Picard Groups of Affine Curves

Victor I. PierceyUniversity 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 thendevelop computational techniques for Picard groups of affine curves. If the curve is non-singular, thetechnique will require some geometry. However, in the singular case, we will find a purely algebraictool. Along the way we will obtain an explicit description of the Picard group for every non-singularaffine curve over C as well as affine curves with a single cusp or node singularity.

Contents1 Introduction 1

2 Invertible Modules and the Picard Group 12.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2 The Picard Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

3 Picard Groups of Non-Singular Curves 73.1 Weil Divisors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.2 Nonsingular Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83.3 The Jacobian of a Riemann Surface and Pic(X) over C . . . . . . . . . . . . . . . . . . . 103.4 Non-Singular Affine Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

4 Singular Affine Curves 124.1 Projective Modules over Fiber Products . . . . . . . . . . . . . . . . . . . . . . . . . . . 124.2 The Conductor Square . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184.3 The Cusp and the Node . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

5 Conclusion 29

References 29

Picard Groups of Affine Curves

Victor I. PierceyMath 518

May 7, 2008

1 IntroductionIn algebraic geometry, one encounters invertible sheaves (or equivalently, line bundles) over an algebraicvariety. Invertible sheaves over a fixed algebraic variety form a group under the tensor product called thePicard 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 algebraicvarieties will be replaced by commutative rings with identity. If the variety is affine, the corresponding ringis the coordinate ring. Invertible sheaves in this context are invertible modules. The invertible modulesform a group under the tensor product which we will call the Picard group. We will proceed to developmethods 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 bycomputing 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 DefinitionsFirst 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 primeideal p ∈ Spec(A), Ip ∼= Ap as Ap-modules.

Remark 2.2. The condition that Ip ∼= Ap is often described by saying that I is locally free of rank 1.

If I is an ideal inA and is invertible as anA-module, we say that I is an invertible ideal. The notation I forthe 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 thefree 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

Suppose I = (x) for some x ∈ A. Let the norm N : A→ Z be given by N(a+ b√−5) = a2 + 5b2. Then

2 ∈ I and N(2) = 4, hence N(x) divides 4. Since 1 +√−5 ∈ 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) 6= ±1. ThusN(x) = ±2, and since N(x) > 0, it must be that N(x) = 2. But a2 + 5b2 = 2 requires b = 0, and there isno solution to a2 = 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 ∼= F2.

Now, if I * p, then I ∩ (A \ p) 6= ∅ which implies Ip = Ap. Therefore we suppose I ⊂ p. But then I = p

since I is maximal. Since 3 6∈ I (otherwise 1 ∈ I), 3 is invertible in Ap. Therefore:

2 =(1 +

√−5)(1−

√−5)

3∈ pAp.

It follows that Ip = (1 +√−5)p which is principal and hence isomorphic to Ap as Ap-modules.

Example 2.6. Let k be an algebraically closed field and let A = k[x, y]/(y2 − x3 + x). The ring A is thecoordinate ring for a non-singular, affine elliptic curve. Since the curve is non-singular, A is a Dedekinddomain. 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 mp = Ap for all p 6= m. Furthermore, since A is a Dedekinddomain, Am is a discrete valuation ring whose unique maximal ideal mAm is principal (see Proposition9.2 in [AM69]). Thus the maximal ideals are invertible. On the other hand, note that these ideals are notprincipal. There is a geometric reason for this. It comes down to the fact that when this affine curve isembedded in projective space P2

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 tocheck that Im ∼= Am for all m ∈ Max(A).

2.2 The Picard GroupThe isomorphism classes of invertible modules from a group, called the Picard group. The operation isgiven by the tensor product. The identity is the isomorphism class of A as a module over itself. If I is aninvertible module, its inverse is the dual module I∗ = HomA(I, A). Hence invertible modules are modulesthat have “inverses” in this sense.

Before proving the theorem that justifies the statements above, we need a couple of preliminaries. First ofall, given an A-module I , there is a natural map µ : I∗ ⊗ I → A defined by ϕ⊗ a 7→ ϕ(a). The map µ isthe 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 anA-module. The A-submodules of K(A) are called fractional ideals of A. If I is a finitely generated frac-tional ideal of A, we write all the generators over a common denominator. This shows that I is isomorphicas an A-module to an ordinary ideal of A. For any set I ⊂ K(A), define I−1 := s ∈ K(A)

∣∣ sI ⊂ A.

2

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 anisomorphism.

2. Every invertible module is isomorphic to a fractional ideal of A. Every invertible fractional idealcontains 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 7→ st isan isomorphism, as is the natural map I−1J → HomA(I, J) given by t 7→ ϕt where ϕt(a) = ta. Inparticular, I−1 ∼= I∗.

4. If I ⊂ K(A) is any A-submodule, then I is invertible if and only if I−1I = A.

PROOF: First, suppose I is invertible. Then for any prime p ∈ Spec(A),

A∗p ⊗Ap Ap = HomAp(Ap, Ap) = Ap ⊗Ap Ap

and µ localizes to the natural isomorphism

µp : Ap ⊗Ap Ap → Ap.

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 ϕi and ai such that

µ

(n∑

i=1

ϕi ⊗ ai

)= 1.

Since µ is an isomorphism, for any prime p we have an isomorphism µp : (I∗ ⊗A I)p = I∗p ⊗Ap Ip → Ap.I claim that Ip ∼= Ap where Ip is generated by some ai.

Since µ (ϕi ⊗ ai) =∑ϕi(ai) = 1, and p is a proper ideal of A, it follows that there is some i such

that ϕi(ai) 6∈ p. Therefore we can define

v =1

ϕi(ai)∈ Ap

and set a = vai ∈ Ap. It follows that (ϕi)p(a) = 1. Note that (ϕi)p(a) = 1 implies a is not a zero devisorin Ap. Therefore if we denote the cyclic Ap-submodule of Ip generated by a by Apa then Apa ∼= Ap asAp-modules and ker(ϕi)p ∩ Apa = 0. Now suppose for some α ∈ Ip, (ϕi)p(α) = β 6= 0 ∈ Ap. Then(ϕi)p(α − βa) = 0, so α − βa ∈ ker(ϕi)p. Thus it follows that Ip = Apa ⊕ ker(ϕi)p. Similarly, we canconsider the dual element ϕa ∈ HomAp(I

∗p , Ap) defined by ϕa(a) = 1 and ϕa(b) = 0 for any b 6∈ Apa. By

the same argument as above, we have I∗p = Apϕi ⊕ ker(ϕa) with Apϕi∼= Ap. Thus:

I∗p ⊗Ap Ip = (Apϕi ⊗Ap Apa)⊕ (Apϕi ⊗ ker(ϕi)p)⊕ · · · .

Sinceµp(Apϕi ⊗Ap ker(ϕi)p) = (ϕi)p(ker(ϕi)p) = 0

and µp is an isomorphism, Apϕi ⊗Ap ker(ϕi)p = 0. Since Apϕi∼= Ap, it follows that ker(ϕi)p = 0 and

(ϕi)p is an isomorphism. Hence Ip = Apa ∼= Ap. Therefore I is locally free of rank 1. Note that ai is

3

mapped to a generator in Ip. It follows that the natural map (a1, . . . , an) → I is locally surjective, hencesurjective. Thus I is generated by a1, . . . , an. Therefore I is finitely generated and locally free of rank1. 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:

I ⊗A K = I ⊗A A(0)∼= I(0) ∼= A(0) = K.

Therefore it suffices to find an embedding

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 Am. Then uα = 0 in Im ∼= Am. SinceAm is a domain, u = 0 or α = 0 in Am. If u = 0 in Am, then there is some element u′ ∈ A \ m such thatu′u = 0. This is not possible, since 0 ∈ m implies u′ 6= 0 while u 6= 0 in A and A being a domain impliesthat u is not a zero divisor. Thus u 6= 0 in Am and therefore α = 0 in Am. Since m ∈ MaxA was arbitrary,it follows that α = 0 in Im 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 ofthe 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 inK. Since K is a field and u 6= 0, it follows that I = (0). Since (0) is not invertible, it follows that I is notinvertible. Therefore, by contrapositive logic, every invertible submodule of K contains a nonzero (hencenonzerodivisor) 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. Themap

I ⊗ J → IJ

given by s⊗ t 7→ st is an A-module homomorphism that is clearly surjective. Therefore it suffices to showthat this map is injective. By exactness of localization, it suffices to show that this map is injective whenlocalized at p for any p ∈ SpecA. Note that Kp = K. Therefore we wish to show that the map

Ip ⊗Ap Jp → (IJ)p ⊂ K(Ap) ⊂ K

is injective. Since Ap is a domain and K(Ap) is its field of fractions, we may assume A is a local ring.

