The Stacks Project


Tag 0BI1

10.155. Japanese rings

In this section we being to discuss finiteness of integral closure.

Definition 10.155.1. Let $R$ be a domain with field of fractions $K$.

  1. We say $R$ is N-1 if the integral closure of $R$ in $K$ is a finite $R$-module.
  2. We say $R$ is N-2 or Japanese if for any finite extension $K \subset L$ of fields the integral closure of $R$ in $L$ is finite over $R$.

The main interest in these notions is for Noetherian rings, but here is a non-Noetherian example.

Example 10.155.2. Let $k$ be a field. The domain $R = k[x_1, x_2, x_3, \ldots]$ is N-2, but not Noetherian. The reason is the following. Suppose that $R \subset L$ and the field $L$ is a finite extension of the fraction field of $R$. Then there exists an integer $n$ such that $L$ comes from a finite extension $k(x_1, \ldots, x_n) \subset L_0$ by adjoining the (transcendental) elements $x_{n + 1}, x_{n + 2}$, etc. Let $S_0$ be the integral closure of $k[x_1, \ldots, x_n]$ in $L_0$. By Proposition 10.156.16 below it is true that $S_0$ is finite over $k[x_1, \ldots, x_n]$. Moreover, the integral closure of $R$ in $L$ is $S = S_0[x_{n + 1}, x_{n + 2}, \ldots]$ (use Lemma 10.36.8) and hence finite over $R$. The same argument works for $R = \mathbf{Z}[x_1, x_2, x_3, \ldots]$.

Lemma 10.155.3. Let $R$ be a domain. If $R$ is N-1 then so is any localization of $R$. Same for N-2.

Proof. These statements hold because taking integral closure commutes with localization, see Lemma 10.35.11. $\square$

Lemma 10.155.4. Let $R$ be a domain. Let $f_1, \ldots, f_n \in R$ generate the unit ideal. If each domain $R_{f_i}$ is N-1 then so is $R$. Same for N-2.

Proof. Assume $R_{f_i}$ is N-2 (or N-1). Let $L$ be a finite extension of the fraction field of $R$ (equal to the fraction field in the N-1 case). Let $S$ be the integral closure of $R$ in $L$. By Lemma 10.35.11 we see that $S_{f_i}$ is the integral closure of $R_{f_i}$ in $L$. Hence $S_{f_i}$ is finite over $R_{f_i}$ by assumption. Thus $S$ is finite over $R$ by Lemma 10.23.2. $\square$

Lemma 10.155.5. Let $R$ be a domain. Let $R \subset S$ be a quasi-finite extension of domains (for example finite). Assume $R$ is N-2 and Noetherian. Then $S$ is N-2.

Proof. Let $K = f.f.(R) \subset L = f.f.(S)$. Note that this is a finite field extension (for example by Lemma 10.121.2 (2) applied to the fibre $S \otimes_R K$, and the definition of a quasi-finite ring map). Let $S'$ be the integral closure of $R$ in $S$. Then $S'$ is contained in the integral closure of $R$ in $L$ which is finite over $R$ by assumption. As $R$ is Noetherian this implies $S'$ is finite over $R$. By Lemma 10.122.15 there exist elements $g_1, \ldots, g_n \in S'$ such that $S'_{g_i} \cong S_{g_i}$ and such that $g_1, \ldots, g_n$ generate the unit ideal in $S$. Hence it suffices to show that $S'$ is N-2 by Lemmas 10.155.3 and 10.155.4. Thus we have reduced to the case where $S$ is finite over $R$.

Assume $R \subset S$ with hypotheses as in the lemma and moreover that $S$ is finite over $R$. Let $M$ be a finite field extension of the fraction field of $S$. Then $M$ is also a finite field extension of $f.f(R)$ and we conclude that the integral closure $T$ of $R$ in $M$ is finite over $R$. By Lemma 10.35.16 we see that $T$ is also the integral closure of $S$ in $M$ and we win by Lemma 10.35.15. $\square$

Lemma 10.155.6. Let $R$ be a Noetherian domain. If $R[z, z^{-1}]$ is N-1, then so is $R$.

Proof. Let $R'$ be the integral closure of $R$ in its field of fractions $K$. Let $S'$ be the integral closure of $R[z, z^{-1}]$ in its field of fractions. Clearly $R' \subset S'$. Since $K[z, z^{-1}]$ is a normal domain we see that $S' \subset K[z, z^{-1}]$. Suppose that $f_1, \ldots, f_n \in S'$ generate $S'$ as $R[z, z^{-1}]$-module. Say $f_i = \sum a_{ij}z^j$ (finite sum), with $a_{ij} \in K$. For any $x \in R'$ we can write $$ x = \sum h_i f_i $$ with $h_i \in R[z, z^{-1}]$. Thus we see that $R'$ is contained in the finite $R$-submodule $\sum Ra_{ij} \subset K$. Since $R$ is Noetherian we conclude that $R'$ is a finite $R$-module. $\square$

Lemma 10.155.7. Let $R$ be a Noetherian domain, and let $R \subset S$ be a finite extension of domains. If $S$ is N-1, then so is $R$. If $S$ is N-2, then so is $R$.

