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Tag 02ML

Chapter 10: Commutative Algebra > Section 10.121: Quasi-finite maps

Lemma 10.121.10. Let $R \to S$ be a ring map of finite type. Let $\mathfrak p \subset R$ be a minimal prime. Assume that there are at most finitely many primes of $S$ lying over $\mathfrak p$. Then there exists a $g \in R$, $g \not \in \mathfrak p$ such that the ring map $R_g \to S_g$ is finite.

Proof. Let $x_1, \ldots, x_n$ be generators of $S$ over $R$. Since $\mathfrak p$ is a minimal prime we have that $\mathfrak pR_{\mathfrak p}$ is a locally nilpotent ideal, see Lemma 10.24.1. Hence $\mathfrak pS_{\mathfrak p}$ is a locally nilpotent ideal, see Lemma 10.31.2. By assumption the finite type $\kappa(\mathfrak p)$-algebra $S_{\mathfrak p}/\mathfrak pS_{\mathfrak p}$ has finitely many primes. Hence (for example by Lemmas 10.60.3 and 10.114.4) $\kappa(\mathfrak p) \to S_{\mathfrak p}/\mathfrak pS_{\mathfrak p}$ is a finite ring map. Thus we may find monic polynomials $P_i \in R_{\mathfrak p}[X]$ such that $P_i(x_i)$ maps to zero in $S_{\mathfrak p}/\mathfrak pS_{\mathfrak p}$. By what we said above there exist $e_i \geq 1$ such that $P(x_i)^{e_i} = 0$ in $S_{\mathfrak p}$. Let $g_1 \in R$, $g_1 \not \in \mathfrak p$ be an element such that $P_i$ has coefficients in $R[1/g_1]$ for all $i$. Next, let $g_2 \in R$, $g_2 \not \in \mathfrak p$ be an element such that $P(x_i)^{e_i} = 0$ in $S_{g_1g_2}$. Setting $g = g_1g_2$ we win. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 28996–29004 (see updates for more information).

    \begin{lemma}
    \label{lemma-generically-finite}
    Let $R \to S$ be a ring map of finite type.
    Let $\mathfrak p \subset R$ be a minimal prime.
    Assume that there are at most finitely many primes of $S$
    lying over $\mathfrak p$. Then there exists a
    $g \in R$, $g \not \in \mathfrak p$ such that the
    ring map $R_g \to S_g$ is finite.
    \end{lemma}
    
    \begin{proof}
    Let $x_1, \ldots, x_n$ be generators of $S$ over $R$.
    Since $\mathfrak p$ is a minimal prime we have that
    $\mathfrak pR_{\mathfrak p}$ is a locally nilpotent ideal, see
    Lemma \ref{lemma-minimal-prime-reduced-ring}.
    Hence $\mathfrak pS_{\mathfrak p}$ is a locally nilpotent ideal, see
    Lemma \ref{lemma-locally-nilpotent}.
    By assumption the finite type $\kappa(\mathfrak p)$-algebra
    $S_{\mathfrak p}/\mathfrak pS_{\mathfrak p}$ has finitely many
    primes. Hence (for example by
    Lemmas \ref{lemma-finite-type-algebra-finite-nr-primes} and
    \ref{lemma-Noether-normalization})
    $\kappa(\mathfrak p) \to S_{\mathfrak p}/\mathfrak pS_{\mathfrak p}$
    is a finite ring map. Thus we may find monic polynomials
    $P_i \in R_{\mathfrak p}[X]$ such that $P_i(x_i)$ maps to zero
    in $S_{\mathfrak p}/\mathfrak pS_{\mathfrak p}$. By what we said
    above there exist $e_i \geq 1$ such that $P(x_i)^{e_i} = 0$
    in $S_{\mathfrak p}$. Let $g_1 \in R$, $g_1 \not \in \mathfrak p$
    be an element such that $P_i$ has coefficients in $R[1/g_1]$ for all $i$.
    Next, let $g_2 \in R$, $g_2 \not \in \mathfrak p$ be an element
    such that $P(x_i)^{e_i} = 0$ in $S_{g_1g_2}$. Setting $g = g_1g_2$
    we win.
    \end{proof}

    Comments (4)

    Comment #954 by JuanPablo on August 23, 2014 a 7:15 pm UTC

    Hi. In this proof noether normalization was used to see that $\kappa(\mathfrak p) \to S_{\mathfrak p}/\mathfrak pS_{\mathfrak p}$ is a finite ring map. From noether normalization I see that the map is finite if and only if $\dim(S_{\mathfrak p}/\mathfrak pS_{\mathfrak p})=0$ but a priori it is not so clear that having finitely many primes implies dimension $0$ (in this case), as there could be proper inclusion among the finite primes.

    So maybe there should be a reference to tag 00GQ.

    Comment #962 by Johan (site) on August 28, 2014 a 5:08 pm UTC

    Yes, you are right. I added a couple of fun lemmas on the dimension of rings with finite nrs of primes and then I referred to one of those. See here. Thanks!

    Comment #2451 by John Smith on March 11, 2017 a 11:13 am UTC

    Typo in the second last line? I assume $P_i$ should be contained in the polynomial(!)-ring over $R[1/g_1]$?

    Comment #2493 by Johan (site) on April 13, 2017 a 11:10 pm UTC

    OK, yes, thanks! Fixed here.

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