If (A,m) is a local ring, then A = Am. Therefore if I and J are invertible, I = Im, J = Jm andA = Am which implies that I ∼= A ∼= J as A-modules. Therefore there are nonzero elements s, t ∈ Ksuch that I = (s), J = (t) and IJ = (st). It follows that the desired map is injective, completing the firstpart of the third statement.

4

Now consider the map I−1J → HomA(I, J) given by t 7→ ϕt where ϕt(a) = ta. Note that t ∈ I−1Jimplies that

t =∑

airisi

with ai ∈ A, ri ∈ I−1 and si ∈ J . Therefore:

ta =(∑

airisi

)a =

∑ai(ria)si.

Since ri ∈ I−1 and a ∈ I , ria ∈ A and therefore ta ∈ J . It is clear that ϕt is a homomorphism. Now ifv ∈ A ∩ I is nonzero (which exists by the second statement of the theorem proved above), then tv 6= 0 inK whenever t 6= 0. Thus ϕt = 0 if and only if t = 0 and our map is injective.

Therefore it suffices to show that the map t 7→ ϕt is surjective. It suffices to show this locally. Letp ∈ Spec(A). Let ϕ ∈ HomAp(Ip, Jp) be any homomorphism. Since I is invertible, Ip ∼= Ap and there-fore Ip = xAp for some x ∈ K(A). Let w = ϕ(x). Note that if α = xa ∈ Ip (where a ∈ Ap), then(1/x)(xa) = a ∈ Ap and therefore 1/x ∈ I−1

p . Then w/x ∈ I−1p Jp.1 The claim is that ϕ = ϕw/x. Note

that ϕw/x(x) = w = ϕ(x). Since x is a generator of Ip, the claim follows. Therefore our map is surjectivelocally, and hence is surjective.

Finally, suppose I ⊂ K is invertible. Then the map µ : I∗ ⊗ I → A is an isomorphism, and by thethird part proved above this map is the same as the multiplication map I−1 ⊗ I → A. It follows thatI−1I ∼= A. Since I−1I ⊂ A it follows that I−1I = A. This establishes one direction of the fourth state-ment.

Now suppose I ⊂ K is an A-submodule with I−1I = A. By localization, we can assume A is a lo-cal ring with maximal ideal m. It then suffices to show under this assumption, I ∼= A. Since I−1I = A,there is an element v ∈ I−1 and an element α ∈ I such that vα 6∈ m. Therefore vI * m. I claim thatthe map v : I → A given by α 7→ vα is an isomorphism. Since vI * m implies v 6= 0, then vα = 0 inA ⊂ K implies α = 0, so multiplication by v is injective. Now since vI * m, there is α ∈ I such thatvα = y /∈ m. Since m is the unique maximal ideal in the local ring A, it follows that y ∈ A×. Thereforewe have vy−1α = 1 with y−1α ∈ I . It follows that if a ∈ A, then v(ay−1α) = a with ay−1α ∈ I . Hencev 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 arepresentative as a submodule of K. Note that distinct invertible submodules of K may be isomorphic! Inany case, the collection of isomorphism classes of invertible modules of A can be realized as a subset ofthe power set of K, and is therefore a set.

Corollary 2.10. The set of isomorphism classes of invertible modules of A is a group under ⊗ withidentity 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 2.11. The Picard group of A, denoted by Pic(A), is the group of all isomorphism classes ofinvertible A-modules.

1I am tacitly asserting that (I−1)p = (Ip)−1, which is true over a Noetherian ring.

5

Before considering Picard groups of rings of affine curves, we should look at the relationship betweenPic(A) and invertible submodules of K. We will exploit this relationship in the next section. Observe thatthe collection of all invertible submodules of K (NOT isomorphism classes!) is a subset of the power setof K and is therefore a set. This set also forms a group with identity element given by A as a module overitself and where the inverse of I is given by I−1 (hence the notation!). All of this follows from Theorem2.8. This allows us to give the following definition.

Definition 2.12. The group of invertible submodules of K is called the group of Cartier divisors anddenoted by C(A).

Definition 2.13. A principal divisor is an element of C(A) of the form Au for u ∈ K×. Denote the setof principal divisors by PC(A).

Remark 2.14. Note that if Au and Av are principal divisors, then Au ⊗ Av = A(uv), and uv ∈ K×. Itfollows that PC(A) is a subgroup of C(A).

Lemma 2.15. PC(A) ∼= K×/A×.

PROOF: We clearly have a surjective map K× → PC(A) given by u 7→ Au. Suppose Au = Av. Thenthere 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 ofour map is A×, completing the proof.

The relationship between C(A) and Pic(A) is given by the following.

Corollary 2.16. Let ϕ : C(A) → Pic(A) be given by ϕ(I) = [I] (where [I] is the isomorphism class ofthe invertible module I).

1. The map ϕ is a surjective group homomorphism and kerϕ = 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 PC(A). It is clear that ifAu ∈ PC(A),then Au ∼= A is an A-module under the map u 7→ 1. Therefore one inclusion is obvious and it sufficesto show the other inclusion. If I is any invertible module and Au ∈ PC(A), then (Au)I = u(AI) ⊂ uIon the one hand, while clearly uI = (1 · u)I ⊂ (Au)I . Therefore (Au)I = uI for any Au ∈ PC(A).Accordingly, it suffices to show that if I, J ∈ C(A) and I ∼= J , then J = uI for some u ∈ K×. But thisis immediate from the third statement in Theorem 2.8, since any isomorphism ψ : I → J is an element ofHomA(I, J) and thus can be realized as multiplication by a nonzero element u ∈ I−1J (since multiplica-tion by zero would not be an isomorphism). Any nonzero element of I−1J is a nonzero element of K andtherefore 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 invertiblefractional ideal by part (4) of Theorem 2.8, and therefore by the second part of Theorem 2.8, there is anonzero 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 2.17. Pic(A) ∼= C(A)/PC(A).

PROOF: This is immediate from Corollaries 2.16 and 2.15.

6

3 Picard Groups of Non-Singular CurvesWe will now determine a method for computing Picard groups of nonsingular curves over an algebraicallyclosed field k. If X is a nonsingular affine curve, its Picard group Pic(X) is the Picard group Pic(A)whereA is the affine coordinate ring k[x1, . . . , xn]/I of the curve. We will be able to describe these Picardgroups explicitly when the ground field is C. To lay the ground work for this discussion, we begin withobjects called Weil divisors. The Weil divisors form a group, and after taking a quotient we obtain a groupcalled the class group of the curve. We will then relate the class group, the group of Cartier divisors, andthe Picard group. This will allow us to write down a split short exact sequence that can be used to computethe Picard group of a nonsingular curve embedded in projective space. We can then pull this program backto obtain the Picard group of the affine curve.

3.1 Weil DivisorsWe 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 theset 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 showsthat when A is a Dedekind domain, they are actually the same. This will suffice for us since affine ringsof nonsingular curves are Dedekind domain.

Lemma 3.2. Every nonzero proper ideal I ⊂ A of a Dedekind domain A can be written uniquely as theproduct 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

(IJ)(IJ)−1 = (IJ)(J−1I−1) = I(JJ−1)I−1 = IAI−1 = II−1 = A.

Therefore IJ is invertible.

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 anyother prime ideal q ⊂ A, pAq = Aq. Since A is a Dedekind domain, Ap is a discrete valuation ring andhence its unique maximal ideal pAp is principal (see [AM69], Proposition 9.2). Therefore as Ap-modules,pAp

∼= 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 setof maximal ideals of A.

7

PROOF: First of all, suppose I ⊂ A is a proper invertible ideal. Then I is nonzero, so I is an elementof the free abelian group generated by the nonzero prime ideals of A. Since any nonzero prime ideal isa maximal ideal in a Dedekind domain, it follows that I is an element of the free abelian group on themaximal ideals of A. Finally, every fractional ideal in C(A) is an element of the free abelian group onmaximal ideals of A by the second part of Corollary 2.16. The reverse inclusion is obvious by Lemma 3.3and 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 maximalideals. 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 inC(A).

Definition 3.8. Let A be a Dedekind domain. Then the group of Weil divisors modulo principal divisorsis called the class group and denoted

Cl(A) ∼= Div(A)/PC(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:

Pic(A) ∼= C(A)/PC(A) = Div(A)/PC(A) = Cl(A)

as claimed.

