Lemma 37.48.3. Let $S$ be a scheme. If $U \subset S$ is open and $V \to U$ is a surjective integral morphism, then there exists a surjective integral morphism $\overline{V} \to S$ with $\overline{V} \times _ S U$ isomorphic to $V$ as schemes over $U$.

Proof. Let $V' \to S$ be the normalization of $S$ in $U$, see Morphisms, Section 29.53. By construction $V' \to S$ is integral. By Morphisms, Lemmas 29.53.6 and 29.53.12 we see that the inverse image of $U$ in $V'$ is $V$. Let $Z$ be the reduced induced scheme structure on $S \setminus U$. Then $\overline{V} = V' \amalg Z$ works. $\square$

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