hcju/stacksqa huggingface dataset:
The second statement is Lemma \ref{lemma-indecomposable-injective-noetherian}. For the first statement, let $I$ be an injective $R$-module. We will use transfinite induction to construct $I_\alpha \subset I$ for ordinals $\alpha$ which are direct sums of indecomposable injective $R$-modules $E_{\beta + 1}$ for $\beta < \alpha$. For $\alpha = 0$ we let $I_0 = 0$. Suppose given an ordinal $\alpha$ such that $I_\alpha$ has been constructed. Then $I_\alpha$ is an injective $R$-module by Lemma \ref{lemma-sum-injective-modules}. Hence $I \cong I_\alpha \oplus I'$. If $I' = 0$ we are done. If not, then $I'$ has an associated prime by Algebra, Lemma \ref{algebra-lemma-ass-zero}. Thus $I'$ contains a copy of $R/\mathfrak p$ for some prime $\mathfrak p$. Hence $I'$ contains an indecomposable submodule $E$ by Lemmas \ref{lemma-injective-hull-unique} and \ref{lemma-injective-hull-indecomposable}. Set $I_{\alpha + 1} = I_\alpha \oplus E_\alpha$. If $\alpha$ is a limit ordinal and $I_\beta$ has been constructed for $\beta < \alpha$, then we set $I_\alpha = \bigcup_{\beta < \alpha} I_\beta$. Observe that $I_\alpha = \bigoplus_{\beta < \alpha} E_{\beta + 1}$. This concludes the proof.
By Lemma \ref{lemma-endos} we have $R = \Hom_R(E, E) = D(E)$. Of course we have $E = \Hom_R(R, E) = D(R)$. Since $E$ is injective the functor $D$ is exact. The result now follows immediately from the description of the categories in Lemma \ref{lemma-describe-categories}.
Assume $E$ is locally constructible. Then there exists an open covering $X = \bigcup U_i$ such that $E \cap U_i$ is constructible in $U_i$ for each $i$. Let $V \subset X$ be any affine open. We can find a finite open affine covering $V = V_1 \cup \ldots \cup V_m$ such that for each $j$ we have $V_j \subset U_i$ for some $i = i(j)$. By Topology, Lemma \ref{topology-lemma-open-immersion-constructible-inverse-image} we see that each $E \cap V_j$ is constructible in $V_j$. Since the inclusions $V_j \to V$ are quasi-compact (see Schemes, Lemma \ref{schemes-lemma-quasi-compact-affine}) we conclude that $E \cap V$ is constructible in $V$ by Topology, Lemma \ref{topology-lemma-collate-constructible}. The converse implication is immediate.
As the complement of a locally constructible subset is locally constructible it suffices to show (2). We may assume $X$ is affine and hence $E$ constructible (Lemma \ref{lemma-locally-constructible}). In this case $X$ is a spectral space (Algebra, Lemma \ref{algebra-lemma-spec-spectral}). Then $\xi \not \in E$ implies $\xi \not \in \overline{E}$ by Topology, Lemma \ref{topology-lemma-constructible-stable-specialization-closed} and the fact that there are no points of $X$ different from $\xi$ which specialize to $\xi$.
If $U$ and $V$ are quasi-compact open then $U \cap V = \Delta^{-1}(U \times V)$, where $\Delta : X \to X \times X$ is the diagonal. As $X$ is quasi-separated we see that $\Delta$ is quasi-compact. Hence we see that $U \cap V$ is quasi-compact as $U \times V$ is quasi-compact (details omitted; use Schemes, Lemma \ref{schemes-lemma-affine-covering-fibre-product} to see $U \times V$ is a finite union of affines). The other assertions follow from the first and Topology, Lemma \ref{topology-lemma-topology-quasi-separated-scheme}.
By Topology, Definition \ref{topology-definition-spectral-space} we have to check that $X$ is sober, quasi-compact, has a basis of quasi-compact opens, and the intersection of any two quasi-compact opens is quasi-compact. This follows from Schemes, Lemma \ref{schemes-lemma-scheme-sober} and \ref{schemes-lemma-basis-affine-opens} and Lemma \ref{lemma-quasi-separated-quasi-compact-open-retrocompact} above.
