Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 113 / 3

Let $K$ be the function field of the integral scheme $X$ and choose a nonzero rational section of a line bundle $s\in {\mathcal{L}}_{\eta }$. Choose an open trivializing cover $U_i$ with generators $e_i$ for the invertible sheaf $\mathcal{L}$. On each nonempty $U_i$, write $s=a_i{e}_i$ with $a_i\in K^{\ast }$. If $e_i=u_{ij}{e}_j$ on an overlap, then $a_j=a_i{u}_{ij}$, so $a_i/a_j=u_{ij}^{-1}$ is a regular unit. Consequently the local rational functions $a_i$ define a Cartier divisor $D$.

The divisor line bundle ${\mathcal{O}}_X(D)$ restricts to $a_i^{-1}{\mathcal{O}}_{U_i}$. Define its local isomorphism from $\mathcal{L}$ by $e_i\mapsto a_i^{-1}$. On an overlap,so the maps glue. HenceThe rational section maps to the rational function $1$, which also fixes the sign convention for $D$. Choosing a different nonzero rational section changes $D$ by a principal Cartier divisor, so its linear equivalence of Cartier divisors class is unchanged. No projectivity or Noetherian hypothesis is required.

For each point $x$, choose a global section $s$ whose germ $s_x$ generates the rank-one free ${\mathcal{O}}_{X,x}$-module ${\mathcal{L}}_x$. The section $s\otimes s$ is global, and tensor products commute with taking stalks. Thereforegenerates $(\mathcal{L}{\otimes }_{{\mathcal{O}}_X}\mathcal{L}{)}_x$. Thus the tensor square is globally generated. More generally, the tensor product of globally generated line bundles is globally generated: choose a generator section for each factor at the point and tensor them. This works on every scheme, including nonreduced schemes.