## 28.13 Japanese and Nagata schemes

The notions considered in this section are not prominently defined in EGA. A “universally Japanese scheme” is mentioned and defined in [IV Corollary 5.11.4, EGA]. A “Japanese scheme” is mentioned in [IV Remark 10.4.14 (ii), EGA] but no definition is given. A Nagata scheme (as given below) occurs in a few places in the literature (see for example [Definition 8.2.30, Liu] and [Page 142, Greco]).

We briefly recall that a domain $R$ is called *Japanese* if the integral closure of $R$ in any finite extension of its fraction field is finite over $R$. A ring $R$ is called *universally Japanese* if for any finite type ring map $R \to S$ with $S$ a domain $S$ is Japanese. A ring $R$ is called *Nagata* if it is Noetherian and $R/\mathfrak p$ is Japanese for every prime $\mathfrak p$ of $R$.

Definition 28.13.1. Let $X$ be a scheme.

Assume $X$ integral. We say $X$ is *Japanese* if for every $x \in X$ there exists an affine open neighbourhood $x \in U \subset X$ such that the ring $\mathcal{O}_ X(U)$ is Japanese (see Algebra, Definition 10.161.1).

We say $X$ is *universally Japanese* if for every $x \in X$ there exists an affine open neighbourhood $x \in U \subset X$ such that the ring $\mathcal{O}_ X(U)$ is universally Japanese (see Algebra, Definition 10.162.1).

We say $X$ is *Nagata* if for every $x \in X$ there exists an affine open neighbourhood $x \in U \subset X$ such that the ring $\mathcal{O}_ X(U)$ is Nagata (see Algebra, Definition 10.162.1).

Being Nagata is the same thing as being locally Noetherian and universally Japanese, see Lemma 28.13.8.

Lemma 28.13.3. A Nagata scheme is locally Noetherian.

**Proof.**
This is true because a Nagata ring is Noetherian by definition.
$\square$

Lemma 28.13.4. Let $X$ be an integral scheme. The following are equivalent:

The scheme $X$ is Japanese.

For every affine open $U \subset X$ the domain $\mathcal{O}_ X(U)$ is Japanese.

There exists an affine open covering $X = \bigcup U_ i$ such that each $\mathcal{O}_ X(U_ i)$ is Japanese.

There exists an open covering $X = \bigcup X_ j$ such that each open subscheme $X_ j$ is Japanese.

Moreover, if $X$ is Japanese then every open subscheme is Japanese.

**Proof.**
This follows from Lemma 28.4.3 and Algebra, Lemmas 10.161.3 and 10.161.4.
$\square$

Lemma 28.13.5. Let $X$ be a scheme. The following are equivalent:

The scheme $X$ is universally Japanese.

For every affine open $U \subset X$ the ring $\mathcal{O}_ X(U)$ is universally Japanese.

There exists an affine open covering $X = \bigcup U_ i$ such that each $\mathcal{O}_ X(U_ i)$ is universally Japanese.

There exists an open covering $X = \bigcup X_ j$ such that each open subscheme $X_ j$ is universally Japanese.

Moreover, if $X$ is universally Japanese then every open subscheme is universally Japanese.

**Proof.**
This follows from Lemma 28.4.3 and Algebra, Lemmas 10.162.4 and 10.162.7.
$\square$

Lemma 28.13.6. Let $X$ be a scheme. The following are equivalent:

The scheme $X$ is Nagata.

For every affine open $U \subset X$ the ring $\mathcal{O}_ X(U)$ is Nagata.

There exists an affine open covering $X = \bigcup U_ i$ such that each $\mathcal{O}_ X(U_ i)$ is Nagata.

There exists an open covering $X = \bigcup X_ j$ such that each open subscheme $X_ j$ is Nagata.

Moreover, if $X$ is Nagata then every open subscheme is Nagata.

**Proof.**
This follows from Lemma 28.4.3 and Algebra, Lemmas 10.162.6 and 10.162.7.
$\square$

Lemma 28.13.7. Let $X$ be a locally Noetherian scheme. Then $X$ is Nagata if and only if every integral closed subscheme $Z \subset X$ is Japanese.

**Proof.**
Assume $X$ is Nagata. Let $Z \subset X$ be an integral closed subscheme. Let $z \in Z$. Let $\mathop{\mathrm{Spec}}(A) = U \subset X$ be an affine open containing $z$ such that $A$ is Nagata. Then $Z \cap U \cong \mathop{\mathrm{Spec}}(A/\mathfrak p)$ for some prime $\mathfrak p$, see Schemes, Lemma 26.10.1 (and Definition 28.3.1). By Algebra, Definition 10.162.1 we see that $A/\mathfrak p$ is Japanese. Hence $Z$ is Japanese by definition.

Assume every integral closed subscheme of $X$ is Japanese. Let $\mathop{\mathrm{Spec}}(A) = U \subset X$ be any affine open. As $X$ is locally Noetherian we see that $A$ is Noetherian (Lemma 28.5.2). Let $\mathfrak p \subset A$ be a prime ideal. We have to show that $A/\mathfrak p$ is Japanese. Let $T \subset U$ be the closed subset $V(\mathfrak p) \subset \mathop{\mathrm{Spec}}(A)$. Let $\overline{T} \subset X$ be the closure. Then $\overline{T}$ is irreducible as the closure of an irreducible subset. Hence the reduced closed subscheme defined by $\overline{T}$ is an integral closed subscheme (called $\overline{T}$ again), see Schemes, Lemma 26.12.4. In other words, $\mathop{\mathrm{Spec}}(A/\mathfrak p)$ is an affine open of an integral closed subscheme of $X$. This subscheme is Japanese by assumption and by Lemma 28.13.4 we see that $A/\mathfrak p$ is Japanese.
$\square$

Lemma 28.13.8. Let $X$ be a scheme. The following are equivalent:

$X$ is Nagata, and

$X$ is locally Noetherian and universally Japanese.

**Proof.**
This is Algebra, Proposition 10.162.15.
$\square$

This discussion will be continued in Morphisms, Section 29.18.

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