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 10.122.11. Let $R$ be a ring. Let $S = R[x]/I$. Let $\mathfrak q \subset S$ be a prime. Assume $R \to S$ is quasi-finite at $\mathfrak q$. Let $S' \subset S$ be the integral closure of $R$ in $S$. Then there exists an element $g \in S'$, $g \not\in \mathfrak q$ such that $S'_ g \cong S_ g$.

Proof. Let $\mathfrak p$ be the image of $\mathfrak q$ in $\mathop{\mathrm{Spec}}(R)$. There exists an $f \in I$, $f = a_ nx^ n + \ldots + a_0$ such that $a_ i \not\in \mathfrak p$ for some $i$. Namely, otherwise the fibre ring $S \otimes _ R \kappa (\mathfrak p)$ would be $\kappa (\mathfrak p)[x]$ and the map would not be quasi-finite at any prime lying over $\mathfrak p$. We conclude there exists a relation $b_ m x^ m + \ldots + b_0 = 0$ with $b_ j \in S'$, $j = 0, \ldots , m$ and $b_ j \not\in \mathfrak q \cap S'$ for some $j$. We prove the lemma by induction on $m$. The base case is $m = 0$ is vacuous (because the statements $b_0 = 0$ and $b_0 \not\in \mathfrak q$ are contradictory).

The case $b_ m \not\in \mathfrak q$. In this case $x$ is integral over $S'_{b_ m}$, in fact $b_ mx \in S'$ by Lemma 10.122.1. Hence the injective map $S'_{b_ m} \to S_{b_ m}$ is also surjective, i.e., an isomorphism as desired.

The case $b_ m \in \mathfrak q$. In this case we have $b_ mx \in S'$ by Lemma 10.122.1. Set $b'_{m - 1} = b_ mx + b_{m - 1}$. Then

\[ b'_{m - 1}x^{m - 1} + b_{m - 2}x^{m - 2} + \ldots + b_0 = 0 \]

Since $b'_{m - 1}$ is congruent to $b_{m - 1}$ modulo $S' \cap \mathfrak q$ we see that it is still the case that one of $b'_{m - 1}, b_{m - 2}, \ldots , b_0$ is not in $S' \cap \mathfrak q$. Thus we win by induction on $m$. $\square$

Comments (2)

Comment #3632 by Brian Conrad on

The treatment of the "base" of the induction in this argument should be handled more clearly: in the case necessarily because otherwise we have which lies in , so the case is vacuous and hence tautologically true. Of course, from the expository viewpoint that is bad writing and instead what is really being shown after noting that necessarily is that we can eventually put ourselves into the case with a possibly smaller . So the final case in the proof should be done first for clearer exposition.

Comment #3731 by on

Well, I tried to improve it following your suggestions, but I think it is essentially the same as before. Here are the changes.

There are also:

  • 3 comment(s) on Section 10.122: Zariski's Main Theorem

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