As $X$ is quasi-compact we can choose a finite affine open covering $X = V_1 \cup \ldots \cup V_m$. As $X$ is quasi-separated each $V_i$ is retrocompact in $X$ by Lemma \ref{lemma-quasi-separated-quasi-compact-open-retrocompact}. Hence by Topology, Lemma \ref{topology-lemma-collate-constructible} we see that $E \subset X$ is constructible in $X$ if and only if $E \cap V_j$ is constructible in $V_j$. Thus we win by Lemma \ref{lemma-locally-constructible}.
See Topology, Definitions \ref{topology-definition-quasi-compact}, \ref{topology-definition-paritition}, and \ref{topology-definition-locally-finite} for the definitions of retrocompact, partition, and locally finite. \medskip\noindent If the partition is locally finite and $U \subset X$ is an affine open, then we see that $U = \coprod_{i \in I} U \cap X_i$ is a finite partition (more precisely, all but a finite number of its parts are empty). Hence $U \cap X_i$ is quasi-compact and its complement is retrocompact in $U$ as a finite union of retrocompact parts. Thus $U \cap X_i$ is constructible by Topology, Lemma \ref{topology-lemma-locally-closed-constructible-image}. It follows that $X_i$ is locally constructible by Lemma \ref{lemma-locally-constructible}. \medskip\noindent Assume the parts are locally constructible. Then for any affine open $U \subset X$ we obtain a covering $U = \coprod X_i \cap U$ by constructible subsets. Since the constructible topology is quasi-compact, see Topology, Lemma \ref{topology-lemma-constructible-hausdorff-quasi-compact}, this covering has a finite refinement, i.e., the partition is locally finite.
Assume (1). By Schemes, Lemma \ref{schemes-lemma-scheme-sober} we see that $X$ has a unique generic point $\eta$. Then $X = \overline{\{\eta\}}$. Hence $\eta$ is an element of every nonempty affine open $U \subset X$. This implies that $U = \overline{\{\eta\}}$ and that any two nonempty affines meet. Thus (1) implies both (2) and (3). \medskip\noindent Assume (2). Suppose $X = Z_1 \cup Z_2$ is a union of two closed subsets. For every $i$ we see that either $U_i \subset Z_1$ or $U_i \subset Z_2$. Pick some $i \in I$ and assume $U_i \subset Z_1$ (possibly after renumbering $Z_1$, $Z_2$). For any $j \in I$ the open subset $U_i \cap U_j$ is dense in $U_j$ and contained in the closed subset $Z_1 \cap U_j$. We conclude that also $U_j \subset Z_1$. Thus $X = Z_1$ as desired. \medskip\noindent Assume (3). Choose an affine open covering $X = \bigcup_{i \in I} U_i$. We may assume that each $U_i$ is nonempty. Since $X$ is nonempty we see that $I$ is not empty. By assumption each $U_i$ is irreducible. Suppose $U_i \cap U_j = \emptyset$ for some pair $i, j \in I$. Then the open $U_i \amalg  U_j = U_i \cup U_j$ is affine, see Schemes, Lemma \ref{schemes-lemma-disjoint-union-affines}. Hence it is irreducible by assumption which is absurd. We conclude that (3) implies (2). The lemma is proved.
If $X$ is irreducible, then every affine open $\Spec(R) = U \subset X$ is irreducible. If $X$ is reduced, then $R$ is reduced, by Lemma \ref{lemma-characterize-reduced} above. Hence $R$ is reduced and $(0)$ is a prime ideal, i.e., $R$ is an integral domain. \medskip\noindent If $X$ is integral, then for every nonempty affine open $\Spec(R) = U \subset X$ the ring $R$ is reduced and hence $X$ is reduced by Lemma \ref{lemma-characterize-reduced}. Moreover, every nonempty affine open is irreducible. Hence $X$ is irreducible, see Lemma \ref{lemma-characterize-irreducible}.
Of course (1) $\Leftrightarrow$ (3) and (2) $\Rightarrow$ (1). If (3) $\Rightarrow$ (2), then the final statement of the lemma holds and it follows easily that (4) is also equivalent to (1). Thus we show (3) $\Rightarrow$ (2). \medskip\noindent Let $X = \bigcup U_i$ be an affine open covering, say $U_i = \Spec(R_i)$. Assume $P(R_i)$. Let $\Spec(R) = U \subset X$ be an arbitrary affine open. By Schemes, Lemma \ref{schemes-lemma-good-subcover} there exists a standard covering of $U = \Spec(R)$ by standard opens $D(f_j)$ such that each ring $R_{f_j}$ is a principal localization of one of the rings $R_i$. By Definition \ref{definition-property-local} (1) we get $P(R_{f_j})$. Whereupon $P(R)$ by Definition \ref{definition-property-local} (2).