Remark 3.10. Suppose A is a Dedekind domain. Then A is a unique factorization domain if and only ifPic(A) is trivial.

PROOF: First of all, it is clear that PC(A) can be identified with the principal ideals of A. Therefore itis clear that in a Dedekind domain, A is a principal ideal domain if and only if PC(A) = Div(A), whichholds if and only if Cl(A) is trivial. In addition, if A is a Dedekind domain, A is a principal ideal domainif 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 factorizationdomain if and only if Cl(A) is trivial if and only if Pic(A) is trivial.

3.2 Nonsingular CurvesWe now start with a nonsingular curve X of genus g embedded into projective space X → PN for someN . This embedding allows the curve to satisfy a “completeness” hypothesis. So far we have only dis-cussed Pic(A) for a ring A and by extension Pic(X) for an affine curve X . A projective curve does notcorrespond to a single coordinate ring. Instead, it has affine “patches” that have corresponding coordinate

8

rings, which may not all be the same.2 Therefore it is necessary to make sense of the Picard group inthis situation. Taking a collection of invertible modules over the affine coordinate rings and gluing themtogether in a consistent manner forms a sheaf of modules that is locally free of rank one. Such a sheafis called an invertible sheaf. Invertible sheaves are precisely line bundles over the curve. In the sameway that invertible modules form a group, the invertible sheaves form a group under tensor product. Theidentity element is the trivial sheaf and the inverse is given by the dual. The group of isomorphism classesof invertible sheaves over X is now the Picard group Pic(X). Restricting to an affine part of the curvegives 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. Todo so, we also need to make sense of Weil divisors on a projective curve. A Weil divisor can be thought ofas a formal (finite) sum of integral multiples of points on the projective curve. The group of Weil divisorsDiv(X) is then the free abelian group on the points of the curve. Linear equivalence is defined a littledifferently from above. One defines a principal divisor to be the formal sum of zeros and poles of certainfunctions on the curve whose coefficients are the corresponding order (where the order of the zero or poleis 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) asabove. The class group is useful here because of the following result.

Lemma 3.11. There is a surjective group homomorphism D : Div(X) → Z that descends to a welldefined group homomorphism d : Cl(X) → Z.

PROOF: For a divisor∑nipi, define

D(∑

nipi

)=∑

ni.

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 3.12. The map d in Lemma 3.11 is called the degree map.

For a nonsingular projective curveX , we have Pic(X) ∼= Cl(X). Hence we have a degree map on Pic(X).

Definition 3.13. For an isomorphism class of invertible sheaves (line bundles) L ∈ Pic(X), the degreeof L is the image under d of the corresponding Weil divisor class. The isomorphism classes of invertiblesheaves of degree zero are the kernel of the degree map and therefore form a subgroup of Pic(X) denotedby Pic0(X).

Proposition 3.14. Pic(X) ∼= Z⊕ Pic0(X)

PROOF: Since the degree map from Pic(X) → Z is surjective and its kernel is Pic0(X), we have a shortexact sequence of abelian groups

0 → Pic0(X) → Pic(X) → Z → 0.

Since Z is a projective Z-module, this short exact sequence splits and we obtain the desired result.2This 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.3The definitions are modeled after meromorphic functions on Riemann surfaces.

9

3.3 The Jacobian of a Riemann Surface and Pic(X) over CBy Proposition 3.14, describing Pic(X) for a nonsingular projective curve X is reduced to describingPic0(X), the group of invertible sheaves (line bundles) of degree zero. This has a nice, explicit descriptionif we assume the underlying ground field is C. Over C, a nonsingular projective curve X is a compactRiemann surface of genus g. To understand Pic0(X) we take a brief detour and discuss the Jacobian of aRiemann surface.

Recall that for a Riemann surface X of genus g, the first singular homology group is given by

H1(X,Z) ∼= Z2g.

Recall that a holomorphic 1-form ω on X is closed. Accordingly, if c = ∂A is a boundary in H1(X,Z),then by Stokes’ theorem: ∫

c

ω =

∫∂A

ω =

∫A

dω = 0.

Accordingly, denoting the vector space of holomorphic one-forms by Ω1(X), for each [c] ∈ H1(X,Z) wehave a well-defined linear functional: ∫

[c]

: Ω1(X) → C.

This gives us an embedding H1(X,Z) → (Ω1(X))∗.

Definition 3.15. The Jacobian of the curve X , denoted by J(X), is defined as:

J(X) = (Ω1(X))∗/H1(X,Z).

Proposition 3.16. If X is a Riemann surface of genus g, then

J(X) ∼= T2g

where the isomorphism is in the category of abelian groups.

PROOF: The dual (Ω1(X))∗ is a complex vector space of dimension g, the genus ofX . Identify (Ω1(X))∗

with Cg. The corresponding embedding H1(X,Z) → Cg gives us H1(X,Z) ∼= Λ where Λ ⊂ Cg is a rank2g lattice. Accordingly, as abelian groups:

J(X) ∼= Cg/Λ ∼= R2g/Z2g ∼= T2g

which is the desired result.

A very deep theorem of Abel and Jacobi shows that Pic0(X) ∼= J(X). See Chapter VIII of [Mir95]for details. This gives us the following immediate corollary.

Corollary 3.17. If X is a nonsingular projective curve over C of genus g, then

Pic(X) ∼= Z⊕ T2g.

PROOF: By Proposition 3.14, the Abel-Jacobi theorem, and Proposition 3.16:

Pic(X) ∼= Z⊕ Pic0(X) ∼= Z⊕ J(X) ∼= Z⊕ T2g.

Remark 3.18. As an interesting observation, note that if X is an elliptic curve, Pic(X) ∼= Z⊕X.

Remark 3.19. A curve of genus zero is called a rational curve. A rational curve over C embedded inprojective space has Picard group Pic(X) ∼= Z. Since X ∼= P1, we can explicitly describe a generator forthe Picard group. The generator is the so-called tautological bundle. This is the line bundle over P1 wherethe fiber over a point ` ∈ P1 is the line in C2 represented by `.

10

3.4 Non-Singular Affine CurvesNow suppose A = k[x1, . . . , xn]/I is the coordinate ring of a non-singular affine curve X . To computePic(A), we will “complete” the curve to a projective curve X ′, obtaining the Picard group Pic(X ′) asabove, and determine how the Picard group responds when restricting to the affine part that we startedwith.

The first observation is that our coordinate ring gives us a closed embedding of our curve X := SpecAinto An = Speck[x1, . . . , xn]. Complete An to Pn. This amounts to adding in the “hyperplane at infinity”,which can be chosen to be the hyperplane in Pn given by the homogeneous equation z0 = 0 where Pn isgiven homogeneous coordinates [z0 : z1 : · · · : zn]. 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 Pn, we canreturn to X by removing finitely many points. The question therefore becomes how the Picard group ofX ′ changes when finitely many points are removed.

To see how this works, we exploit again the isomorphism between the Picard group Pic(X ′) and theclass group Cl(X ′).

Proposition 3.20. Let X ′ be a nonsingular curve embedded in projective space Pn. Let

Z = p1, . . . , pk ⊂ X ′

be a finite subset of the points of X and let U = X ′ \ Z. Then there is an exact sequence:

Zk → Cl(X ′) → Cl(U) → 0.

PROOF: Map Zk → Cl(X ′) by the map sending the ith generator of Zk to the divisor pi. Then theimage is the residue of the subgroup Zp1 + Zp2 + . . . + Zpk modulo principal divisors. Define the mapCl(X ′) → Cl(U) by mapping a divisor of the form

∑niqi on X to a divisor

∑niqi where ni = ni

if qi ∈ U and zero otherwise. Then this map is surjective, and the kernel is precisely the image of Zk.Therefore this sequence is exact.

Corollary 3.21. Let A be the coordinate ring of an affine non-singular curve X of genus g over C. Let kbe the number of points added in its embedding into projective space. Then:

Pic(A) ∼= T2g/Zk−1.

PROOF: Let X ′ denote the image of X in projective space. By Corollary 3.17 we have

Pic(X ′) ∼= Z⊕ T2g.

Let Z be the set of points of X ′ \X. Then (since Pic(A) = Pic(X)) by Proposition 3.20, we have an exactsequence:

Zk → Pic(X ′) → Pic(A) → 0

and hencePic(A) ∼= Pic(X ′)/Zk ∼= (Z⊕ T2g)/Zk ∼= T2g/Zk−1

as desired.

