Lemma 59.43.3. Let $f : X \to Y$ be a finite 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 finite over $Y$, this also implies that $X$ is affine. During the proof we may (finitely often) replace $(V, v)$ by an étale neighbourhood $(V', v')$ and correspondingly the covering $\{ U_ i \to X \times _ Y V\} $ by $\{ V' \times _ V U_ i \to X \times _ Y V'\} $.

Since $X \times _ Y V \to V$ is finite there exist finitely many (pairwise distinct) points $x_1, \ldots , x_ n \in X \times _ Y V$ mapping to $v$. We may apply More on Morphisms, Lemma 37.40.5 to $X \times _ Y V \to V$ and the points $x_1, \ldots , x_ n$ lying over $v$ and find an étale neighbourhood $(V', v') \to (V, v)$ such that

with $T_ a \to V'$ finite with exactly one point $p_ a$ lying over $v'$ and moreover $\kappa (v') \subset \kappa (p_ a)$ purely inseparable, and such that $R \to V'$ has empty fibre over $v'$. Because $X \to Y$ is finite, also $R \to V'$ is finite. Hence after shrinking $V'$ we may assume that $R = \emptyset $. Thus we may assume that $X \times _ Y V = X_1 \amalg \ldots \amalg X_ n$ with exactly one point $x_ l \in X_ l$ lying over $v$ with moreover $\kappa (v) \subset \kappa (x_ l)$ purely inseparable. Note that this property is preserved under refinement of the étale neighbourhood $(V, v)$.

For each $l$ choose an $i_ l$ and a point $u_ l \in U_{i_ l}$ mapping to $x_ l$. Now we apply property (A) for the finite morphism $X \times _ Y V \to V$ and the étale morphisms $U_{i_ l} \to X \times _ Y V$ and the points $u_ l$. This is permissible by Lemma 59.42.3 This gives produces an étale neighbourhood $(V', v') \to (V, v)$ and decompositions

and $X$-morphisms $a_ l : W_ l \to U_{i_ l}$ whose image contains $u_{i_ l}$. Here is a picture:

After replacing $(V, v)$ by $(V', v')$ we conclude that each $x_ l$ is contained in an open and closed neighbourhood $W_ l$ such that the inclusion morphism $W_ l \to X \times _ Y V$ factors through $U_ i \to X \times _ Y V$ for some $i$. Replacing $W_ l$ by $W_ l \cap X_ l$ we see that these open and closed sets are disjoint and moreover that $\{ x_1, \ldots , x_ n\} \subset W_1 \cup \ldots \cup W_ n$. Since $X \times _ Y V \to V$ is finite we may shrink $V$ and assume that $X \times _ Y V = W_1 \amalg \ldots \amalg W_ n$ as desired. $\square$

## 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.

## Comments (0)