Picard Groups of Affine Curves

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1 Picard Groups of Affine Curves Victor I. Piercey University of Arizona Math 518 May 7, 2008 Abstract We will develop a purely algebraic definition for the Picard group of an affine variety. We will then develop computational techniques for Picard groups of affine curves. If the curve is non-singular, the technique will require some geometry. However, in the singular case, we will find a purely algebraic tool. Along the way we will obtain an explicit description of the Picard group for every non-singular affine curve over C as well as affine curves with a single cusp or node singularity.

2 Contents 1 Introduction 1 2 Invertible Modules and the Picard Group Basic Definitions The Picard Group Picard Groups of Non-Singular Curves Weil Divisors Nonsingular Curves The Jacobian of a Riemann Surface and Pic(X) over C Non-Singular Affine Curves Singular Affine Curves Projective Modules over Fiber Products The Conductor Square The Cusp and the Node Conclusion 29 References 29

3 Picard Groups of Affine Curves 1 Introduction Victor I. Piercey Math 518 May 7, 2008 In algebraic geometry, one encounters invertible sheaves (or equivalently, line bundles) over an algebraic variety. Invertible sheaves over a fixed algebraic variety form a group under the tensor product called the Picard group. The elements of the Picard group are inherently geometric objects. We will begin with a purely algebraic definition of invertible sheaves and the Picard group. The algebraic varieties will be replaced by commutative rings with identity. If the variety is affine, the corresponding ring is the coordinate ring. Invertible sheaves in this context are invertible modules. The invertible modules form a group under the tensor product which we will call the Picard group. We will proceed to develop methods to compute the Picard groups of coordinate rings of affine curves. When the curve is nonsingular, we will have to use some geometry. We will have an explicit description if the underlying field is C. When the curve is singular, the technique is purely algebraic. It involves constructing a Meier-Veitoris sequence out of a commutative diagram called the conductor square. We will illustrate this method by computing the Picard group explicitly for a curve with a cusp and a curve with a node. 2 Invertible Modules and the Picard Group 2.1 Basic Definitions First we define what it means for a module over a ring to be invertible. Definition 2.1. For a ring A, an A-module I is invertible if I is finitely generated and for every prime ideal p Spec(A), I p = Ap as A p -modules. Remark 2.2. The condition that I p = Ap is often described by saying that I is locally free of rank 1. If I is an ideal in A and is invertible as an A-module, we say that I is an invertible ideal. The notation I for the modules is chosen because we will see that every invertible module is isomorphic to an invertible ideal. Before proceeding, we consider a couple of examples. Example 2.3. If I is a free module of rank 1, then I is invertible. Note that the isomorphism class of the free module of rank 1 will play a special role shortly. Example 2.4. Any principal ideal generated by a nonzero divisor is invertible. Example 2.5. Let A = Z[ 5] and let I = (2, 1 + 5). I claim I is not principle, but I is locally free of rank 1. 1

4 Suppose I = (x) for some x A. Let the norm N : A Z be given by N(a + b 5) = a 2 + 5b 2. Then 2 I and N(2) = 4, hence N(x) divides 4. Since I and N(1 + 5) = 6, N(x) divides 6. Therefore N(x) = ±1 or ±2. Since I is a proper ideal, x is not a unit and therefore N(x) ±1. Thus N(x) = ±2, and since N(x) > 0, it must be that N(x) = 2. But a 2 + 5b 2 = 2 requires b = 0, and there is no solution to a 2 = 2 in Z. Therefore I is not principle. Now we will show that I is locally free of rank 1. Note that I is a maximal ideal, since Z[ 5]/I = F 2. Now, if I p, then I (A \ p) which implies I p = A p. Therefore we suppose I p. But then I = p since I is maximal. Since 3 I (otherwise 1 I), 3 is invertible in A p. Therefore: 2 = (1 + 5)(1 5) 3 pa p. It follows that I p = (1 + 5) p which is principal and hence isomorphic to A p as A p -modules. Example 2.6. Let k be an algebraically closed field and let A = k[x, y]/(y 2 x 3 + x). The ring A is the coordinate ring for a non-singular, affine elliptic curve. Since the curve is non-singular, A is a Dedekind domain. I claim every maximal ideal m Max(A) is invertible. Since the only prime ideal p in Spec(A) with m p is p = m, it follows that m p = A p for all p m. Furthermore, since A is a Dedekind domain, A m is a discrete valuation ring whose unique maximal ideal ma m is principal (see Proposition 9.2 in [AM69]). Thus the maximal ideals are invertible. On the other hand, note that these ideals are not principal. There is a geometric reason for this. It comes down to the fact that when this affine curve is embedded in projective space P 2 k, the result is a curve of genus 1. If all maximal ideals were principal, then all ideals would be principal (since every ideal in a Dedekind domain is a finite product of primes). This would imply that the curve has genus zero. We will see a sketch of the last implication below. Remark 2.7. Note that to check whether or not a finitely generated module I is invertible, it suffices to check that I m = Am for all m Max(A). 2.2 The Picard Group The isomorphism classes of invertible modules from a group, called the Picard group. The operation is given by the tensor product. The identity is the isomorphism class of A as a module over itself. If I is an invertible module, its inverse is the dual module I = Hom A (I, A). Hence invertible modules are modules that have inverses in this sense. Before proving the theorem that justifies the statements above, we need a couple of preliminaries. First of all, given an A-module I, there is a natural map µ : I I A defined by ϕ a ϕ(a). The map µ is the unique homomorphism induced by the duality pairing on I I. Let K = K(A) denote the total ring of fractions. Since there is a natural map A K(A), K(A) is an A-module. The A-submodules of K(A) are called fractional ideals of A. If I is a finitely generated fractional ideal of A, we write all the generators over a common denominator. This shows that I is isomorphic as an A-module to an ordinary ideal of A. For any set I K(A), define I 1 := {s K(A) si A}. 2