Remark 3.22. Over an arbitrary field k, the same argument shows that under the same hypotheses,Pic(A) ∼= Pic0(X ′)/Zk−1.

11

One consequence is the following.

Corollary 3.23. Let A be the coordinate ring of an affine non-singular curve X . Then A is a uniquefactorization 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 gbe the genus. Then it follows that

Cl(A) ∼= Pic(X) ∼= T2g/Zk−1.

If g 6= 0, then T2g is uncountable and Zk−1 is countable and hence Cl(A) is nontrivial and hence A isnot a unique factorization domain. Conversely, if g = 0 then Cl(A) is trivial and hence A is a uniquefactorization domain by Remark 3.10.

The proof over an arbitrary field k requires showing that for the completion X ′ of X in projective space,Pic0(X) is uncountable.

4 Singular Affine CurvesWe now turn to the case where X is a singular affine curve. At the end of this discussion we willcompute Pic(X) where X = SpecA for A = k[x, y]/(y2 − x3) and Pic(Y ) where Y = SpecB forB = k[x, y]/(y2 − x2(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 coordi-nate rings for singular curves are Noetherian domains, but they are not Dedekind domains. Therefore alot of the work we did above fails. To compute Picard groups in this new context, we will be able to mapPic(A) to Pic(A1), where A1 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 overa fiber product.

4.1 Projective Modules over Fiber ProductsAs 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 Ap-modules

0 →Mp → Np → Ip → 0.

Since I is invertible, Ip ∼= Ap and is therefore a projective Ap-module. Accordingly, for any p ∈ SpecA,

Ip ∼= Mp ⊕Np = (M ⊕N)p.

12

Accordingly I ∼= M ⊕N which implies the original exact sequence splits. It follows that I is a projectiveA-module.

Proposition 4.1 is the motivation for the work we will do to construct projective modules over fiber prod-ucts. For the next several results, we assume the following is a fiber-product diagram of commutative ringswith identity:

Aα1 //

α2

A1

β1

A2 β2

// S

We will also assume β1 is surjective. The construction of fiber products in the category of commutativerings with identity provides an explicit description of A:

A = (a1, a2) ∈ A1 × A2

∣∣ β1(a1) = β2(a2).

Suppose Pi is anAi-module for i = 1, 2 and we have an S-module isomorphism ϕ : P1⊗A1S → P2⊗A2S.Define:

M(P1, P2, ϕ) := (p1, p2) ∈ P1 × P2

∣∣ ϕ(p1 ⊗ 1) = p2 ⊗ 1.

Lemma 4.2. M(P1, P2, ϕ) is an A-module.

PROOF: For (p1, p2) ∈M(P1, P2, ϕ) and a ∈ A, define

a · (p1, p2) = (α1(a) · p1, α2(a) · p2).

Thenϕ(α1(a) · p1︸ ︷︷ ︸

A1−action

⊗1) = β1(α1(a))ϕ(p1 ⊗ 1)︸ ︷︷ ︸S−action

= β2(α2(a))ϕ(p1 ⊗ 1) = α2(a) · p2 ⊗ 1.

It is clear that the module axioms are satisfied.

Our first goal is to show that if Pi are Ai-modules, then M(P1, P2, ϕ) is a projective A-module. Wewill 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 ofA-modules to the category of A′-modules. It is clear that if M is projective, free, or finitely generated thenf]M is projective, free or finitely generated respectively. There is a natural A-linear map f∗ : M → f]Mdefined by

f∗(m) = 1⊗m.

In this notation, the module constructed above is

M = M(P1, P2, ϕ)

where ϕ : (β1)]P1 → (β2)]P2 is an isomorphism.

13

Remark 4.3. In this notation, in the category of A-modules, M(P1, P2, ϕ) is the fiber product of P1 andP2 over (β2)]P2.

PROOF: For i = 1, 2, define pi : M(P1, P2, ϕ) → Pi to be the natural projections. If we consider thefollowing diagram:

M(P1, P2, ϕ)p1 //

p2

P1

ϕ(β1)∗

P2 (β2)∗

// (β2)]P2

the result is clear from the definition ofM(P1, P2, ϕ) and the construction of fiber products in the categoryof A-modules.

We begin by assuming that Pi is free and finitely generated over Ai for i = 1, 2. Let xi be a basisfor P1 over A1 and let yj be a basis for P2 over A2. Then clearly (β1)∗xi is a basis for (β1)]P1 and(β2)∗yj is a basis for (β2)]P2, both over S. Thus there are elements s′ij ∈ S such that

ϕ((β1)∗xi) =∑

j

s′ij(β2)∗yj.

Since ϕ is an isomorphism, the matrixT ′ = (s′ij)

is invertible. Let (T ′)−1 = (sij) denote the inverse, which represents ϕ−1. Then:

ϕ−1((β2)∗yi) =∑

j

sij(β1)∗xj.

Lemma 4.4. If the matrix T is the image under β1 of an invertible matrix over A1, then M(P1, P2, ϕ) isfree.

PROOF: Suppose sij = β1(cij) (recall β1 is assumed to be surjective) where the matrix (cij) is invertible.Define:

x′i =∑

j

cijxj ∈ A1.

Since (cij) is invertible, x′i forms a basis for P1 over A1. Observe that

(β1)∗(x′i) = (β1)∗

(∑j

cijxj

)=

∑j

β1(cij)(β1)∗xj

=∑

j

sij(β1)∗xj

= ϕ−1((β2)∗yi).

14

Therefore ϕ((β1)∗x′i) = (β2)∗yi and thus

zi = (x′i, yi) ∈ P1 × P2

are elements of M(P1, P2, ϕ). I claim that zi forms a basis for M(P1, P2, ϕ) over A.

Suppose q = (p1, p2) ∈ M(P1, P2, ϕ). Then p1 ∈ A1 and p2 ∈ A2. Thus there are elements bi ∈ A1

and ci ∈ A2 such that

p1 =∑

i

bix′i

p2 =∑

i

ciyi.

Moreover:

∑i

β1(bi)ϕ((β1)∗x′i) = ϕ((β1)∗p1) = (β2)∗p2 = (β2)∗

(∑i

ciyi

)=∑

i

β2(ci)(β2)∗yi.

Sinceϕ((β1)∗x

′i) = (β2)∗yi

and moreover (β2)∗yi is a basis for (β2)]P2 as an S-module, it follows that for each i, β1(bi) = β2(ci).Therefore we have

ai = (bi, ci) ∈ A

and we can writeq =

∑aizi.

Uniqueness is an immediate consequence of the fact that x′i forms a basis for P1 and yi forms a basisfor P2. Therefore zi forms a basis for M(P1, P2, ϕ) over A and M(P1, P2, ϕ) is free over A.

Corollary 4.5. Under the hypothesis of Lemma 4.4,

rank(P1) = rank(P2) = rank(M(P1, P2, ϕ)).

PROOF: Since for i = 1, 2, Pi is free and finitely generated overAi, Pi⊗AiS is free and finitely generated

over S with the same rank as Pi. Since P1 ⊗A1 S∼= P2 ⊗A2 S:

rank(P1) = rank(P1 ⊗A1 S) = rank(P2 ⊗A2 S) = rank(P2).

The basis zi constructed for M(P1, P2, ϕ) in the proof of Lemma 4.4 has the same cardinality as anybasis for P1 or P2 and the desired result follows.

If P1 and P2 are free, but ϕ does not come from an invertible matrix over A1, M(P1, P2, ϕ) is not neces-sarily free, but it is at least projective.

Lemma 4.6. Suppose P1 and P2 are free. Then M(P1, P2, ϕ) is a projective A-module.

15

PROOF: First we define Q1 to be a free module over A1 with a basis uj in one-to-one correspondencewith the basis yj for P2. Similarly, defineQ2 to be a free module overA2 with a basis vi in one-to-onecorrespondence with the basis xi. Let

ψ : (β1)]Q1 → (β2)]Q2

be the isomorphism of S-modules given by T defined above Lemma 4.4. It is clear that

M(P1, P2, ϕ)⊕M(Q1, Q2, ψ) ∼= M(P1 ⊕Q1, P2 ⊕Q2, ϕ⊕ ψ)

and that the isomorphism ϕ⊕ ψ corresponds with the matrix(T ′ 00 T

).

Observe that since T ′ = T−1 and vice versa:(T ′ 00 T

)=

(I T0 I

)(I 0−T ′ I

)(I T0 I

)(0 −II 0

).

