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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

  1. an étale ring map $R \to R'$,
  2. a prime $\mathfrak p' \subset R'$ lying over $\mathfrak p$.
  3. a product decomposition $$ R' \otimes_R S = A \times B $$

with the following properties

  1. $R' \to A$ is surjective, and
  2. $\mathfrak p'A$ 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. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 39739–39759 (see updates for more information).

    \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)

    Comment #3276 by Dario WeiƟmann on April 15, 2018 a 3:10 pm UTC

    Typo: by a etale ring map -> by an etale ring map.

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