5 Theorem 2.8. Let A be a Noetherian domain. 1. If I is an A-module, then I is invertible if and only if the natural map µ : I I A is an isomorphism. 2. Every invertible module is isomorphic to a fractional ideal of A. Every invertible fractional ideal contains a non-zero-divisor of A. 3. If I, J K(A) are invertible modules, then the natural map I J IJ given by s t st is an isomorphism, as is the natural map I 1 J Hom A (I, J) given by t ϕ t where ϕ t (a) = ta. In particular, I 1 = I. 4. If I K(A) is any A-submodule, then I is invertible if and only if I 1 I = A. PROOF: First, suppose I is invertible. Then for any prime p Spec(A), and µ localizes to the natural isomorphism A p Ap A p = Hom Ap (A p, A p ) = A p Ap A p µ p : A p Ap A p A p. Therefore, since the property of being an isomorphism is local, it follows that µ is an isomorphism. Suppose now that µ is an isomorphism. Then there are some ϕ i and a i such that ( n ) µ ϕ i a i = 1. i=1 Since µ is an isomorphism, for any prime p we have an isomorphism µ p : (I A I) p = I p Ap I p A p. I claim that I p = Ap where I p is generated by some a i. Since µ (ϕ i a i ) = ϕ i (a i ) = 1, and p is a proper ideal of A, it follows that there is some i such that ϕ i (a i ) p. Therefore we can define v = 1 ϕ i (a i ) A p and set a = va i A p. It follows that (ϕ i ) p (a) = 1. Note that (ϕ i ) p (a) = 1 implies a is not a zero devisor in A p. Therefore if we denote the cyclic A p -submodule of I p generated by a by A p a then A p a = A p as A p -modules and ker(ϕ i ) p A p a = 0. Now suppose for some α I p, (ϕ i ) p (α) = β 0 A p. Then (ϕ i ) p (α βa) = 0, so α βa ker(ϕ i ) p. Thus it follows that I p = A p a ker(ϕ i ) p. Similarly, we can consider the dual element ϕ a Hom Ap (I p, A p ) defined by ϕ a (a) = 1 and ϕ a (b) = 0 for any b A p a. By the same argument as above, we have I p = A p ϕ i ker(ϕ a ) with A p ϕ i = Ap. Thus: Since I p Ap I p = (A p ϕ i Ap A p a) (A p ϕ i ker(ϕ i ) p ). µ p (A p ϕ i Ap ker(ϕ i ) p ) = (ϕ i ) p (ker(ϕ i ) p ) = 0 and µ p is an isomorphism, A p ϕ i Ap ker(ϕ i ) p = 0. Since A p ϕ i = Ap, it follows that ker(ϕ i ) p = 0 and (ϕ i ) p is an isomorphism. Hence I p = A p a = A p. Therefore I is locally free of rank 1. Note that a i is 3

6 mapped to a generator in I p. It follows that the natural map (a 1,..., a n ) I is locally surjective, hence surjective. Thus I is generated by {a 1,..., a n }. Therefore I is finitely generated and locally free of rank 1. Thus I is invertible. This proves the first statement. For the second statement, suppose I is an invertible module of A. Note that since A is a domain, (0) SpecA and K = A (0). Then by definition and the assumption that I is invertible: Therefore it suffices to find an embedding I A K = I A A (0) = I(0) = A(0) = K. I I A K = I (0). I claim that the natural map η : I I (0) is a monomorphism. First observe: η(α) = 0 α 1 = 0 uα = 0 for some u A \ {0}. Let m MaxA be arbitrary. Localize at m and view u and α in A m. Then uα = 0 in I m = Am. Since A m is a domain, u = 0 or α = 0 in A m. If u = 0 in A m, then there is some element u A \ m such that u u = 0. This is not possible, since 0 m implies u 0 while u 0 in A and A being a domain implies that u is not a zero divisor. Thus u 0 in A m and therefore α = 0 in A m. Since m MaxA was arbitrary, it follows that α = 0 in I m for every m and thus α = 0. Therefore η is injective and thus is an embedding. It follows that I = η(a) K. This proves the first part of the second statement. Now suppose I K is a finitely generated fractional ideal such that I A = {0}. Let u be the product of the denominators of the generators of I. Then u A and ui A I = {0}. Therefore, for every α I, uα = 0. Since A K = A (0) canonically, and I K, then uα = 0 for every α I is an equation in K. Since K is a field and u 0, it follows that I = (0). Since (0) is not invertible, it follows that I is not invertible. Therefore, by contrapositive logic, every invertible submodule of K contains a nonzero (hence nonzerodivisor) element of A. This completes the proof of the second statement. Now if I and J are invertible modules, by the second statement we can assume that I, J K. The map I J IJ given by s t st is an A-module homomorphism that is clearly surjective. Therefore it suffices to show that this map is injective. By exactness of localization, it suffices to show that this map is injective when localized at p for any p SpecA. Note that K p = K. Therefore we wish to show that the map I p Ap J p (IJ) p K(A p ) K is injective. Since A p is a domain and K(A p ) is its field of fractions, we may assume A is a local ring. If (A, m) is a local ring, then A = A m. Therefore if I and J are invertible, I = I m, J = J m and A = A m which implies that I = A = J as A-modules. Therefore there are nonzero elements s, t K such that I = (s), J = (t) and IJ = (st). It follows that the desired map is injective, completing the first part of the third statement. 4

7 Now consider the map I 1 J Hom A (I, J) given by t ϕ t where ϕ t (a) = ta. Note that t I 1 J implies that t = a i r i s i with a i A, r i I 1 and s i J. Therefore: ( ) ta = ai r i s i a = a i (r i a)s i. Since r i I 1 and a I, r i a A and therefore ta J. It is clear that ϕ t is a homomorphism. Now if v A I is nonzero (which exists by the second statement of the theorem proved above), then tv 0 in K whenever t 0. Thus ϕ t = 0 if and only if t = 0 and our map is injective. Therefore it suffices to show that the map t ϕ t is surjective. It suffices to show this locally. Let p Spec(A). Let ϕ Hom Ap (I p, J p ) be any homomorphism. Since I is invertible, I p = Ap and therefore I p = xa p for some x K(A). Let w = ϕ(x). Note that if α = xa I p (where a A p ), then (1/x)(xa) = a A p and therefore 1/x Ip 1. Then w/x Ip 1 J p. 1 The claim is that ϕ = ϕ w/x. Note that ϕ w/x (x) = w = ϕ(x). Since x is a generator of I p, the claim follows. Therefore our map is surjective locally, and hence is surjective. Finally, suppose I K is invertible. Then the map µ : I I A is an isomorphism, and by the third part proved above this map is the same as the multiplication map I 1 I A. It follows that I 1 I = A. Since I 1 I A it follows that I 1 I = A. This establishes one direction of the fourth statement. Now suppose I K is an A-submodule with I 1 I = A. By localization, we can assume A is a local ring with maximal ideal m. It then suffices to show under this assumption, I = A. Since I 1 I = A, there is an element v I 1 and an element α I such that vα m. Therefore vi m. I claim that the map v : I A given by α vα is an isomorphism. Since vi m implies v 0, then vα = 0 in A K implies α = 0, so multiplication by v is injective. Now since vi m, there is α I such that vα = y / m. Since m is the unique maximal ideal in the local ring A, it follows that y A. Therefore we have vy 1 α = 1 with y 1 α I. It follows that if a A, then v(ay 1 α) = a with ay 1 α I. Hence v is surjective. This completes the proof of the fourth statement and hence the theorem. Corollary 2.9. The collection of isomorphism classes of invertible modules of A form a set. PROOF: By the second part of Theorem 2.8, each isomorphism class of invertible modules of A has a representative as a submodule of K. Note that distinct invertible submodules of K may be isomorphic! In any case, the collection of isomorphism classes of invertible modules of A can be realized as a subset of the power set of K, and is therefore a set. Corollary The set of isomorphism classes of invertible modules of A is a group under with identity element given by A (as a module over itself) and the inverse of I given by I. PROOF: This is an immediate consequence of Theorem 2.8. Definition The Picard group of A, denoted by Pic(A), is the group of all isomorphism classes of invertible A-modules. 1 I am tacitly asserting that (I 1 ) p = (I p ) 1, which is true over a Noetherian ring. 5

