Lemma 59.42.3. Let $f : X \to Y$ be an integral morphism of schemes. Then property (A) holds.

Proof. Let $U \to X$ be étale, and let $u \in U$ be a point. We have to find $V \to Y$ étale, a disjoint union decomposition $X \times _ Y V = W \amalg W'$ and an $X$-morphism $W \to U$ with $u$ in the image. We may shrink $U$ and $Y$ and assume $U$ and $Y$ are affine. In this case also $X$ is affine, since an integral morphism is affine by definition. Write $Y = \mathop{\mathrm{Spec}}(A)$, $X = \mathop{\mathrm{Spec}}(B)$ and $U = \mathop{\mathrm{Spec}}(C)$. Then $A \to B$ is an integral ring map, and $B \to C$ is an étale ring map. By Algebra, Lemma 10.143.3 we can find a finite $A$-subalgebra $B' \subset B$ and an étale ring map $B' \to C'$ such that $C = B \otimes _{B'} C'$. Thus the question reduces to the étale morphism $U' = \mathop{\mathrm{Spec}}(C') \to X' = \mathop{\mathrm{Spec}}(B')$ over the finite morphism $X' \to Y$. In this case the result follows from Lemma 59.42.2. $\square$

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