 Solution

Since $X$ is an irreducible variety, every nonempty open subset is irreducible and the sheaf of nonzero rational functions on an irreducible variety has value $k(X{)}^{\ast }$ there. All its restriction maps between nonempty opens are identities, and restriction to the empty set is surjective. Thus ${\mathcal{K}}_X^{\ast }$ is a flabby sheaf, and the introductory argument gives $H^1(X,{\mathcal{K}}_X^{\ast })=0$.

Apply the long exact sequence in sheaf cohomology to the short exact sequence of sheavesIts relevant portion isThe connecting homomorphism is therefore surjective, with kernel the image of $k(X{)}^{\ast }$. The computation in part (a) now givesConcretely, local rational equations $f_i$ of a divisor on an algebraic curve differ by regular units on overlaps. Their ratios give the gluing data of an invertible sheaf; changing the local equations by units changes those data by a coboundary. This is the transition-function interpretation of the identification with the Picard group. Reversing all transition ratios replaces a class by its inverse; either consistent convention gives the same group isomorphism.