Lemma 38.28.7. In Situation 38.28.1 let $K$ be as in Lemma 38.28.2. Let $W \subset X$ be as in Lemma 38.28.6. 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 support of $\mathcal{F}$. This is a closed subset as $\mathcal{F}$ is of finite presentation. By part (3) of Lemma 38.28.6 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, Lemma 37.52.9 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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