8 Before considering Picard groups of rings of affine curves, we should look at the relationship between Pic(A) and invertible submodules of K. We will exploit this relationship in the next section. Observe that the collection of all invertible submodules of K (NOT isomorphism classes!) is a subset of the power set of K and is therefore a set. This set also forms a group with identity element given by A as a module over itself and where the inverse of I is given by I 1 (hence the notation!). All of this follows from Theorem 2.8. This allows us to give the following definition. Definition The group of invertible submodules of K is called the group of Cartier divisors and denoted by C(A). Definition A principal divisor is an element of C(A) of the form Au for u K. Denote the set of principal divisors by P C(A). Remark Note that if Au and Av are principal divisors, then Au Av = A(uv), and uv K. It follows that P C(A) is a subgroup of C(A). Lemma P C(A) = K /A. PROOF: We clearly have a surjective map K P C(A) given by u Au. Suppose Au = Av. Then there is a A such that u = av. Similarly, there is b A such that v = bu. It follows that (ab)u = u. Since u is a unit in K, it follows that ab = 1. Therefore u and v differ by a unit of A. Hence the kernel of our map is A, completing the proof. The relationship between C(A) and Pic(A) is given by the following. Corollary Let ϕ : C(A) Pic(A) be given by ϕ(i) = [I] (where [I] is the isomorphism class of the invertible module I). 1. The map ϕ is a surjective group homomorphism and ker ϕ = K /A. 2. The group C(A) is generated by the set of invertible ideals of A. PROOF: It is clear that ϕ is a group homomorphism and surjectivity follows by part (2) of Theorem 2.8. Therefore it suffices by Lemma 2.15 to show that the kernel of ϕ is P C(A). It is clear that if Au P C(A), then Au = A is an A-module under the map u 1. Therefore one inclusion is obvious and it suffices to show the other inclusion. If I is any invertible module and Au P C(A), then (Au)I = u(ai) ui on the one hand, while clearly ui = (1 u)i (Au)I. Therefore (Au)I = ui for any Au P C(A). Accordingly, it suffices to show that if I, J C(A) and I = J, then J = ui for some u K. But this is immediate from the third statement in Theorem 2.8, since any isomorphism ψ : I J is an element of Hom A (I, J) and thus can be realized as multiplication by a nonzero element u I 1 J (since multiplication by zero would not be an isomorphism). Any nonzero element of I 1 J is a nonzero element of K and therefore a unit of K. Thus J = ψ(i) = ui with u K. This proves the first statement. For the second statement, suppose I K is an invertible fractional ideal. Then I 1 is an invertible fractional ideal by part (4) of Theorem 2.8, and therefore by the second part of Theorem 2.8, there is a nonzero element a A I 1. Since a I 1, ai A. We now have I = ai (a) 1, where ai and (a) 1 are invertible ideals of A. Corollary Pic(A) = C(A)/P C(A). PROOF: This is immediate from Corollaries 2.16 and

9 3 Picard Groups of Non-Singular Curves We will now determine a method for computing Picard groups of nonsingular curves over an algebraically closed field k. If X is a nonsingular affine curve, its Picard group Pic(X) is the Picard group Pic(A) where A is the affine coordinate ring k[x 1,..., x n ]/I of the curve. We will be able to describe these Picard groups explicitly when the ground field is C. To lay the ground work for this discussion, we begin with objects called Weil divisors. The Weil divisors form a group, and after taking a quotient we obtain a group called the class group of the curve. We will then relate the class group, the group of Cartier divisors, and the Picard group. This will allow us to write down a split short exact sequence that can be used to compute the Picard group of a nonsingular curve embedded in projective space. We can then pull this program back to obtain the Picard group of the affine curve. 3.1 Weil Divisors We begin with the definition of a Weil divisor. Definition 3.1. The group of Weil divisors of a ring A, denoted Div(A), is the free abelian group on the set of codimension one prime ideals of A. A Weil divisor is an element of Div(A). Weil divisors and Cartier divisors are generally speaking very different. However, the following shows that when A is a Dedekind domain, they are actually the same. This will suffice for us since affine rings of nonsingular curves are Dedekind domain. Lemma 3.2. Every nonzero proper ideal I A of a Dedekind domain A can be written uniquely as the product of finitely many prime ideals. PROOF: See Theorem 3.7 in [J.S08]. Lemma 3.3. A product of invertible ideals in a Noetherian ring A is invertible. PROOF: Let I, J A be invertible. View I and J as fractional ideals in K(A). Then Therefore IJ is invertible. (IJ)(IJ) 1 = (IJ)(J 1 I 1 ) = I(JJ 1 )I 1 = IAI 1 = II 1 = A. Lemma 3.4. Every nonzero prime ideal in a Dedekind domain A is invertible. PROOF: If p A is a nonzero prime ideal, since A is a Dedekind domain, p is maximal. Hence for any other prime ideal q A, pa q = A q. Since A is a Dedekind domain, A p is a discrete valuation ring and hence its unique maximal ideal pa p is principal (see [AM69], Proposition 9.2). Therefore as A p -modules, pa p = Ap. Hence p is invertible. Lemma 3.5. Every nonzero ideal in a Dedekind domain A is invertible. PROOF: This is an immediate consequence of Lemma 3.2, Lemma 3.3 and Lemma 3.4. Proposition 3.6. If A is a Dedekind domain, the Cartier divisors C(A) is the free abelian group on the set of maximal ideals of A. 7

