# Picard group and cohomology

It's an easy but boring exercise (Hartshorne Ex. III.4.5 or Liu 5.2.7) that the group $Pic(X)$ of isomorphism classes of invertible sheaves on a ringed topological space (well, maybe we can restrict to schemes) is isomorphic to $H^{1}(X, \mathcal{O}_X^{*})$, where $\mathcal{O}_X^{*}$ denotes the sheaf whose sections over an open set $U$ are the units in the ring $\mathcal{O}_X(U)$.

The proof that I know (that uses the hint given by Hartshorne) uses heavily Cech cohomology: basically the idea is that given an invertible sheaf $\mathcal{L}$ and an affine open covering $\mathcal{U}=(U_i)$ on which $\mathcal{L}$ is free, we can construct an element in $\check{C}^1(\mathcal{U},\mathcal{O}_X^{*})$ using the restriction of the local isomorphism to the intersections $U_i\cap U_j$. The cocycle condition on triple intersection implies that we have a well defined element in $\check{H}^{1}(X, \mathcal{O}_X^{*})$. Then one proves that the map is an isomorphism of groups.

My question is the following: this approach is not very enlightening. Is there a more intrinsic proof of the isomorphism between $Pic(X)$ and $H^{1}(X, \mathcal{O}_X^{*})$, without Cech cohomology?

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I'm puzzled. The proof via Čech cohomology is the most enlightening way of seeing the isomorphism! There is no a priori reason why a cohomology group computed via injective resolutions or other abstract nonsense should have anything to do with the classification of line bundles. – Zhen Lin Mar 17 '12 at 18:36

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I agree with @Zhen: thinking about line bundles in the "Čech way" seems pretty valuable. Not that other proofs wouldn't be cool to see. – Dylan Moreland Mar 17 '12 at 18:37
Of course the "Čech way" is pretty valuable. I was just asking for a different approach. – FedeB Mar 18 '12 at 11:57

## 1 Answer

Suppose for simplicity that $X$ is integral. Consider the exact sequence of groups $$1\to O_X^* \to K_X^* \to K_X^*/O_X^* \to 1$$ where $K_X$ is the constant sheaf of rational functions on $X$. Taking the cohomology will give $$K(X)^* \to H^0(X, K_X^*/O_X^*) \to H^1(X, O_X^*) \to H^1(X, K_X^*).$$ Now the last term vanishes because $K_X^*$ is a flasque sheaf, and the cokernel of the left arrow is by definition the group of Cartier divisors on $X$ up to linear equivalence. As $X$ is integral, this cokernel is known to be isomorphic to $\mathrm{Pic}(X)$.

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Ok, this works in the integral case. But what about the general case? The isomorphism between $H^1(X,\mathcal{O}_X^*)$ and Pic(X) can be defined even in the non integral case, also when you don't have the isomorphism between Cartier divisors and Picard group. – FedeB Mar 18 '12 at 12:10
For example if $X$ is not Noetherian and if you have embedded points, then the canonical map $CaCl(X)\to Pic(X)$ is not surjective. – FedeB Mar 18 '12 at 12:11
@FedeB, the above proof works for Noetherian schemes without embedded points. Unfortunately, in general I don't have an alternative proof to the usual one. – QiL'8 Mar 18 '12 at 20:43
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