To show this it suffices to show that being Noetherian is a local property of rings, see Lemma \ref{lemma-locally-P}. Any localization of a Noetherian ring is Noetherian, see Algebra, Lemma \ref{algebra-lemma-Noetherian-permanence}. By Algebra, Lemma \ref{algebra-lemma-cover} we see the second property to Definition \ref{definition-property-local}.
A closed immersion is clearly quasi-compact. A composition of quasi-compact morphisms is quasi-compact, see Topology, Lemma \ref{topology-lemma-composition-quasi-compact}. Hence it suffices to show that an open immersion into a locally Noetherian scheme is quasi-compact. Using Schemes, Lemma \ref{schemes-lemma-quasi-compact-affine} we reduce to the case where $X$ is affine. Any open subset of the spectrum of a Noetherian ring is quasi-compact (for example combine Algebra, Lemma \ref{algebra-lemma-Noetherian-topology} and Topology, Lemmas \ref{topology-lemma-Noetherian} and \ref{topology-lemma-Noetherian-quasi-compact}).
By Schemes, Lemma \ref{schemes-lemma-characterize-quasi-separated} we have to show that the intersection $U \cap V$ of two affine opens of $X$ is quasi-compact. This follows from Lemma \ref{lemma-immersion-into-noetherian} above on considering the open immersion $U \cap V \to U$ for example. (But really it is just because any open of the spectrum of a Noetherian ring is quasi-compact.)
The second assertion follows from the first (using Schemes, Lemma \ref{schemes-lemma-reduced-closed-subscheme} and Lemma \ref{lemma-locally-closed-in-Noetherian}). Consider any nonempty affine open $U \subset X$. Let $x \in U$ be a closed point. If $x$ is a closed point of $X$ then we are done. If not, let $X_0 \subset X$ be the reduced induced closed subscheme structure on $\overline{\{x\}}$. Then $U_0 = U \cap X_0$ is an affine open of $X_0$ by Schemes, Lemma \ref{schemes-lemma-closed-subspace-scheme} and $U_0 = \{x\}$. Let $y \in X_0$, $y \not = x$ be a specialization of $x$. Consider the local ring $R = \mathcal{O}_{X_0, y}$. This is a Noetherian local ring as $X_0$ is Noetherian by Lemma \ref{lemma-locally-closed-in-Noetherian}. Denote $V \subset \Spec(R)$ the inverse image of $U_0$ in $\Spec(R)$ by the canonical morphism $\Spec(R) \to X_0$ (see Schemes, Section \ref{schemes-section-points}.) By construction $V$ is a singleton with unique point corresponding to $x$ (use Schemes, Lemma \ref{schemes-lemma-specialize-points}). By Algebra, Lemma \ref{algebra-lemma-Noetherian-local-domain-dim-2-infinite-opens} we see that $\dim(R) = 1$. In other words, we see that $y$ is an immediate specialization of $x$ (see Topology, Definition \ref{topology-definition-dimension-function}). In other words, any point $y \not = x$ such that $x \leadsto y$ is an immediate specialization of $x$. Clearly each of these points is a closed point as desired.
Let $x' \leadsto x$ be a specialization in $X$, and let $\kappa(x') \subset K$ be a finitely generated extension of fields. By Schemes, Lemma \ref{schemes-lemma-specialize-points} and the discussion following Schemes, Lemma \ref{schemes-lemma-characterize-points} this leads to ring maps $\mathcal{O}_{X, x} \to \kappa(x') \to K$. Let $R \subset K$ be any discrete valuation ring whose field of fractions is $K$ and which dominates the image of $\mathcal{O}_{X, x} \to K$, see Algebra, Lemma \ref{algebra-lemma-exists-dvr}. The ring map $\mathcal{O}_{X, x} \to R$ induces the morphism $f : \Spec(R) \to X$, see Schemes, Lemma \ref{schemes-lemma-morphism-from-spec-local-ring}. This morphism has all the desired properties by construction.
