Lemma 37.30.4. Let $f : X \to S$ and $\mathcal{E}$ be as in Lemma 37.30.1 and in addition assume $\mathcal{E}$ is an invertible $\mathcal{O}_ X$-module. If moreover the geometric fibres of $f$ are integral, then $Z$ is closed in $S$.

Proof. Since $j : Z \to S$ is of finite presentation, it suffices to show: for any morphism $g : \mathop{\mathrm{Spec}}(A) \to S$ where $A$ is a valuation ring with fraction field $K$ such that $g(\mathop{\mathrm{Spec}}(K)) \in j(Z)$ we have $g(\mathop{\mathrm{Spec}}(A)) \subset j(Z)$. See Morphisms, Lemma 29.6.5. This follows from Lemma 37.30.3 and the characterization of $j : Z \to S$ in Lemma 37.30.1. $\square$

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