The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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 10.31.3. 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. $\square$


Comments (4)

Comment #954 by JuanPablo on

Hi. In this proof noether normalization was used to see that is a finite ring map. From noether normalization I see that the map is finite if and only if but a priori it is not so clear that having finitely many primes implies dimension (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 on

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

Typo in the second last line? I assume should be contained in the polynomial(!)-ring over ?


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