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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