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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.2012-03-17
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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.2012-03-17
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    Of course the "Čech way" is pretty valuable. I was just asking for a different approach.2012-03-18

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