The Stacks project

Proof. Since $s$ is a regular section, so is $s^ n \in \Gamma (X, \mathcal{L}^{\otimes n})$ for all $n > 1$. Moreover, the inclusion morphism $Z(s) \to Z(s^ n)$ is a bijection on underlying topological spaces. Hence if $Z(s)$ is disconnected, so is $Z(s^ n)$. Now consider the canonical short exact sequence

Consider the $k$-algebra $R_ n = \Gamma (X, \mathcal{O}_{Z(s^ n)})$. If $Z(s)$ is disconnected, i.e., $Z(s^ n)$ is disconnected, then either $R_ n$ is zero in case $Z(s^ n) = \emptyset$ or $R_ n$ contains a nontrivial idempotent in case $Z(s^ n) = U \amalg V$ with $U, V \subset Z(s^ n)$ open and nonempty (the reader may wish to consult Lemma 33.9.3). Thus the map $\Gamma (X, \mathcal{O}_ X) \to R_ n$ cannot be an isomorphism. It follows that either $H^0(X, \mathcal{L}^{\otimes -n})$ or $H^1(X, \mathcal{L}^{\otimes -n})$ is nonzero for infinitely many positive $n$. This contradicts Lemma 33.48.1 or 33.48.2 and the proof is complete. $\square$