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 37.53.9 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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