Proof. Omitted. (Hint: Integral closures of $R$ in extension fields are contained in integral closures of $S$ in extension fields.) $\square$

Lemma 10.155.8. Let $R$ be a Noetherian normal domain with fraction field $K$. Let $K \subset L$ be a finite separable field extension. Then the integral closure of $R$ in $L$ is finite over $R$.

Proof. Consider the trace pairing (Fields, Definition 9.20.6) $$ L \times L \longrightarrow K, \quad (x, y) \longmapsto \langle x, y\rangle := \text{Trace}_{L/K}(xy). $$ Since $L/K$ is separable this is nondegenerate (Fields, Lemma 9.20.7). Moreover, if $x \in L$ is integral over $R$, then $\text{Trace}_{L/K}(x)$ is in $R$. This is true because the minimal polynomial of $x$ over $K$ has coefficients in $R$ (Lemma 10.37.6) and because $\text{Trace}_{L/K}(x)$ is an integer multiple of one of these coefficients (Fields, Lemma 9.20.3). Pick $x_1, \ldots, x_n \in L$ which are integral over $R$ and which form a $K$-basis of $L$. Then the integral closure $S \subset L$ is contained in the $R$-module $$ M = \{y \in L \mid \langle x_i, y\rangle \in R, ~i = 1, \ldots, n\} $$ By linear algebra we see that $M \cong R^{\oplus n}$ as an $R$-module. Hence $S \subset R^{\oplus n}$ is a finitely generated $R$-module as $R$ is Noetherian. $\square$

Example 10.155.9. Lemma 10.155.8 does not work if the ring is not Noetherian. For example consider the action of $G = \{+1, -1\}$ on $A = \mathbf{C}[x_1, x_2, x_3, \ldots]$ where $-1$ acts by mapping $x_i$ to $-x_i$. The invariant ring $R = A^G$ is the $\mathbf{C}$-algebra generated by all $x_ix_j$. Hence $R \subset A$ is not finite. But $R$ is a normal domain with fraction field $K = L^G$ the $G$-invariants in the fraction field $L$ of $A$. And clearly $A$ is the integral closure of $R$ in $L$.

The following lemma can sometimes be used as a substitute for Lemma 10.155.8 in case of purely inseparable extensions.

Lemma 10.155.10. Let $R$ be a Noetherian normal domain with fraction field $K$ of characteristic $p > 0$. Let $a \in K$ be an element such that there exists a derivation $D : R \to R$ with $D(a) \not = 0$. Then the integral closure of $R$ in $L = K[x]/(x^p - a)$ is finite over $R$.

Proof. After replacing $x$ by $fx$ and $a$ by $f^pa$ for some $f \in R$ we may assume $a \in R$. Hence also $D(a) \in R$. We will show by induction on $i \leq p - 1$ that if $$ y = a_0 + a_1x + \ldots + a_i x^i,\quad a_j \in K $$ is integral over $R$, then $D(a)^i a_j \in R$. Thus the integral closure is contained in the finite $R$-module with basis $D(a)^{-p + 1}x^j$, $j = 0, \ldots, p - 1$. Since $R$ is Noetherian this proves the lemma.

If $i = 0$, then $y = a_0$ is integral over $R$ if and only if $a_0 \in R$ and the statement is true. Suppose the statement holds for some $i < p - 1$ and suppose that $$ y = a_0 + a_1x + \ldots + a_{i + 1} x^{i + 1},\quad a_j \in K $$ is integral over $R$. Then $$ y^p = a_0^p + a_1^p a + \ldots + a_{i + 1}^pa^{i + 1} $$ is an element of $R$ (as it is in $K$ and integral over $R$). Applying $D$ we obtain $$ (a_1^p + 2a_2^p a + \ldots + (i + 1)a_{i + 1}^p a^i)D(a) $$ is in $R$. Hence it follows that $$ D(a)a_1 + 2D(a) a_2 x + \ldots + (i + 1)D(a) a_{i + 1} x^i $$ is integral over $R$. By induction we find $D(a)^{i + 1}a_j \in R$ for $j = 1, \ldots, i + 1$. (Here we use that $1, \ldots, i + 1$ are invertible.) Hence $D(a)^{i + 1}a_0$ is also in $R$ because it is the difference of $y$ and $\sum_{j > 0} D(a)^{i + 1}a_jx^j$ which are integral over $R$ (since $x$ is integral over $R$ as $a \in R$). $\square$

Lemma 10.155.11. A Noetherian domain of characteristic zero is N-1 if and only if it is N-2 (i.e., Japanese).

