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Tag 032P

Chapter 10: Commutative Algebra > Section 10.155: Japanese rings

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. $R \cong \mathop{\rm lim}\nolimits_n R/x^nR$ 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 10.45.6) 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. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 42892–42903 (see updates for more information).

    \begin{lemma}[Tate]
    \label{lemma-tate-japanese}
    Let $R$ be a ring.
    Let $x \in R$.
    Assume
    \begin{enumerate}
    \item $R$ is a normal Noetherian domain,
    \item $R/xR$ is a domain and N-2,
    \item $R \cong \lim_n R/x^nR$ is complete with respect to $x$.
    \end{enumerate}
    Then $R$ is N-2.
    \end{lemma}
    
    \begin{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 \ref{lemma-domain-char-zero-N-1-2}. Hence we may assume
    that the characteristic of $K$ is $p > 0$, and we may apply
    Lemma \ref{lemma-domain-char-p-N-1-2}. 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 $R$.
    
    \medskip\noindent
    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 \ref{lemma-p-ring-map})
    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 \ref{lemma-characterize-dvr}.
    By Lemma \ref{lemma-finite-extension-residue-fields-dimension-1} 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 \ref{lemma-intersect-powers-ideal-module-zero}).
    Thus we may apply Lemma \ref{lemma-finite-over-complete-ring} to conclude
    that $S$ is finite over $R$, and we win.
    \end{proof}

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