Let $T_0 \subset T$ be the set of $t \in T$ which do not specialize to another point of $T$. If $T_0$ is infinite, then $T' = T_0$ works. Hence we may and do assume $T_0$ is finite. Inductively, for $i > 0$, consider the set $T_i \subset T$ of $t \in T$ such that \begin{enumerate} \item $t \not \in T_{i - 1} \cup T_{i - 2} \cup \ldots \cup T_0$, \item there exist a nontrivial specialization $t \leadsto t'$ with $t' \in T_{i - 1}$, and \item for any nontrivial specialization $t \leadsto t'$ with $t' \in T$ we have $t' \in T_{i - 1} \cup T_{i - 2} \cup \ldots \cup T_0$. \end{enumerate} Again, if $T_i$ is infinite, then $T' = T_i$ works. Let $d$ be the maximum of the dimensions of the local rings $\mathcal{O}_{S, t}$ for $t \in T_0$; then $d$ is an integer because $T_0$ is finite and the dimensions of the local rings are finite by Algebra, Proposition \ref{algebra-proposition-dimension}. Then $T_i = \emptyset$ for $i > d$. Namely, if $t \in T_i$ then we can find a sequence of nontrivial specializations $t = t_i \leadsto t_{i - 1} \leadsto \ldots \leadsto t_0$ with $t_0 \in T_0$. As the points $t = t_i, t_{i - 1}, \ldots, t_0$ are in $\Spec(\mathcal{O}_{S, t_0})$ (Schemes, Lemma \ref{schemes-lemma-specialize-points}), we see that $i \leq d$. Thus $\bigcup T_i = T_d \cup \ldots \cup T_0$ is a finite subset of $T$. \medskip\noindent Suppose $t \in T$ is not in $\bigcup T_i$. Then there must be a specialization $t \leadsto t'$ with $t' \in T$ and $t' \not \in \bigcup T_i$. (Namely, if every specialization of $t$ is in the finite set $T_d \cup \ldots \cup T_0$, then there is a maximum $i$ such that there is some specialization $t \leadsto t'$ with $t' \in T_i$ and then $t \in T_{i + 1}$ by construction.) Hence we get an infinite sequence $$ t \leadsto t' \leadsto t'' \leadsto \ldots $$ of nontrivial specializations between points of $T \setminus \bigcup T_i$. This is impossible because the underlying topological space of $S$ is Noetherian by Lemma \ref{lemma-locally-Noetherian-quasi-separated}.
Recall that $\dim(\mathcal{O}_{S, s}) < \infty$ for any $s \in S$, see Algebra, Proposition \ref{algebra-proposition-dimension}. Let $t \in T$. If $t' \leadsto t$, then by dimension theory $\dim(\mathcal{O}_{S, t'}) \leq \dim(\mathcal{O}_{S, t})$ with equality if and only if $t' = t$. Thus if we pick $t' \leadsto t$ with $\dim(\mathcal{O}_{T, t'})$ minimal, then $t' \in T_0$. In other words,  every $t \in T$ is the specialization of an element of $T_0$.
Let $T'$ be the set of points $s \in S$ such that $\overline{\{s\}} \cap T$ contains a countable subset whose closure is $\overline{\{s\}}$. Since a finite set is countable we have $T \subset T'$. For $s \in T'$ choose such a countable subset $E_s \subset \overline{\{s\}} \cap T$. Let $E' = \{s_1, s_2, s_3, \ldots\} \subset T'$ be a countable subset. Then the closure of $E'$ in $S$ is the closure of the countable subset $\bigcup_n E_{s_n}$ of $T$. It follows that if $Z$ is an irreducible component of the closure of $E'$, then the generic point of $Z$ is in $T'$. \medskip\noindent Denote $T'_0 \subset T'$ the subset of $t \in T'$ such that there is no nontrivial specialization $t' \leadsto t$ with $t' \in T'$ as in Lemma \ref{lemma-maximal-points} whose results we will use without further mention. If $T'_0$ is infinite, then we choose a countable subset $E' \subset T'_0$. By the argument in the first paragraph, the generic points of the irreducible components of the closure of $E'$ are in $T'$. However, since one of these points specializes to infinitely many distinct elements of $E' \subset T'_0$ this is a contradiction. Thus $T'_0$ is finite, say $T'_0 = \{s_1, \ldots, s_m\}$. Then it follows that $S$, which is the closure of $T$, is contained in the closure of $\{s_1, \ldots, s_m\}$, which in turn is contained in the closure of the countable subset $E_{s_1} \cup \ldots \cup E_{s_m} \subset T$ as desired.