Proof. This is clear from Lemma 10.155.8 since every field extension in characteristic zero is separable. $\square$

Lemma 10.155.12. Let $R$ be a Noetherian domain with fraction field $K$ of characteristic $p > 0$. Then $R$ is N-2 if and only if for every finite purely inseparable extension $K \subset L$ the integral closure of $R$ in $L$ is finite over $R$.

Proof. Assume the integral closure of $R$ in every finite purely inseparable field extension of $K$ is finite. Let $K \subset L$ be any finite extension. We have to show the integral closure of $R$ in $L$ is finite over $R$. Choose a finite normal field extension $K \subset M$ containing $L$. As $R$ is Noetherian it suffices to show that the integral closure of $R$ in $M$ is finite over $R$. By Fields, Lemma 9.27.3 there exists a subextension $K \subset M_{insep} \subset M$ such that $M_{insep}/K$ is purely inseparable, and $M/M_{insep}$ is separable. By assumption the integral closure $R'$ of $R$ in $M_{insep}$ is finite over $R$. By Lemma 10.155.8 the integral closure $R''$ of $R'$ in $M$ is finite over $R'$. Then $R''$ is finite over $R$ by Lemma 10.7.3. Since $R''$ is also the integral closure of $R$ in $M$ (see Lemma 10.35.16) we win. $\square$

Lemma 10.155.13. Let $R$ be a Noetherian domain. If $R$ is N-1 then $R[x]$ is N-1. If $R$ is N-2 then $R[x]$ is N-2.

Proof. Assume $R$ is N-1. Let $R'$ be the integral closure of $R$ which is finite over $R$. Hence also $R'[x]$ is finite over $R[x]$. The ring $R'[x]$ is normal (see Lemma 10.36.8), hence N-1. This proves the first assertion.

For the second assertion, by Lemma 10.155.7 it suffices to show that $R'[x]$ is N-2. In other words we may and do assume that $R$ is a normal N-2 domain. In characteristic zero we are done by Lemma 10.155.11. In characteristic $p > 0$ we have to show that the integral closure of $R[x]$ is finite in any finite purely inseparable extension of $f.f.(R[x]) = K(x) \subset L$ with $K = f.f.(R)$. Clearly there exists a finite purely inseparable field extension $K \subset L'$ and $q = p^e$ such that $L \subset L'(x^{1/q})$. As $R[x]$ is Noetherian it suffices to show that the integral closure of $R[x]$ in $L'(x^{1/q})$ is finite over $R[x]$. And this integral closure is equal to $R'[x^{1/q}]$ with $R \subset R' \subset L'$ the integral closure of $R$ in $L'$. Since $R$ is N-2 we see that $R'$ is finite over $R$ and hence $R'[x^{1/q}]$ is finite over $R[x]$. $\square$

Lemma 10.155.14. Let $R$ be a Noetherian domain. If there exists an $f \in R$ such that $R_f$ is normal then $$ U = \{\mathfrak p \in \mathop{\rm Spec}(R) \mid R_{\mathfrak p} \text{ is normal}\} $$ is open in $\mathop{\rm Spec}(R)$.

Proof. It is clear that the standard open $D(f)$ is contained in $U$. By Serre's criterion Lemma 10.151.4 we see that $\mathfrak p \not \in U$ implies that for some $\mathfrak q \subset \mathfrak p$ we have either

  1. Case I: $\text{depth}(R_{\mathfrak q}) < 2$ and $\dim(R_{\mathfrak q}) \geq 2$, and
  2. Case II: $R_{\mathfrak q}$ is not regular and $\dim(R_{\mathfrak q}) = 1$.

This in particular also means that $R_{\mathfrak q}$ is not normal, and hence $f \in \mathfrak q$. In case I we see that $\text{depth}(R_{\mathfrak q}) = \text{depth}(R_{\mathfrak q}/fR_{\mathfrak q}) + 1$. Hence such a prime $\mathfrak q$ is the same thing as an embedded associated prime of $R/fR$. In case II $\mathfrak q$ is an associated prime of $R/fR$ of height 1. Thus there is a finite set $E$ of such primes $\mathfrak q$ (see Lemma 10.62.5) and $$ \mathop{\rm Spec}(R) \setminus U = \bigcup\nolimits_{\mathfrak q \in E} V(\mathfrak q) $$ as desired. $\square$

Lemma 10.155.15. Let $R$ be a Noetherian domain. Assume

  1. there exists a nonzero $f \in R$ such that $R_f$ is normal, and
  2. for every maximal ideal $\mathfrak m \subset R$ the local ring $R_{\mathfrak m}$ is N-1.

Then $R$ is N-1.

