** 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

an étale ring map $R \to R'$,

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

a product decomposition

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

with the following properties

$R' \to A_ i$ is surjective,

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

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