The Stacks project

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. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 04DR. Beware of the difference between the letter 'O' and the digit '0'.