Lemma 31.24.1. Let $X$ be a quasi-compact scheme. Let $h \in \Gamma (X, \mathcal{O}_ X)$ and $f \in \Gamma (X, \mathcal{K}_ X)$ such that $f$ restricts to zero on $X_ h$. Then $h^ n f = 0$ for some $n \gg 0$.
Proof. We can find a covering of $X$ by affine opens $U$ such that $f|_ U = s^{-1}a$ with $a \in \mathcal{O}_ X(U)$ and $s \in \mathcal{S}(U)$. Since $X$ is quasi-compact we can cover it by finitely many affine opens of this form. Thus it suffices to prove the lemma when $X = \mathop{\mathrm{Spec}}(A)$ and $f = s^{-1}a$. Note that $s \in A$ is a nonzerodivisor hence it suffices to prove the result when $f = a$. The condition $f|_{X_ h} = 0$ implies that $a$ maps to zero in $A_ h = \mathcal{O}_ X(X_ h)$ as $\mathcal{O}_ X \subset \mathcal{K}_ X$. Thus $h^ na = 0$ for some $n > 0$ as desired. $\square$
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