Lemma 31.29.4. Assumptions and notation as in Lemma 31.29.3. If $s$ is nonzero, then every irreducible component of $X \setminus U$ has codimension $1$ in $X$.
Proof. Let $\xi \in X$ be a generic point of an irreducible component $Z$ of $X \setminus U$. After replacing $X$ by an open neighbourhood of $\xi $ we may assume that $Z = X \setminus U$ is irreducible. Since $s : \mathcal{O}_ U \to \mathcal{F}|_ U$ is an isomorphism, if the codimension of $Z$ in $X$ is $\geq 2$, then $s : \mathcal{O}_ X \to \mathcal{F}$ is an isomorphism by Lemma 31.12.12 and Serre's criterion Properties, Lemma 28.12.5. This would mean that $Z = \emptyset $, a contradiction. $\square$
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