∗ k ∗ • ∗ • 0 0 k H (X , g˜ F) = h (˜g C (X)). But clearly g˜ C (X ) = R ⊗R C (X) and since the 0 functor R ⊗R − commutes with taking cohomology, the assertion follows.

THEOREM 14.4. Let S be a noetherian scheme and F a coherent O r -module. For PS ˜∗ ˜ r r s ∈ S, denote by Ps ∈ Q[T ] the Hilbert polynomial of isF(s), where is : Ps ⊂ PS. Then F is ﬂat over S if and only if the Hilbert polynomial deﬁnes a locally constant function P : |S| → Q[T ] (in which case π∗F(n) is locally free for n 0).

PROOF. Without loss of generality we may assume that S is affine with R noe- r therian. We first show that for every s ∈ S the natural map κ(s) ⊗ Γ(PS, F(n)) → r ∗ Γ(Pκ, isF(n)) is an isomorphism when n 0, so that P is in fact computable from r the graded R-module ⊕nΓ(PS, F(n)). k Choose a finite set of generators of mS,s so that have an exact sequence R → R → κ(s) → 0. This yields an exact sequence k ˜ ˜∗ F → F → is∗isF → 0 which remains exact after Serre twisting. For n 0, H1 of F(n)k and the image of F(n)k → F(n) vanish, so that we then have an exact sequence r k r r ˜∗ Γ(PS, F(n)) → Γ(PS, F(n)) → Γ(Ps, isF(n)) → 0. r k The first map is also obtained by tensoring Γ(PS, F(n)) with R → R and so the r r ˜∗ natural map κ(s) ⊗ Γ(PS, F(n)) → Γ(Pκ, isF(n)) is an isomorphism indeed. In view of the fact that for a finitely generated R-module M, M is flat if and only if the function s ∈ S 7→ dimκ(s) κ(s) ⊗ M is locally constant, it therefore r suffices to show that F is flat if and only if Γ(PS, F(n)) is flat for n 0. Assume that F is flat over S. Let 0 → F → C0 → C1 → · · · be an exact sequence of quasi-coherent flat O r -modules that are Γ-exact (as for instance obtained in PS remark 14.2). This remains so after Serre twisting. For n 0, we have by the r r 0 r 1 Serre vanishing property that 0 → Γ(PS, F(n)) → Γ(PS, C (n)) → Γ(PS, C (n)) r is also exact so that Γ(PS, F(n)), being the kernel of a homomorphism of flat R- modules, is a flat R-module. r Conversely, assume that Γ(PS, F(n)) is a flat R-module for n ≥ n0. This means r that the graded R-module ⊕n≥n0 Γ(PS, F(n)) is flat. This implies that F as the associated coherent sheaf is flat over R. 15. SERRE DUALITY 141

r REMARK 14.5. We may apply the preceding to a closed subscheme X ⊂ PS with S is connected and with F = OX . We then find that X is flat over S if and r only if the Hilbert polynomials of the fibers Xs of X → PS are constant. Since the dimension of Xs is the degree of Ps and the degree deg(Xs) of Xs is dim(Xt)! times the leading coefficient of Pt, it follows that then both s 7→ dim(Xs) and s 7→ deg(Xs) are constant functions on S.

15. Serre duality In this section X denotes a scheme of dimension n that is proper over the algebraically closed field k. The main result only concerns the case when X is i projective over k. For a coherent OX -module F, H (X, F) is a finite dimensional i vector space. So we have defined a functor H (X, −): Coh(X) → Vectk and by i ∨ ◦ dualizing, H (X, −) : Coh(X) → Vectk.

DEFINITION 15.1.A weak dualizing sheaf for X is a representation of the func- n ∨ ◦ tor H (X, −) : Coh(X) → Vectk (recall that n = dim(X)). In other words, n ∨ this is a pair (ω, t), where ω is a coherent OX -module and t ∈ H (X, ω) = n Homk(H (X, ω), k) (often called the trace map) such that for every coherent OX - module F, the map

n ∨ (H ) n ∨ n ∨ n ∨ HomX (F, ω) −−−−→ Homk(H (X, ω) , H (X, F) ) −→ H (X, F) (where the last map is evaluation in t) is an isomorphism. 7 We say that the pair (ω, t) is a dualizing sheaf if for every coherent OX -module F, and every i, the composite

i n−i n ExtX (F, ω) → Homk(H (X, F), H (X, ω)) = n ∨ n−i ∨ n−i ∨ = Homk(H (X, ω) , H (X, F) ) → H (X, F) (with the ﬁrst map deﬁned as in the appendix and the last map evaluating in t so that for i = 0 we get the map appearing above) is an isomorphism. By the Yoneda lemma a weak dualizing sheaf is unique up to unique isomor- phism8. We will show that it exists when X is projective.

0 ◦ REMARK 15.2. The functors Tj,Tj : Coh(X) → Vectk defined by Tj(F) := j 0 n−j ∨ Ext (F, ωX ) resp. Tj(F) := H (X, F) are δ-functors with the first one being universal. (The contravariant nature leads us to use a homological indexing by subscripts rather than superscripts.) The homomorphisms j n−j n n−j ∨ 0 Tj(F) = ExtX (F, ω) → Homk(H (X, F), H (X, ω)) → H (X, F) = Tj(F) defined above constitute in fact a morphism of δ-functors. To say that (ω, t) is a 0 weak dualizing sheaf means that T0 ⇒ T0 is an isomorphism. So to say that such a 0 weak dualizing sheaf is dualizing means that T• is also universal. We first treat the case of a projective space.

7Hartshorne calls a dualizing sheaf what we call a weak dualizing sheaf. 8Beware that this does not imply that for a given ω, Hn(X, ω)∨ has a canonical element t: we may just as well replace t by t0 = λt, where λ is a nonzero scalar and then multiplication by λ maps the pair (ω, t) isomorphically onto the pair (ω, t0). 142 3. SCHEMES

r PROPOSITION 15.3. For P := Pk, the pair (ωP := OP (−r − 1), tP ), where tP is r ∨ a generator of the 1-dimensional k-vector space H (P, OP (−n − 1)) , is a dualizing sheaf.

PROOF. Proposition 11.1 shows that (ω = OP (−r − 1), t) has all the desired properties for the OP -modules F of the form OP (n) (the dualizing property is trivially fulfilled in degrees 6= 0, r). It then also holds for any OP -module that ⊕s is a finite direct sum of such sheaves. In particular, when P = OP (−d) with d > 0, then both members of the asserted identity vanish unless i = 0 (in which case they are naturally isomorphic). As observed in Remark 15.2, F 7→ Ti(F) := i 0 n−i ∨ ExtP (F, ω) and F 7→ Ti (F) := H (P, F) are δ-functors. By 11.3 there exists a left resolution P• → F if F in Coh(P ) by modules P as above (this is of course a ◦ right resolution in Coh(P ) ). The Pi are acyclic for both δ-functors and so by the 0 0 abstract De Rham theorem, Ti(F) = hi(T0(P•)) and Ti (F) = hi(T0(P•)). Since we ∼ 0 have T0(P•) = T0(P•) as chain complexes, the proposition follows. r THEOREM 15.4. Let i : X ⊂ Pk =: P be a closed subscheme of dimension n and j let (ωP , tP ) be a dualizing sheaf for P . Then Ext P (i∗OX , ωP ) = 0 for j < r − n and ∗ r−n we have a weak dualizing sheaf (ωX , tX ) for X with ωX = i Ext P (i∗OX , ωP ). This j is a dualizing sheaf if also Ext P (i∗OX , ωP ) = 0 for j > r − n. j PROOF. We first show that Ext P (i∗OX , ωP ) = 0 for j < r − n. Since OP (q) is invertible, it is equivalent to prove this property for some Serre j twist Ext P (i∗OX , ωP )(q)). When q 0, then by the Serre’s vanishing theorem 11.2, j Ext P (i∗OX , ωP )(q) is generated by its global sections and is Γ-acyclic. So it suffices j to prove that Γ(P, Ext P (i∗OX , ωP )(q)) = 0 for q 0. Since OP (q) is locally free, 3.18-iv of the appendix implies that j ∼ j ∼ j Ext P (i∗OX , ωP )(q) = Ext P (i∗OX , ωP (q)) = Ext P (i∗OX (−q), ωP )). j In particular, Ext P (i∗OX (−q), ωP ) is Γ-acyclic for q 0. We invoke 3.18-v of j the appendix and find a natural isomorphism from Γ(P, Ext P (i∗OX (−q), ωP )) onto j ExtP (i∗OX (−q), ωP ). By Serre duality on P , j ∼ r−j ∨ ∼ r−j ∨ ExtP (i∗OX (−q), ωP ) = H (P, i∗OX (−q)) = H (X, OX (−q)) and the last vector space is zero when r − j > n, i.e., when j < n − r. 9 We now fix a coherent OX -module F and consider the functor G : Coh(P ) → ∗ Vectk which assigns to a coherent OX -module G the k-vector space HomX (F, i G) = j j HomP (i∗F, G). We have R G(G) = ExtP (i∗F, G) by definition. We now regard G as the composite of two functors, to be precise, we write G = GF with F : G 7→ j j HomP (i∗OX , G). Then R F (G) = Ext P (i∗OX , G) and so if we take G = ωP , then by the above, these derived functors vanish for j < n − r. Hence by Proposition 2.21 there is a natural isomorphism n−r ∼ n−r HomP (i∗F, Ext P (i∗OX , ωP )) = ExtP (i∗F, ωP ). ∗ n−j The left hand side can be identified with HomX (F, i Ext P (i∗OX , ωP )) and the n right hand side is by Serre duality on P isomorphic to the k-dual of H (P, i∗F) = n ∗ n−j H (X, F). Thus ωX = i Ext P (i∗OX , ωP ) serves as a weak dualizing sheaf.

9You may ﬁnd it strange that we proved a local property by means of global arguments. A purely local (commutative algebra) proof is indeed possible. 15. SERRE DUALITY 143

j Now assume that we also have that Ext P (i∗OX , ωP ) = 0 for j > n − r. Then the stronger part of Proposition 2.21 says that there is a natural isomorphism j n−r ∼ n−r+j ExtP (i∗F, Ext P (i∗OX , ωP )) = ExtP (i∗F, ωP ), j ≥ 0. r−j r−j The right hand side can be identified with the k-dual of H (P, i∗F) = H (X, F). In order to identify the left hand side, we make the observation that for any OP - j • module G, Ext P (i∗F, G) is a i∗OX -module (choose an injective resolution G → I • and note that Hom(i∗F, I ) is a complex of i∗O-modules) so that the natural map j ∗ j Ext P (i∗FX , G) → i∗i Ext P (i∗FX , G) is an isomorphism. If G is a OX -module, then j ∼ j ∗ j ∗ Ext P (i∗F, G) = Ext P (i∗F, i∗i G) = i∗ Ext X (F, i G) (for i∗ is exact) and similarly, j j ∗ ExtP (i∗F, G) = ExtX (F, i G). If we apply this to the present situation we find that j n−r j ∗ n−r j ExtP (i∗F, Ext P (i∗OX , ωP )) = ExtP (i∗F, i∗i Ext P (i∗OX , ωP )) = ExtX (F, ωX ). j ∼ r−j ∨ We have thus shown that ExtX (F, ωX ) = H (X, F) . The Cohen-Macaulay property. The vanishing condition in Theorem 15.4 (of j certain Ext P (i∗OX , ωP )) that gives us a dualizing sheaf is remains elusive, as long as we have no way of establishing it. At this point it is also unclear to what extent this depends on the immersion of X in a projective space. Let us first show that this is an issue of a local nature:

PROPOSITION 15.5. Let F be a coherent sheaf on a noetherian scheme Y and let y ∈ Y . Then for every O -module G, we have Ext j (F, G) = Extj (F , G ). Y Y y OY,y y y j PROOF. For F = OY this is obvious: we have Ext Y (OY , G) = 0 for j 6= 0 0 and Ext Y (OY , G) = G. For the same reason this is true when F is a free OX - module of finite rank. In general, choose an affine open neighborhood U of y in Y and a right resolution of F|U by free OU of finite rank: L• → F|U. Then j j Ext Y (F, G)|U = Ext U (F|U, G|U) = hj(HomU L•, G|U)) by 3.18-(ii) and (vi) and so Ext j (L, N ) = h (Hom (L , G|U)) = h Hom (L , G ) = Extj (F , G ). U y j U • y j U y,• y OY,y y y

CHAPTER 4

Appendix

1. The basics of category theory In this part of mathematics one quickly runs into foundational aspects of mathematics and one must be careful about which kind of set theory one adopts (otherwise we will be soon be talking about a nonentity like ‘the set of all sets). We leave these issues aside. Let us just say that we at least accept the axiom of choice.

