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
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