Proof. Set $K = f.f.(R)$. Suppose that $R \subset R' \subset K$ is a finite extension of $R$ contained in $K$. Note that $R_f = R'_f$ since $R_f$ is already normal. Hence by Lemma 10.155.14 the set of primes $\mathfrak p' \in \mathop{\rm Spec}(R')$ with $R'_{\mathfrak p'}$ non-normal is closed in $\mathop{\rm Spec}(R')$. Since $\mathop{\rm Spec}(R') \to \mathop{\rm Spec}(R)$ is closed the image of this set is closed in $\mathop{\rm Spec}(R)$. For such a ring $R'$ denote $Z_{R'} \subset \mathop{\rm Spec}(R)$ this image.

Pick a maximal ideal $\mathfrak m \subset R$. Let $R_{\mathfrak m} \subset R_{\mathfrak m}'$ be the integral closure of the local ring in $K$. By assumption this is a finite ring extension. By Lemma 10.35.11 we can find finitely many elements $r_1, \ldots, r_n \in K$ integral over $R$ such that $R_{\mathfrak m}'$ is generated by $r_1, \ldots, r_n$ over $R_{\mathfrak m}$. Let $R' = R[x_1, \ldots, x_n] \subset K$. With this choice it is clear that $\mathfrak m \not \in Z_{R'}$.

As $\mathop{\rm Spec}(R)$ is quasi-compact, the above shows that we can find a finite collection $R \subset R'_i \subset K$ such that $\bigcap Z_{R'_i} = \emptyset$. Let $R'$ be the subring of $K$ generated by all of these. It is finite over $R$. Also $Z_{R'} = \emptyset$. Namely, every prime $\mathfrak p'$ lies over a prime $\mathfrak p'_i$ such that $(R'_i)_{\mathfrak p'_i}$ is normal. This implies that $R'_{\mathfrak p'} = (R'_i)_{\mathfrak p'_i}$ is normal too. Hence $R'$ is normal, in other words $R'$ is the integral closure of $R$ in $K$. $\square$

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. $\square$

Lemma 10.155.17. Let $R$ be a ring. If $R$ is Noetherian, a domain, and N-2, then so is $R[[x]]$.

Proof. Observe that $R[[x]]$ is Noetherian by Lemma 10.30.2. Let $R' \supset R$ be the integral closure of $R$ in its fraction field. Because $R$ is N-2 this is finite over $R$. Hence $R'[[x]]$ is finite over $R[[x]]$. By Lemma 10.36.9 we see that $R'[[x]]$ is a normal domain. Apply Lemma 10.155.16 to the element $x \in R'[[x]]$ to see that $R'[[x]]$ is N-2. Then Lemma 10.155.7 shows that $R[[x]]$ is N-2. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 42450–42921 (see updates for more information).

    \section{Japanese rings}
    \label{section-japanese}
    
    \noindent
    In this section we being to discuss finiteness of integral closure.
    
    \begin{definition}
    \label{definition-N}
    Let $R$ be a domain with field of fractions $K$.
    \begin{enumerate}
    \item We say $R$ is {\it N-1} if the integral closure of $R$ in $K$
    is a finite $R$-module.
    \item We say $R$ is {\it N-2} or {\it Japanese} if for any finite
    extension $K \subset L$ of fields the integral closure of $R$ in $L$
    is finite over $R$.
    \end{enumerate}
    \end{definition}
    
    \noindent
    The main interest in these notions is for Noetherian rings,
    but here is a non-Noetherian example.
    
    \begin{example}
    \label{example-Japanese-not-Noetherian}
    Let $k$ be a field. The domain $R = k[x_1, x_2, x_3, \ldots]$ is N-2,
    but not Noetherian. The reason is the following. Suppose that $R \subset L$
    and the field $L$ is a finite extension of the fraction field of $R$.
    Then there exists an integer $n$ such that $L$ comes from a finite
    extension $k(x_1, \ldots, x_n) \subset L_0$ by adjoining
    the (transcendental) elements $x_{n + 1}, x_{n + 2}$, etc.
    Let $S_0$ be the integral
    closure of $k[x_1, \ldots, x_n]$ in $L_0$. By
    Proposition \ref{proposition-ubiquity-nagata} below
    it is true that $S_0$ is finite over $k[x_1, \ldots, x_n]$.
    Moreover, the integral closure of $R$ in $L$ is
    $S = S_0[x_{n + 1}, x_{n + 2}, \ldots]$ (use
    Lemma \ref{lemma-polynomial-domain-normal}) and
    hence finite over $R$. The same argument works for
    $R = \mathbf{Z}[x_1, x_2, x_3, \ldots]$.
    \end{example}
    
    \begin{lemma}
    \label{lemma-localize-N}
    Let $R$ be a domain.
    If $R$ is N-1 then so is any localization of $R$.
    Same for N-2.
    \end{lemma}
    
