LINE BUNDLES ON PROJECTIVE SPACE

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1 LINE BUNDLES ON PROJECTIVE SPACE DANIEL LITT We wish to show that any line bundle over P n k is isomorphic to O(m) for some m; we give two proofs below, one following Hartshorne, and the other assuming some knowledge of sheaf cohomology. Throughout, let X be an integral scheme. Definition 1 (Cartier Divisors). We define O X to be the sheaf of groups which is the subsheaf of O X consisting of units, and K X be the sheaf associated to the presheaf K X : U Frac(Γ(U, O X )), which as we will see is actually just this presheaf in the integral case. (This can be generalized to non-integral schemes by taking the total ring of fractions instead of the fraction field). Let K X be the sheaf of groups consisting of the units of K X. Then there is a short exact sequence 0 O X K X K X/O X 0; a Cartier Divisor is a global section of K X /O X. Elements of the image of the map Γ(X, K X ) Γ(X, K X /O X ) are principal divisors. Remark 1 (Explicitly writing down a Cartier Divisor). By our explicit construction of the quotient sheaf, a Cartier Divisor is an equivalence class [{U i, f i }] where the U i are an open cover of X, f i Γ(U i, K X ), and for each i, j, we have that f i /f j (where it is defined) is in the image of the map Γ(U i U j, O X) Γ(U i U j, K X). The equivalence relation here is that {U i, f i } {V i, g i } if the two collections have a common refinement, or, equivalently, if {U i V j, f i Ui V j } {U i V j, g j Ui V j } is a Cartier divisor. In this set-up, we may write the product of two Cartier divisors {U i, f i }, {V i, g i } as {U i V j, f i g j }. We will soon motivate this definition. Before continuing, we wish to note quickly that Lemma 1. The group of Cartier divisors modulo principal divisors, that is is isomorphic to H 1 (X, O X ). coker(γ(x, K X) Γ(X, K X/O X)), Proof. By the long exact sequence in sheaf cohomology, it suffices to show that H 1 (X, K X ) = 0. We claim in fact that the sheaf K X is simply the constant sheaf S defined by Γ(U, S) = K, where K is the field of rational functions on X (the residue field of the generic point of X). This is a sheaf as X is irreducible. Indeed, let Spec(A), Spec(B) X be any two affine opens of X. As X is irreducible, both Spec(A), Spec(B) contain the generic point η of X; thus Frac(A) O X,η Frac(B). So K X agrees with S on basic opens, and thus agrees with S. But then K X is a flasque sheaf, and thus has vanishing cohomology. 1

2 We may now motivate Cartier divisors as a tool in our study of line bundles. Definition 2. We have noted before that isomorphism classes of line bundles over a scheme for a group, with the ring of regular functions as the identity, tensor product as the operation, and dualizing as the inverse. We call this group the Picard group of a scheme X, or Pic(X). Lemma 2 (Line Bundles are Cartier Divisors). There is a natural isomorphism L : Γ(X, K X/O X) {Invertible subsheaves of K X } Proof. We first define the desired map L. Given a Cartier divisor D = {U i, f i }, let L(D) be the subsheaf of K X generated by f 1 i on U i. Note that on intersections U i U j, f i /f j is invertible, so f 1 i, f 1 j generate the same sub-module. The map is trivially well-defined across equivalence classes. Furthermore, note that given an invertible subsheaf of K X, say L K X, we may recover a divisor by letting U i be a cover of X on which L is trivial, and letting f i be the inverse of a local generator on U i, giving a Cartier divisor {U i, f i }. These maps are mutually inverse by construction. We must now check that the map is a homomorphism, e.g. that given two Cartier divisors D 1, D 2, we have that L(D 1 D2 1 ) L(D 1) L(D 2 ) 1, with the natural embedding in K X. But indeed, if D 1 = {U i, f i } and D 2 = {U i, g i }, then both sides are locally generated by f 1 i g i. Furthermore, it is clear that {X, 1} maps to O X K X, completing the proof. Lemma 3 (The Picard Group is the Cartier Class Group). We claim that the map above induces an isomorphism H 1 (X, O X) Pic(X). Proof. We first check that the map is well-defined and injective. This is exactly the claim that the kernel of the map Γ(X, K X/O X) {Invertible subsheaves of K X } Pic(X) is exactly the image of Γ(X, K X ) Γ(X, K X /O X ), where the last map sends an invertible subsheaf of K X to its isomorphism class. Equivalently, we wish to show that a divisor D is principal if and only if L(D) O X. If D is principal and given by some f Γ(X, K ), L(D) is simply f 1 O X, so sending 1 to f 1 gives an isomorphism O X L(D). Similarly, given such an isomorphism, the inverse of the image of the element 1 gives the desired principal divisor. To see surjectivity, it suffices to show that the map {Invertible subsheaves of K X } Pic(X) is surjective that is, that every line bundle on X admits an embedding into K X. Let L be a line bundle. Note that by our proof of Lemma 1, the sheaf K X is the constant sheaf S. Furthermore, for U with L trivial, we have Γ(U, L K X ) Γ(U, K X ), so L K X is the constant sheaf S as well, giving a map L L K X K X, where the first map is injective by integrality. This completes the proof. So to characterize line bundles on projective space P n k, it would suffice to compute the group Pic(Pn k ), and to find which element corresponds to, say, O(1). However, we need more computational tools before we can do this. From here on we assume that X is separated, noetherian and normal. Furthermore, for convenience we work in slightly less generality than Hartshorne; rather than assuming that the scheme is locally factorial, we assume it admits an open cover by the spectra of UFDs (all of these conditions are satisfied by P n k ). 2

