Tag 00UX
Chapter 10: Commutative Algebra > Section 10.147: Unramified ring maps
Lemma 10.147.10. 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 a 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 10.141.23. 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 10.147.6). Say $\mathfrak q = \mathfrak r_1$. By Lemma 10.147.5 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 10.40.11 (which applies to $R \to A_1$ because a finite ring map satisfies going up by Lemma 10.35.22) we have $(A_1)_{\mathfrak r_1} = (A_1)_{\mathfrak p}$. It follows from Nakayama's Lemma 10.19.1 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. $\square$
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\begin{lemma}
\label{lemma-etale-makes-unramified-closed-at-prime}
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
\begin{enumerate}
\item an \'etale ring map $R \to R'$,
\item a prime $\mathfrak p' \subset R'$ lying over $\mathfrak p$.
\item a product decomposition
$$
R' \otimes_R S = A \times B
$$
\end{enumerate}
with the following properties
\begin{enumerate}
\item $R' \to A$ is surjective, and
\item $\mathfrak p'A$ is a prime of $A$ lying over $\mathfrak p'$ and
over $\mathfrak q$.
\end{enumerate}
\end{lemma}
\begin{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 a \'etale 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.
\medskip\noindent
The assumption that $R \to S$ is of finite type means that we may apply
Lemma \ref{lemma-etale-makes-quasi-finite-finite-variant}. 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 \ref{lemma-unramified-quasi-finite}).
Say $\mathfrak q = \mathfrak r_1$.
By Lemma \ref{lemma-unramified-at-prime} 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 \ref{lemma-unique-prime-over-localize-below}
(which applies to $R \to A_1$ because a finite ring map satisfies going up by
Lemma \ref{lemma-integral-going-up})
we have $(A_1)_{\mathfrak r_1} = (A_1)_{\mathfrak p}$.
It follows from Nakayama's Lemma \ref{lemma-NAK}
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.
\end{proof}Comments (1)
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