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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