Since β1 is surjective, there exist matrices T1 and T2 such that β1(T1) = T and β1(T2) = T ′ (where themap acts component-wise). It follows that the above equation can lift to the following:(

T2 00 T1

)=

(I T1

0 I

)(I 0−T2 I

)(I T1

0 I

)(0 −II 0

).

Since the matrices on the right are invertible, so is the matrix on the left. Therefore the matrix(T ′ 00 T

)= β1

(T2 00 T1

)and is the image of an invertible matrix. It follows from Lemma 4.4 that M(P1 ⊕Q1, P2 ⊕Q2, ϕ⊕ ψ) isa free A-module. Therefore, since

M(P1 ⊕Q1, P2 ⊕Q2, ϕ⊕ ψ) ∼= M(P1, P2, ϕ)⊕M(Q1, Q1, ψ)

it follows that M(P1, P2, ϕ) is a projective A-module.

Now we generalize to the case where P1 and P2 are merely projective. We will also assume that P1

and P2 are finitely generated.

Lemma 4.7. There are projective modulesQi overAi such that Pi⊕Qi are finitely generated free modulesover Ai, for i = 1, 2, and so that (β1)]Q1

∼= (β2)]Q2.

PROOF: Since Pi are finitely generated and projective over Ai, there exist projective modules Ni over Ai

such that Pi⊕Ni are finitely-generated free A-modules. Therefore there are positive integers r and s suchthat

P1 ⊕N1∼= Ar

1

P2 ⊕N2∼= As

2.

16

Define P ′ = (β1)]P1. Then P ′ ∼= (β2)]P2. Since (β1)] and (β2)] preserve direct sums, it follows that:

P ′ ⊕ (β1)]N1 = (β1)]P1 ⊕ (β1)]N1∼= (β1)](P1 ⊕N1) ∼= (β1)](A

r1)∼= Sr

andP ′ ⊕ (β2)]N2

∼= (β2)]P2 ⊕ (β2)]N2∼= (β2)](P2 ⊕N2) ∼= (β2)](A

s2)∼= Ss.

Define the following:

Q1 = N1 ⊕ As1

Q2 = N2 ⊕ Ar1.

Then it is clear that Pi⊕Qi are finitely generated, freeAi-modules for i = 1, 2. Therefore,Qi are projectiveAi-modules. Moreover, by associativity and commutativity of the direct sum:

(β1)]Q1∼= ((β1)]N1)⊕ Ss ∼= (β1)]N1 ⊕ P ′ ⊕ (β2)]N2

∼= ((β2)]N2)⊕ Sr ∼= (β2)]Q2.

This is the desired result.

We are now in a position to prove the results we need to derive the Meyer-Vietoris sequence for Picardgroups.

Theorem 4.8. If Pi are finitely-generated, projectiveAi modules, thenM(P1, P2, ϕ) is a finitely-generated,projective A-module.

PROOF: Choose Qi for i = 1, 2 as in Lemma 4.7 and choose an isomorphism

ψ : (β1)]Q1∼= (β2)]Q2.

By Lemma 4.6, since Pi ⊕Qi are free Ai-modules, the module

M(P1, P2, ϕ)⊕M(Q1, Q2, ψ) ∼= M(P1 ⊕Q1, P2 ⊕Q2, ϕ⊕ ψ)

is projective and therefore M(P1, P2, ϕ) is projective. Since we can choose Qi such that Pi ⊕ Qi arefinitely generated Ai-modules, by the proof of Lemma 4.4 and Lemma 4.6 it follows that M(P1, P2, ϕ) isfinitely generated.

Theorem 4.9. Every projective A-module is isomorphic to M(P1, P2, ϕ) for some P1, P2 and ϕ.

PROOF: Let P be a projective A-module. Define Pi = (αi)]P for i = 1, 2. Then each Pi is a projectiveAi-module. Since β1α1 = β2α2, there is a natural isomorphism

ϕ : (β1)]P1 → (β2)]P2

such thatϕ(β1)∗(α1)∗ = (β2)∗(α1)∗.

Therefore we have the following commutative diagram (of abelian groups):

P(α1)∗ //

(α2)∗

P1

(ϕβ1)∗

P2

(β2)∗// (β2)]P2

17

We can identify P with the subgroup of P1 × P2 such that

(ϕβ1)∗(p1) = (β2)∗(p2).

Accordingly, P is the fiber product of P1 and P2 over (β2)]P2. By Remark 4.3, P ∼= M(P1, P2, ϕ).

Theorem 4.10. Given projective modules P1 and P2 and an isomorphism ϕ : (β1)]P1 → (β2)]P2, letM = M(P1, P2, ϕ). Then Pi

∼= (αi)]M .

PROOF: Recall the definition of M(P1, P2, ϕ):

M(P1, P2, ϕ) := (p1, p2) ∈ P1 × P2

∣∣ ϕ(p1 ⊗ 1) = p2 ⊗ 1.

Viewing P1 as an A-module via the map α1, there is a natural A-linear map f : M → P1 given byprojection. This induces a natural A1-linear map g : (α1)]M → P1 given by g(1⊗m) = f(m). Since theargument does not depend on the index 1 or 2, it suffices to prove g is an isomorphism. In the special casewhere the hypotheses of Lemma 4.4 are satisfied, this is clear since both (α1)]M and P1 are free over A1

with the same number of generators and g takes one basis to the other. In the general case, choosing Q1

and Q2 as in Lemma 4.7, we obtain the A-module

M(P1 ⊕Q1, P2 ⊕Q2, g ⊕ h)

which satisfies the hypotheses of Lemma 4.4, hence g ⊕ h is an isomorphism. It follows that g is anisomorphism.

4.2 The Conductor SquareTo apply this machinery to come up with a computational technique for Picard groups, we begin witha lemma that determines when M(P1, P2, ϕ) and M(P ′

1, P′2, ϕ

′) are isomorphic. We continue to use thesame setup.

Lemma 4.11. Suppose Pi and P ′i are invertibleAi modules for i = 1, 2. ThenM(P1, P2, ϕ) ∼= M(P ′

1, P′2, ϕ

′)if and only if there are isomorphisms ψi : Pi → P ′

i for i = 1, 2 such that

ϕ = (ψ−12 ⊗ IdS) ϕ′ (ψ1 ⊗ IdS).

PROOF: Define the following notation:

M = M(P1, P2, ϕ)

M ′ = M(P ′1, P

′2, ϕ

′)

Suppose we have isomorphisms ψi as in the statement. Define f : M →M ′ by

f(p1, p2) = (ψ1(p1), ψ2(p2)).

Then we have the following:

ϕ′(ψ1(p1)⊗ 1) = (ψ2 ⊗ IdS) ϕ(p1 ⊗ 1) = (ψ2 ⊗ IdS)(p2 ⊗ 1) = ψ2(p2)⊗ 1.

This shows that f(M) ⊂ M ′. Since ψi are isomorphisms, it is clear that f is bijective and A-linear andhence f is an isomorphism.

18

Now suppose we have an isomorphism f : M → M ′. First of all, I claim that given p1 ∈ P1, thereis a unique p2 ∈ P2 such that (p1, p2) ∈ M . Existence is clear, since given p1 ∈ P1, there is p2 ∈ P2 suchthat ϕ(p1⊗ 1) = p2⊗ 1 (since ϕ is an isomorphism of S-modules). Now suppose there is p2, q2 ∈ P2 suchthat ϕ(p1 ⊗ 1) = p2 ⊗ 1 = q2 ⊗ 1. Then by bilinearity we have

(p2 − q2)⊗ 1 = 0.

We can localize this equation. Then we have for any prime p, in the localization since (P2)p∼= (A2)p,

0 = (p2 − q2)⊗ 1 = (p2 − q2)(1⊗ 1)

which implies p2 − q2 = 0. Since this holds for all primes, it follows that p2 − q2 = 0 in P2 and hencep2 = q2. This establishes uniqueness. Observe that this holds for P2 (using ϕ−1) as well as P ′

1 and P ′2.

It follows that we can define an inclusion ιi : Pi → M and a projection π′i : M ′ → P ′i and that the

composition ψi = π′i f ιi is an isomorphism Pi → P ′i .