Categories. The following deﬁnition is modeled after the coarse structure per- ceived in various mathematical areas.

DEFINITION 1.1.A category C consists of (i) a collection (a class, to be precise) called objects Obj(C), (ii) for any pair X,Y ∈ Obj(C) a (possibly) empty set, called the set of C- morphisms from X to Y and denoted MorC(X,Y ) and for (iii) any triple X,Y,Z ∈ Obj(C) a map, called composition

(g, f) ∈ MorC(Y,Z) × MorC(X,Y ) 7→ g ◦ f ∈ MorC(X,Z). These data must obey: Associativity: (h ◦ g) ◦ f = h ◦ (g ◦ f) whenever that makes sense and Identity element: for every X ∈ Obj(C) there exists a 1X ∈ MorC(X,X) such that for all f ∈ MorC(X,Y ) and g ∈ MorC(Y,X), f ◦ 1X = f and 1X ◦ g = g.

A standard argument shows that there is only one element in MorC(X,X) satisfying the last property. It is called the identity element of X. We often write f : X → Y instead of f ∈ MorC(X,Y ) and we call X the source and Y the target of f. We say that f : X → Y is an isomorphism if it has a 2-sided inverse, i.e., if there exists a g : Y → X such that g ◦ f = 1X and f ◦ g = 1Y . Again a standard argument shows that g is then unique; it is called the inverse of f and denoted f −1. Two objects of C are said to be isomorphic if an isomorphism between them exists. The generality of this notion is both its power and its limitation.

EXAMPLES 1.2. Here are just a few. Set: the category Set of sets and maps between sets, Top: the category of topological spaces and continuous maps, Htp: the category of topological spaces and homotopy classes of maps, Grp: the category of groups and homomorphisms, Ab: its subcategory of abelian groups, Ring: the category of rings and ring homomorphisms, VectK : the category of ﬁnite dimensional K-vector spaces and linear maps (K a ﬁeld), and more generally,

145 146 4. APPENDIX

ModR: the category ModR of R-modules and R-homomorphisms (R a ring), AbX : the category of abelian sheaves of on a space X, ModO: the category of sheaves of O-modules, where O is a sheaf of rings on a space X (to be discussed later). The definition shows that reversal of the arrows of C produces another category, called the dual of C and denoted C◦: it has the same objects: Obj(C◦) := Obj(C), but MorC◦ (X,Y ) := MorC(Y,X) (if we view f ∈ MorC(Y,X) as an element of 0 ◦ ◦ MorC◦ (X,X ), we may denote it by f ), the composition in C being the one of C: 0 0 00 ◦ ◦ ◦ if f ∈ MorC(X,X ) and g ∈ MorC(X ,X ), then f ◦ g := (g ◦ f) . Functors. Categories can be related to each other with the help of the notion of functor. DEFINITION 1.3. Let C and C0 be categories. A functor from C to C0 is a map F which assigns to each object X of C an object F (X) of C0 and to each morphism f : X → Y a morphism F (f): F (X) → F (Y ) which sends identity elements to identity elements: F (1X ) = 1F (X) and respects composition: F (g◦f) = F (g)◦F (f). This is also called a covariant functor in order to distinguish it from a contravariant functor from C to C0, which is simply a functor from C◦ to C0 (so it reverses arrows). It is clear that functors F : C → C0 and F 0 : C0 → C00 can be composed to produce a functor F 0 ◦ F : C → C00. An important class of functors form the ones that ‘forget’ structure (and for this reason often referred to as forgetful functors). For instance, we have the forgetful functors Top → Set (forgets the topology of a space), Grp → Set (forgets the group structure of a group). Other examples of functors are π0 : Top → Set and irr : Top → Set (associate to a space its set of connected resp. irreducible components). We have an evident functor Top → Htp and we have encountered the functor Ring◦ → Top which assigns to a ring its prime ideal spectrum. We ◦ also note that we have a contravariant functor ModR → ModR which assigns to an R-module M, HomR(M,R). Remarkably often a functor is of the type we describe next. An object A of a category C determines two functors. The first one is contravariant: it is the functor ◦ hA : C → Set which assigns to an object X of C, the set hA(X) := MorC(X,A) and 0 0 to a morphism ξ : X → X the induced map hA(ξ) : MorC(X ,A) → MorC(X,A) (given by precomposition with ξ). The other, hA : C → Set, is covariant and assigns A 0 to an object X of C, the set h (X) := MorC(A, X) and to a morphism ξ : X → X 0 the induced map hA(ξ) : MorC(A, X) → MorC(A, X ) (given by composition with A ξ). A functor F that is naturally isomorphic to one of these (i.e., to hA or h according to whether F if contravariant or covariant) is said to be representable (we give a more precise definition later). EXAMPLES 1.4. The forgetful functor Top → Set is representable by the singleton space ∗, for assigning to a space X the set of continuous maps ∗ → X (which is of course the set of all maps) gives us back X as a set. Similarly, the forgetful functor Grp → Set is representable by Z. EXERCISE 90. Show that the forgetful functor Ring → Set is representable. Our somewhat off-hand definition of a representable functor is incomplete because we did not say what we mean by two functors being naturally isomorphic. This notion is a special case of that of a natural transformation. 1. THE BASICS OF CATEGORY THEORY 147

DEFINITION 1.5. Let F,G : C → C0 be functors between categories. A natural transformation from F to G (denoted T : F ⇒ G) assigns to every object X of C a C0-morphism T (X): F (X) → G(X) with the property that for every C-morphism φ : X → Y the diagram F (φ) F (X) −−−−→ F (Y )     T (X)y T (Y )y G(φ) G(X) −−−−→ G(Y ) commutes. We say that T is a natural isomorphism (and that F and G are naturally isomorphic) if every T (X) is an isomorphism; we then write T : F =∼ G. This is a very fundamental concept and ideally the adjective ‘natural’ when used in mathematical sense, should always be understood in this way. As the following exercise shows, a natural isomorphism need not be unique. ◦ EXERCISE 91. Let D : VectK → VectK be the functor which assigns to a K- ∗ vector space V its dual V . Construct a natural isomorphism DD ∼ 1 ◦ : = VectK VectK → VectK . Construct two such when K is not of characteristic 2. Two categories C and C0 are for all practical purposes the ‘same’ if they are equivalent in the sense of the following deﬁnition.

DEFINITION 1.6. A functor F : C → C0 is said to be an equivalence (of cate- 0 ∼ ∼ gories) is there exist a functor G : C → C such that GF = 1C and FG = 1C0 . In other words, there must exist for every object X of C isomorphisms T (X): X → GF (X) and T 0(X): X → FG(X) such that for every C-morphism φ : X → Y we have GF (φ) ◦ T (X) = T (Y ) ◦ φ and FG(φ) ◦ T 0(X) = T 0(Y ) ◦ φ. The following example may help to illustrate the subtleties involved here.

EXAMPLE 1.7. Fix a field K and let MatK be the category whose objects are the nonnegative integers 0, 1, 2,... . A morphism m → n is given by a m × n-matrix (which has m rows and n columns) with entries in K. Composition of morphisms is composition of matrices. This is in fact a subcategory of VectK : we have an inclusion functor F : MatK ⊂ VectK . We claim that this is an equivalence of categories. The inverse functor G : VectK → MatK will on objects be clearly the map which assigns to every finite dimensional K-vector space V its dimension dim V . But defining it on morphisms requires us to choose an ordered basis of dim V ∼ V , or equivalently, an isomorphism eV : K = V for every V (so here we need the axiom of choice). Then for a K-linear map φ : V → W we let G(φ) be the matrix of φ in terms of the chosen bases of V and W , in other words, of −1 dim V dim W eW φeV : K → K . Now a natural isomorphism T : 1MatK ⇒ GF is defined by letting T (n) be the matrix of eKn ∈ GL(n, K) (so had we been careful to choose each eKn to be the identity, then T (n) would have been the identity matrix) 0 0 −1 and a natural isomorphism T : 1VectK ⇒ FG is given by letting T (V ) := eV . This example shows that a given functor F : C → C0 may define an equivalence without there being an obvious choice for its ‘inverse’ G. But the idea behind this example generalizes well to a criterion for F being an equivalence: it is necessary and sufficient that

F is full: for every X,Y ∈ C, F : MorC(X,Y ) → MorC0 (F (X),F (Y )) is surjective, F is faithful: for every X,Y ∈ C, F : MorC(X,Y ) → MorC0 (F (X),F (Y )) is injective, 148 4. APPENDIX

F is essentially surjective: every object of C0 is isomorphic to one of the form F (X) with X ∈ C. The basic example in algebraic geometry is the functor Spec : Ring◦ → Affsch. In this case an inverse functor is also given, for it is the global section functor Γ: Affsch → Ring◦ which assigns to an afﬁne scheme the ring of global sections of its structure sheaf. For a ring R resp. an afﬁne scheme X, we found natural isomorphisms R → Γ Spec(R) resp. X → Spec Γ(X), but it is not always a good idea to regard these as identities.

Given categories C and C0, then the functors F : C → C0 are the objects of a new category whose morphisms are the natural transformations. We denote this 0 category Func(C, C ), so that MorFunc(C◦,Set)(F,G) = Nat(F,G). We encounter an example in the Yoneda lemma that we discuss next. The Yoneda Lemma. Notice that a C-morphism α : A → A0 defines a natural transformation hA ⇒ hA0 (postcomposition with α). In this way we can regard ◦ the map A 7→ hA as a functor h∗ : C → Func(C , Set) from the category C to the category Func(C◦, Set) whose objects are the contravariant functors C → Set and whose morphisms are the natural transformations between them. The Yoneda lemma implies that this embeds C in Func(C◦, Set) as a full subcategory. In particular, A ∈ C is determined (up to unique isomorphism) by its associated functor hA. LEMMA 1.8 (The Yoneda Lemma). Let F : C◦ → Set be a contravariant functor from C to Set. Then the map which assigns to any natural transformation T : hA ⇒ F the element sT := T (A)(1A) ∈ F (A) defines a bijection between the set Nat(hA,F ) of natural transformations hA ⇒ F and F (A). Its inverse assigns to s ∈ F (A) the natural transformation Ts : hA ⇒ F defined by

Ts(X): φ ∈ hA(X) = MorC(X,A) 7→ F (φ)(s) ∈ F (X). ∼ In particular (take F = hA0 ), this yields a bijection Nat(hA, hA0 ) = hA0 (A) = 0 ◦ MorC(A, A ) and thus h∗ : C → Func(C , Set) is fully faithful. The proof is best left as an exercise. If we replace in the preceding the category C by its opposite, then we find that A also defines a covariant functor hA, which assigns to an object X of C, A 0 the set h (X) := MorC(A, X) and to a morphism ξ : X → X the induced map A 0 0 h (ξ) : MorC(A, X) → MorC(A, X ). The morphism α : A → A defines a natural A ∗ ◦ transformation hA0 → hA so that A 7→ h defines a functor h : C → Func(C, Set). In this case the Yoneda lemma asserts that for every functor G : C◦ → Set there is a natural bijection between Nat(hA,G) and G(A).