    \begin{proof}
    These statements hold because taking integral closure commutes
    with localization, see Lemma \ref{lemma-integral-closure-localize}.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-Japanese-local}
    Let $R$ be a domain. Let $f_1, \ldots, f_n \in R$ generate the
    unit ideal. If each domain $R_{f_i}$ is N-1 then so is $R$.
    Same for N-2.
    \end{lemma}
    
    \begin{proof}
    Assume $R_{f_i}$ is N-2 (or N-1).
    Let $L$ be a finite extension of the fraction field of $R$ (equal to
    the fraction field in the N-1 case). Let $S$ be the integral
    closure of $R$ in $L$. By Lemma \ref{lemma-integral-closure-localize}
    we see that $S_{f_i}$ is the integral closure of $R_{f_i}$ in $L$.
    Hence $S_{f_i}$ is finite over $R_{f_i}$ by assumption.
    Thus $S$ is finite over $R$ by Lemma \ref{lemma-cover}.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-quasi-finite-over-Noetherian-japanese}
    Let $R$ be a domain.
    Let $R \subset S$ be a quasi-finite extension of domains
    (for example finite).
    Assume $R$ is N-2 and Noetherian.
    Then $S$ is N-2.
    \end{lemma}
    
    \begin{proof}
    Let $K = f.f.(R) \subset L = f.f.(S)$. Note that this is
    a finite field extension (for example by
    Lemma \ref{lemma-isolated-point-fibre} (2)
    applied to the fibre $S \otimes_R K$, and the definition of a
    quasi-finite ring map).
    Let $S'$ be the integral closure of $R$ in $S$.
    Then $S'$ is contained in the integral closure of $R$ in $L$
    which is finite over $R$ by assumption. As $R$ is Noetherian this
    implies $S'$ is finite over $R$.
    By Lemma \ref{lemma-quasi-finite-open-integral-closure}
    there exist elements $g_1, \ldots, g_n \in S'$
    such that $S'_{g_i} \cong S_{g_i}$ and such that $g_1, \ldots, g_n$
    generate the unit ideal in $S$. Hence it suffices to show that
    $S'$ is N-2 by Lemmas \ref{lemma-localize-N} and \ref{lemma-Japanese-local}.
    Thus we have reduced to the case where $S$ is finite over $R$.
    
    \medskip\noindent
    Assume $R \subset S$ with hypotheses as in the lemma and moreover
    that $S$ is finite over $R$. Let $M$ be a finite field extension
    of the fraction field of $S$. Then $M$ is also a finite field extension
    of $f.f(R)$ and we conclude that the integral closure $T$ of $R$ in
    $M$ is finite over $R$. By Lemma \ref{lemma-integral-closure-transitive}
    we see that $T$ is also the integral closure of $S$ in $M$ and we win by
    Lemma \ref{lemma-integral-permanence}.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-Laurent-ring-N-1}
    Let $R$ be a Noetherian domain.
    If $R[z, z^{-1}]$ is N-1, then so is $R$.
    \end{lemma}
    
    \begin{proof}
    Let $R'$ be the integral closure of $R$ in its field of fractions $K$.
    Let $S'$ be the integral closure of $R[z, z^{-1}]$ in its field of fractions.
    Clearly $R' \subset S'$.
    Since $K[z, z^{-1}]$ is a normal domain we see that $S' \subset K[z, z^{-1}]$.
    Suppose that $f_1, \ldots, f_n \in S'$ generate $S'$ as $R[z, z^{-1}]$-module.
    Say $f_i = \sum a_{ij}z^j$ (finite sum), with $a_{ij} \in K$.
    For any $x \in R'$ we can write
    $$
    x = \sum h_i f_i
    $$
    with $h_i \in R[z, z^{-1}]$. Thus we see that $R'$ is contained in the
    finite $R$-submodule $\sum Ra_{ij} \subset K$. Since $R$ is Noetherian
    we conclude that $R'$ is a finite $R$-module.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-finite-extension-N-2}
    Let $R$ be a Noetherian domain, and let $R \subset S$ be a
    finite extension of domains. If $S$ is N-1, then so is $R$.
    If $S$ is N-2, then so is $R$.
    \end{lemma}
    
    \begin{proof}
    Omitted. (Hint: Integral closures of $R$ in extension fields
    are contained in integral closures of $S$ in extension fields.)
    \end{proof}
    
    \begin{lemma}
    \label{lemma-Noetherian-normal-domain-finite-separable-extension}
    Let $R$ be a Noetherian normal domain with fraction field $K$.
    Let $K \subset L$ be a finite separable field extension.
    Then the integral closure of $R$ in $L$ is finite over $R$.
    \end{lemma}
    
