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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