10 PROOF: First of all, suppose I A is a proper invertible ideal. Then I is nonzero, so I is an element of the free abelian group generated by the nonzero prime ideals of A. Since any nonzero prime ideal is a maximal ideal in a Dedekind domain, it follows that I is an element of the free abelian group on the maximal ideals of A. Finally, every fractional ideal in C(A) is an element of the free abelian group on maximal ideals of A by the second part of Corollary The reverse inclusion is obvious by Lemma 3.3 and Lemma 3.4. We now have the following immediate corollary. Corollary 3.7. In A is a Dedekind domain, C(A) = Div(A). PROOF: In a Dedekind domain, by definition the codimension 1 prime ideals are precisely the maximal ideals. The result therefore follows from Proposition 3.6. As above, in the case where A is a Dedekind domain, principal divisors in Div(A) are the same as in C(A). Definition 3.8. Let A be a Dedekind domain. Then the group of Weil divisors modulo principal divisors is called the class group and denoted Cl(A) = Div(A)/P C(A). Two elements that are equivalent modulo PC(A) are said to be linearly equivalent. We now have the following corollary. Corollary 3.9. If A is a Dedekind domain, Pic(A) = Cl(A). PROOF: By the first part of Corollary 2.16, and by Corollary 3.7 since A is a Dedekind domain, we have: as claimed. Pic(A) = C(A)/P C(A) = Div(A)/P C(A) = Cl(A) Remark Suppose A is a Dedekind domain. Then A is a unique factorization domain if and only if Pic(A) is trivial. PROOF: First of all, it is clear that P C(A) can be identified with the principal ideals of A. Therefore it is clear that in a Dedekind domain, A is a principal ideal domain if and only if P C(A) = Div(A), which holds if and only if Cl(A) is trivial. In addition, if A is a Dedekind domain, A is a principal ideal domain if and only if A is a unique factorization domain. See Proposition 3.18 on page 45 of [J.S08] for the proof. Finally, since Pic(A) = Cl(A) for a Dedekind domain A, it follows that A is a uniform factorization domain if and only if Cl(A) is trivial if and only if Pic(A) is trivial. 3.2 Nonsingular Curves We now start with a nonsingular curve X of genus g embedded into projective space X P N for some N. This embedding allows the curve to satisfy a completeness hypothesis. So far we have only discussed Pic(A) for a ring A and by extension Pic(X) for an affine curve X. A projective curve does not correspond to a single coordinate ring. Instead, it has affine patches that have corresponding coordinate 8

11 rings, which may not all be the same. 2 Therefore it is necessary to make sense of the Picard group in this situation. Taking a collection of invertible modules over the affine coordinate rings and gluing them together in a consistent manner forms a sheaf of modules that is locally free of rank one. Such a sheaf is called an invertible sheaf. Invertible sheaves are precisely line bundles over the curve. In the same way that invertible modules form a group, the invertible sheaves form a group under tensor product. The identity element is the trivial sheaf and the inverse is given by the dual. The group of isomorphism classes of invertible sheaves over X is now the Picard group Pic(X). Restricting to an affine part of the curve gives the Picard of the corresponding coordinate ring. We will relate the Picard group Pic(X) to the divisor class group Cl(X) when X is nonsingular. To do so, we also need to make sense of Weil divisors on a projective curve. A Weil divisor can be thought of as a formal (finite) sum of integral multiples of points on the projective curve. The group of Weil divisors Div(X) is then the free abelian group on the points of the curve. Linear equivalence is defined a little differently from above. One defines a principal divisor to be the formal sum of zeros and poles of certain functions on the curve whose coefficients are the corresponding order (where the order of the zero or pole is determined by a certain discrete valuation on a certain function field on the curve). See Chapter II, Lemma 6.1 and definitions that follow in [Har77] for details. 3 We can then form the quotient Cl(X) as above. The class group is useful here because of the following result. Lemma There is a surjective group homomorphism D : Div(X) Z that descends to a well defined group homomorphism d : Cl(X) Z. PROOF: For a divisor n i p i, define ( ) D ni p i = n i. It is clear that this defines a group homomorphism. Surjectivity is also clear, since for any n Z, D(np) = n. Finally, if (f) is a principal divisor, D(f) = 0. See Proposition 6.4 in Chapter II of [Har77]. Therefore if two divisors are linearly equivalent, they have the same image under D. Definition The map d in Lemma 3.11 is called the degree map. For a nonsingular projective curve X, we have Pic(X) = Cl(X). Hence we have a degree map on Pic(X). Definition For an isomorphism class of invertible sheaves (line bundles) L Pic(X), the degree of L is the image under d of the corresponding Weil divisor class. The isomorphism classes of invertible sheaves of degree zero are the kernel of the degree map and therefore form a subgroup of Pic(X) denoted by Pic 0 (X). Proposition Pic(X) = Z Pic 0 (X) PROOF: Since the degree map from Pic(X) Z is surjective and its kernel is Pic 0 (X), we have a short exact sequence of abelian groups 0 Pic 0 (X) Pic(X) Z 0. Since Z is a projective Z-module, this short exact sequence splits and we obtain the desired result. 2 This actually describes the space of a scheme. The other data involved in a scheme involves keeping track of the coordinate rings of open (and hence affine) subsets. These coordinate rings form a sheaf, called the structure sheaf. 3 The definitions are modeled after meromorphic functions on Riemann surfaces. 9

12 3.3 The Jacobian of a Riemann Surface and Pic(X) over C By Proposition 3.14, describing Pic(X) for a nonsingular projective curve X is reduced to describing Pic 0 (X), the group of invertible sheaves (line bundles) of degree zero. This has a nice, explicit description if we assume the underlying ground field is C. Over C, a nonsingular projective curve X is a compact Riemann surface of genus g. To understand Pic 0 (X) we take a brief detour and discuss the Jacobian of a Riemann surface. Recall that for a Riemann surface X of genus g, the first singular homology group is given by H 1 (X, Z) = Z 2g. Recall that a holomorphic 1-form ω on X is closed. Accordingly, if c = A is a boundary in H 1 (X, Z), then by Stokes theorem: ω = ω = dω = 0. c A Accordingly, denoting the vector space of holomorphic one-forms by Ω 1 (X), for each [c] H 1 (X, Z) we have a well-defined linear functional: : Ω 1 (X) C. This gives us an embedding H 1 (X, Z) (Ω 1 (X)). Definition The Jacobian of the curve X, denoted by J(X), is defined as: [c] J(X) = (Ω 1 (X)) /H 1 (X, Z). Proposition If X is a Riemann surface of genus g, then A J(X) = T 2g where the isomorphism is in the category of abelian groups. PROOF: The dual (Ω 1 (X)) is a complex vector space of dimension g, the genus of X. Identify (Ω 1 (X)) with C g. The corresponding embedding H 1 (X, Z) C g gives us H 1 (X, Z) = Λ where Λ C g is a rank 2g lattice. Accordingly, as abelian groups: which is the desired result. J(X) = C g /Λ = R 2g /Z 2g = T 2g A very deep theorem of Abel and Jacobi shows that Pic 0 (X) = J(X). See Chapter VIII of [Mir95] for details. This gives us the following immediate corollary. Corollary If X is a nonsingular projective curve over C of genus g, then Pic(X) = Z T 2g. PROOF: By Proposition 3.14, the Abel-Jacobi theorem, and Proposition 3.16: Pic(X) = Z Pic 0 (X) = Z J(X) = Z T 2g. Remark As an interesting observation, note that if X is an elliptic curve, Pic(X) = Z X. Remark A curve of genus zero is called a rational curve. A rational curve over C embedded in projective space has Picard group Pic(X) = Z. Since X = P 1, we can explicitly describe a generator for the Picard group. The generator is the so-called tautological bundle. This is the line bundle over P 1 where the fiber over a point l P 1 is the line in C 2 represented by l. 10