    \begin{proof}
    Consider the trace pairing
    (Fields, Definition \ref{fields-definition-trace-pairing})
    $$
    L \times L \longrightarrow K,
    \quad (x, y) \longmapsto \langle x, y\rangle := \text{Trace}_{L/K}(xy).
    $$
    Since $L/K$ is separable this is nondegenerate
    (Fields, Lemma \ref{fields-lemma-separable-trace-pairing}).
    Moreover, if $x \in L$ is integral over $R$, then
    $\text{Trace}_{L/K}(x)$ is in $R$. This is true because the
    minimal polynomial of $x$ over $K$ has coefficients in $R$
    (Lemma \ref{lemma-minimal-polynomial-normal-domain})
    and because $\text{Trace}_{L/K}(x)$ is an
    integer multiple of one of these coefficients
    (Fields, Lemma \ref{fields-lemma-trace-and-norm-from-minimal-polynomial}).
    Pick $x_1, \ldots, x_n \in L$ which are integral over $R$
    and which form a $K$-basis of $L$. Then the integral closure
    $S \subset L$ is contained in the $R$-module
    $$
    M = \{y \in L \mid \langle x_i, y\rangle \in R, \ i = 1, \ldots, n\}
    $$
    By linear algebra we see that $M \cong R^{\oplus n}$ as an $R$-module.
    Hence $S \subset R^{\oplus n}$ is a finitely generated $R$-module
    as $R$ is Noetherian.
    \end{proof}
    
    \begin{example}
    \label{example-bad-invariants}
    Lemma \ref{lemma-Noetherian-normal-domain-finite-separable-extension}
    does not work if the ring is not Noetherian.
    For example consider the action of $G = \{+1, -1\}$ on
    $A = \mathbf{C}[x_1, x_2, x_3, \ldots]$ where $-1$ acts by
    mapping $x_i$ to $-x_i$. The invariant ring $R = A^G$ is
    the $\mathbf{C}$-algebra generated by all $x_ix_j$. Hence
    $R \subset A$ is not finite. But $R$ is a normal domain
    with fraction field $K = L^G$ the $G$-invariants in the fraction field
    $L$ of $A$. And clearly $A$ is the integral closure of $R$ in
    $L$.
    \end{example}
    
    \noindent
    The following lemma can sometimes be used as a substitute for
    Lemma \ref{lemma-Noetherian-normal-domain-finite-separable-extension}
    in case of purely inseparable extensions.
    
    \begin{lemma}
    \label{lemma-Noetherian-normal-domain-insep-extension}
    Let $R$ be a Noetherian normal domain with fraction field $K$
    of characteristic $p > 0$.
    Let $a \in K$ be an element such that there exists a derivation
    $D : R \to R$ with $D(a) \not = 0$. Then the integral closure
    of $R$ in $L = K[x]/(x^p - a)$ is finite over $R$.
    \end{lemma}
    
    \begin{proof}
    After replacing $x$ by $fx$ and $a$ by $f^pa$ for some $f \in R$
    we may assume $a \in R$. Hence also $D(a) \in R$. We will show
    by induction on $i \leq p - 1$ that if
    $$
    y = a_0 + a_1x + \ldots + a_i x^i,\quad a_j \in K
    $$
    is integral over $R$, then $D(a)^i a_j \in R$. Thus the integral
    closure is contained in the finite $R$-module with basis
    $D(a)^{-p + 1}x^j$, $j = 0, \ldots, p - 1$. Since $R$ is Noetherian
    this proves the lemma.
    
    \medskip\noindent
    If $i = 0$, then $y = a_0$ is integral over $R$ if and only if $a_0 \in R$
    and the statement is true. Suppose the statement holds for some $i < p - 1$
    and suppose that
    $$
    y = a_0 + a_1x + \ldots + a_{i + 1} x^{i + 1},\quad a_j \in K
    $$
    is integral over $R$. Then
    $$
    y^p = a_0^p + a_1^p a + \ldots + a_{i + 1}^pa^{i + 1}
    $$
    is an element of $R$ (as it is in $K$ and integral over $R$). Applying
    $D$ we obtain
    $$
    (a_1^p + 2a_2^p a + \ldots + (i + 1)a_{i + 1}^p a^i)D(a)
    $$
    is in $R$. Hence it follows that
    $$
    D(a)a_1 + 2D(a) a_2 x + \ldots + (i + 1)D(a) a_{i + 1} x^i
    $$
    is integral over $R$. By induction we find $D(a)^{i + 1}a_j \in R$
    for $j = 1, \ldots, i + 1$. (Here we use that $1, \ldots, i + 1$
    are invertible.) Hence $D(a)^{i + 1}a_0$ is also in $R$ because it
    is the difference of $y$ and $\sum_{j > 0} D(a)^{i + 1}a_jx^j$ which
    are integral over $R$ (since $x$ is integral over $R$ as $a \in R$).
    \end{proof}
    