We say that a contravariant functor F : C◦ → Set is representable if there exists a pair (A, s), where A is an object of C and s ∈ F (A), such that the associated natural transformation Ts : hA ⇒ F is an isomorphism. In other words, if for every object X of C, the map

φ ∈ MorC(X,A) 7→ F (φ)(s) ∈ F (X) is a bijection. We then call (A, s) a universal element for F . The pair (A, s) is unique up to unique isomorphism. For if F is also represented 0 0 0 −1 ∼ by the pair (A , s ∈ F (A )), then we get a natural isomorphism Ts0 Ts : hA = hA0 1. THE BASICS OF CATEGORY THEORY 149 and by the Yoneda Lemma this must come from an unique isomorphism A =∼ A0. It is straightforward to check that it takes s to s0. Over and under category; initial and final object. Given a category C, then every object A of C defines a category denoted C/A: its objects are morphisms with target A: f : X → A; a morphism from f : X → A to f 0 : X0 → A is then simply a C-morphism g : X → X0 such that f 0g = f. It is often called the category over A. We have a forgetful functor C/A → C (which tells us to forget about the morphism to A). This is the identity precisely if A is a final object of C, i.e., if for every object X of C, Mor(X,A) is a singleton. Such an object is unique up to unique isomorphism. If we reverse all arrows we arrive at the dual notion, the category under A, denoted A/C (whose objects are morphisms A → X). The forgetful functor A/C → C is the identity precisely if A is an initial object of C: it has the property that Mor(A, X) is a singleton for every object X of C. EXERCISE 92. Determine of each category listed in 1.2 whether it has a final resp. initial object and if it exists, describe it. Prove that C/A resp. A/C has a final object resp. initial object. Adjoint functors. The notion of an adjoint functor encodes (and sometimes even suggests) in a categorical way universal constructions such as the formation of the free group generated by a set, the field of fractions of a ring, the largest group that is the quotient of a semigroup etc. DEFINITION 1.9. Given categories C and C0, then functors F : C → C0 and F 0 : C0 → C are said to make up an adjoint pair if for every object X of C and object 0 0 0 0 ∼ 0 X of C we are given a bijection MorC(F (X ),X) = MorC0 (X ,F (X)) which is natural1 in X and X0 (covariant in X and contravariant in X0). We say that F 0 is a left adjoint of F and that F is a right adjoint of F 0. Before giving examples, let us note that if we take X = F 0(X0) and take the 0 0 0 0 0 0 0 0 image of 1 ∈ MorC(F (X ),F (X )) we get a C -morphism X → FF (X ). This 0 0 defines a natural transformation η : 1C0 ⇒ FF . Similarly, by taking X = F (X), we get a C-morphism F 0F (X) → F giving rise to a natural transformation ε : 0 F F ⇒ 1C. We also observe that Yoneda’s lemma implies a right adjoint (or left adjoint) of a functor is unique up to natural isomorphism. EXAMPLES 1.10. (ii) A standard example is provided by the forgetful functor Forget : Grp → Set. We are looking for left adjoint of it, that is, a functor F : Set → Grp such that for every set X and every group G, Hom(F (X),G) = Map(X, Forget(G)). This problem has a solution indeed: given a set map X → G, we can always extend it (uniquely) to a group homomorphism from the free group Free(X) generated by X to G (by convention Free(∅) is the trivial group). We thus find that F := Free does the job. We have indeed a map of sets X → Forget Free(X) and a natural group homomorphism Free Forget(G) → G. (ii) The inclusion Ab ⊂ Grp has as left adjoint the abelianization functor Grp → Ab, which assigns to a group G, G/[G, G]. (iii) Let Dom be the category of integral domains whose morphisms are injective homomorphisms and let Field ⊂ Dom be its subcategory of fields. Then

1 0 0 0 The use of ‘natural’ in this definition is explicated as follows: both (X,X ) 7→ MorC(F (X ),X) 0 0 0 ◦ and (X,X ) 7→ MorC0 (X ,F (X)) define functors from the product category C × (C ) to the category Set of sets and we ask here for a natural equivalence between the two. 150 4. APPENDIX the inclusion has a left adjoint Frac : Dom → Field given by the formation of the fraction field. Product and sum. Let C be a category. Given objects A and B of C, then α β form the category whose objects are the diagrams A ←− X −→ B in C and where α β α0 β0 a morphism from A ←− X −→ B in C to A ←− X0 −→ B in C is the given by a morphism f : X → X0 so that the resulting diagram commutes: α = α0f and β = β0f. If this category has a final object, then we call it ‘the’ product of A and B and, denoted A u B (the definite article is justified since a final object is unique up to unique isomorphism). So A u B comes with morphisms πA : A u B → A and πB : A u B → B and has the property that whenever we are given a diagram α β A ←− X −→ B, then there is a unique morphism f u g : X → A u B such that we get a commutative diagram. If we apply this definition to an over category C/C, then we get in C the notion of a fibered product. So we are then given a diagram α β A −→ C ←− B and we form a category whose objects are commutative squares A ←−−−− X   α  y y β C ←−−−− B and whose morphisms are defined in an obvious manner. If an final object of this category exists, then we call it the fibered product of the pair (α, β), denoted

πA/C A ←−−−− A uC B   α πB/C y y β C ←−−−− B In other words, if we are given morphisms f : X → A and X → B such that αf = βg, then there exists a unique a morphism f u g : X → A uC B over which f and g both factor. We say that C has finite products if every pair of its objects has a product. If this is the case for every overcategory C/C, then we say that C has finite fibered products. In many categories (such as Set, see Example 1.11) it is more common to write A ×C B instead of A uC B and will in fact use this notation in the category of schemes Sch. i Reversal of arrows gives the notion of the sum of A and B, denoted A −→A i A t B −→B B. When applied to the undercategory C/C, we get the notion of the iC/A iC/B cofibered sum A −−−→ A tC B ←−−− B.

EXAMPLE 1.11. In the category Set, fibered products and cofibered sums exist: the product is the ordinary product of sets and the sum is the disjoint union. The α β fibered product of A −→ C ←− B is a subset of A × B and given as the set of (a, b) ∈ A × B with α(a) = β(b) (and is usually denoted A ×C B). The cofibered φ ψ sum of A ←− C −→ B is the quotient of A t B and obtained by dividing out by the equivalence relation generated by φ(c) ∼ ψ(c) (and is usually denoted A ∪C B). In Grp fibered products and cofibered sums also exist: for two groups A, B, A u B = A × B (the direct product) and A t B = A ∗ B (the free product). 2. ABELIAN CATEGORIES AND DERIVED FUNCTORS 151

Furthermore, A uC B is the subgroup A ×C B ⊂ A × B of (a, b) ∈ A × B with α(a) = β(b), and A tC B is the amalgamated product A ∗C B (the quotient of A ∗ B by the normal divisor generated by the elements φ(c)ψ(c−1), c ∈ C).

EXERCISE 93. Prove the category Ring has ﬁbered products and coﬁbered sums. More precisely, show that A u B = A ⊗ B, A uC B = A ⊗C B, A t B = A × B and A tC B is A × B modulo the ideal generated by (φ(c), ψ(−c)). Monomorphism and epimorphism. The notion of an injective resp. surjective map in the category of sets has the following analogue in a categorical setting.

DEFINITION 1.12. We say that a morphism f : A → B in a category C is a monomorphism if for two morphisms g1, g2 ∈ MorC(X,A) are equal if and only if fg1 and fg2 are. Dually, we say that f : A → B is an epimorphism if if for two morphisms g1, g2 ∈ MorC(B,X) are equal if and only if g1f and g2f are.

In the categories Grp, Ring, Top, ModR these also amount to injection resp. surjection.

EXERCISE 94. Prove that in the category AbX , a morphism of sheaves φ : A → B is a monomorphism resp. epimorphism if and only if for every x ∈ X the homomorphism of stalks φx : Ax → Bx has this property. 2. Abelian categories and derived functors Abelian categories. These are categories to which the standard techniques of homological algebra extend in an immediate manner. But the advantage of the categorical approach goes beyond the generalization: it offers also a different, ‘higher’ point of view. The categories C under consideration will all have an object that is both ﬁnal and initial. It is called the zero object and will be denoted 0. This determines in every MorC(A, B) a distinguished element, namely the composite A → 0 → B; we call this the zero morphism and denote it also by 0. That enables us to deﬁne the notion of kernel and cokernel: given a morphism f : A → B in C, then a morphism i : K → A is called a kernel of f if fi = 0 and i is universal for this property: any morphism g : X → A in C such that fg = 0, factors over i be means of a unique morphism g˜ : X → K (g = ig˜). Notice that the uniqueness property implies i has to be a monomorphism. We also note that a kernel is unique up to unique isomorphism. We denote a kernel of f often by Ker(f) → A. If f happens to be a monomorphism, then it has a kernel, namely the zero map 0 → A. For if g : X → A is such that gf = 0, then it has that property in common with the zero morphism 0 : X → A, and so the monomorphism property implies that we must have g = 0. This just means that g factors (uniquely) as X → 0 → A and so 0 → A may serve as the kernel of f. Dually, a morphism p : B → C is called a cokernel of f if pf = 0 and p is universal for this property: any morphism g : B → X in C such that gf = 0, factors over p be means of a unique morphism g˜ : C → X (g =gp ˜ ). We denote a cokernel of f often by B → Coker(f). If f happens to be an epimorphism, then B → 0 is the cokernel of f.

EXERCISE 95. Describe the kernel of f ∈ MorC(A, B) as a representative object of a functor C◦ → Set and describe the cokernel of f as a representative object of a functor C → Set. 152 4. APPENDIX

EXERCISE 96. Prove that the kernel of A → 0 and the cokernel of 0 → A are both represented by the identity 1A : A → A.

We now also assume that any two objects A1,A2 of C have a sum that at the same time serves as product (this is called in the literature a biproduct). We denote it by A1 ⊕ A2. This structure turns MorC(A1,A2) into an abelian semigroup which has 0 as zero, as follows. The product role of ⊕ make that A1 ⊕ A2 comes with the projections p1 : A1 ⊕ A2 → A1, p2 : A1 ⊕ A2 → A2, whereas its sum role endows it with morphisms k1 : A1 → A1 ⊕ A2 and k2 : A2 → A1 ⊕ A2. Let us then note that if we have fi ∈ MorC(Ai,Bi) (i = 1, 2), then the two composites fi ki Ai −→ Bi −→ B1 ⊕ B2 combine to deﬁne a morphism

f1 ⊕ f2 : A1 ⊕ A2 → B1 ⊕ B2.

We have a diagonal ∆ : A → A ⊕ A characterized by the property that p1∆ and p1∆ are the identity and a codiagonal ∇ : A⊕A → A characterized by the property that ∇k1 and ∇k2 are the identity. We then deﬁne the sum of f1, f2 ∈ MorC(A, B) as the composite

∆ f1⊕f2 ∇ f1 + f2 : A −−−−→ A ⊕ A −−−−→ B ⊕ B −−−−→ B. It is a (not entirely trivial) exercise to check that the sum is commutative, associative and has 0 as its zero. The construction shows that the identity of A⊕B is equal to k1p1 + k2p2. If we are also given g ∈ MorC(X,A) resp. h ∈ MorC(B,Y ), then you may check that (f1 + f2)g = f1g + f2g and h(f1 + f2) = hf1 + hf2. In other words, the composition map

MorC(A, B) × MorC(B,C) → MorC(A, C) is bilinear.

DEFINITION 2.1. A category C is said to be an abelian category if it is as above, that is, possesses zero object and any two objects have a biproduct, and has in addition the following properties

(i) the semigroup structure on MorC(A, B) is in a fact a group structure (so that MorC(A, B) is an (additively written) abelian group, (ii) every morphism has a kernel and a cokernel, (iii) every morphism f : A → B admits a factorization A → I → B, where A → I is the cokernel of Ker(f) → A and I → B is the kernel of B → Coker(φ). The factorization in (iii) is called the formation of the image of f, and I is often denoted Im(f). Notice that A → Im(f) is an epimorphism and Im(f) → B is a monomorphism. Let us observe that the dual of an abelian category is also abelian. We further A note that for an object A of an abelian category, the functors hA and h takes values in the category Ab.

LEMMA 2.2. Let f : A → B be a morphism in an abelian category. Then f is a monomorphism if and only if it has trivial kernel and in that case A → Im(f) is an isomorphism. Dually, f is an epimorphism if and only if it has trivial cokernel and in that case Im(f) → B is an isomorphism. 2. ABELIAN CATEGORIES AND DERIVED FUNCTORS 153

In particular, f is an isomorphism if and only if both its kernel and its cokernel are trivial.

PROOF. With the help Exercise 96 we see that Ker(f) = 0 implies that A → I is an isomorphism. This also shows that f is monomorphism, as is then the composite of the isomorphism A =∼ I and the monomorphism I → B (a kernel). We already observed that a monomorphism has trivial kernel and so the ﬁrst assertion follows. The proof of the second assertion is similar. EXAMPLES 2.3. The most basic example is that of the category of abelian groups, Grp. Other examples are ModR and (as we will show later), ModO. But Ring is not one (why?). We have now all the ingredients in place for the notions related to exactness. f g For instance, if we have a diagram A −→ B −→ C in C with gf = 0, then the composite Im(f) → B → C is zero, hence must factor uniquely through a morphism Im(f) → Ker(g). This is a monomorphism of which we can form the cokernel. This cokernel is called the (co)homology of the diagram and denoted h(A −→ B −→ C). We say that the diagram is exact if this is the zero object.

DEFINITION 2.4. A complex in C is a chain of morphisms

k−2 k−1 k k+1 A• ... −−−→d Ak−1 −−−→d Ak −→d Ak+1 −−−→d ... such that dkdk−1 = 0 for all k. The k-th cohomology object hk(A•) is deﬁned as the cokernel of the natural morphism Im(dk−1) → Ker(dk). If all of these are zero, then we say that the complex is exact. Given two complexes A• and B• in C, then a chain morphism φ• : A• → B• is a collection of morphisms φk : Ak → Bk, k ∈ Z, that make a commutative ladder. Two such chain morphisms φ•, ψ• : A• → B• are said to be (chain) homotopic if there exist morphisms ki : Ai → Bi−1 such that di−1ki + ki+1di = φi − ψi for all i ∈ Z. The same proofs as in the category Ab yield the following two lemmas.

