Lemma 37.41.4. Let $f : X \to S$ be a morphism of schemes. Let $s \in S$. Let $x_1, \ldots , x_ n \in X_ s$. Assume that

$f$ is locally of finite type,

$f$ is separated, and

$x_1, \ldots , x_ n$ are pairwise distinct isolated points of $X_ s$.

Then there exists an elementary étale neighbourhood $(U, u) \to (S, s)$ and a decomposition

\[ U \times _ S X = W \amalg V_1 \amalg \ldots \amalg V_ n \]

into open and closed subschemes such that the morphisms $V_ i \to U$ are finite, the fibres of $V_ i \to U$ over $u$ are singletons $\{ v_ i\} $, each $v_ i$ maps to $x_ i$ with $\kappa (x_ i) = \kappa (v_ i)$, and the fibre of $W \to U$ over $u$ contains no points mapping to any of the $x_ i$.

**Proof.**
Choose $(U, u) \to (S, s)$ and $V_ i \subset X_ U$ as in Lemma 37.41.2. Since $X_ U \to U$ is separated (Schemes, Lemma 26.21.12) and $V_ i \to U$ is finite hence proper (Morphisms, Lemma 29.44.11) we see that $V_ i \subset X_ U$ is closed by Morphisms, Lemma 29.41.7. Hence $V_ i \cap V_ j$ is a closed subset of $V_ i$ which does not contain $v_ i$. Hence the image of $V_ i \cap V_ j$ in $U$ is a closed set (because $V_ i \to U$ proper) not containing $u$. After shrinking $U$ we may therefore assume that $V_ i \cap V_ j = \emptyset $ for all $i, j$. This gives the decomposition as in the lemma.
$\square$

## Comments (0)

There are also: