## 27.13 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 27.13.1. Let $X$ be a scheme.

1. 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 10.155.1).

2. 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 10.156.1).

3. 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 10.156.1).

Being Nagata is the same thing as being locally Noetherian and universally Japanese, see Lemma 27.13.8.

Remark 27.13.2. In a (locally Noetherian) scheme $X$ is called Japanese if for every $x \in X$ and every associated prime $\mathfrak p$ of $\mathcal{O}_{X, x}$ the ring $\mathcal{O}_{X, x}/\mathfrak p$ is Japanese. We do not use this definition since there exists a one dimensional noetherian domain with excellent (in particular Japanese) local rings whose normalization is not finite. See [Example 1, Hochster-loci] or or [Exposé XIX, Traveaux]. On the other hand, we could circumvent this problem by calling a scheme $X$ Japanese if for every affine open $\mathop{\mathrm{Spec}}(A) \subset X$ the ring $A/\mathfrak p$ is Japanese for every associated prime $\mathfrak p$ of $A$.

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

Lemma 27.13.4. Let $X$ be an integral scheme. The following are equivalent:

1. The scheme $X$ is Japanese.

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

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

4. 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.

Lemma 27.13.5. Let $X$ be a scheme. The following are equivalent:

1. The scheme $X$ is universally Japanese.

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

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

4. 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.

Lemma 27.13.6. Let $X$ be a scheme. The following are equivalent:

1. The scheme $X$ is Nagata.

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

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

4. 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.

Lemma 27.13.7. 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 25.10.1 (and Definition 27.3.1). By Algebra, Definition 10.156.1 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 27.5.2). 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 25.12.4. 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 27.13.4 we see that $A/\mathfrak p$ is Japanese. $\square$

Lemma 27.13.8. Let $X$ be a scheme. The following are equivalent:

1. $X$ is Nagata, and

2. $X$ is locally Noetherian and universally Japanese.

Proof. This is Algebra, Proposition 10.156.15. $\square$

This discussion will be continued in Morphisms, Section 28.17.

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