LEMMA 2.5. Homotopy is an equivalence relation on the collection of chain morphisms which behaves well with respect to composition: a homotopy class of chain morphisms A• → B• can be composed with a homotopy class of chain morphisms B• → C • to obtain a homotopy class of chain morphisms A• → C •. A chain morphism f • : A• → B• induces morphisms hk(f •): hk(A•) → hk(B•) for all k ∈ Z and these morphisms only depend on its homotopy class. Two chain morphisms f • : A• → B• and g• : B• → C • are said to deﬁne a short exact sequence of chain morphisms, often written as

• • • f • g 0 → A −→ B −→ C • → 0, when each sequence 0 → Ak → Bk → Ck → 0 is exact (so each fk is then a monomorphism and each gk is an epimorphism).

LEMMA 2.6. A short exact sequence of chain morphisms as above generates a long exact sequence

hk(f •) hk(g•) k · · · → hk(A•) −−−−→ hk(B•) −−−−→ hk(C •) −→δ hk+1(A•) → · · · . 154 4. APPENDIX

d0 dk−1 dk REMARKS 2.7. Often we are given a complex as 0 → A0 −→ ... −−−→ Ak −→ ... . It is then understood that all the terms Ck with k < 0 are zero. Sometimes we are given a complex with decreasing indexing (the index is then almost always written as a subscript):

dk+1 dk dk−1 A• ... −−−→ Ak −→ Ak−1 −−−→ ..., k but one can always pass to an increasing indexing by putting A := A−k and k d := d−k. Derived functors. In what follows F : C → C0 is a functor of abelian categories. It is an exercise to show that F (A ⊕ B) is canonically isomorphic to F (A) ⊕ F (B) (there are natural maps between them in each direction) and that F takes diagonal to diagonal and codiagonal to codiagonal. It then follows that F is additive in the sense that for any two objects A, B of C the map MorC(A, B) → MorC0 (F (A),F (B)) is a homomorphisms of abelian groups: F (φ1 + φ2) = F (φ1) + F (φ2) for any pair φ1, φ2 ∈ MorC(A, B).

DEFINITION 2.8. We say that F is left exact if it takes an exact sequence of the form 0 → A → B → C to a sequence that is still exact (so 0 → F (A) → F (B) → F (C) is exact); in particular, F takes monomorphisms to monomorphisms. Similarly, we say that F is right exact if it takes an exact sequence of the form A → B → C → 0 to sequence that is still exact (so that F takes epimorphisms to epimorphisms). We say that F is exact if it is both left and right exact. This deﬁnition (and the subsequent discussion) is justiﬁed by the fact that many important functors are only half exact, witness the following examples.

EXAMPLES 2.9. (a) Given a ring R and an R-module M, then the functor ⊗RM : ModR → ModR, A 7→ A ⊗R M is right exact. It is not always left ex- ×5 act: if R = Z, then tensoring the exact sequence 0 → Z −−→ Z → Z/5 with Z/5 ×0 ×1 yields the non-exact sequence 0 → Z/5 −−→ Z/5 −−→ Z/5. (b) For R and M as above, we can consider HomR(M, −): ModR → ModR instead. This functor is left exact, but need not be right exact: taking R = Z and ×5 M = Z/5 and applying the functor HomZ(Z/5, −) to the exact sequence Z −−→ Z → Z/5 → 0 yields 0 → 0 → Z/5 → 0 which is evidently not exact. (c) For a space X we have a functor Γ: AbX → Ab which assigns to an abelian sheaf F its group of sections Γ(X, G) (which we often prefer to write H0(X, F)). This functor is left exact, but not always right exact. For example, lat X be the 0 ∞ 1 circle (with its usual topology). Let ES1 be the sheaf of C -functions on S and 1 0 ∞ 1 ES1 = ES1 dθ be the sheaf of C -differentials on S . Then we have a short exact sequence of abelian sheaves 0 d 1 0 → RS1 → ES1 −→ES1 → 0 (the differential of a function is zero if and only if it is locally constant and any differential on an open subset of S1 can locally be integrated to a function). But if we apply Γ(S1, −) we get

0 1 d 1 1 0 → R → E (S ) −→E (S ) → 0, Ei(S1) stands for the functions (i = 0) resp. differentials (i = 1) on S1. Exactness fails at E1(S1), for the differential dθ can not be integrated to function on S1: θ 2. ABELIAN CATEGORIES AND DERIVED FUNCTORS 155 is only defined up to an integral multiple of 2π and is not globally definable as a C∞-function, although dθ is globally defined. (d) Suppose f : X → Y is a continuous map of topological spaces. Then −1 0 00 f : AbY → AbX is exact, for if 0 → F → F → F → 0 is an exact sequence of abelian sheaves on Y , then 0 → f −1F 0 → f −1F → f −1F 00 → 0 is also exact: this property needs only be tested for stalks, and the stalk sequence at x is just 0 00 0 → Ff(x) → Ff(x) → Ff(x) → 0. (e) In the above situation, f∗ : AbX → AbY is left exact, but need not be right exact. In case Y is a singleton, we recover Example (c). (f) There are plenty variations and combinations of the previous examples. For example, we can start with a ringed space (X, O) and put R := Γ(X, O). Then Γ defines a functor ModO → ModR. This functor is left exact. If we are also given an O-module M, then A 7→ HomO(A, O) defines a left exact functor HomO(−, M): ModO → ModO. And getting a bit ahead of ourselves, we can also consider the functor HomO(−, M): ModO → ModR defined by A 7→ HomO(A, M). This is left exact as the composite of two left exact functors, namely as

HomO (−,M) Γ HomO(−, M): ModO −−−−−−−−→ ModO −−−−→ ModR. In hindsight, one may say that the desire to thoroughly understand and mea- sure how the half exact functors mentioned in examples (a) and (b) fail to be exact gave rise to the birth of homological algebra. It was observed in the 1950’s that the tools developed for this purpose can render the same services in the present setting. We go brieﬂy through the relevant notions.

DEFINITION 2.10. We say that an object I of C is injective if the functor hI = ◦ MorC(−,I): C → Ab is exact. We say that C has enough injectives, if for every object A of C there exists a monomorphism from A to an injective object. Dually, an object P of C is said to be projective if the functor MorC(P, −): C → Ab is exact. In particular, if we are given an epimorphism A → B, then then any morphism P → B lifts to a morphism P → A. We say that C has enough projectives, if for every object A of C there exists a epimorphism from a projective object to A.

Since MorC(−,I) is always left exact, the injectivity property amounts to: for any monomorphism A → B, any morphism A → I extends to a morphism B → I. Given an object A of C, then a resolution of A is simply a long exact exact sequence which begins with A:

0 1 2 0 → A −→α C0 −→d C1 −→d C2 −→·d · ·

We prefer to think of this a morphism of complexes: α : A → C •, where A is identiﬁed with the complex that has A placed in degree zero and zeroes elsewhere. The exactness property then amounts to the statement that α induces an isomorphism on cohomology: A =∼ h0(C •) and hk(C •) = 0 for k 6= 0. A complex with the property that it is exact on all nonzero degrees is alcso called an acyclic complex. We omit the proof of the following proposition, as it follows a standard pattern.

PROPOSITION 2.11. Suppose C has enough injective objects. Then (i) for every object A of C, there exists an injective resolution α : A → I • (by which we mean that each Ik is injective). 156 4. APPENDIX

(ii) Given a morphism φ : A → B in C, an arbitrary resolution α : A → C • and an injective resolution β : B → J •. Then φ can be extended to a morphism φ• of chain maps:

0 1 2 −−−−→ 0 −−−−→ A −−−−→α C0 −−−−→d C1 −−−−→d C2 −−−−→d · · ·        0 1 2 y φy φ y φ y φ y β 0 1 2 −−−−→ 0 −−−−→ B −−−−→ J 0 −−−−→d J 1 −−−−→d J 2 −−−−→d · · · and any two such extensions φ•, ψ• are chain homotopic.

In the remainder of this subsection, F : C → C0 is a left exact functor between abelian categories and it is assumed that C has enough injectives. With the help of the previous proposition we can now define a functor RiF : C → C0 as follows. Given an object A, choose an injective resolution A → I • and put RiF (A) := hi(F (I •)). Before we prove that this is well-defined, we first define RiF on morphisms. Given a morphism φ : A → B in C, choose injective resolutions A → I • and B → J • and lift φ to chain morphism φ• : I • → J •. This induces a chain morphism F (φ•): F (I •) → F (J •) and hence morphisms hi(F (I •)) → hi(F (J •)). Given our choice of injective resolutions, the homotopy class of φ• only depends on φ and hence the same is true for the homotopy class of F (φ•). We may therefore denote hi(F (I •)) → hi(F (J •)) provisionally by hi(φ, F ). If we have another morphism ψ : B → C and an injective resolution C → K •, then ψ can be lifted to a chain morphism ψ• : J • → K •. So ψ•φ• defines a chain a morphism I • → K •. We thus find that hi(ψφ, F ) = hi(ψ, F )hi(φ, F ). If we apply this to the case where φ is an isomorphism with inverse ψ, we find that hi(ψ, F ) is a two sided inverse of hi(φ, F ). In particular, if we take A = B and φ the identity, then it follows that the object hi(F (I •)) is unique up to unique isomorphism. We denote this object by RiF (A) and we denote the mor- i i i phism R F (A) → R F (B) induced by φ ∈ MorC(A, B) by R F (φ). We have thus constructed a functor RiF : C → C0, called the ith right derived functor of F .

PROPOSITION-DEFINITION 2.12. We have R0F = F and when A is injective, then φ ψ Ri(A) = 0 for i 6= 0. Moreover any short exact sequence 0 → A −→ B −→ C → 0 in C in C gives rise to a long exact sequence in C0:

RiF (φ) RiF (ψ) i · · · → RiF (A) −−−−−→ RiF (B) −−−−−→ RiF (C) −→δ Ri+1F (A) → · · · , which is functorial: we have a functor from the category of short exact sequences in C to the category of long exact sequences in C0. For any resolution A → C • of A we have for every n ≥ 0 a natural morphism hn(F (C •)) → RnF (C).

PROOF. Let A be an arbitrary object of C and let A → I • be an injective resolution. Since F is left exact, the exactness of 0 → A → I0 → I1 implies the exactness of 0 → F (A) → F (I0) → F (I1). This shows that R0F (A) = Ker(F (I0) → F (I1)) =∼ F (A). It is easily checked that under these isomorphisms, R0F (φ) = φ for any morphism in C. If A is injective, then we can regard A as its own injective resolution (put 0 in degrees 6= 0) and we ﬁnd that Ri(A) = 0 for i 6= 0. 2. ABELIAN CATEGORIES AND DERIVED FUNCTORS 157

The long exact sequence follows by choosing our injective resolutions compat- ibly with the short exact sequence: φ ψ 0 −−−−→ A −−−−→ B −−−−→ C −−−−→ 0    α  γ y yβ y φ• ψ• 0 −−−−→ I • −−−−→ J • −−−−→ K • −−−−→ 0 in the sense that α, β, γ are injective resolutions as part of a short sequence of complexes (show that you can do this). Since F is exact on a short sequences of injective objects,

F (φ•) F (ψ•) 0 −−−−→ F (I •) −−−−→ F (J •) −−−−→ K • −−−−→ 0 is also a short exact sequence of complexes. The associated long exact cohomology sequence is the exact sequence with the asserted properties. The last assertion follows from Proposition 2.11.

EXAMPLE 2.13. Let us return to the Examples 2.9-a,b. Let R be a ring and M an R-module. Since R has enough projectives, the right exact functor ⊗RM : R ModR → ModR has left derived functors; they are denoted Tori (M, −) familiar from homological algebra. Similarly, since R has enough injectives, the left exact functor HomR(M, −): ModR → ModR has right derived functors; they are denoted i ExtR(M, −). Injective resolutions have in general only a theoretical interest: they serve to define left derived functors, but in practice they are unfit to do computations with. Proposition 2.14 below implies that for calculations it suffices to have at our dis- posal a large collection of objects on which the Rk for k > 0 vanish.

Derived functors without injective objects. Let C and C0 be abelian categories. We have seen that in case C has enough injective objects, then any left exact i i 0 ∞ functor F yields a collection of functors {T = R F : C → C}i=0 and natural trans- φ ψ formations from the category of short exact sequences S : 0 → A −→ B −→ C → 0 in C to the category of long exact exact sequences in C0 that are of the form

T i(φ) T i(ψ) δi · · · → T i(A) −−−→ T i(B) −−−−→ T i(C) −→S T i+1(A) → · · · . The data needed to write down this sequence are of course just the morphisms i i i+1 i i δS : T (C) → T (A). Let us agree to call this system of data (the T and the δ ) a δ-package (or δ-functor). Notice that T 0 is then automatically left exact. We have an abstract De Rham theorem for such packages:

i i PROPOSITION 2.14 (Abstract De Rham theorem). Let (T , δ )i be δ-package for (C, C0) and let be A → C • is a right resolution in C. Then we have for every n ≥ 0 a natural morphism hn(T 0(C •)) → T n(A) and this is an isomorphism when T i(Cj) = 0 for 0 < i ≤ n − j.

