Lemma 33.48.3. Let $k$ be a field. Let $X$ be a proper scheme over $k$. Let $\mathcal{L}$ be an ample invertible $\mathcal{O}_ X$-module. Let $s \in \Gamma (X, \mathcal{L})$. Assume

$s$ is a regular section (Divisors, Definition 31.14.6),

for every closed point $x \in X$ we have $\text{depth}(\mathcal{O}_{X, x}) \geq 2$, and

$X$ is connected.

Then the zero scheme $Z(s)$ of $s$ is connected.

**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

\[ 0 \to \mathcal{L}^{\otimes -n} \xrightarrow {s^ n} \mathcal{O}_ X \to \mathcal{O}_{Z(s^ n)} \to 0 \]

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.
$\square$

## Comments (3)

Comment #5449 by Johan on

Comment #5450 by Laurent Moret-Bailly on

Comment #5669 by Johan on