Next, by construction, given p1 ∈ P2 there is a unique p2 ∈ P2 and (since f is an isomorphism) aunique (p′1, p

′2) ∈M ′ such that:

ψ1(p1) = π′1fι1(p1) = π′1f(p1, p2)π′1(p

′1, p

′2) = p′1

and moreover ψ2(p2) = p′2. Accordingly:

(ψ−12 ⊗ IdS) ϕ′ (ψ1 ⊗ IdS)(p1 ⊗ 1) = (ψ−1

2 ⊗ IdS) ϕ′(ψ1(p1)⊗ 1)

= (ψ−12 ⊗ IdS) ϕ′(p′1 ⊗ 1)

= (ψ−12 ⊗ IdS)(p′2 ⊗ 1)

= p2 ⊗ 1

= ϕ(p1 ⊗ 1).

Since p⊗ 1 for p ∈ P1 generates P1 ⊗A S as an S-module, it follows that

ϕ = (ψ−12 ⊗ IdS) ϕ′ (ψ1 ⊗ IdS)

as desired.

Lemma 4.12. If Pi,Qi are projectiveAi-modules for i = 1, 2 and ϕ : P1⊗AS → P2⊗AS, ψ : Q1⊗AS →Q2 ⊗A S are isomorphisms, then

M(P1, P2, ϕ)⊗A M(Q1, Q2, ψ) ∼= M(P1 ⊗A1 Q1, P2 ⊗A2 Q2, ϕ⊗ ψ).

PROOF: Given ((p1, p2), (q1, q2)) ∈M(P1, P2, ϕ)×M(Q1, Q2, ψ), define the following map:

ξ((p1, p2), (q1, q2)) = (p1 ⊗ q1, p2 ⊗ q2).

Then note that(ϕ⊗ ψ)((p1 ⊗ q1)⊗ 1S) = (p2 ⊗ q2)⊗ 1S

19

and therefore ξ is a map

ξ : M(P1, P2, ϕ)×M(Q1, Q2, ψ) →M(P1 ⊗A1 Q1, P2 ⊗A2 Q2, ϕ⊗ ψ).

It is clear that ξ is bilinear, hence ξ induces a unique linear map

M(P1, P2, ϕ)⊗A M(Q1, Q2, ψ) →M(P1 ⊗A1 Q1, P2 ⊗A2 Q2, ϕ⊗ ψ)

that is clearly bijective.

In order to define the sequence below, we need a little bit of algebraic K-theory.

Definition 4.13. For a commutative ring A with identity, K0(A) is the quotient of the free abelian groupon isomorphism classes of finitely generated projective modules over A modulo the relation [P ] + [Q] =[P ⊕Q].

The group K0(A) can be made into a commutative ring with identity where multiplication is given by thetensor product and the identity is given by [A] (viewed as a module over itself).

Proposition 4.14. The Picard group Pic(A) embeds into K0(A) as the group of units.

PROOF: By Lemma 4.1, Pic(A) embeds intoK0(A) as a set and by part 1 of Theorem 2.8, Pic(A) actuallyembeds into the group of units K0(A)×. Now suppose [M ] ∈ K0(A)×. Then there is [N ] ∈ K0(A) suchthat

M ⊗A N ∼= A.

Let p ∈ SpecA be arbitrary. Localizing at p gives

Mp ⊗Ap Np∼= Ap.

Since M and N are projective, they are locally free. Therefore there are positive integers k and l such that

Mp∼= Ak

p

andNp

∼= Alp.

Accordingly,Mp ⊗Ap Np

∼= Aklp∼= Ap

which implies k = l = 1. Therefore M and N are locally free of rank 1. Since M and N are finitelygenerated, 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 anyhomomorphism of commutative rings with identity (requiring f(1) = 1) and a ∈ A× is a unit, then f(a)is a unit. Consequently, αi (for i = 1, 2) induces a map A× → A×

i which (by an abuse of notation) wewill also denote by αi. The same goes for the βi’s. Define

ι : A× → A×1 ⊕ A×

2

byι(a) = (α1(a), α2(a)).

20

Definej : A×

1 ⊕ A×2 → S×

byj(a1, a2) = β1(a1)(β2(a2))

−1.

Finally, for each i = 1, 2, define a map δi : Pic(A) → Pic(Ai) by

δi(I) = Ai ⊗A I.

Then δi(I) is a finitely generated Ai-module. By properties of localization, given a prime p ∈ SpecAi andq ∈ SpecA such that α−1

i (p) = q, since I is invertible we have:

(Ai ⊗A I)p = (Ai)p ⊗Aq Iq∼= (Ai)p ⊗Aq Aq

∼= (Ai)p.

It follows that δi(I) really is an element of Pic(Ai). It is clear that δi is a group homomorphism, since ifI, J ∈ Pic(A), by associativity of the tensor product:

δi(I)δi(J) = [(I ⊗A Ai)⊗Ai(J ⊗A Ai)]

= [I ⊗A (Ai ⊗Ai(J ⊗A Ai))]

= [I ⊗A (J ⊗A Ai)]

= [(I ⊗A J)⊗A Ai]

= δ(IJ).

Finally, defineδ : Pic(A) → Pic(A1)⊕ Pic(A2)

by δ = (δ1, δ2).

Finally, I add the additional assumption on the fiber square that α1 is injective (this has been unneces-sary thusfar, but will be satisfied in the conductor square below).

Theorem 4.15. There is a homomorphism

∂ : S× → Pic(A)

such that following sequence is exact:

0 // A× ι // A×1 ⊕ A×

2

j // S×∂ // Pic(A) δ // Pic(A1)⊕ Pic(A2) .

PROOF: First, since α1 is injective it is clear that ι is injective. Next I show that ker j = Imi. Onecontainment is obvious. Now suppose j(a1, a2) = 1. This means that β1(a1)(β2(a2))

−1 = 1. Thusβ1(a1) = β2(a2). Since A is the fiber product of A1 and A2 over S, there is a ∈ A such that αi(a) = ai

which precisely says that ι(a) = (a1, a2). Hence ker j = Imj.

Now we define the connecting homomorphism ∂. Let Pi = Ai viewed as modules over Ai for i = 1, 2.Let s ∈ S×. Then s induces an S-module isomorphism ϕs : A1 ⊗A1 S → A2 ⊗A2 S given byϕ(1A1 ⊗ 1S) = s(1A2 ⊗ 1S) (since 1Ai

⊗ 1S generates Ai ⊗A S as an S-module). Therefore we canform the module Ms = M(A1, A2, ϕs). By Theorem 4.8, [Ms] ∈ K0(A).

21

First I claim thatMst

∼= Ms ⊗A Mt.

Starting from the right hand side, by Lemma 4.12, we have :

Ms ⊗A Mt∼= M(A1 ⊗A1 A1, A2 ⊗A2 A2, ϕs ⊗ ϕt) ∼= M(A1, A2, ϕs ⊗ ϕt) ∼= M(A1, A2, ϕst) = Mst

(since it is clear that ϕs ⊗ ϕt = ϕst). In addition, note that

M1 = M(A1, A2, ϕ1S) ∼= A.

Therefore:Ms ⊗Ms−1

∼= Mss−1∼= M1

∼= A.

Therefore in K0(A) as a ring, [Ms][Ms−1 ] = [A] and [Ms] is a unit. Accordingly, by Proposition 4.14,[Ms] ∈ Pic(A) for every s ∈ S×. Define ∂ : S× → Pic(A) by ∂(s) = [Ms]. Then we have

∂(st) = [Mst] = [Ms ⊗A Mt] = [Ms]⊗A [Mt] = ∂(s)∂(t).

Thus ∂ is a group homomorphism.

Now note that if s = j(a1, a2) = β1(a1)β2(a2)−1 we have

Ms = Mβ1(a1)β2(a2)−1∼= Mβ1(a1) ⊗A Mβ2(a2).

By the proof of Lemma 4.4, since A1 and A2 are free modules of rank one and ϕβ(a1) as a one-by-onematrix is the image of the one-by-one matrix over A1 given by (a1) (and likewise for ϕβ(a−1

2 )), it followsthat Mβ1(a1)

∼= A and Mβ2(a2)−1 = Mβ2(a−12 )

∼= A.4 Accordingly, if s ∈ Im(j), then

∂(s) = [Ms] = [A⊗A A] = [A]

which implies s ∈ ker ∂.

On the other hand, suppose ∂(s) = [Ms] = [A]. This means, in particular, that we have

M(A1, A2, ϕs) ∼= M(A1, A2, ϕ1).

By Lemma 4.11, there are isomorphisms ψi : Ai → Ai such that:

ϕ1 = (ψ−12 ⊗ IdS) ϕs (ψ1 ⊗ IdS). (4.1)

Note that for each i = 1, 2, 1Ai⊗ 1S generates Ai ⊗Ai

S as an S-module. Therefore, we plug 1A1 ⊗ 1S

into equation (4.1). On the right hand side, we have

ϕ1(1A1 ⊗ 1S) = 1A2 ⊗ 1S.

