## 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 $K$.

- We say $R$ is
N-1if the integral closure of $R$ in $K$ is a finite $R$-module.- We say $R$ is
N-2orJapaneseif 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

- Case I: $\text{depth}(R_{\mathfrak q}) < 2$ and $\dim(R_{\mathfrak q}) \geq 2$, and
- 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

- there exists a nonzero $f \in R$ such that $R_f$ is normal, and
- 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

- $R$ is a normal Noetherian domain,
- $R/xR$ is a domain and N-2,
- $R \cong \mathop{\rm lim}\nolimits_n R/x^nR$ 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 42498–42969 (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 (3)

## 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.