## 10.161 Japanese rings

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

Definition 10.161.1. 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 $L/K$ 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 10.161.2. 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 $L_0/k(x_1, \ldots , x_ n)$ 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 10.162.16 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 10.37.8) and hence finite over $R$. The same argument works for $R = \mathbf{Z}[x_1, x_2, x_3, \ldots ]$.

Lemma 10.161.3. 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 10.36.11. $\square$

Lemma 10.161.4. 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 10.36.11 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 10.161.5. 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 $L/K$ be the induced extension of fraction fields. Note that this is a finite field extension (for example by Lemma 10.122.2 (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 10.123.14 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 10.161.3 and 10.161.4. 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 $K$ and we conclude that the integral closure $T$ of $R$ in $M$ is finite over $R$. By Lemma 10.36.16 we see that $T$ is also the integral closure of $S$ in $M$ and we win by Lemma 10.36.15. $\square$

Lemma 10.161.6. 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 10.161.7. 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 10.161.8. Let $R$ be a Noetherian normal domain with fraction field $K$. Let $L/K$ 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 10.38.6) 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 10.161.9. Lemma 10.161.8 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 10.161.8 in case of purely inseparable extensions.

Lemma 10.161.10. 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 10.161.11. A Noetherian domain whose fraction field has characteristic zero is N-1 if and only if it is N-2 (i.e., Japanese).

Proof. This is clear from Lemma 10.161.8 since every field extension in characteristic zero is separable. $\square$

Lemma 10.161.12. 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 $L/K$ 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 $L/K$ 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 $M/K$ 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 $M/M_{insep}/K$ 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 10.161.8 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 10.36.16) we win. $\square$

Lemma 10.161.13. 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 10.37.8), hence N-1. This proves the first assertion.

For the second assertion, by Lemma 10.161.7 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 10.161.11. In characteristic $p > 0$ we have to show that the integral closure of $R[x]$ is finite in any finite purely inseparable extension of $L/K(x)$ where $K$ is the fraction field of $R$. There exists a finite purely inseparable field extension $L'/K$ and $q = p^ e$ such that $L \subset L'(x^{1/q})$; some details omitted. 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 10.161.14. Let $R$ be a Noetherian domain. If there exists an $f \in R$ such that $R_ f$ is normal then

$U = \{ \mathfrak p \in \mathop{\mathrm{Spec}}(R) \mid R_{\mathfrak p} \text{ is normal}\}$

is open in $\mathop{\mathrm{Spec}}(R)$.

Proof. It is clear that the standard open $D(f)$ is contained in $U$. By Serre's criterion Lemma 10.157.4 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 10.63.5) and

$\mathop{\mathrm{Spec}}(R) \setminus U = \bigcup \nolimits _{\mathfrak q \in E} V(\mathfrak q)$

as desired. $\square$

Lemma 10.161.15. Let $R$ be a Noetherian domain. Then $R$ is N-1 if and only if the following two conditions hold

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.

Proof. First assume $R$ is N-1. Let $R'$ be the integral closure of $R$ in its field of fractions $K$. By assumption we can find $x_1, \ldots , x_ n$ in $R'$ which generate $R'$ as an $R$-module. Since $R' \subset K$ we can find $f_ i \in R$ nonzero such that $f_ i x_ i \in R$. Then $R_ f \cong R'_ f$ where $f = f_1 \ldots f_ n$. Hence $R_ f$ is normal and we have (1). Part (2) follows from Lemma 10.161.3.

Assume (1) and (2). Let $K$ be the fraction field of $R$. 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 10.161.14 the set of primes $\mathfrak p' \in \mathop{\mathrm{Spec}}(R')$ with $R'_{\mathfrak p'}$ non-normal is closed in $\mathop{\mathrm{Spec}}(R')$. Since $\mathop{\mathrm{Spec}}(R') \to \mathop{\mathrm{Spec}}(R)$ is closed the image of this set is closed in $\mathop{\mathrm{Spec}}(R)$. For such a ring $R'$ denote $Z_{R'} \subset \mathop{\mathrm{Spec}}(R)$ 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 10.36.11 we can find finitely many elements $x_1, \ldots , x_ n \in K$ integral over $R$ such that $R_{\mathfrak m}'$ is generated by $x_1, \ldots , x_ n$ 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 $\mathop{\mathrm{Spec}}(R)$ 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 10.161.16 (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 10.161.11. Hence we may assume that the characteristic of $K$ is $p > 0$, and we may apply Lemma 10.161.12. Thus given $L/K$ 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 10.46.7) 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 $y^ q = x$. Observe that $R_{\mathfrak p}$ is a discrete valuation ring by Lemma 10.119.7. Then $S_{\mathfrak q}$ is Noetherian by Krull-Akizuki (Lemma 10.119.12). Whereupon we conclude $S_{\mathfrak q}$ is a discrete valuation ring by Lemma 10.119.7 once again. By Lemma 10.119.10 we see that $\kappa (\mathfrak q)/\kappa (\mathfrak p)$ 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 10.51.4). Thus we may apply Lemma 10.96.12 to conclude that $S$ is finite over $R$, and we win. $\square$

Lemma 10.161.17. 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 10.31.2. 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 10.37.9 we see that $R'[[x]]$ is a normal domain. Apply Lemma 10.161.16 to the element $x \in R'[[x]]$ to see that $R'[[x]]$ is N-2. Then Lemma 10.161.7 shows that $R[[x]]$ is N-2. $\square$

Comment #2546 by Ashwin Iyengar on

Where did the name "Japanese" come from?

Comment #2547 by sdf on

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

Comment #6452 by Yijin Wang on

A minor lapses.In lemma 10.161.15,the third paragraph,R' should be R[r_1,...r_n] instead of R[x_1,...x_n]

Comment #6461 by on

Thanks and fixed here by renaming the $r_i$ instead.

Comment #6691 by nhw on

In the first sentence should "being" be "begin"?

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