4Lemma 4.4 could just as easily been proved using β2 instead of β1. In fact, this is how Milnor proves the lemma in [Mil71].

22

Therefore:

1A2 ⊗ 1S = (ψ−12 ⊗ IdS) ϕs (ψ1 ⊗ IdS)(1A1 ⊗ 1S)

= (ψ−12 ⊗ IdS) ϕs(ψ1(1A1)⊗ 1S)

= β1(ψ1(1A1))(ψ−12 ⊗ IdS) ϕs(1A1 ⊗ 1S)

= sβ1(ψ1(1A1))(ψ−12 ⊗ IdS)(1A2 ⊗ 1S)

= sβ1(ψ1(1A1))(ψ−12 (1A2)⊗ 1S)

= sβ1(ψ1(1A2))β2(ψ−12 (1A2))(1A2 ⊗ 1S).

Since 1A2 ⊗ 1S is a generator, it follows that

sβ1(ψ1(1A2))β2(ψ−12 (1A2)) = 1S. (4.2)

Note that since ψi is an isomorphism for each i, ψi(1Ai) must be a generator for Ai as an Ai-module, and

therefore must be a unit. Therefore, let a1 = (ψ1(1A1))−1 and let a2 = ψ−1

2 (1A2). Then equation (4.2)reads:

sβ1(a−11 )β2(a2) = 1

which implies (since βi is a ring homomorphism for each i = 1, 2) that

s = (β1(a−11 ))−1(β2(a2))

−1 = β1(a1)(β2(a2))−1 = j(a1, a2).

It follows that s ∈ Im(j). Therefore ker ∂ = Im(j).

Finally we show that ker δ = Im(∂). Suppose [I] ∈ Im(∂). Then [I] = [Ms] for some s ∈ S×. ThereforeI ∼= M(A1, A2, ϕs). By Theorem 4.10, I ⊗Ai

Ai∼= Ai, it follows that δ([I]) = ([A1], [A2]) which implies

[I] ∈ ker δ.

On the other hand, suppose δ([I]) = ([A1], [A2]). By Theorem 4.9, there are projective Ai-modulesPi and an S-module isomorphism ϕ : P1 ⊗A S → P2 ⊗A S such that [I] = [M(P1, P2, ϕ)]. Therefore, byTheorem 4.10 and by the hypothesis on δ([I]):

Pi∼= I ⊗A Ai

∼= Ai.

ThereforeI ∼= M(A1, A2, ϕ)

for some S-module isomorphism ϕ : A1 ⊗A S → A2 ⊗A S. As an S-module, Ai ⊗A S is generated by1Ai

⊗1S . Therefore there is some s ∈ S such that ϕ(1A1⊗1S) = s(1A2⊗1S). Since ϕ is an isomorphism,it follows that s ∈ S×. Therefore ϕ = ϕs. Hence

[I] = [M(A1, A2, ϕs)] = [Ms] = ∂(s)

which implies [I] ∈ Im(∂). Thus Im(∂) = ker δ. This completes the proof.

Finally, we put together a special fiber square that is useful in computations for affine singular curves.Given the coordinate ring of an affine singular curve A, let A1 be the normalization of A. Then A → A1

and A1/A is an A-module.

Definition 4.16. The annihilator in A of the A-module A1/A is called the conductor of A and denoted c.

23

Lemma 4.17. Let ι : A → A1 be the natural inclusion. Then ce = c.

PROOF: By definition, cA1 ⊂ A. If f ∈ ce = (ι(c)) = (c), then there are fi ∈ c and gi ∈ A1 such thatf =

∑figi. For each i, figi ∈ A, so figi ∈ A ∩ ce = c for each i. Hence f ∈ c.

Let A2 = A/c and S = A1/c. Let α1 : A → A1 and β2 : A2 → S be the natural inclusions and letα2 : A→ A2 and β1 : A1 → S be the natural projections.

Proposition 4.18. The map β1 is surjective and the following diagram:

Aα1 //

α2

A1

β1

A/c

β2 // A1/c

is a fiber square.

PROOF: It is clear that the square above commutes, and it is obvious that β1 is surjective. Note as wellthat α2 is surjective, α1 is injective.

Let A′ denote the fiber product of A1 and A/c over β1 and β2 (with maps α′1 : A′ → A1 andα′2 : A′ → A/c). Then A′ is explicitly described as

A′ = (a1, [a]) ∈ A1 × A/c∣∣ β1(a1) = β2([a]).

Since the diagram above commutes, by the universal property of fiber products there is a unique ringhomomorphism ϕ : A→ A′ such that the following diagram commutes:

A

α2

α1

%%

ϕ

!!CC

CC

A′α′1 //

α′2

A1

β1

A/c

β2

// A1/c

This map can be described explicitly:

ϕ(a) = (α1(a), α2(a)).

I claim that ϕ is an isomorphism. Since ϕ is a ring homomorphism, it suffices to show that ϕ is bijective.Since α1 is injective, ϕ must be injective. Now suppose (a1, [a]) ∈ A′. Then β1(a1) = β2([a]), whichimplies that:

a1 ≡ a mod ce

which, by Lemma 4.17 implies thata1 − a ∈ ce = c ⊂ A.

This implies that a1 = ι(a) and hence (a1, [a]) = ϕ(a). Therefore ϕ is surjective.

24

Definition 4.19. The square in Proposition 4.18 is called the conductor square.

The following corollary is an immediate consequence of Theorem 4.15 and Proposition 4.18.

Corollary 4.20. The following is an exact sequence:

0 → A× → A×1 ⊕ (A/c)× → (A1/c)

× → Pic(A) → Pic(A1)⊕ Pic(A/c).

4.3 The Cusp and the NodeFirst we consider the cusped curve. Let k be a field and consider the ring

A = k[t2, t3] = k[x, y]/(y2 − x3).

This is a singular curve in A2k with a cusp at the origin. When k = R, the picture as follows:

Lemma 4.21. The normalization ofA is given byA1 = k[t] and the conductor is given by c = (t2, t3) ⊂ A.

PROOF: Denote A = k[x, y]. Then x, y are clearly integral over A. Moreover, since x2 = y3, in the fieldof fractions K(A), we have

(x/y)2 = y. (4.3)

Therefore x/y is integral over A. It follows that A is not integrally closed. Consider the ring

C = k[x, y, x/y].

Sincex = y (x/y)

it follows thatC = k[y, x/y].

On the other hand, by equation (4.3) we have

C = k[x/y] = k[t].

Thus C is a minimal integrally closed subring of K(A) that contains A, and therefore is the integral clo-sure of A. Hence A1 = k[t]. Note that t2 = y and t3 = x.

It is clear that y ∈ c since yt = x ∈ A. Moreover,

xt = x2/y = y3/y = y2 ∈ A.

Therefore x ∈ c. Hence (x, y) ⊂ c. On the other hand, by the correspondence principle (x, y) is a maximalideal in A. The entire ring A cannot be the conductor, since otherwise 1 · A1 ⊂ A which implies A isintegral. Therefore c = (x, y) = (t2, t3) as claimed

25

Lemma 4.22. For the conductor square for A, we have A×1 = k×, (A/c)× = k×. There is an embedding

of the additive group of k, k+, into (A1/c)× such that (A1/c)

× = k× ⊕ k+.

PROOF: First of all, by Lemma 4.21, A1 = k[t] and therefore (k[t])× = k×. Again, by Lemma 4.21,A/c = k and hence (A/c)× = k×.

Finally, note that in k[t], t3 ∈ (t2) and therefore c = (t2) in A1. Thus A1/c = k[t]/(t2). Therefore if[f ] ∈ A1/c, there is a representative f such that f(t) = a + bt, b ∈ k. Suppose [f ] is a unit with inverseg(t) = c+ dt. Since t2 = 0 we have:

1 = (a+ bt)(c+ dt) = ac+ (bc+ da)t.

Therefore a, c ∈ k× and bc = −da. Hence we can rewrite f as f = a(1 + b/at). If we let u = b/a, wehave f = a(1 + ut). Therefore

(A1/c)× = a(1 + ut)

∣∣ a ∈ k×, u ∈ k.Define a map ι : k+ → (A1/c)

× by ι(u) = 1 + ut (which is clearly injective). Then:

ι(u)ι(v) = (1 + ut)(1 + vt) = 1 + (u+ v)t = ι(u+ v).