PROOF. Denote by Zk the kernel of dk : Ck → Ck+1. We have short exact sequences 0 → A → C0 → Z1 → 0 and 0 → Zk → Ck → Zk+1 → 0 (for k ≥ 0). The associated long exact T -sequences yield for n > 0 a string of maps: T n(A) ←−δ T n−1(Z1) ←−·δ · · ←−δ T 1(Zn−1) ←−δ Coker(T 0(Cn−1) → T 0(Zn)). 158 4. APPENDIX

Since T 0 is left exact, we have T 0(Zn) = ker(T 0(Cn) → T 0(Cn+1) and so we can identify Coker(T 0(Cn−1) → T 0(Zn)) with hn(T 0(A)). The maps δ in our string are i j all isomorpisms when T (C ) = 0 for 0 < i ≤ n − j. Note that the preceding proposition applies in particular to right derived functors.

REMARK 2.15. The name comes from the fact that its content and proof are inspired by a key step in the proof of De Rham theorem. The Poincar´e lemma asserts that on a C∞-manifold M of dimension m, the complex of sheaves of differential forms

0 d 1 d d m 0 → RM → EM −→EM −→· · · −→EM → 0 • is exact. We think of this as a resolution RM → EM of the constant sheaf RM . With the help k of partitions of unity one can show that each EM is acyclic for the functor Γ: AbM → Ab. It k k follows that R Γ(M, RM ) (which we usually write as H (M, RM )) is computed as the kth • cohomology group of the De Rham complex Γ(M, EM ). There may exist several δ-packages for given C and C0. (That is for instance the case, if C has enough injectives and there are several left exact functors C → C0.) The δ-packages for the pair (C, C0) are objects of a category: a morphism i i ∞ 0i 0i ∞ from on such package (T , δ )i=0 to another such (T , δ )i=0 consists of natural transformations T i ⇒ T 0i (i = 0, 1,... ) such that we get for every morphism of short exact sequences in C a morphism of long exact exact sequences in C0. i i ∞ 0 DEFINITION 2.16. We say that the δ-package (T , δ )i=0 for (C, C ) is universal if 0i 0i ∞ 0 0 00 for every δ-package (T , δ )i=0 for (C, C ), every natural transformation T ⇒ T i i ∞ 0i 0i ∞ extends uniquely to morphism of δ-packages from (T , δ )i=0 to (T , δ )i=0. 0i 0i ∞ 0 In particular, if (T , δ )i=0 is a package with T0 = T0, then such a morphism 0i 0i ∞ of δ-packages always exists and is unique. So if (T , δ )i=0 is also universal, then this morphism has a two-sided inverse. Hence there is, up to unique isomorphism, i i ∞ 0 at most one universal δ-package (T , δ )i=0 for a given initial functor T . Here is a useful criterion for a δ-package to be universal. i i ∞ n PROPOSITION 2.17. A δ-package (T , δ )i=0 is universal if each T with n > 0 is effaceable, meaning that every object of C is the source of a monomorphism u in C with T n(u) = 0.

PROOF. We only deﬁne for every object A a morphism F i(A): T i(A) → T 0i(A), leaving to you the proof that they are functors. Suppose n > 0 and that the natural transformations F i : T i ⇒ T 0i have been constructed for i < n. Given A ∈ C, then choose a monomorphism u : A → B such that T n(u) = 0. Denote by B → C the cokernel of u so that we have a short exact sequence 0 → A → B → C → 0. Now consider the commutative diagram with exact rows T n−1(B) −−−−→ T n−1(C) −−−−→ T n(A) −−−−→0 T n(B)    n−1  n−1 yF (B) yF (C) T 0n−1(B) −−−−→ T 0n−1(C) −−−−→ T 0n(A) So T n−1(C) → T n(A) is an epimorphism whose kernel is contained in the kernel of the composite T n−1(C) → T 0n−1(C) → T 0n(A). It follows that there is unique morphism T n(A) → T 0n(A) which makes this diagram commute. To prove independence of u, let u0 : A → B0 be a second monomorphism with T n(u0) = 0. If 2. ABELIAN CATEGORIES AND DERIVED FUNCTORS 159

u we have a factorization u0 : A −→ B → B0, then it easily checked that we get the same morphism T n(A) → T 0n(A). But we can always reduce to this case by factoring u and u0 through the monomorphism (u, u0): A → B ⊕ B0 (observe that n 0 n 0n T (U, u ) = 0). So we have a well-deﬁned morphism T (A) → T (A). We get nothing new when C has enough injectives:

COROLLARY 2.18. Assume C has enough injectives. Then for every left exact func- 0 i i tor F : C → C the system of its right derived functors (R F, δ )i is a universal δ- package and conversely, any universal δ-package is of this form (so that T i is the ith derived functor of T 0).

PROOF. Every object A admits a monomorphism u : A → I with injective. Then for n > 0, RnF (u): RnF (A) → RnF (I) = 0 and so by Proposition 2.17, i i (R F, δ )i is universal as a δ-package. i i 0 Conversely, if (T , δ )i is a universal δ-package, then T is left exact, and so gives rise to a δ-package. Since this package is universal and has the same functor i i in degree zero, it must be isomorphic to (T , δ )i. Here are two applications.

PROPOSITION 2.19. Let f : X → Y be a continuous map of topological spaces so that we have defined the left exact functor f∗ : AbX → AbY . Then the right derived k functor R f∗ : AbX → AbY applied to an abelian sheaf F on X is the sheafification of the presheaf on Y which assigns to an open subset V ⊂ Y , Hk(f −1V, F).

k PROOF. The sheafification process defines for any k ≥ 0 a functor T : AbX → AbY . For k = 0 this functor is simply f∗. The {Tk} are part of a delta package: for a fixed open subset V ⊂ Y , the functor F 7→ Hk(f −1V, F) is clearly part of one and this then remains true if we let V vary by passing to Tk. For an injective sheaf I on k X, T (I) is zero for k > 0 and hence is universal. Now apply Corollary 2.18.

Let R be ring and L a free R-module. Then the functor HomR(L, −) is exact i and so ExtR(L, −) = 0 for all i > 0.

PROPOSITION 2.20. For every R-module M, we have natural R-homomorphisms

i ∼= i ExtR(M ⊗ L, N) −→ ExtR(M, HomR(L, N)). and these are isomorphisms when L is of ﬁnite rank.

PROOF. We have a natural transformation of functors

ModR ⇒ ModR : HomR(M ⊗ L, −) ⇒ HomR(M, HomR(L, −)) and this is an isomorphism in case L is of ﬁnite rank. Both functors are part of δ- i i i packages with T being represented by ExtR(M⊗L, −) resp. ExtR(M, HomR(L, −)). According to Corollary 2.18 the former is universal, and so the natural transforma- i i tion extends to one of δ-packages ExtR(M ⊗ L, −) ⇒ ExtR(M, HomR(L, −)) and this extension is unique. To see that the latter is also universal we check that the i ExtR(M, HomR(L, −)) are effaceable (and apply Proposition 2.17). If N is an R- module, and N,→ I embeds N in an injective module, then HomR(L, I), being i a direct sum of copies of I, is also injective and so ExtR(M, HomR(L, I)) = 0 for i > 0. So when L is of ﬁnite rank, the transformations above are isomorphisms. 160 4. APPENDIX

The following proposition is usually derived with the help of a spectral sequence.

PROPOSITION 2.21. Let F : C → C0 and G : C0 → C00 be left exact functors between abelian categories of which the source categories C and C0 have enough injectives. Then for every object A of C and every integer k ≥ 0 there are natural homomorphisms R(GF )k(A) → G(RkF (A)). Assume that every object of C admits a monomorphism to an injective object I with the property that F (I) is G-acyclic. This is an isomorphism when RjF (A) is G-acyclic for j < k. If RjF (A) = 0 for j 6= k, then we have even for every integer l a natural isomorphism R(GF )l(A) =∼ RGk+l(RcF (A)).

PROOF. Choose an injective right resolution A → I • and put C • := F (I •), so that RjF (A) = hj(C •) and Rj(GF )(A) = hj(G(C •)). Let Zj := ker(Cj → Cj+1) and Bj = Im(Cj−1 → Cj). Since hk(C •) is the cokernel of the monomorphism Bk → Zk we have a long exact sequence 0 → G(Bk) → G(Zk) → G(hk(C •)) → R1G(Bk) → R1G(Zk) → · · · As hk(G(C •)) is the cokernel of G(Ck−1) → G(Zk) (which factors as G(Ck−1) → G(Bk) → G(Zk)) is follows that we have a natural morphism Rk(GF )(A) = hk(G(C •)) → G(hk(C •)) = G(RkF (A)). We also see that this is an isomorphism if R1G(Bk) = 0 and R1G(Zk−1) = 0 (for G(Ck−1) → G(Bk) is then an epimorphism). We verify that this is so under the given hypotheses. We then may assume that each Cj is G-acyclic. If we feed this in the long exact sequence associated with the short exact sequence 0 → Zj−1 → Cj−1 → Bj → 0 we ﬁnd that we have for i ≥ 1 isomorphisms ∼ RiF (Bj) −→= Ri+1F (Zj−1). Consider the strings ∼ ∼ R1G(Bk)) −→= R2G(Zk−1) ←R2G(Bk−1) −→·= · · ← Rk+1G(B0) = 0, ∼ ∼ R1G(Zk−1) ← R1G(Bk−1)) −→= R2G(Zk−2) ← R2G(Bk−2) −→·= · · ← RkG(B0) = 0.

Since R2G(hk−1(C)) = ··· = Rk+1G(h0(C)) = 0, the left headed arrows in the ﬁrst line are all isomorphisms so that R1G(Bk)) = 0. Similarly, since R1G(hk−2(C)) = ··· = RkG(h0(C)) = 0, the left headed arrows in the second line are ismorphisms, except that R1G(Zk−1) ← R1G(Bk−1)) may only be just onto. The conclusion is still that R1G(Zk−1) = 0. The ﬁrst assertion follows. Assume now that RjF (A) = 0 for j 6= k. Then the above argument shows that for l > 0, RlG(Bk) = 0. This implies that RlG(Zk) → RlG(hk(C •)) = RlG(RkF (A)) is an isomorphism for l > 0. On the other hand, the sequence 0 → Zk → Ck → Ck+1 → · · · is exact and so we may regard this as a G-acyclic resolu- k l k k+l • k+l tion of Z . It follows that for l > 0, R G(Z ) = h (G(C )) = R (GF )(A).

3. Sheaf cohomology The category of modules over a ringed space. In this subsection we ﬁx a ringed space (X, O). 3. SHEAF COHOMOLOGY 161

DEFINITION 3.1. Let (X, O) be ringed space. An O-module is an abelian sheaf A on X such that for every open U ⊂ X, A(U) comes with the structure of an O(U)- module and for U 0 ⊂ U the restriction map A(U) → F(U 0) is a homomorphism of O(U)-modules (where O(U) acts on A(U 0) via the ring homomorphism O(U) → O(U 0)). We say that a sheaf-homomorphism between abelian sheaves φ : A → B that are O-modules is a O-homomorphism if for ever open U ⊂ X, φ(U): A(U) → B(U) is a homomorphism of O(U)-modules.

It is clear that we have thus deﬁned a category of O-modules ModO. Notice that when X is a singleton, then O is given a by a single ring R := O(X) and ModO is then simply the category of R-modules. The goal of this subsection is to show that in general the category ModO has many properties in common with a category of modules over a ring.

PROPOSITION 3.2. The category ModO is abelian. For its structure as an abelian category, a sequence of O-homomorphisms · · · → Ak → Ak+1 → · · · is exact if and k k+1 only for every x ∈ X, the stalk sequence · · · → Ax → Ax → · · · is so. PROOF. The category has an evident zero object: given an open U ⊂ X, then this assigns to U the zero module. The biproduct is also obvious: given O-modules A and B, then A ⊕ B assigns to U the direct sum A(U) ⊕ B(U). The kernel of φ : A → B assigns to U ker φ(U): A(U) → B(U). But if we assign to U the cokernel of φ(U) we get in general only a presheaf and we must sheaﬁfy to get Coker(φ). The last assertion is left as an exercise.

