Lemma 10.152.2. Let $R \to S$ be a ring map. Let $\mathfrak q$ be a prime of $S$ lying over $\mathfrak p \subset R$. Assume that $R \to S$ is of finite type and unramified at $\mathfrak q$. Then there exist

an étale ring map $R \to R'$,

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

a product decomposition

\[ R' \otimes _ R S = A \times B \]

with the following properties

$R' \to A$ is surjective, and

$\mathfrak p'A$ is a prime of $A$ lying over $\mathfrak p'$ and over $\mathfrak q$.

**Proof.**
We may replace $(R \to S, \mathfrak p, \mathfrak q)$ with any base change $(R' \to R'\otimes _ R S, \mathfrak p', \mathfrak q')$ by an étale ring map $R \to R'$ with a prime $\mathfrak p'$ lying over $\mathfrak p$, and a choice of $\mathfrak q'$ lying over both $\mathfrak q$ and $\mathfrak p'$. Note also that given $R \to R'$ and $\mathfrak p'$ a suitable $\mathfrak q'$ can always be found.

The assumption that $R \to S$ is of finite type means that we may apply Lemma 10.145.4. Thus 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 $\mathfrak p$. Then clearly $\mathfrak q = \mathfrak r_ i$ for some $i$, since an unramified morphism is quasi-finite (see Lemma 10.151.6). Say $\mathfrak q = \mathfrak r_1$. By Lemma 10.151.5 we see that $\kappa (\mathfrak p) \subset \kappa (\mathfrak r_1)$ is separable hence the trivial field extension, and that $\mathfrak p(A_1)_{\mathfrak r_1}$ is the maximal ideal. Also, by Lemma 10.41.11 (which applies to $R \to A_1$ because a finite ring map satisfies going up by Lemma 10.36.22) we have $(A_1)_{\mathfrak r_1} = (A_1)_{\mathfrak p}$. It follows from Nakayama's Lemma 10.20.1 that the map of local rings $R_{\mathfrak p} \to (A_1)_{\mathfrak p} = (A_1)_{\mathfrak r_1}$ is surjective. Since $A_1$ is finite over $R$ we see that there exists a $f \in R$, $f \not\in \mathfrak p$ such that $R_ f \to (A_1)_ f$ is surjective. After replacing $R$ by $R_ f$ we win.
$\square$

## Comments (2)

Comment #3260 by Dario Weißmann on

Comment #3356 by Johan on