The final assertion of the lemma holds by Topology, Lemma \ref{topology-lemma-jacobson-inherited}. The equivalence of (5) and (1) is Topology, Lemma \ref{topology-lemma-jacobson-local}. Hence, using Lemma \ref{lemma-affine-jacobson}, we see that (1) $\Leftrightarrow$ (2). To finish proving the lemma it suffices to show that ``Jacobson'' is a local property of rings, see Lemma \ref{lemma-locally-P}. Any localization of a Jacobson ring at an element is Jacobson, see Algebra, Lemma \ref{algebra-lemma-Jacobson-invert-element}. Suppose $R$ is a ring, $f_1, \ldots, f_n \in R$ generate the unit ideal and each $R_{f_i}$ is Jacobson. Then we see that $\Spec(R) = \bigcup D(f_i)$ is a union of open subsets which are all Jacobson, and hence $\Spec(R)$ is Jacobson by Topology, Lemma \ref{topology-lemma-jacobson-local} again. This proves the second property of Definition \ref{definition-property-local}.
Proof of (3). Observe that $\Spec(R) - V(I)$ has a covering by the affine opens $\Spec(R_f)$ for $f \in I$. The rings $R_f$ are Jacobson by More on Algebra, Lemma \ref{more-algebra-lemma-noetherian-zariski-jacobson-complement}. Hence $\Spec(R) \setminus V(I)$ is Jacobson by Lemma \ref{lemma-locally-jacobson}. Parts (1) and (2) are special cases of (3). \medskip\noindent Direct proof of case (1). Since $\Spec(R)$ is a Noetherian scheme, $S$ is a Noetherian scheme (Lemma \ref{lemma-locally-closed-in-Noetherian}). Hence $S$ is a sober, Noetherian topological space (use Schemes, Lemma \ref{schemes-lemma-scheme-sober}). Assume $S$ is not Jacobson to get a contradiction. By Topology, Lemma \ref{topology-lemma-non-jacobson-Noetherian-characterize} there exists some non-closed point $\xi \in S$ such that $\{\xi\}$ is locally closed. This corresponds to a prime $\mathfrak p \subset R$ such that (1) there exists a prime $\mathfrak q$, $\mathfrak p \subset \mathfrak q \subset \mathfrak m$ with both inclusions strict, and (2) $\{\mathfrak p\}$ is open in $\Spec(R/\mathfrak p)$. This is impossible by Algebra, Lemma \ref{algebra-lemma-Noetherian-local-domain-dim-2-infinite-opens}.
It is immediate from the definitions that (2) implies (1). Let $X$ be a normal scheme such that every quasi-compact open has a finite number of irreducible components. If $X$ is affine then $X$ satisfies (2) by Algebra, Lemma \ref{algebra-lemma-characterize-reduced-ring-normal}. For a general $X$, let $X = \bigcup X_i$ be an affine open covering. Note that also each $X_i$ has but a finite number of irreducible components, and the lemma holds for each $X_i$. Let $T \subset X$ be an irreducible component. By the affine case each intersection $T \cap X_i$ is open in $X_i$ and an integral normal scheme. Hence $T \subset X$ is open, and an integral normal scheme. This proves that $X$ is the disjoint union of its irreducible components, which are integral normal schemes.
Set $R = \Gamma(X, \mathcal{O}_X)$. It is clear that $R$ is a domain. Suppose $f = a/b$ is an element of its fraction field which is integral over $R$. Say we have $f^d + \sum_{i = 0, \ldots, d - 1} a_i f^i = 0$ with $a_i \in R$. Let $U \subset X$ be affine open. Since $b \in R$ is not zero and since $X$ is integral we see that also $b|_U \in \mathcal{O}_X(U)$ is not zero. Hence $a/b$ is an element of the fraction field of $\mathcal{O}_X(U)$ which is integral over $\mathcal{O}_X(U)$ (because we can use the same polynomial $f^d + \sum_{i = 0, \ldots, d - 1} a_i|_U f^i = 0$ on $U$). Since $\mathcal{O}_X(U)$ is a normal domain (Lemma \ref{lemma-locally-normal}), we see that $f_U = (a|_U)/(b|_U) \in \mathcal{O}_X(U)$. It is easy to see that $f_U|_V = f_V$ whenever $V \subset U \subset X$ are affine open. Hence the local sections $f_U$ glue to a global section $f$ as desired.