We will usually write HomO(A, B) (or just Hom(A, B) for the set of sheaf homomorphisms A → B; the structure of an abelian group that is intrinsic to fact that ModO is an abelian category is of course the obvious one. But there is also an object HomO(A, B) of ModO which assigns to U, the O(U)-module of sheaf homomorphisms HomO|U (A|U, B|U) (this is a sheaf indeed). In particular, an O- homomorphism φ ∈ HomO(A, B) is a section of HomO(A, B) over all of X. We further have a tensor product of two objects A and B of ModO (which should not be confused with the biproduct: it is not a product or a sum in ModO): this is the sheaf associated to the presheaf U 7→ A(U) ⊗O(U) B(U). The category ModR of modules over a ring R has enough injectives: for any R-module A there exists a monomorphism A → I with I injective. The same is true for ModO:

PROPOSITION 3.3. The category ModO has enough injectives: for every O-module A there exists a monomorphism of A to an injective object I of O. This I has the property that every local section is the restriction of a global section.

PROOF. Given x ∈ X, then the category ModOx is injective and so there exists a monomorphism of Ox-modules αx : Ax → Ix to an injective Ox-module Ix. We the deﬁne a sheaf I on X as construction which ignores the topology on X: it assigns Q to an open U ⊂ X the product x∈U Ix, so that a section s of I over U is given by a collection (sx ∈ Ix)x∈U . There is an evident O-homomorphism α : A → I: if a ∈ A(U), then α(a) = (αx(ax))x∈U . This is clearly a monomorphism. It remains to see that I is injective, that is, we must show that HomO(−, I) is exact. If F is an arbitrary O-module, then to give a homomorphism φ : HomO(F, I) amounts to giving a collection of stalk homomorphisms (φx ∈ HomOx (Fx → Ix)x∈X . 162 4. APPENDIX

If 0 → F 0 → F → F 00 → 0 is an exact sequence of O-modules, then so is every stalk 0 00 0 → Fx → Fx → Fx → 0 of this sequence. Since Ix is injective, this remains so if we apply HomOx (−,Ix). But this tells us that HomO(−, I) is exact when applied to our exact sequence.

If we take for O the constant sheaf ZX , then we ﬁnd that the functor Γ: AbX → Ab which assigns to an abelian sheaf A the abelian group Γ(X, O) is left exact. We will denote its right derived functors by Hi(X, −) and call it sheaf cohomology: Hi(X, F) = RiΓ(F).

REMARK 3.4. One can prove that for every paracompact Hausdorff space X, the singular cohomology of X with values in M is naturally isomorphic with the sheaf cohomology of X of the constant sheaf ZX . But for X a scheme endowed with its structure sheaf OX , the use of these coholomology modules will be of a different kind. Flasque sheaves. An abelian sheaf F on X is said be flasque if for every open subset U ⊂ X, the restriction map F(X) → F(U) is onto. In other words, every section of F over U must extend to one over X. The injective objects of ModO constructed in Proposition 3.3 have evidently this property. (Indeed, one can show that any injective object of ModO is flasque.) It is clear that if F is flasque, then for any open subset U ⊂ X, F|U is also flasque and that so is the sheaf which assigns to an open V ⊂ X, F(U ∩ V ) (if the −1 −1 inclusion is denoted j : U ⊂ X, then these are j F and j∗j F respectively).

LEMMA 3.5. Let F be a sheaf on a space X and let F 0 ⊂ F a ﬂasque subsheaf. Then for every open subset U ⊂ X, the map F(U) → (F/F 0)(U) is onto.

PROOF. Put F := F/F 0 and let s¯ ∈ F(U) for some open U ⊂ X. Consider the collection of pairs (U 0, s ∈ F(U 0)) with the property that U 0 is an open subset of U and s maps to s¯|U 0. This collection is partially ordered in an evident manner. By Zorn’s lemma there exists an maximal element (U 0, s ∈ F(U 0)). We show that 0 0 U = U (which evidently suffices). If not, then choose x ∈ U − U and let sx ∈ Fx map to s¯x ∈ F x. We then have a neighborhood Ux of x in U on which sx is defined 0 0 0 0 and maps to s¯|Ux. Then s|U ∩Ux−sx|U ∩Ux is an element of F (U ∩Ux) and hence 0 0 0 extends to all of X. If we denote this extension s and put sx := sx + s |Ux, then we 0 0 0 0 see that now s|U ∩Ux = sx|U ∩Ux. So this defines an element of F(U ∪Ux) which 0 0 0 maps to s¯|U ∪ Ux. This contradicts the maximality property of (U , s ∈ I(U )) and 0 we conclude that U = U.

PROPOSITION 3.6. Let F be a ﬂasque abelian sheaf F on a space X. Then F is acyclic for the global section functor: Hi(X, F) = 0 for all i > 0. More generally, if f : X → Y is a continuous map between topological spaces, then F is acyclic for the i direct image functor: R f∗F = 0 for all i > 0.

PROOF. We only prove the first assertion; the second then follows from Propo- sition 2.19. Let F → I be a monomorphism to an injective sheaf that we constructed in Proposition 3.3: first choose for every x ∈ X, an injective monomorphism Fx ,→ Ix and then let I(U) consist of the collections (sx ∈ Ix)x∈U . We put G := I/F. In view of Lemma 3.5, the sequence 0 → Γ(X, F) → Γ(X, I) → Γ(X, G) → 0, 3. SHEAF COHOMOLOGY 163 is then exact. Since I is acyclic for Γ, the long exact sequence yields: H1(X, F) = 0 and Hi(X, G) → Hi+1(X, F) for all i ≥ 1. We now prove with induction on n ≥ 1 that for any flasque sheaf H, Hi(X, H) = 0 for i ≤ n (we just checked the case n = 1). We do this by proving that G is flasque as well: if Hi(X, G) = 0 for i ≤ n, it then follows that Hi(X, F) = 0 for i ≤ n + 1. Any s¯ ∈ G(U) is by Lemma 3.5 the image of some s = (sx)x∈U ∈ I(U). But it is clear that we can extend s to an element s˜ ∈ I(X) by putting s˜x = 0 when x∈ / U (I is flasque). Then the image of s˜ in F(X) restricts to s¯ and so G is flasque. The advantage of the flasqueness property over injectivity is that it is one that it ignores any additional structure the abelian sheaf might have. We exploit this in the following corollary.

COROLLARY 3.7. For a ringed space (X, O), denote by ΓO : ModO → ModO(X) the global section functor for O-modules. Then for any O-module M, the abelian i i group underlying R ΓO(M) is H (X, M).

PROOF. We noted that the injective O-resolution M → I • produced in Propo- sition 3.3 is also one by flasque sheaves. By definition RkΓ(M) = Hk(Γ(X, I •)). According to Proposition 3.6, each Ik is acyclic for the global section functor AbX → Ab for abelian sheaves and so by the Abstract De Rham theorem 2.14, i this also computes H (X, M). i k This justifies our writing H (X, M) for R ΓO(M); it then should be understood that Hi(X, M) comes with the structure of a O(X)-module. In particular, if R is a i ring and we take for O the constant sheaf RX , then H (X, M) is a natural manner a RX (X)-module, and hence a R-module.

Comparison with Cechˇ cohomology. In this subsection X is a space and F is an abelian sheaf on X. Let A be a set. A k-simplex of A is simply an element k+1 of A , that is a sequence of (k + 1)-elements σ = (α0, . . . , αk). We have for k+1 k i = 0, . . . , k a map ∂i : A → A which forgets the element in slot αi. If Ck(A) denotes the free Z-module spanned by Ak+1, then we have a boundary operator d : Ck(A) → Ck−1(A) deﬁned by k X i d(σ) = (−1) ∂i(σ). i=0

This defines a complex C•(A) that can be identified with the simplicial (homological) complex of the abstract simplex spanned by A. If A 6= ∅, then this simplex is contractible, so has only homology in degree zero, namely Z, but it is also fairly straightforward to prove this directly (for A = ∅ we get of course the zero complex). Suppose now that A indexes an open covering U = {Uα}α∈A of X. We then k define find for every k-simplex σ = (α0, . . . , αk) of A an open subset Uσ = ∩i=0Uαi , which may or may not be empty. We put k Y C (U, F) := F(Uσ)

σ∈Ck(A)

(note that when Uσ = ∅, then F(Uσ) = {0} and so such a factor can be omitted). k k+1 In other words, an element of C (U, F) is given by a map s : σ ∈ A 7→ sσ ∈ 164 4. APPENDIX

• k k F(Uσ). We make it a (cohomological) complex C (U, F) by letting d : C (U, F) → Ck+1(U, F) be deﬁned by k+1 k k+2 X i d s : σ ∈ A 7→ (−1) s∂i(σ)|Uσ ∈ F(Uσ). i=0 This is called the Cechˇ complex of U with values in F. Its cohomology is called the • Cechˇ cohomology of F with respect to the covering U and is denoted by Hˇ (U, F). It is immediate from the deﬁnitions that Hˇ 0(U, F) = Γ(X, F) = H0(X, F).

k k EXERCISE 97. Let Calt(U, F) consist of the elements of C (U, F) that transform under permutations according to the sign character, i.e., the set of (sσ ∈ F(Uσ))σ such that sτ(σ) = sign(τ)sσ for all τ ∈ Sk+1. Show that this deﬁnes a subcomplex • • Calt(U, F) ⊂ C (U, F) and that this inclusion is a chain homotopy equivalence, hence yields the same cohomology. (This subcomplex is often more practical for doing computations.)

REMARK 3.8. This construction makes sense in much greater generality. Note that A deﬁnes a category: its objects are the ﬁnite sequences in A and its morphisms are the maps ∂i. Any contravariant functor from A to an abelian category C gives rise to such a complex. Such a functor is called a semisimplical object in C.

We also have a sheaf analogue. For any open jU : U ⊂ X consider the sheaf −1 jU∗jU F on X. So this sheaf assigns to the open subset V ⊂ X the group F(U ∩V ). We put Ck(U, F) := ⊕ j j−1F σ∈Ck(A) Uσ ∗ Uσ and deﬁne the sheaf homomorphism dk : Ck(U, F) → Ck+1(U, F) in the obvious way. Notice that then Γ(X, Ck(U, F)) = Ck(U, F).

PROPOSITION 3.9. This deﬁnes a resolution of F. In other words, the complex

0 1 2 0 → F → C0(U, F) −→Cd 1(U, F) −→Cd 2(U, F) −→·d · · is exact. In particular, we have natural homomorphisms Hˇ k(U, F) → Hk(X, F).

PROOF. Exactness is tested on stalks. For x ∈ X, let Ax ⊂ A be the set of α ∈ A for which x ∈ Uα. Then the stalk at x of the complex in question is Hom(C•(A), Fx).

This is acyclic, because C•(A) is. The last assertion is an application of Proposition 2.14.

COROLLARY 3.10. If F is flasque, then Hˇ k(U, F) = 0 for k 6= 0. −1 PROOF. In that case jU∗jU F is flasque for any open subset j : U ⊂ X. Hence each Ck(U, F), being a direct sum of flasque sheaves, is also flasque. Now apply the abstract De Rham theorem 2.14 and conclude that Hˇ k(U, F) = Hk(X, F). Accord- ing to Proposition 3.6 the latter is zero when k 6= 0. k COROLLARY 3.11. Suppose that H (Uσ, F) = 0 for all k > 0 and all σ. Then the natural maps Hˇ k(U, F) → Hk(X, F) are isomorphisms.

