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 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. $\square$
Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #5795 by James A. Myer on
Comment #5808 by Johan on