The Stacks project

In an unramified ring map, one can separate the points in a fiber by passing to an étale neighbourhood.

Lemma 10.152.3. Let $R \to S$ be a ring map. Let $\mathfrak p$ be a prime of $R$. If $R \to S$ is unramified then there exist

  1. an étale ring map $R \to R'$,

  2. a prime $\mathfrak p' \subset R'$ lying over $\mathfrak p$.

  3. a product decomposition

    \[ R' \otimes _ R S = A_1 \times \ldots \times A_ n \times B \]

with the following properties

  1. $R' \to A_ i$ is surjective,

  2. $\mathfrak p'A_ i$ is a prime of $A_ i$ lying over $\mathfrak p'$, and

  3. there is no prime of $B$ lying over $\mathfrak p'$.

Proof. We may apply Lemma 10.145.4. Thus, after an étale base change, we may assume that $S = A_1 \times \ldots \times A_ n \times B$, that each $R \to A_ i$ is finite with exactly one prime $\mathfrak r_ i$ lying over $\mathfrak p$ such that $\kappa (\mathfrak p) \subset \kappa (\mathfrak r_ i)$ is purely inseparable, and that $R \to B$ is not quasi-finite at any prime lying over $\mathfrak p$. Since $R \to S$ is quasi-finite (see Lemma 10.151.6) we see there is no prime of $B$ lying over $\mathfrak p$. By Lemma 10.151.5 we see that $\kappa (\mathfrak r_ i)/\kappa (\mathfrak p)$ is separable hence the trivial field extension, and that $\mathfrak p(A_ i)_{\mathfrak r_ i}$ is the maximal ideal. Also, by Lemma 10.41.11 (which applies to $R \to A_ i$ because a finite ring map satisfies going up by Lemma 10.36.22) we have $(A_ i)_{\mathfrak r_ i} = (A_ i)_{\mathfrak p}$. It follows from Nakayama's Lemma 10.20.1 that the map of local rings $R_{\mathfrak p} \to (A_ i)_{\mathfrak p} = (A_ i)_{\mathfrak r_ i}$ is surjective. Since $A_ i$ is finite over $R$ we see that there exists a $f \in R$, $f \not\in \mathfrak p$ such that $R_ f \to (A_ i)_ f$ is surjective. After replacing $R$ by $R_ f$ we win. $\square$

Comments (1)

Comment #1118 by Simon Pepin Lehalleur on

Suggested slogan: In an unramified ring map, one can separate the points in a fiber by passing to an étale neighbourhood.

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