13 3.4 Non-Singular Affine Curves Now suppose A = k[x 1,..., x n ]/I is the coordinate ring of a non-singular affine curve X. To compute Pic(A), we will complete the curve to a projective curve X, obtaining the Picard group Pic(X ) as above, and determine how the Picard group responds when restricting to the affine part that we started with. The first observation is that our coordinate ring gives us a closed embedding of our curve X := SpecA into A n = Speck[x 1,..., x n ]. Complete A n to P n. This amounts to adding in the hyperplane at infinity, which can be chosen to be the hyperplane in P n given by the homogeneous equation z 0 = 0 where P n is given homogeneous coordinates [z 0 : z 1 : : z n ]. The curve X is then completed using this embedding, and it requires the addition of finitely many points. Therefore, given our complete curve X in P n, we can return to X by removing finitely many points. The question therefore becomes how the Picard group of X changes when finitely many points are removed. To see how this works, we exploit again the isomorphism between the Picard group Pic(X ) and the class group Cl(X ). Proposition Let X be a nonsingular curve embedded in projective space P n. Let Z = {p 1,..., p k } X be a finite subset of the points of X and let U = X \ Z. Then there is an exact sequence: Z k Cl(X ) Cl(U) 0. PROOF: Map Z k Cl(X ) by the map sending the i th generator of Z k to the divisor p i. Then the image is the residue of the subgroup Zp 1 + Zp Zp k modulo principal divisors. Define the map Cl(X ) Cl(U) by mapping a divisor of the form n i q i on X to a divisor n i q i where n i = n i if q i U and zero otherwise. Then this map is surjective, and the kernel is precisely the image of Z k. Therefore this sequence is exact. Corollary Let A be the coordinate ring of an affine non-singular curve X of genus g over C. Let k be the number of points added in its embedding into projective space. Then: Pic(A) = T 2g /Z k 1. PROOF: Let X denote the image of X in projective space. By Corollary 3.17 we have Pic(X ) = Z T 2g. Let Z be the set of points of X \ X. Then (since Pic(A) = Pic(X)) by Proposition 3.20, we have an exact sequence: Z k Pic(X ) Pic(A) 0 and hence as desired. Pic(A) = Pic(X )/Z k = (Z T 2g )/Z k = T 2g /Z k 1 Remark Over an arbitrary field k, the same argument shows that under the same hypotheses, Pic(A) = Pic 0 (X )/Z k 1. 11

14 One consequence is the following. Corollary Let A be the coordinate ring of an affine non-singular curve X. Then A is a unique factorization domain if and only if the curve is rational. PROOF: I will prove this only over C. Let k be the number of points at infinity for the curve X and let g be the genus. Then it follows that Cl(A) = Pic(X) = T 2g /Z k 1. If g 0, then T 2g is uncountable and Z k 1 is countable and hence Cl(A) is nontrivial and hence A is not a unique factorization domain. Conversely, if g = 0 then Cl(A) is trivial and hence A is a unique factorization domain by Remark The proof over an arbitrary field k requires showing that for the completion X of X in projective space, Pic 0 (X) is uncountable. 4 Singular Affine Curves We now turn to the case where X is a singular affine curve. At the end of this discussion we will compute Pic(X) where X = SpecA for A = k[x, y]/(y 2 x 3 ) and Pic(Y ) where Y = SpecB for B = k[x, y]/(y 2 x 2 (x + 1)). The first curve is a curve with a cusp and the second curve has a node. The method used for non-singular curves no longer works in the singular case. In the non-singular case, the ring A is a Dedekind domain and we have isomorphisms with the divisor class group. Affine coordinate rings for singular curves are Noetherian domains, but they are not Dedekind domains. Therefore a lot of the work we did above fails. To compute Picard groups in this new context, we will be able to map Pic(A) to Pic(A 1 ), where A 1 is the normalization of A and corresponds to an affine non-singular curve. We will be able to do this by constructing the beginning of a Meyer-Vietoris sequence for Picard groups. This sequence will be the conclusion of a series of lemmata involving constructing projective curves over a fiber product. 4.1 Projective Modules over Fiber Products As a preliminary step, I will show that an invertible A-module is projective. Proposition 4.1. If I is an invertible A-module, then I is a projective A-module. PROOF: Suppose M and N are A-modules and that 0 M N I 0 is a short exact sequence. Localize at an arbitrary prime p to give us an exact sequence of A p -modules 0 M p N p I p 0. Since I is invertible, I p = Ap and is therefore a projective A p -module. Accordingly, for any p SpecA, I p = Mp N p = (M N) p. 12

15 Accordingly I = M N which implies the original exact sequence splits. It follows that I is a projective A-module. Proposition 4.1 is the motivation for the work we will do to construct projective modules over fiber products. For the next several results, we assume the following is a fiber-product diagram of commutative rings with identity: A α 1 A 1 α 2 β 1 A 2 β 2 S We will also assume β 1 is surjective. The construction of fiber products in the category of commutative rings with identity provides an explicit description of A: A = {(a 1, a 2 ) A 1 A 2 β1 (a 1 ) = β 2 (a 2 )}. Suppose P i is an A i -module for i = 1, 2 and we have an S-module isomorphism ϕ : P 1 A1 S P 2 A2 S. Define: M(P 1, P 2, ϕ) := {(p 1, p 2 ) P 1 P 2 ϕ(p1 1) = p 2 1}. Lemma 4.2. M(P 1, P 2, ϕ) is an A-module. PROOF: For (p 1, p 2 ) M(P 1, P 2, ϕ) and a A, define a (p 1, p 2 ) = (α 1 (a) p 1, α 2 (a) p 2 ). Then ϕ(α 1 (a) p }{{} 1 1) = β 1 (α 1 (a))ϕ(p 1 1) = β }{{} 2 (α 2 (a))ϕ(p 1 1) = α 2 (a) p 2 1. A 1 action S action It is clear that the module axioms are satisfied. Our first goal is to show that if P i are A i -modules, then M(P 1, P 2, ϕ) is a projective A-module. We will then be able to use this result to derive the desired exact sequence. The arguments below are from [Mil71]. Suppose f : A A is a ring homomorphism and M is an A-module. Define the following notation (inspired by [Mil71]): f M = A A M. Then f M is naturally an A -module. In fact, f induces an additive covariant functor from the category of A-modules to the category of A -modules. It is clear that if M is projective, free, or finitely generated then f M is projective, free or finitely generated respectively. There is a natural A-linear map f : M f M defined by f (m) = 1 m. In this notation, the module constructed above is where ϕ : (β 1 ) P 1 (β 2 ) P 2 is an isomorphism. M = M(P 1, P 2, ϕ) 13

