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\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*}

[Theorem 23.1.3, EGA]

Lemma 10.155.16 (Tate). Let $R$ be a ring. Let $x \in R$. Assume

  1. $R$ is a normal Noetherian domain,

  2. $R/xR$ is a domain and N-2,

  3. $R \cong \mathop{\mathrm{lim}}\nolimits _ n R/x^ nR$ is complete with respect to $x$.

Then $R$ is N-2.

Proof. We may assume $x \not= 0$ since otherwise the lemma is trivial. Let $K$ be the fraction field of $R$. If the characteristic of $K$ is zero the lemma follows from (1), see Lemma 10.155.11. Hence we may assume that the characteristic of $K$ is $p > 0$, and we may apply Lemma 10.155.12. Thus given $K \subset L$ be a finite purely inseparable field extension we have to show that the integral closure $S$ of $R$ in $L$ is finite over $R$.

Let $q$ be a power of $p$ such that $L^ q \subset K$. By enlarging $L$ if necessary we may assume there exists an element $y \in L$ such that $y^ q = x$. Since $R \to S$ induces a homeomorphism of spectra (see Lemma 10.45.7) there is a unique prime ideal $\mathfrak q \subset S$ lying over the prime ideal $\mathfrak p = xR$. It is clear that

\[ \mathfrak q = \{ f \in S \mid f^ q \in \mathfrak p\} = yS \]

since $y^ q = x$. Hence $R_{\mathfrak p}$ and $S_{\mathfrak q}$ are discrete valuation rings, see Lemma 10.118.7. By Lemma 10.118.10 we see that $\kappa (\mathfrak p) \subset \kappa (\mathfrak q)$ is a finite field extension. Hence the integral closure $S' \subset \kappa (\mathfrak q)$ of $R/xR$ is finite over $R/xR$ by assumption (2). Since $S/yS \subset S'$ this implies that $S/yS$ is finite over $R$. Note that $S/y^ nS$ has a finite filtration whose subquotients are the modules $y^ iS/y^{i + 1}S \cong S/yS$. Hence we see that each $S/y^ nS$ is finite over $R$. In particular $S/xS$ is finite over $R$. Also, it is clear that $\bigcap x^ nS = (0)$ since an element in the intersection has $q$th power contained in $\bigcap x^ nR = (0)$ (Lemma 10.50.4). Thus we may apply Lemma 10.95.12 to conclude that $S$ is finite over $R$, and we win. $\square$


Comments (2)

Comment #2749 by Ariyan Javanpeykar on

A reference: EGA IV_0, Theorem 23.1.3

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  • 3 comment(s) on Section 10.155: Japanese rings

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