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

1. $\pi$ is finite,

2. $\pi$ is unramified,

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

4. $\kappa (s) \subset \kappa (x)$ is purely inseparable1.

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 10.41.11 we have $B_{\mathfrak q} = B_{\mathfrak p}$ (note that a finite ring map satisfies going up, see Algebra, Section 10.41.) 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 10.20.1) 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. $\square$

[1] In view of condition (2) this is equivalent to $\kappa (s) = \kappa (x)$.

Comment #5795 by James A. Myer on

I think "By Conditions (2), (3), and (4) imply that $B_\mathfrak{q}/\mathfrak{p}B_\mathfrak{q}=\kappa(\mathfrak{p})$. Algebra, Lemma..." should read instead "Conditions (2), (3), and (4) imply that $B_\mathfrak{q}/\mathfrak{p}B_\mathfrak{q}=\kappa(\mathfrak{p})$. By Algebra, Lemma..." -- I guess the "By" got cycled around.

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