    \begin{lemma}
    \label{lemma-domain-char-zero-N-1-2}
    A Noetherian domain of characteristic zero is N-1 if and only if
    it is N-2 (i.e., Japanese).
    \end{lemma}
    
    \begin{proof}
    This is clear from
    Lemma \ref{lemma-Noetherian-normal-domain-finite-separable-extension}
    since every field extension in characteristic zero is separable.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-domain-char-p-N-1-2}
    Let $R$ be a Noetherian domain with fraction field $K$ of
    characteristic $p > 0$. Then $R$ is N-2 if and only if
    for every finite purely inseparable extension $K \subset L$ the integral
    closure of $R$ in $L$ is finite over $R$.
    \end{lemma}
    
    \begin{proof}
    Assume the integral closure of $R$ in every finite purely inseparable
    field extension of $K$ is finite.
    Let $K \subset L$ be any finite extension. We have to show the
    integral closure of $R$ in $L$ is finite over $R$.
    Choose a finite normal field extension $K \subset M$
    containing $L$. As $R$ is Noetherian it suffices to show that
    the integral closure of $R$ in $M$ is finite over $R$.
    By Fields, Lemma \ref{fields-lemma-normal-case}
    there exists a subextension $K \subset M_{insep} \subset M$
    such that $M_{insep}/K$ is purely inseparable, and $M/M_{insep}$
    is separable. By assumption the integral closure $R'$ of $R$ in
    $M_{insep}$ is finite over $R$. By
    Lemma \ref{lemma-Noetherian-normal-domain-finite-separable-extension}
    the integral
    closure $R''$ of $R'$ in $M$ is finite over $R'$. Then $R''$ is finite
    over $R$ by Lemma \ref{lemma-finite-transitive}.
    Since $R''$ is also the integral closure
    of $R$ in $M$ (see Lemma \ref{lemma-integral-closure-transitive}) we win.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-polynomial-ring-N-2}
    Let $R$ be a Noetherian domain.
    If $R$ is N-1 then $R[x]$ is N-1.
    If $R$ is N-2 then $R[x]$ is N-2.
    \end{lemma}
    
    \begin{proof}
    Assume $R$ is N-1. Let $R'$ be the integral closure of $R$
    which is finite over $R$. Hence also $R'[x]$ is finite over
    $R[x]$. The ring $R'[x]$ is normal (see
    Lemma \ref{lemma-polynomial-domain-normal}), hence N-1.
    This proves the first assertion.
    
    \medskip\noindent
    For the second assertion, by Lemma \ref{lemma-finite-extension-N-2}
    it suffices to show that $R'[x]$ is N-2. In other words we may
    and do assume that $R$ is a normal N-2 domain. In characteristic zero
    we are done by Lemma \ref{lemma-domain-char-zero-N-1-2}.
    In characteristic $p > 0$ we have to show that the integral
    closure of $R[x]$ is finite in any finite purely inseparable extension
    of $f.f.(R[x]) = K(x) \subset L$ with $K = f.f.(R)$. Clearly there
    exists a finite purely inseparable field extension $K \subset L'$
    and $q = p^e$ such that $L \subset L'(x^{1/q})$. As $R[x]$ is
    Noetherian it suffices to show that the integral closure of $R[x]$
    in $L'(x^{1/q})$ is finite over $R[x]$. And this integral closure
    is equal to $R'[x^{1/q}]$ with $R \subset R' \subset L'$ the integral
    closure of $R$ in $L'$.
    Since $R$ is N-2 we see that $R'$ is finite over $R$ and hence
    $R'[x^{1/q}]$ is finite over $R[x]$.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-openness-normal-locus}
    Let $R$ be a Noetherian domain.
    If there exists an $f \in R$ such that $R_f$ is normal
    then
    $$
    U = \{\mathfrak p \in \Spec(R) \mid R_{\mathfrak p} \text{ is normal}\}
    $$
    is open in $\Spec(R)$.
    \end{lemma}
    
