Lemma 76.12.7. In Situation 76.12.1 let $K$ be as in Lemma 76.12.2. Let $W \subset X$ be as in Lemma 76.12.6. Set $\mathcal{F} = H^0(K)|_ W$. Then, after possibly shrinking the open $W$, the support of $\mathcal{F}$ is proper over $A$.

Proof. Fix $n \geq 1$. Let $I_ n = \mathop{\mathrm{Ker}}(A \to A_ n)$. By More on Algebra, Lemma 15.11.3 the pair $(A, I_ n)$ is henselian. Let $Z \subset W$ be the scheme theoretic support of $\mathcal{F}$. This is a closed subspace as $\mathcal{F}$ is of finite presentation. By part (3) of Lemma 76.12.6 we see that $Z \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A_ n)$ is equal to the support of $\mathcal{F}_ n$ and hence proper over $\mathop{\mathrm{Spec}}(A/I)$. By More on Morphisms of Spaces, Lemma 75.36.10 we can write $Z = Z_1 \amalg Z_2$ with $Z_1, Z_2$ open and closed in $Z$, with $Z_1$ proper over $A$, and with $Z_1 \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A/I_ n)$ equal to the support of $\mathcal{F}_ n$. In other words, $|Z_2|$ does not meet $X_0$. Hence after replacing $W$ by $W \setminus Z_2$ we obtain the lemma. $\square$

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