16 Remark 4.3. In this notation, in the category of A-modules, M(P 1, P 2, ϕ) is the fiber product of P 1 and P 2 over (β 2 ) P 2. PROOF: For i = 1, 2, define p i : M(P 1, P 2, ϕ) P i to be the natural projections. If we consider the following diagram: M(P 1, P 2, ϕ) p 1 P 1 p 2 ϕ(β 1 ) P 2 (β 2 ) (β 2 ) P 2 the result is clear from the definition of M(P 1, P 2, ϕ) and the construction of fiber products in the category of A-modules. We begin by assuming that P i is free and finitely generated over A i for i = 1, 2. Let {x i } be a basis for P 1 over A 1 and let {y j } be a basis for P 2 over A 2. Then clearly {(β 1 ) x i } is a basis for (β 1 ) P 1 and {(β 2 ) y j } is a basis for (β 2 ) P 2, both over S. Thus there are elements s ij S such that ϕ((β 1 ) x i ) = j s ij(β 2 ) y j. Since ϕ is an isomorphism, the matrix T = (s ij) is invertible. Let (T ) 1 = (s ij ) denote the inverse, which represents ϕ 1. Then: ϕ 1 ((β 2 ) y i ) = j s ij (β 1 ) x j. Lemma 4.4. If the matrix T is the image under β 1 of an invertible matrix over A 1, then M(P 1, P 2, ϕ) is free. PROOF: Suppose s ij = β 1 (c ij ) (recall β 1 is assumed to be surjective) where the matrix (c ij ) is invertible. Define: x i = c ij x j A 1. j Since (c ij ) is invertible, {x i} forms a basis for P 1 over A 1. Observe that ( ) (β 1 ) (x i) = (β 1 ) c ij x j j = j β 1 (c ij )(β 1 ) x j = j s ij (β 1 ) x j = ϕ 1 ((β 2 ) y i ). 14

17 Therefore ϕ((β 1 ) x i) = (β 2 ) y i and thus z i = (x i, y i ) P 1 P 2 are elements of M(P 1, P 2, ϕ). I claim that {z i } forms a basis for M(P 1, P 2, ϕ) over A. Suppose q = (p 1, p 2 ) M(P 1, P 2, ϕ). Then p 1 A 1 and p 2 A 2. Thus there are elements b i A 1 and c i A 2 such that p 1 = i p 2 = i b i x i c i y i. Moreover: ( β 1 (b i )ϕ((β 1 ) x i) = ϕ((β 1 ) p 1 ) = (β 2 ) p 2 = (β 2 ) i i c i y i ) = i β 2 (c i )(β 2 ) y i. Since ϕ((β 1 ) x i) = (β 2 ) y i and moreover (β 2 ) y i is a basis for (β 2 ) P 2 as an S-module, it follows that for each i, β 1 (b i ) = β 2 (c i ). Therefore we have a i = (b i, c i ) A and we can write q = a i z i. Uniqueness is an immediate consequence of the fact that {x i} forms a basis for P 1 and {y i } forms a basis for P 2. Therefore {z i } forms a basis for M(P 1, P 2, ϕ) over A and M(P 1, P 2, ϕ) is free over A. Corollary 4.5. Under the hypothesis of Lemma 4.4, rank(p 1 ) = rank(p 2 ) = rank(m(p 1, P 2, ϕ)). PROOF: Since for i = 1, 2, P i is free and finitely generated over A i, P i Ai S is free and finitely generated over S with the same rank as P i. Since P 1 A1 S = P 2 A2 S: rank(p 1 ) = rank(p 1 A1 S) = rank(p 2 A2 S) = rank(p 2 ). The basis {z i } constructed for M(P 1, P 2, ϕ) in the proof of Lemma 4.4 has the same cardinality as any basis for P 1 or P 2 and the desired result follows. If P 1 and P 2 are free, but ϕ does not come from an invertible matrix over A 1, M(P 1, P 2, ϕ) is not necessarily free, but it is at least projective. Lemma 4.6. Suppose P 1 and P 2 are free. Then M(P 1, P 2, ϕ) is a projective A-module. 15

18 PROOF: First we define Q 1 to be a free module over A 1 with a basis {u j } in one-to-one correspondence with the basis {y j } for P 2. Similarly, define Q 2 to be a free module over A 2 with a basis {v i } in one-to-one correspondence with the basis {x i }. Let ψ : (β 1 ) Q 1 (β 2 ) Q 2 be the isomorphism of S-modules given by T defined above Lemma 4.4. It is clear that M(P 1, P 2, ϕ) M(Q 1, Q 2, ψ) = M(P 1 Q 1, P 2 Q 2, ϕ ψ) and that the isomorphism ϕ ψ corresponds with the matrix ( ) T 0. 0 T Observe that since T = T 1 and vice versa: ( ) ( ) ( T 0 I T I 0 = 0 T 0 I T I ) ( I T 0 I ) ( 0 I I 0 ). Since β 1 is surjective, there exist matrices T 1 and T 2 such that β 1 (T 1 ) = T and β 1 (T 2 ) = T (where the map acts component-wise). It follows that the above equation can lift to the following: ( ) ( ) ( ) ( ) ( ) T2 0 I T1 I 0 I T1 0 I =. 0 T 1 0 I T 2 I 0 I I 0 Since the matrices on the right are invertible, so is the matrix on the left. Therefore the matrix ( ) ( ) T 0 T2 0 = β 0 T 1 0 T 1 and is the image of an invertible matrix. It follows from Lemma 4.4 that M(P 1 Q 1, P 2 Q 2, ϕ ψ) is a free A-module. Therefore, since M(P 1 Q 1, P 2 Q 2, ϕ ψ) = M(P 1, P 2, ϕ) M(Q 1, Q 1, ψ) it follows that M(P 1, P 2, ϕ) is a projective A-module. Now we generalize to the case where P 1 and P 2 are merely projective. We will also assume that P 1 and P 2 are finitely generated. Lemma 4.7. There are projective modules Q i over A i such that P i Q i are finitely generated free modules over A i, for i = 1, 2, and so that (β 1 ) Q 1 = (β2 ) Q 2. PROOF: Since P i are finitely generated and projective over A i, there exist projective modules N i over A i such that P i N i are finitely-generated free A-modules. Therefore there are positive integers r and s such that P 1 N 1 = A r 1 P 2 N 2 = A s 2. 16