Therefore ι embeds k+ into (A1/c)×. It remains to show that

(A1/c)× ∼= k× ⊕ k+.

Defineζ : (A1/c)

× → k× ⊕ k+

by ζ(a(1 + ut)) = (a, u). Then:

ζ(a(1 + ut)b(1 + vt)) = ζ(ab(a+ (u+ v)t)) = (ab, u+ v) = (a, u) · (b, v) = ζ(a(1 + ut))ζ(b(1 + vt)).

Hence ζ is a group homomorphism. It is clearly bijective and hence an isomorphism. This completes theproof.

Proposition 4.23. Pic(A) = k+.

PROOF: Using the Meyer-Vietoris sequence, the following is exact:

0 → A× → k× × k× → k× ⊕ k+ → Pic(A) → Pic(k[t])⊕ Pic(k).

Clearly Pic(k) = 0 and since k[t] is a principal ideal domain, Pic(k[t]) = 0. Therefore we have thefollowing exact sequence:

0 → A× → k× × k× → k× ⊕ k+ → Pic(A) → 0.

The image of k× × k× in k× ⊕ k+ is the factor k×, since the map takes units in k to themselves sittinginside the unit group of k[x]/(x2), which gives the factor of k×, whereas k+ sits in (k[x]/(x2))× as degreeone polynomials of the form 1+ut. Hence the image lies in the k×-factor. On the other hand, given a units ∈ k×, s is the image of (s, 1). So the entire factor is the image. Since the map ∂ : k× ⊕ k+ → Pic(A) issurjective, and its kernel is the image of k× × k×, it follows that

Pic(A) ∼= (k× ⊕ k+)/ ker ∂ = (k× ⊕ k+)/k× = k+.

26

This proves the proposition.

Now we consider a nodal curve. Consider the ring

B = k[t2 − 1, t3 − t] = k[x, y]/(y2 − x2(x+ 1)).

This is a singular curve in A2k with a node at the origin. For simplicity (in one step below), I will assume

char(k) 6= 2. When k = R, the picture as follows:

Lemma 4.24. The normalization of B is given by B1 = k[t] and the conductor is given by c = (t2 −1, t3 − t).

PROOF: Denote B = k[x, y]. Then x, y are clearly integral over B. Moreover, in the field of fractionsK(B), we have

(y/x)2 = x+ 1. (4.4)

Therefore y/x is integral over B. It follows that B is not integrally closed. Consider the ring

C = k[x, y, y/x].

Sincey = x(y/x)

it follows thatC = k[x, y/x].

On the other hand, by equation (4.4) we have

C = k[y/x] = k[t].

Thus C is a minimal integrally closed subring of K(B) that contains B, and therefore is the integral clo-sure of B. Hence B1 = k[t]. Note that t2 − 1 = x and t3 − t = xt = y.

It is clear that x ∈ c since xt = y ∈ B. Moreover,

yt = y2/x = x2(x+ 1)/x = x(x+ 1) ∈ B.

Therefore y ∈ c. Hence (x, y) ⊂ c. Therefore since by maximality of the ideal (x, y), c = (x, y) =(t2 − 1, t3 − t) as claimed.

Lemma 4.25. For the conductor square for B, we have B×1 = k×, (B/c)× = k× and (B1/c)

× = k×⊕k×.

27

PROOF: Since B1 = k[t], (B1)× = k×. Since B/c = k[t2− 1, t3− t]/(t2− 1, t3− t) = k, (B/c)× = k×.

Since t ∈ B1, t3− t ∈ (t2− 1) and therefore c = (t2− 1) in B1. Since I assumed char(k) 6= 2, (t− 1) and(t+ 1) are coprime in B1. Hence (t2 − 1) = (t− 1) ∩ (t+ 1) and by the Chinese Remainder Theorem:

k[t]/(t2 − 1) = k[t]/[(t− 1) ∩ (t+ 1)] ∼= k[t]/(t− 1)× k[t]/(t+ 1) ∼= k × k.

Since (R× S)× ∼= R× × S× for any rings R and S, it follows that (B1/c)× ∼= k× ⊕ k× as claimed.

Proposition 4.26. Pic(B) = k×.

PROOF: Using the Meyer-Vietoris sequence, the following is exact:

0 → B× → k× ⊕ k× → k× ⊕ k× → Pic(B) → Pic(k[t])⊕ Pic(k).

Again, Pic(k) = 0 and Pic(k[t]) = 0. Therefore we have the following exact sequence:

0 → B× → k× ⊕ k× → k× ⊕ k× → Pic(B) → 0.

The image of k× ⊕ k× in k× ⊕ k× is one factor of k×. This is because the units in both B1 and B/c arethe respective images under α1 and α2 of the units in k sitting in B. Hence since the conductor squarecommutes, their images inB1/c must be the same. Moreover, any unit (s, 1) ∈ k×⊕k× is the image underj of (s, 1). Hence the image of j is one factor of k×. Since the map ∂ : k× ⊕ k× → Pic(B) is surjective,and its kernel is the image of k× ⊗ k×, it follows that

Pic(B) ∼= (k× ⊕ k×)/ ker ∂ = (k× ⊕ k×)/k× = k×.

This proves the proposition.

The geometric story for the node can be described intuitively. Think of the elements of the Picardgroup as line bundles. The normalization of B corresponds to an affine, non-singular rational curveY1 ≈ A1

k = Spec(k[t]). Since Pic(Y1) is trivial, the only line bundle over Y1 is the trivial line bundleY1 × k. The map from B to B1 corresponds to a map from the corresponding curve Y to Y1 that sepa-rates the node on X into two distinct points p1, p2 on Y1. Every line bundle over Y corresponds to someidentification of the fibers through p1 and p2 of the trivial line bundle over Y1. There is a one-to-onecorrespondence between the possible identifications of the fibers and the transition maps in a local triv-ialization. The transition maps are in a one-to-one correspondence with maps from the trivializing openneighborhood of the node into the general linear group over k with rank one. This is precisely k×. In otherwords, every transition map is given by multiplication by some unit u ∈ k×, hence Pic(Y ) ∼= k×.

Remark 4.27. If one completes the cusped curve in projective space, the resulting Picard group is Z⊕k+,whereas the completion of the nodal curve has Picard group Z⊕k×. The interested reader can see part (b)of Exercise 6.9 in Chapter II, Section 6 of [Har77]. In this exercise, one proves that if X is the completionof the cusped curve, there is a short exact sequence

0 → k+ → Pic(X) → Z → 0

and in the case where Y is the completion of the nodal curve there is a short exact sequence

0 → k× → Pic(Y ) → Z → 0.

Since both short exact sequences end in Z, which is a projective Z-module, they split giving the descrip-tions of Pic(X) and Pic(Y ) above.

28

5 ConclusionWe have developed a purely algebraic definition for the Picard group of an affine variety and developedtechniques for its computation in the case of affine curves. When the curve is non-singular, we had to deferto some geometry. However, in the singular case, the Meier-Vietoris sequence and the conductor squareprovided a purely algebraic technique for computation.

The kernel of the natural map Pic(A) → Pic(A1), where A1 is the normalization of A, is an invariantof the singularities of a singular affine curve. When the field is an infinite, perfect domain, triviality ofthis kernel is equivalent to direct-sum cancelation holding for finitely-generated, torsion free A-modules.When k is algebraically closed, triviality of the kernel implies A is a Dedekind domain and the curveis non-singular. When k is not algebraically closed, if the kernel is trivial and A is not a Dedekind do-main then the curve has exactly one singularity. See [Wie89] for proofs of these results and for furtherdiscussion.

References[AM69] M.F. Aatiyah and I.G. McDonald. Introduction to Commutative Algebra. Westview Press, 1969.

[BL04] Christina Birkenhake and Herbert Lanbge. Complex Abelian Varieties. Springer, 2004.

[Eis04] David Eisenbud. Commutative Algebra with a View Toward Algebraic Geometry. Spring, 2004.

[Har77] Robin Hartshorne. Algebraic Geometry. Springer, 1977.

[J.S08] J.S.Milne. Algebraic number theory course notes, February 2008.

[Mil71] John Milnor. Introduction to Algebraic K-Theory. Number 72 in Annals of Mathematics Studies.Princeton University Press, 1971.

[Mir95] Rick Miranda. Algebraic Curves and Riemann Surfaces. American Mathematical Society, 1995.

[Wie89] Roger Wiegand. Picard groups of singular affine curves over a perfect field. MathematischeZeitschrift, 200:301–311, 1989.

29