 Solution

For a divisor on an algebraic curve $D=\sum n_{P}P$, define its line bundle associated to a divisor inside the sheaf of rational functions byLocally choose a rational equation $f_i$ for $D$, as in the construction of the quotient sheaf above. ThenConsequently this is an invertible sheaf. Multiplication gives an isomorphismas is checked on the displayed local generators. If $D=\mathrm{div}(f)$, multiplication by $f$ identifies ${\mathcal{O}}_X(D)$ with ${\mathcal{O}}_X$. Thus $D\mapsto [{\mathcal{O}}_X(D)]$ induces a group homomorphism from the divisor class group to the Picard group.

For surjectivity, let $\mathcal{L}$ be an invertible sheaf. Trivialize its one-dimensional rational fibre over the function field $k(X)$; equivalently, choose a nonzero rational section and identify $\mathcal{L}$ with a subsheaf of the rational-function sheaf. A finite trivializing cover givesOn overlaps $a_i/a_j$ is a regular unit. The integers $-v_P(a_i)$ therefore agree wherever two descriptions apply and define a finite divisor on an algebraic curve $D$. Its local description gives $\mathcal{L}={\mathcal{O}}_X(D)$. A different rational trivialization multiplies all $a_i$ by the same rational function, changing $D$ by a principal divisor, so the resulting divisor class is intrinsic.

For injectivity, an isomorphism ${\mathcal{O}}_X(D)\to {\mathcal{O}}_X(E)$ becomes multiplication by some $g\in k(X{)}^{\ast }$ on the rational fibre. At each closed point its effect on the local free generators givesHence $D-E=\mathrm{div}(g)$. Conversely this equality gives the required isomorphism by multiplication by $g$. We have proved the group isomorphism