19 Define P = (β 1 ) P 1. Then P = (β2 ) P 2. Since (β 1 ) and (β 2 ) preserve direct sums, it follows that: and Define the following: P (β 1 ) N 1 = (β 1 ) P 1 (β 1 ) N 1 = (β1 ) (P 1 N 1 ) = (β 1 ) (A r 1) = S r P (β 2 ) N 2 = (β2 ) P 2 (β 2 ) N 2 = (β2 ) (P 2 N 2 ) = (β 2 ) (A s 2) = S s. Q 1 = N 1 A s 1 Q 2 = N 2 A r 1. Then it is clear that P i Q i are finitely generated, free A i -modules for i = 1, 2. Therefore, Q i are projective A i -modules. Moreover, by associativity and commutativity of the direct sum: (β 1 ) Q 1 = ((β1 ) N 1 ) S s = (β1 ) N 1 P (β 2 ) N 2 = ((β2 ) N 2 ) S r = (β2 ) Q 2. This is the desired result. We are now in a position to prove the results we need to derive the Meyer-Vietoris sequence for Picard groups. Theorem 4.8. If P i are finitely-generated, projective A i modules, then M(P 1, P 2, ϕ) is a finitely-generated, projective A-module. PROOF: Choose Q i for i = 1, 2 as in Lemma 4.7 and choose an isomorphism ψ : (β 1 ) Q 1 = (β2 ) Q 2. By Lemma 4.6, since P i Q i are free A i -modules, the module M(P 1, P 2, ϕ) M(Q 1, Q 2, ψ) = M(P 1 Q 1, P 2 Q 2, ϕ ψ) is projective and therefore M(P 1, P 2, ϕ) is projective. Since we can choose Q i such that P i Q i are finitely generated A i -modules, by the proof of Lemma 4.4 and Lemma 4.6 it follows that M(P 1, P 2, ϕ) is finitely generated. Theorem 4.9. Every projective A-module is isomorphic to M(P 1, P 2, ϕ) for some P 1, P 2 and ϕ. PROOF: Let P be a projective A-module. Define P i = (α i ) P for i = 1, 2. Then each P i is a projective A i -module. Since β 1 α 1 = β 2 α 2, there is a natural isomorphism such that ϕ : (β 1 ) P 1 (β 2 ) P 2 ϕ(β 1 ) (α 1 ) = (β 2 ) (α 1 ). Therefore we have the following commutative diagram (of abelian groups): P (α 1 ) P 1 (α 2 ) (ϕβ 1 ) P 2 (β 2 ) (β 2 ) P 2 17

20 We can identify P with the subgroup of P 1 P 2 such that (ϕβ 1 ) (p 1 ) = (β 2 ) (p 2 ). Accordingly, P is the fiber product of P 1 and P 2 over (β 2 ) P 2. By Remark 4.3, P = M(P 1, P 2, ϕ). Theorem Given projective modules P 1 and P 2 and an isomorphism ϕ : (β 1 ) P 1 M = M(P 1, P 2, ϕ). Then P i = (αi ) M. (β 2 ) P 2, let PROOF: Recall the definition of M(P 1, P 2, ϕ): M(P 1, P 2, ϕ) := {(p 1, p 2 ) P 1 P 2 ϕ(p 1 1) = p 2 1}. Viewing P 1 as an A-module via the map α 1, there is a natural A-linear map f : M P 1 given by projection. This induces a natural A 1 -linear map g : (α 1 ) M P 1 given by g(1 m) = f(m). Since the argument does not depend on the index 1 or 2, it suffices to prove g is an isomorphism. In the special case where the hypotheses of Lemma 4.4 are satisfied, this is clear since both (α 1 ) M and P 1 are free over A 1 with the same number of generators and g takes one basis to the other. In the general case, choosing Q 1 and Q 2 as in Lemma 4.7, we obtain the A-module M(P 1 Q 1, P 2 Q 2, g h) which satisfies the hypotheses of Lemma 4.4, hence g h is an isomorphism. It follows that g is an isomorphism. 4.2 The Conductor Square To apply this machinery to come up with a computational technique for Picard groups, we begin with a lemma that determines when M(P 1, P 2, ϕ) and M(P 1, P 2, ϕ ) are isomorphic. We continue to use the same setup. Lemma Suppose P i and P i are invertible A i modules for i = 1, 2. Then M(P 1, P 2, ϕ) = M(P 1, P 2, ϕ ) if and only if there are isomorphisms ψ i : P i P i for i = 1, 2 such that PROOF: Define the following notation: ϕ = (ψ 1 2 Id S ) ϕ (ψ 1 Id S ). M = M(P 1, P 2, ϕ) M = M(P 1, P 2, ϕ ) Suppose we have isomorphisms ψ i as in the statement. Define f : M M by Then we have the following: f(p 1, p 2 ) = (ψ 1 (p 1 ), ψ 2 (p 2 )). ϕ (ψ 1 (p 1 ) 1) = (ψ 2 Id S ) ϕ(p 1 1) = (ψ 2 Id S )(p 2 1) = ψ 2 (p 2 ) 1. This shows that f(m) M. Since ψ i are isomorphisms, it is clear that f is bijective and A-linear and hence f is an isomorphism. 18

21 Now suppose we have an isomorphism f : M M. First of all, I claim that given p 1 P 1, there is a unique p 2 P 2 such that (p 1, p 2 ) M. Existence is clear, since given p 1 P 1, there is p 2 P 2 such that ϕ(p 1 1) = p 2 1 (since ϕ is an isomorphism of S-modules). Now suppose there is p 2, q 2 P 2 such that ϕ(p 1 1) = p 2 1 = q 2 1. Then by bilinearity we have (p 2 q 2 ) 1 = 0. We can localize this equation. Then we have for any prime p, in the localization since (P 2 ) p = (A2 ) p, 0 = (p 2 q 2 ) 1 = (p 2 q 2 )(1 1) which implies p 2 q 2 = 0. Since this holds for all primes, it follows that p 2 q 2 = 0 in P 2 and hence p 2 = q 2. This establishes uniqueness. Observe that this holds for P 2 (using ϕ 1 ) as well as P 1 and P 2. It follows that we can define an inclusion ι i : P i M and a projection π i : M P i and that the composition ψ i = π i f ι i is an isomorphism P i P Next, by construction, given p 1 P 2 there is a unique p 2 P 2 and (since f is an isomorphism) a unique (p 1, p 2) M such that: i. and moreover ψ 2 (p 2 ) = p 2. Accordingly: ψ 1 (p 1 ) = π 1fι 1 (p 1 ) = π 1f(p 1, p 2 )π 1(p 1, p 2) = p 1 (ψ2 1 Id S ) ϕ (ψ 1 Id S )(p 1 1) = (ψ2 1 Id S ) ϕ (ψ 1 (p 1 ) 1) = (ψ2 1 Id S ) ϕ (p 1 1) = (ψ2 1 Id S )(p 2 1) = p 2 1 = ϕ(p 1 1). Since {p 1} for p P 1 generates P 1 A S as an S-module, it follows that as desired. ϕ = (ψ 1 2 Id S ) ϕ (ψ 1 Id S ) Lemma If P i, Q i are projective A i -modules for i = 1, 2 and ϕ : P 1 A S P 2 A S, ψ : Q 1 A S Q 2 A S are isomorphisms, then M(P 1, P 2, ϕ) A M(Q 1, Q 2, ψ) = M(P 1 A1 Q 1, P 2 A2 Q 2, ϕ ψ). PROOF: Given ((p 1, p 2 ), (q 1, q 2 )) M(P 1, P 2, ϕ) M(Q 1, Q 2, ψ), define the following map: ξ((p 1, p 2 ), (q 1, q 2 )) = (p 1 q 1, p 2 q 2 ). Then note that (ϕ ψ)((p 1 q 1 ) 1 S ) = (p 2 q 2 ) 1 S 19

