Lemma 59.43.4. Let $f : X \to Y$ be an integral morphism of schemes. Then property (B) holds.

Proof. Consider $V \to Y$ étale, $\{ U_ i \to X \times _ Y V\}$ an étale covering, and $v \in V$. We have to find a $V' \to V$ and decomposition and maps as in Lemma 59.43.2. We may shrink $V$ and $Y$, hence we may assume that $V$ and $Y$ are affine. Since $X$ is integral over $Y$, this also implies that $X$ and $X \times _ Y V$ are affine. We may refine the covering $\{ U_ i \to X \times _ Y V\}$, and hence we may assume that $\{ U_ i \to X \times _ Y V\} _{i = 1, \ldots , n}$ is a standard étale covering. Write $Y = \mathop{\mathrm{Spec}}(A)$, $X = \mathop{\mathrm{Spec}}(B)$, $V = \mathop{\mathrm{Spec}}(C)$, and $U_ i = \mathop{\mathrm{Spec}}(B_ i)$. Then $A \to B$ is an integral ring map, and $B \otimes _ A C \to B_ i$ are étale ring maps. By Algebra, Lemma 10.143.3 we can find a finite $A$-subalgebra $B' \subset B$ and an étale ring map $B' \otimes _ A C \to B'_ i$ for $i = 1, \ldots , n$ such that $B_ i = B \otimes _{B'} B'_ i$. Thus the question reduces to the étale covering $\{ \mathop{\mathrm{Spec}}(B'_ i) \to X' \times _ Y V\} _{i = 1, \ldots , n}$ with $X' = \mathop{\mathrm{Spec}}(B')$ finite over $Y$. In this case the result follows from Lemma 59.43.3. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).