PROOF. The complex of sheaves C•(U, F) is a resolution of F. It is acyclic for the global section functor, for if l > 0, then for every k l k l H (X, C (U, F)) = ⊕σ∈Ak+1 H (Uσ, F) 3. SHEAF COHOMOLOGY 165 and this is 0 by assumption. Hence the natural map from hk(Γ(X, C•(U, F)) = ˇ k k H (U, F) to H (X, F) is an isomorphism. Absolute Cechˇ cohomology. Let v : A → A0 be a map of sets. Then v extends to a map between the simplices spanned by A and A0 and thus gives rise to map on 0 0 chain complexes v• : C•(A) → C•(A ). Suppose now that A indexes the members 0 0 of another open covering U = (Uα0 )α0∈A0 of X and that v is a ‘refiner map’ in the 0 0 sense that Uα ⊂ Uv(α). Then for any k-simplex σ of A, we have Uσ ⊂ Uv(σ) so that 0 we have a well-defined restriction homomorphism F(Uv(σ)) → F(Uσ). This gives rise to a homomorphism k 0 Y 0 Y k C (U , F) := F(Uσ0 ) → F(Uσ) = C (U, F) 0 0 σ ∈Ck(A ) σ∈Ck(A) 0 which sends (sσ0 ∈ F(Uσ0 ))σ0 to (sv(σ)|Uσ ∈ F(Uσ))σ. This is in fact a chain map C •(U0, F) → C •(U, F) so that v induces a maps on Cechˇ cohomology groups Hˇ k(U0, F) → Hˇ k(U, F). We claim that this last map is independent of v: if w : A → A0 is another refiner function, then we use the homotopies defined as follows: for k+1 σ = (α0, . . . , αk) ∈ A , and j = 0, . . . , k, put

(v ∗j w)(σ) := (v(α0), . . . , v(αj), w(αj), . . . , w(αk)). 0 Observe that then Uσ ⊂ U(v∗j w)(σ). If we deﬁne hk : Ck(A) → Ck+1(A ) by k k+1 X j k+2 σ ∈ A 7→ (−1) (v ∗j w)(σ) ∈ Ck+2(A) , j=0 then hk−1dk + dk+1hk = w − v. In addition hk induces a map k k+1 k+2 0 X j k+1 0 0 h (U, F):(sσ )σ ∈ C (U , F) 7→ (−1) s(v∗j w)|Uσ ∈ C (U, F) σ j=0

k k k k • We have hk+1d + dk−1h = w (U, F) − v (U, F) and hence h (U, F) defines a homotopy between the chain maps defined by v and w. In particular, they induce the same map Hˇ k(U0, F) → Hˇ k(U, F). Since every two open coverings U and U0 admit a common refinement (namely the collection U ∩ U 0 with U a member of U and U 0 a member of U0), we can form the direct limit: Hˇ k(X, F) := lim Hˇ k(U, F). −→ U This is the (absolute) Cechˇ cohomology of X with values in F. So an element of Hˇ k(X, F) is given by some c ∈ Hˇ k(U, F) with the understanding that c0 ∈ Hˇ k(U0, F) represents the same element if c and c0 have the same image in Hˇ k(U00, F) for some common refinement U00 of U and U0. The natural maps Hˇ k(U, F) → Hk(X, F) are compatible under refinement: if U0 is refined by U, then the natural map Hˇ k(U0, F) → Hk(X, F) is just the composition Hˇ k(U0, F) → Hˇ k(U, F) → Hk(X, F). Hence we have also natural maps Hˇ k(X, F) → Hk(U, F). We know that this is an isomorphism for k = 0. This is also the case for k = 1: 1 PROPOSITION 3.12. The natural map Hˇ 1(X, F) → H (X, F) is an isomorphism. 166 4. APPENDIX

PROOF. Choose a monomorphism F ,→ I with I flasque and denote by G its cokernel. Then the long exact sequence for cohomology produces an isomorphism 0 ∼ 1 H (X, G) = H (X, F). On the other hand, for every open covering U = (Uα)α of X, C •(U, I) is an acyclic complex and so the short exact sequence of complexes 0 → C •(U, F) → C •(U, I) → C •(U, I)/C •(U, I) → 0 yields an isomorphism h0(C •(U, I)/C •(U, I)) =∼ Hˇ 1(U, F). Notice that the group 0 • • h (C (U, I)/C (U, I)) is simpy the set of (sα ∈ I(Uα)/F(Uα))α with the property that sα and sβ have the same image in I(Uα ∩ Uβ)/F(Uα ∩ Uβ) for all α, β. In particular, such a system defines a section of G. Indeed, the resulting map h0(C •(U, I)/C •(U, F)) → H0(X, G). corresponds to the natural map Hˇ 1(U, F) → H1(X, F). The image consists of the sections s of G with the property that for every α, s|Uα lifts to a section of I|Uα. But for any given s there always exists an open covering of X for which this is the case. On the other hand, if (sα ∈ I(Uα)/F(Uα))α lies in the kernel, then there 0 is a refinement U of U such that (sα ∈ I(Uα)/F(Uα))α maps already to zero in h0(C •(U0, I)/C •(U0, F)). So if we pass to the direct limit over all the coverings we ˇ 1 1 find that H (X, F) → H (X, F) is an isomorphism. Morphisms between ringed spaces. We first recall an elementary fact from commutative algebra. If we are given a ring homomorphism φ : R0 → R, then 0 any R-module M can be regarded as R -module. This defines a functor ModR → ModR0 , called restriction. But we can also go in the opposite direction: from an 0 0 0 0 R -module M we get an R-module: R ⊗R0 M . Moreover, any R -homomorphism 0 0 α : M → M extends in an obvious manner to R-homomorphism R ⊗R0 M → M 0 0 by (r ⊗R0 m ) 7→ φ(r)α(m ) and this in fact establishes an bijection

0 ∼= 0 HomR0 (M ,M) −→ HomR(R ⊗R0 M ,M) 0 (the inverse is obtained by composing an R-homomorphism R ⊗R0 M → M with 0 0 0 0 0 the obvious R -homomorphism m ∈ M → 1 ⊗R m ∈ R ⊗R0 M ). In categorical language we may express this by saying that the functor R ⊗R0 − : ModR0 → ModR (which is called induction) is a left adjoint of the functor ModR → ModR0 . This generalizes in a straightforward manner to the sheaf setting: let f : (X, O) → (X0, O0) be a morphism between ringed spaces. We recall that this amounts to giving a continuous map X → X0 (also denoted f) and a sheaf homomorphism of rings f −1O0 → O on X, or equivalently, a sheaf homomorphism 0 0 O → f∗O of rings on X . If M is an O-module, then f∗M is in a natural man- 0 0 ner a f∗O-module and hence via O → f∗O a O -module. This deﬁnes a functor ModO → ModO0 . It has a left adjoint ModO0 → ModO, namely the functor which 0 0 −1 0 ∗ 0 assigns to an O -module M the O-module O ⊗f −1O0 f M . We shall write f M for this O-module so that we obtain a bijection

0 ∼= ∗ 0 HomO0 (M , f∗M) −→ HomO(f M , M). Notice that on stalks this is just the construction recalled above: we have ∗ 0 (f M )x = Ox ⊗ 0 M . Of(x) f(x) Corollary 3.7 and its proof (including its use of Proposition 3.6) extend in a straightforward manner to this setting: 3. SHEAF COHOMOLOGY 167

PROPOSITION 3.13. Let f :(X, O) → (X0, O0) be a morphism of ringed spaces and denote by f¯ : X → X0 the underlying (continuous) map. Then for any O-module 0 k k ¯ M, the abelian sheaf underlying the O -module R f(M) is R f∗(M). Sheaf cohomology modules and their relative versions behave contravariantly in the following sense:

PROPOSITION 3.14. Let be given a commutative square of ringed spaces:

0 g˜ (X , OX0 ) −−−−→ (X, OX )   0  f y fy 0 g (Y , OY 0 ) −−−−→ (Y, OY )

Then any OX -module M gives rise to natural OY 0 -homomorphisms ∗ j j 0 ∗ g R f∗(M) → R f∗(˜g M), j = 0, 1, 2,...

PROOF. We first do the case for sheaf cohomology. Let M → I • be an injective −1 −1 • −1 resolution. Then g˜ M → g˜ I is a right resolution by g˜ OX -modules (that need not be injective). By Proposition-definition 2.12, we therefore have a natural homomorphism hj(Γ(X0, g˜−1I •)) → Hj(X0, g˜−1M). The obvious chain map Γ(X, I •) → Γ(X0, g˜−1I •) gives rise to homomorphisms of O(X)-modules Hj(X, M) = hj(Γ(X, I •)) → hj(Γ(X0, g˜−1I •)) → Hj(X0, g˜−1M). This is easily checked to be independent of the choice of the injective resolution. Now compose this with the homomorphism Hj(X0, g˜−1M) → Hj(X0, g˜∗M) induced by the sheaf homomorphism g˜−1M → g˜∗M. This yields natural homomorphism of O(X)-modules Hj(X, M) → Hj(X0, g˜∗M) and hence one of O(X0)- 0 j j 0 ∗ modules O(X ) ⊗O(X) H (X, M) → H (X , g˜ M). The general case then follows from Proposition 2.19, for we have shown that for every open V 0 ⊂ Y 0 and every open V ⊂ Y with V 0 ⊂ g−1V , we have natural 0 0 j −1 j 0−1 0 ∗ O(V )-homomorphisms O(V ) ⊗O(V ) H (f V, M) → H (f V , g˜ M). Vector bundles and invertible sheaves over a ringed space. We fix a ringed space (X, O).

DEFINITION 3.15. An O-module F on X is said to be free of rank n if F =∼ On and is said to be locally free of rank n if we can cover X by open subsets U such that F|U =∼ On|U. An invertible sheaf on X is locally free O-module of rank one. Let F be a locally free on X of rank n. We may think of F as a sheaf of sections of an object that may be thought as a vector bundle: by deﬁnition X admits an open covering U = (Uα)α so that there exists an isomorphism of OUα -modules ∼ n hα : F|Uα = O |Uα (in vector bundle parlance, a local trivialization). We then obtain on an intersection Uα ∩ Uβ the isomorphism

α n hα hβ n gβ : O |Uα ∩ Uβ ←−−−− F|Uα ∩ Uβ −−−−→ O |Uα ∩ Uβ. ∼= ∼= α Note that gβ is given by an n × n-matrix with values in Rαβ := O(Uα ∩ Uβ). Since α gβ is invertible, we may regard it as an element of GLn(Rαβ) (for the zero ring this α is by deﬁnition the trivial group). The collection (gβ ∈ GLn(Rαβ))α,β satisﬁes the following two properties: besides the obvious fact that 168 4. APPENDIX

α (i) gα = 1n ∈ GLn(Rα) for all α, it satisfies α β γ (ii) the cocycle condition: gβ gγ gα = 1n ∈ GLn(Rαβγ ) for all α, β, γ. where Rα := O(Uα) and Rαβγ = O(Uα ∩ Uβ ∩ Uγ ). α The collection (gβ ) ∈ GLn(Rαβ))α,β can be understood as the system of glueing data needed to reconstruct F (in the context of vector bundles called the transition functions). The sheaf F can be reconstructed from this system: a section s of F n over an open subset U ⊂ X corresponds to a collection (sα ∈ O(U ∩ Uα) )α with α n the property that gβ |U ∩ Uα ∩ Uβ takes sα|U ∩ Uα ∩ Uβ ∈ Γ(U ∩ Uα ∩ Uβ, O) to α sβ|U ∩ Uα ∩ Uβ. In fact, given U, then any system (gβ ∈ GLn(Rαβ))α,β satisfying the two properties above thus defines a locally free O-module of rank n over X that is trivial on each member of U. We can express this in the language of sheaf theory: we have a sheaf GLn(O) with values in a group (nonabelian when n > 1) and we then refer to such a system as a Cechˇ 1-cocycle with values in this sheaf. We denote the collection of 1 × such systems by Zˇ (U, GLn(O)). Note that for n = 1, GLn(O) = O is a sheaf of abelian groups, and then this is an ordinary Cechˇ 1-cocycle. If we choose for 0 n a given U a different set of local trivializations hα : F|Uα → O |Uα, then we can 0 write hα = gαhα for some gα ∈ GLn(Rα) and so the associated cocycle is given by 0α α −1 α g β := (gβ|Uα ∩Uβ)gβ (gα |Uα ∩Uβ). We then say that the two Cech 1-cocycles (gβ ) 0α and (g β ) are cohomologous. In particular, when F is trivial, then we can choose a ∼ n 0 global trivialization h : F = O and use hα to be the restriction of h to Uα. Then 0α g β is the identity matrix and so it follows that there exist (gα ∈ GLn(Rα))α such α −1 that gβ = (gα|Uα ∩ Uβ)(gβ |Uα ∩ Uβ) for all α, β. So the set of cohomology classes, 1 denoted Hˇ (U, GLn(O)), can be understood as the set of isomorphism classes of locally free sheaves on X of rank n that are trivial on each member of U. For n > 1 this is just a set with a distinguished element (represented by the triv- 0α α −1 ial sheaf), but when n = 1, being cohomologous indeed means that g β (gβ ) = −1 ˇ 1 (gβ|Uα ∩ Uβ)(gα |Uα ∩ Uβ) is a coboundary and so H (U, GLn(O)) is then the usual Cechˇ cohomology group Hˇ 1(U, O×). The group structure has then a simple interpretation: if L and L0 are invertible sheaves on X that are both trivial on every 0 member of U, then L ⊗O L is an invertible sheaves with the same property. In fact, local trivializations for both invertible sheaves relative to U determine local trivializations for their tensor product and the Cechˇ 1-cocycle for the latter is just the product of the Cechˇ 1-cocycles of L and L0. Thus the group structure on Hˇ 1(U, O×) comes from the tensor product. In particular, if we take the direct limit over all coverings, we find

PROPOSITION-DEFINITION 3.16. The isomorphism classes of the invertible sheaves on X make up a group under tensor product and this group may be identiﬁed with H1(X, O×). This group is called the Picard group of X and usually denoted Pic(X) (although Pic(X, O) would be more correct).

