Lemma 41.7.3. Let $\pi : X \to S$ be a morphism of schemes. Let $s \in S$. Assume that

$\pi $ is finite,

$\pi $ is unramified,

$\pi ^{-1}(\{ s\} ) = \{ x\} $, and

$\kappa (s) \subset \kappa (x)$ is purely inseparable^{1}.

Then there exists an open neighbourhood $U$ of $s$ such that $\pi |_{\pi ^{-1}(U)} : \pi ^{-1}(U) \to U$ is a closed immersion.

**Proof.**
The question is local on $S$. Hence we may assume that $S = \mathop{\mathrm{Spec}}(A)$. By definition of a finite morphism this implies $X = \mathop{\mathrm{Spec}}(B)$. Note that the ring map $\varphi : A \to B$ defining $\pi $ is a finite unramified ring map. Let $\mathfrak p \subset A$ be the prime corresponding to $s$. Let $\mathfrak q \subset B$ be the prime corresponding to $x$. Conditions (2), (3) and (4) imply that $B_{\mathfrak q}/\mathfrak pB_{\mathfrak q} = \kappa (\mathfrak p)$. By Algebra, Lemma 10.40.11 we have $B_{\mathfrak q} = B_{\mathfrak p}$ (note that a finite ring map satisfies going up, see Algebra, Section 10.40.) Hence we see that $B_{\mathfrak p}/\mathfrak pB_{\mathfrak p} = \kappa (\mathfrak p)$. As $B$ is a finite $A$-module we see from Nakayama's lemma (see Algebra, Lemma 10.19.1) that $B_{\mathfrak p} = \varphi (A_{\mathfrak p})$. Hence (using the finiteness of $B$ as an $A$-module again) there exists a $f \in A$, $f \not\in \mathfrak p$ such that $B_ f = \varphi (A_ f)$ as desired.
$\square$

## Comments (2)

Comment #5795 by James A. Myer on

Comment #5808 by Johan on