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