PROOF. The preceding argument shows that the group if question can be iden- tiﬁed with lim Hˇ k(U, O×) = Hˇ 1(X, O×) and according to Proposition 3.12 the −→U 1 × latter maps isomorphically onto H (X, O ). Let f :(X, O) → (X0, O0) be a morphism of ringed spaces. If L0 is an invertible 0 ∗ 0 0 O -module, then f L = O ⊗f −1O0 L is an invertible O-module. Its isomorphism 3. SHEAF COHOMOLOGY 169 class only depends on the isomorphism class of L0 and indeed, we thus have deﬁned a group homomorphism Pic(X0) → Pic(X).

Cohomological dimension of a noetherian space. Let be given a X is a space. Suppose Y ⊂ X is a closed subset with complement U. We denote by i : Y ⊂ X and j : U ⊂ X the inclusions. For any abelian sheaf F on X, we have an −1 evident adjunction homomorphism F → i∗i F (we recall that i∗ : AbY → AbX −1 is a left adjoint of i : AbX → AbY ). On the level of stalks, this is the identity Fx = Fx when x ∈ Y and the zero map 0x → Fx if x ∈ U. There is a complementary construction involving U as follows. For any abelian sheaf G on U, we denote by j!G the sheaf on X obtained by extension by zero: it is the sheaf associated to the presheaf that assigns to an open subset V ⊂ U, G(V ) in case V ⊂ U and zero otherwise. In particular, the stalk of j!G at x is Gx when x ∈ U and is 0 if x ∈ Y . This sheaf is more directly characterized by saying that j!G(V ) is a subgroup of G(V ∩ U), namely the group of s ∈ G(V ∩ U) for which there exists neighborhood N of V ∩ Y in V such s|N ∩ U = 0. This is in fact a left −1 adjoint j! : AbU → AbX for the functor j : AbX → AbU : we have an evident ∼ −1 bijection Hom(j!G, F) = Hom(G, j F). In particular we have a natural adjunction −1 homomorphism j!j F → F. On the level of stalks, this is identity Fx = Fx when x ∈ U and 0 → Fx if x ∈ Y . We thus have obtained an exact sequence

−1 −1 0 → j!j F → F → i∗i F → 0.

It is clear from the definition that i∗ sends flasque sheaves to flasque sheaves. In −1 −1 k −1 particular, it sends a flasque resolution of i F to one of i∗i F so that H (Y, i F) = k −1 H (X, i∗i F) for all k. Hence we have a long exact sequence

k −1 k k −1 k+1 −1 · · · → H (X, j!j F) → H (X, F) → H (Y, i F) → H (X, j!j F) → · · ·

We will use this exact sequence in the proof of Grothendieck’s theorem below. Recall that a topological space X is said to be noetherian of dimension ≤ n if for any strictly descending chain X ⊃ X0 ) X1 ) ··· ) Xm of closed irreducible subsets we have m ≤ n.

PROPOSITION 3.17 (Grothendieck). If X is a noetherian space of dimension ≤ n, then for every abelian sheaf F on X we have Hk(X, F) = 0 for k > n.

PROOF. We first give the proof in case F is locally finitely generated in the sense that we can cover X by open subsets U with the property that there exists an N epimorphism ZU F|U for some N. Note that such sheaves make up an abelian lfg subcategory ModX of ModX . We then use induction on the dimension on X. The induction starts trivially when n = −1, for then X = ∅. Let F be locally finitely generated sheaf on X. If C1,...,Cr are the irreducible components of X, then each Cs contains an open-dense subset Us such that F|Ui is a constant and hence equal to As , where As := F(U ). We may of course assume Us s that the Us’s are pairwise disjoint. Let U := ∪sUs and i : Y = X − U ⊂ X. The exact sequence above becomes · · · → ⊕ Hk(C , j As ) → Hk(X, F) → Hk(Y, i−1F) → · · · , s s s! Us 170 4. APPENDIX

k −1 where js : Us ⊂ Cs is the inclusion. By induction assumption H (Y, i F) = 0 for k > n (even for k ≥ n) and so it remains to see that Hk(C , j As ) = 0 for k > n. s s! Us For this we use the short exact sequence 0 → j As → As → i As → 0, s! Us Cs s∗ Ys where is : Ys := Cs − Us ⊂ Cs. It gives rise to an exact sequence · · · → Hk−1(Y ,As ) → Hk(C , j As ) → Hk(C ,As ) → · · · s Ys s s! Us s Cs A constant sheaf on an irreducible space is clearly ﬂasque and so Hk(C ,As ) = 0 s Cs when k 6= 0. By the induction hypotheses Hk−1(Y ,As ) = 0 for k − 1 > n − 1 and s Ys so the desired vanishing follows. The general result then follows from the observation that an arbitrary abelian sheaf is the union of its locally ﬁnitely generated subsheaves and the fact that for abelian sheaves on a noetherian space, taking cohomology commutes with forming a direct limit (see Hartshorne [7], Ch. III Prop. 2.9). Ext modules. Let (X, O) be a ringed space as before and write R for the ring Γ(X, O). For a O-module M, we have covariant functors

HomO(M, −): ModO → ModO,

HomO(M, −) = Γ(X, HomO(M, −)) : ModO → ModR.

Both are left exact functors, which, since ModO possesses enough injectives, have right derived functors. We call these the local resp. global Ext functors and in- i i stead of writing R HomO(M, −) resp. R HomO(M, −) we usually denote them i i Ext O(M, −) resp. ExtO(M, −). REMARKS 3.18. We make some observations (which sometimes boil down to simple exercises) concerning these modules. (i) In case X is a singleton, then the datum of O just amounts to giving a ring R and then ModO = ModR. If M corresponds to the R-module M, then both functors coincide with HomR(M, −) whose right derived functors are the familiar i modules ExtR(M, −). (ii) If we take M = O, then these two functors are just the identity functor N 7→ N resp. the global section functor N 7→ Γ(X, N ) and hence we ﬁnd that i i i Ext O(O, N ) is zero unlessi = 0, whereas ExtO(O, N ) = H (X, N ). It also follows i that if M is a free O-module of ﬁnite rank, then Ext O(M, N ) is zero unless i = 0 i (in which case we get HomO(M, N )). i ◦ (iii) For a given O-module N , the Ext (−, N ) make up a δ-functor from ModO • to ModR. For let N → I be an injective right resolution. Then for any short exact sequence of O-modules 0 → M0 → M → M00 → 0, the short sequence of 00 • • 0 • complexes 0 → HomO(M , I ) → HomO(M, I ) → HomO(M , I ) → 0 is still exact and so gives rise to a long exact cohomology sequence. This is just

00 0 1 0 0 → HomO(M , N ) → HomO(M, N ) → HomO(M , N ) → ExtO(M , N ) → · · · i 00 i i 0 i+1 0 · · · → ExtO(M , N ) → ExtO(M, N ) → ExtO(M , N ) → ExtO (M , N ) → · · · It is however not always possible to interpret the Exti(−, N ) as derived functors of HomO(−, N ). (iv) Let L be a locally free O-module. Then HomO(L, −) is an exact functor and hence Ext i(L, −) = 0 for all i > 0. This has the following consequence. Let 3. SHEAF COHOMOLOGY 171

M and N be O-modules and let L• → M be a left resolution by locally free O- modules. Then the Abstract De Rham theorem 2.14 (or rather its dual form) shows i i that Ext O(M, N ) = h (HomO(L•, N )). Proceeding as in the proof of Proposition 2.20 we also ﬁnd for every O-module M natural transformations

i i Ext O(M ⊗O L, −) ⇒ Ext O(M, HomO(L, −)), i i ExtO(M ⊗O L, −) ⇒ ExtO(M, HomO(L, −)) that are isomorphisms when L is locally of ﬁnite rank. (v) Let j : U ⊂ X be an open subset and let M, N be O-modules. Then we ∗ ∼ ∗ ∗ have a natural isomorphism j HomO(M, N ) = Homj∗O(j M, j N ), which we regard as an isomorphism of functors ∗ ∼ ∗ ∗ j HomO(M, −) = Homj∗O(j M, j −): ModO → Modj∗O,

i ∗ i i ∗ ∗ that are parts of δ-packages with T equal to j Ext O(M, −) resp. Ext j∗O(j M, j −). These δ-packages are universal, for if I is an injective O-module, then for i > 0, i ∗ i ∗ ∗ Ext O(M, I) = 0 and, since j I is also injective, Ext j∗O(j M, j I) = 0. Hence for all i and all O-modules N we have a natural isomorphism ∗ i ∼ i ∗ ∗ j Ext O(M, N ) = Ext j∗O(j M, j N ).

k (vi) Let M and N be pair O-modules. We can regard ExtO(M, −) as the kth derived functor of the composite functor Γ Hom(M, −) and so by Proposition 2.21 k k we have a natural R-homomorphism ExtO(M, N ) → Γ(X, Ext O(M, N )). That l same proposition tell us that this is an isomorphism if Ext O(M, N ) is Γ-acyclic for l < k, once we observe that an injective resolution N → I • as constructed in k Proposition 3.3 has the property that HomO(M, I ) is ﬂasque for every k. If we combine this with the observations made in (ii), then it also follows that j when L is a locally free O-module of ﬁnite rank, then Ext O(L, N ) = 0 unless j = 0, j in which case we get HomO(L, N ). So if the O-module M admits a right resolution by such modules L• → M, then (iii) combined with the abstract De Rham theorem j implies that Ext O(M, N ) = hj(HomO(L•, N )). (vii) Let L, M, N be O-modules. Then for all i, j ≥ 0 there are natural homomorphisms

i j i+j Ext O(L, M) ⊗O Ext O(M, N ) → Ext O (L, N ), i j i+j ExtO(L, M) ⊗R ExtO(M, N ) → ExtO (L, N ). In particular, we ﬁnd (by taking L = O) natural R-homomorphisms

i j i+j H (X, M) ⊗R ExtO(M, N ) → H (X, N ). This is a special case of a much more general result, which is best presented in the setting of derived categories (which we here avoid). Let us just give the proof for the global Ext. Let M → I •. The point is that replacing M by I • yields the same Ext modules, even when M appears as the ﬁrst variable: let N → J • be a resolution by injec- • • tive O-modules and form the complex of O-modules HomO(I , J ): an element of • • • • • • HomO(I , J ) of degree k is a chain map φ : I → J (k) and d(φ): I → J (k+1) is the chain map which sends a ∈ Is to dφ(a) − φ(da). Then the chain map 172 4. APPENDIX

• • • of O-modules M → I induces a chain map of R-modules HomO(I , J ) → • HomO(M, J ) which induces an isomorphism on cohomology, so that j • • ∼ j • j h (HomO(I , J )) = h (HomO(M, J )) = ExtO(M, N ). Now observe that we have an evident chain map of R-modules

• • • • HomO(L, I ) ⊗R HomO(I , J ) → HomO(L, J ). (the source is a tensor product of complexes of R-modules) and as is well-known, we then have an induced Kunneth¨ map on cohomology:

i • j • • i+j • h (HomO(L, I )) ⊗R h (HomO(I , J )) → h (HomO(L, J )).

i • i • By deﬁnition, we may identify h (HomO(L, I )) with ExtO(L, I ) and likewise i+j • i+j h (HomO(L, J )) with ExtO (L, N ). The assertion follows. Bibliography

[1] M.F. Atiyah, I.G. Macdonald: Introduction to Commutative Algebra, Addison-Wesley (1969). 41, 101, 139 [2] D. Eisenbud: Commutative Algebra with a view toward Algebraic Geometry, GTM 150, Springer Verlag (1995) (there exists a Chinese edition). 139 [3] D. Eisenbud and J. Harris: The Geometry of Schemes, GTM 197, Springer Verlag (2000) (there exists a Chinese edition). [4] A. Grothendieck and J. Dieudonné: El´ éments de Géométrie Algébrique, Ch. 0-IV Publications Mathématiques de l’IHES (1960-1967); a new edition of Ch. I was separately published as vol- ume 166 in the Springer Grundlehren series (1971). [5] A.J. de Jong et alii: The stacks project, available at www.stacks.math.columbia.edu. [6] Fu Lei: Algebraic Geometry, Mathematics Series for Graduate Students, Tsinghua UP (2006) 100 [7] R. Hartshorne: Algebraic Geometry, GTM 52, Springer Verlag (1978) (there exists a Chinese edition). 94, 170 [8] Liu Qing: Algebraic Geometry and Arithmetic curves, Oxford Science Publications (2002) (there exists a Chinese edition). 89 [9] D. Mumford: The Red Book of Varieties and Schemes, Lecture Notes in Mathematics 1358, Springer Verlag (1988). [10] R. Vakil: MATH 216: Foundations of Algebraic Geometry, available on Vakil’s website as a wordpress blog.

173