3 Definition 3 (Weil Divisors). Prime divisors of X are closed integral subschemes of codimension one. The group of Weil divisors, denoted Div(X), is the free Abelian group generated by such primes. Let K = Γ(X, K X ). If p is a prime divisor of X with generic point η, the local ring O η,x is a discrete valuation ring (as e.g. v p : K Z. it is an integrally closed noetherian local ring of dimension one) with valuation Then there is a natural map K Div(X) given by sending f K to v p (f)p. We must show that this sum is finite. Choose an open affine U = Spec(A) on which f is regular (e.g. localize an open affine at the denominator of f); its complement is closed and thus by Noetherianness contains only finitely many prime divisors of X. By our choice of U, v p (f) 0 for p U; those p for which it is nonzero are precisely those containing Af, of which there are finitely many. It is clear that this map is a homomorphism. Divisors in the image of the map K Div(X) are called principal divisors. The cokernel of this map is the class group, denoted Cl(X). Before proceeding, we need a lemma. Lemma 4. Let A be a UFD with field of fractions K, and f K an element such that v p (f) = 0 for all p. Then f is a unit in A. Proof. It suffices to show that f is in A, as v p (1/f) = v p (f). Let f = u f ni i /u g j with f i, g j irreducible and f i g j for all i, j, and u, u units. Up to units, there is a unique way to do this, as A is a UFD. But then the product in the denominator must be empty, as otherwise v (gj)(f) 0 for any g j. But then f A, as desired. It is as yet unclear why Weil divisors should be a useful tool; the next lemma should clear this up. Lemma 5 (Cartier Divisors are the same as Weil Divisors). There is a natural isomorphism Div(X) Γ(X, K X/O X) which induces an isomorphism Cl(X) H 1 (X, O X). Proof. Let D be a Weil divisor, and let Spec(A i ) be an open affine cover of X where the A i are UFDs. Restricting D to each A i and writing it as n i,j p i,j, we may pick generators f i,j of each prime p i,j of A i appearing in D, and send D to the Cartier divisor {Spec(A i ), j f ni,j i,j }. This is a Cartier divisor as the ratios on intersections Spec(A i ) Spec(A j ) are units, by lemma 4 above. To map in the other direction, let {U i, f i } be a Cartier divisor; we send it to v p (f i ), where we choose an f i defined on an open U i containing p. The value is independent of this choice by the definition of a Cartier divisor. To see that the sum is finite, note that each f i has finitely many non-zero valuations by a previous argument we may assume there are only finitely many f i by Noetherianness. It is clear that these maps are mutually inverse. Furthermore, principal divisors map to principal divisors, giving the second isomorphism. We can now directly compute the class group of projective space. Lemma 6 (Computing Class Group using Weil Divisors). Let the degree of a Weil divisor D = n i p i on P n k be given by deg(d) = n i deg(p i ), where deg(p i ) is the degree of the homogeneous prime of k[x 0,..., x n ] 3