    \begin{proof}
    It is clear that the standard open $D(f)$ is contained in $U$.
    By Serre's criterion Lemma \ref{lemma-criterion-normal} we see that
    $\mathfrak p \not \in U$ implies that for some
    $\mathfrak q \subset \mathfrak p$ we have
    either
    \begin{enumerate}
    \item Case I: $\text{depth}(R_{\mathfrak q}) < 2$
    and $\dim(R_{\mathfrak q}) \geq 2$, and
    \item Case II: $R_{\mathfrak q}$ is not regular
    and $\dim(R_{\mathfrak q}) = 1$.
    \end{enumerate}
    This in particular also means that $R_{\mathfrak q}$ is not
    normal, and hence $f \in \mathfrak q$. In case I we see that
    $\text{depth}(R_{\mathfrak q}) =
    \text{depth}(R_{\mathfrak q}/fR_{\mathfrak q}) + 1$.
    Hence such a prime $\mathfrak q$ is the same thing as an embedded
    associated prime of $R/fR$. In case II $\mathfrak q$ is an associated
    prime of $R/fR$ of height 1. Thus there is a finite set $E$
    of such primes $\mathfrak q$ (see Lemma \ref{lemma-finite-ass}) and
    $$
    \Spec(R) \setminus U
    =
    \bigcup\nolimits_{\mathfrak q \in E} V(\mathfrak q)
    $$
    as desired.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-characterize-N-1}
    Let $R$ be a Noetherian domain.
    Assume
    \begin{enumerate}
    \item there exists a nonzero $f \in R$ such that $R_f$ is normal, and
    \item for every maximal ideal $\mathfrak m \subset R$
    the local ring $R_{\mathfrak m}$ is N-1.
    \end{enumerate}
    Then $R$ is N-1.
    \end{lemma}
    
    \begin{proof}
    Set $K = f.f.(R)$. Suppose that $R \subset R' \subset K$ is a finite
    extension of $R$ contained in $K$. Note that $R_f = R'_f$ since
    $R_f$ is already normal. Hence by Lemma \ref{lemma-openness-normal-locus}
    the set of primes
    $\mathfrak p' \in \Spec(R')$ with $R'_{\mathfrak p'}$ non-normal
    is closed in $\Spec(R')$. Since $\Spec(R') \to \Spec(R)$
    is closed the image of this set is closed in $\Spec(R)$.
    For such a ring $R'$ denote $Z_{R'} \subset \Spec(R)$ this image.
    
    \medskip\noindent
    Pick a maximal ideal $\mathfrak m \subset R$.
    Let $R_{\mathfrak m} \subset R_{\mathfrak m}'$ be the integral
    closure of the local ring in $K$. By assumption this is
    a finite ring extension. By Lemma \ref{lemma-integral-closure-localize}
    we can find finitely
    many elements $r_1, \ldots, r_n \in K$ integral over $R$ such that
    $R_{\mathfrak m}'$ is generated by $r_1, \ldots, r_n$ over $R_{\mathfrak m}$.
    Let $R' = R[x_1, \ldots, x_n] \subset K$. With this choice it is clear
    that $\mathfrak m \not \in Z_{R'}$.
    
    \medskip\noindent
    As $\Spec(R)$ is quasi-compact, the above shows that we can
    find a finite collection $R \subset R'_i \subset K$ such that
    $\bigcap Z_{R'_i} = \emptyset$. Let $R'$ be the subring of $K$
    generated by all of these. It is finite over $R$. Also $Z_{R'} = \emptyset$.
    Namely, every prime $\mathfrak p'$ lies over a prime $\mathfrak p'_i$
    such that $(R'_i)_{\mathfrak p'_i}$ is normal. This implies
    that $R'_{\mathfrak p'} = (R'_i)_{\mathfrak p'_i}$ is normal too.
    Hence $R'$ is normal, in other words
    $R'$ is the integral closure of $R$ in $K$.
    \end{proof}
    
    \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}
    
    \begin{lemma}
    \label{lemma-power-series-over-N-2}
    Let $R$ be a ring.
    If $R$ is Noetherian, a domain, and N-2, then so is $R[[x]]$.
    \end{lemma}
    
    \begin{proof}
    Observe that $R[[x]]$ is Noetherian by
    Lemma \ref{lemma-Noetherian-power-series}.
    Let $R' \supset R$ be the integral closure of $R$ in its fraction
    field. Because $R$ is N-2 this is finite over $R$. Hence $R'[[x]]$
    is finite over $R[[x]]$. By
    Lemma \ref{lemma-power-series-over-Noetherian-normal-domain}
    we see that $R'[[x]]$ is a normal domain.
    Apply Lemma \ref{lemma-tate-japanese} to the
    element $x \in R'[[x]]$ to see that $R'[[x]]$ is N-2. Then
    Lemma \ref{lemma-finite-extension-N-2} shows that $R[[x]]$ is N-2.
    \end{proof}

    Comments (2)

    Comment #2546 by Ashwin Iyengar on May 17, 2017 a 2:23 pm UTC

    Where did the name "Japanese" come from?

    Comment #2547 by sdf on May 17, 2017 a 6:17 pm UTC

    I think this name was first used by Grothendieck in EGA. My guess is it in honor of the strong Tokyo school, Nakayama, Takagi, Nagata and the many others there doing things around representation theory around 1920s-1960s

    Add a comment on tag 0BI1

    Your email address will not be published. Required fields are marked.

    In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).

    All contributions are licensed under the GNU Free Documentation License.




    In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

    This captcha seems more appropriate than the usual illegible gibberish, right?