22 and therefore ξ is a map ξ : M(P 1, P 2, ϕ) M(Q 1, Q 2, ψ) M(P 1 A1 Q 1, P 2 A2 Q 2, ϕ ψ). It is clear that ξ is bilinear, hence ξ induces a unique linear map that is clearly bijective. M(P 1, P 2, ϕ) A M(Q 1, Q 2, ψ) M(P 1 A1 Q 1, P 2 A2 Q 2, ϕ ψ) In order to define the sequence below, we need a little bit of algebraic K-theory. Definition For a commutative ring A with identity, K 0 (A) is the quotient of the free abelian group on isomorphism classes of finitely generated projective modules over A modulo the relation [P ] + [Q] = [P Q]. The group K 0 (A) can be made into a commutative ring with identity where multiplication is given by the tensor product and the identity is given by [A] (viewed as a module over itself). Proposition The Picard group Pic(A) embeds into K 0 (A) as the group of units. PROOF: By Lemma 4.1, Pic(A) embeds into K 0 (A) as a set and by part 1 of Theorem 2.8, Pic(A) actually embeds into the group of units K 0 (A). Now suppose [M] K 0 (A). Then there is [N] K 0 (A) such that M A N = A. Let p SpecA be arbitrary. Localizing at p gives M p Ap N p = Ap. Since M and N are projective, they are locally free. Therefore there are positive integers k and l such that and Accordingly, M p = A k p N p = A l p. M p Ap N p = A kl p = A p which implies k = l = 1. Therefore M and N are locally free of rank 1. Since M and N are finitely generated, they are invertible A-modules. Thus [M] Pic(A). Let us now define the maps that will be used in our exact sequence. First of all, if f : A B is any homomorphism of commutative rings with identity (requiring f(1) = 1) and a A is a unit, then f(a) is a unit. Consequently, α i (for i = 1, 2) induces a map A A i which (by an abuse of notation) we will also denote by α i. The same goes for the β i s. Define ι : A A 1 A 2 by ι(a) = (α 1 (a), α 2 (a)). 20

23 Define by j : A 1 A 2 S j(a 1, a 2 ) = β 1 (a 1 )(β 2 (a 2 )) 1. Finally, for each i = 1, 2, define a map δ i : Pic(A) Pic(A i ) by δ i (I) = A i A I. Then δ i (I) is a finitely generated A i -module. By properties of localization, given a prime p SpecA i and q SpecA such that α 1 i (p) = q, since I is invertible we have: (A i A I) p = (A i ) p Aq I q = (Ai ) p Aq A q = (Ai ) p. It follows that δ i (I) really is an element of Pic(A i ). It is clear that δ i is a group homomorphism, since if I, J Pic(A), by associativity of the tensor product: δ i (I)δ i (J) = [(I A A i ) Ai (J A A i )] = [I A (A i Ai (J A A i ))] = [I A (J A A i )] = [(I A J) A A i ] = δ(ij). Finally, define by δ = (δ 1, δ 2 ). δ : Pic(A) Pic(A 1 ) Pic(A 2 ) Finally, I add the additional assumption on the fiber square that α 1 is injective (this has been unnecessary thusfar, but will be satisfied in the conductor square below). Theorem There is a homomorphism such that following sequence is exact: : S Pic(A) 0 A ι A 1 A 2 j S Pic(A) δ Pic(A 1 ) Pic(A 2 ). PROOF: First, since α 1 is injective it is clear that ι is injective. Next I show that ker j = Imi. One containment is obvious. Now suppose j(a 1, a 2 ) = 1. This means that β 1 (a 1 )(β 2 (a 2 )) 1 = 1. Thus β 1 (a 1 ) = β 2 (a 2 ). Since A is the fiber product of A 1 and A 2 over S, there is a A such that α i (a) = a i which precisely says that ι(a) = (a 1, a 2 ). Hence ker j = Imj. Now we define the connecting homomorphism. Let P i = A i viewed as modules over A i for i = 1, 2. Let s S. Then s induces an S-module isomorphism ϕ s : A 1 A1 S A 2 A2 S given by ϕ(1 A1 1 S ) = s(1 A2 1 S ) (since 1 Ai 1 S generates A i A S as an S-module). Therefore we can form the module M s = M(A 1, A 2, ϕ s ). By Theorem 4.8, [M s ] K 0 (A). 21

24 First I claim that M st = Ms A M t. Starting from the right hand side, by Lemma 4.12, we have : M s A M t = M(A1 A1 A 1, A 2 A2 A 2, ϕ s ϕ t ) = M(A 1, A 2, ϕ s ϕ t ) = M(A 1, A 2, ϕ st ) = M st (since it is clear that ϕ s ϕ t = ϕ st ). In addition, note that Therefore: M 1 = M(A 1, A 2, ϕ 1S ) = A. M s M s 1 = M ss 1 = M 1 = A. Therefore in K 0 (A) as a ring, [M s ][M s 1] = [A] and [M s ] is a unit. Accordingly, by Proposition 4.14, [M s ] Pic(A) for every s S. Define : S Pic(A) by (s) = [M s ]. Then we have Thus is a group homomorphism. (st) = [M st ] = [M s A M t ] = [M s ] A [M t ] = (s) (t). Now note that if s = j(a 1, a 2 ) = β 1 (a 1 )β 2 (a 2 ) 1 we have M s = M β1 (a 1 )β 2 (a 2 ) 1 = M β1 (a 1 ) A M β2 (a 2 ). By the proof of Lemma 4.4, since A 1 and A 2 are free modules of rank one and ϕ β(a1 ) as a one-by-one matrix is the image of the one-by-one matrix over A 1 given by (a 1 ) (and likewise for ϕ β(a 1 2 )), it follows that M β1 (a 1 ) = A and M β2 (a 2 ) 1 = M β 2 (a 1 2 ) = A. 4 Accordingly, if s Im(j), then which implies s ker. (s) = [M s ] = [A A A] = [A] On the other hand, suppose (s) = [M s ] = [A]. This means, in particular, that we have M(A 1, A 2, ϕ s ) = M(A 1, A 2, ϕ 1 ). By Lemma 4.11, there are isomorphisms ψ i : A i A i such that: ϕ 1 = (ψ 1 2 Id S ) ϕ s (ψ 1 Id S ). (4.1) Note that for each i = 1, 2, 1 Ai 1 S generates A i Ai S as an S-module. Therefore, we plug 1 A1 1 S into equation (4.1). On the right hand side, we have ϕ 1 (1 A1 1 S ) = 1 A2 1 S. 4 Lemma 4.4 could just as easily been proved using β 2 instead of β 1. In fact, this is how Milnor proves the lemma in [Mil71]. 22

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