4 corresponding to p i. It is clear that deg is a homomorphism Div(X) Z. The degree function induces an isomorphism Cl(P n k) Z. Proof. First, we must show that the map is well-defined that is, that the degree of any principal divisor is zero. But this is clear a rational function on P n k is exactly an element of Frac(k[x 0,..., x n ]) of degree zero, and thus its divisor has degree zero. Furthermore, the map is clearly surjective, as e.g. the divisor H = (x 0 ) has degree 1. To see that the map is injective, we must check that any divisor of degree zero is principal. Indeed, let D = n i p i be a divisor of degree zero, and let f i be a generator of p i, which has degree deg(p i ) (such a generator exists as p i is a prime of height one in a UFD). Then f ni i divisor D. is a rational function inducing the So we now see that Z Cl(P n k ) Pic(Pn k ); all that remains is to see that O(1) corresponds to 1 Z. But this is easy trivializing O(1) locally on U x0, we see that the divisor of O(1) is (x 0 ), which has degree 1. So we ve shown the following: Theorem 1. Every line bundle on P n k is isomorphic to O(m) for some m Z. We give another proof of this claim, which doesn t use Lemma 6. Lemma 7 (Codimension 2 doesn t matter). Let U be an open subscheme of X. Then there is a natural surjective map Cl(X) Cl(U); if the codimension of X U is greater than or equal to 2, this map is an isomorphism. Proof. We have a diagram K Div(X) Cl(X) K Div(U) Cl(U) where the middle arrow is given by quotienting Div(X) by the free abelian group generated by prime divisors contained in X U, and the map Cl(X) Cl(U) is given by the functoriality of the cokernel. Surjectivity of this map follows immediately from the diagram. To see that the map is an isomorphism if X U has codimension greater than or equal to 2, note that primes of height one are unaffected by quotienting by a prime of greater height, so in fact the map Div(X) Div(U) is an isomorphism; principal divisors are the same as K X is a constant sheaf. Lemma 8. Let A be a Noetherian domain. Then if A is a unique factorization domain, Cl(Spec(A)) = 0. Proof. As A is a UFD, every prime of height one is principal. Let p be a prime divisor; then it is generated by some f, which clearly induces the divisor p. From here on let X = P n k. We may now directly compute H1 (P n k, O X ) Pic(Pn k ). Lemma 9 (Computing Class Group using Cohomology). We claim that H 1 (P n k, O X) Z. 4

5 Proof. Let P n k = Proj(k[x 0,..., x n ]). Consider the open subscheme given by U = U x0 U x1. We claim that X U has co-dimension 2; indeed, it is the complement of the closed subscheme cut out by (x 0, x 1 ), which is a prime of height 2. So by lemma 7, it suffices to compute H 1 (U, O X ) Cl(U) Cl(X) H1 (X, O X ). By lemma 8, we have that H 1 (U xi, O X ) Cl(U x i ) 0, so we may use {U x0, U x1 } as a Cech cover of U (as it is an acyclic cover for O X ). This gives the ordered Cech complex 0 k k k Z 0 as Γ(U xi, O X ) (k[x 0,..., x n ] xi ) 0 k, and Γ(U x0 U x1, O X ) (k[x 0,..., x n ] x0,x 1 ) 0 k {(x 0 /x 1 ) n } k Z. It is clear that the map k k k Z is surjective on the k coordinate and does not touch the Z coordinate, giving H 1 (P n k, O X ) Z as desired. To see that O(1) corresponds to 1 Z through this argument, we must analyze the transition map γ 01 : Γ(U x0 U x1, O(1)) Γ(U x0 U x1, O(1)) induced by the two trivializations on U x0, U x1. But in the problem set on Proj, we showed that this is induced by multiplying by x 0 /x 1 in one direction, and x 1 /x 0 in the other unwinding the Cech computation above, this corresponds exactly to 1 Z, as desired. So we have another proof of the claim. 5

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