Algebra, Lemma \ref{algebra-lemma-localize-CM} says that the localization of a Cohen-Macaulay local ring is Cohen-Macaulay. The lemma follows by combining this with Lemma \ref{lemma-locally-Noetherian}, with the existence of closed points on locally Noetherian schemes (Lemma \ref{lemma-locally-Noetherian-closed-point}), and the definitions.
Note that given a chain of specializations $$ \xi_n \leadsto \xi_{n - 1} \leadsto \ldots \leadsto \xi_0 $$ of points of $X$ all of the points $\xi_i$ correspond to prime ideals of the local ring of $X$ at $\xi_0$ by Schemes, Lemma \ref{schemes-lemma-specialize-points}. Hence we see that the dimension of $X$ is the supremum of the dimensions of its local rings. In particular $\dim_x(X) \geq \dim(\mathcal{O}_{X, x})$ as $\dim_x(X)$ is the minimum of the dimensions of open neighbourhoods of $x$. Thus $\sup_{x \in X} \dim_x(X) \geq \dim(X)$. On the other hand, it is clear that $\sup_{x \in X} \dim_x(X) \leq \dim(X)$ as $\dim(U) \leq \dim(X)$ for any open subset of $X$.
As the dimension of $X$ is zero, we see that for any affine open $U \subset X$ the space $U$ is profinite and satisfies a bunch of other properties which we will use freely below, see Algebra, Lemma \ref{algebra-lemma-ring-with-only-minimal-primes}. We choose an affine open covering $X = \bigcup U_i$. \medskip\noindent If (4) holds, then $U_i \cap U_j$ is a closed subset of $U_i$, hence quasi-compact, hence $X$ is quasi-separated, by Schemes, Lemma \ref{schemes-lemma-characterize-quasi-separated}, hence (1) holds. \medskip\noindent If (1) holds, then $U_i \cap U_j$ is a quasi-compact open of $U_i$ hence closed in $U_i$. Then $U_i \cap U_j \to U_i$ is an open immersion whose image is closed, hence it is a closed immersion. In particular $U_i \cap U_j$ is affine and $\mathcal{O}(U_i) \to \mathcal{O}_X(U_i \cap U_j)$ is surjective. Thus $X$ is separated by Schemes, Lemma \ref{schemes-lemma-characterize-quasi-separated}, hence (2) holds. \medskip\noindent Assume (2) and let $x, y \in X$. Say $x \in U_i$. If $y \in U_i$ too, then we can find disjoint open neighbourhoods of $x$ and $y$ because $U_i$ is Hausdorff. Say $y \not \in U_i$ and $y \in U_j$. Then $y \not \in U_i \cap U_j$ which is an affine open of $U_j$ and hence closed in $U_j$. Thus we can find an open neighbourhood of $y$ not meeting $U_i$ and we conclude that $X$ is Hausdorff, hence (3) holds. \medskip\noindent Assume (3). Let $U \subset X$ be affine open. Then $U$ is closed in $X$ by Topology, Lemma \ref{topology-lemma-quasi-compact-in-Hausdorff}. This proves (4) holds. \medskip\noindent We omit the proof of the final statement.
Assume $X$ is catenary. Let $x \in X$. By Lemma \ref{lemma-catenary-local} we may replace $X$ by an affine open neighbourhood of $x$, and then $\Gamma(X, \mathcal{O}_X)$ is a catenary ring. By Algebra, Lemma \ref{algebra-lemma-localization-catenary} any localization of a catenary ring is catenary. Whence $\mathcal{O}_{X, x}$ is catenary. \medskip\noindent Conversely assume all local rings of $X$ are catenary. Let $Y \subset Y'$ be an inclusion of irreducible closed subsets of $X$. Let $\xi \in Y$ be the generic point. Let $\mathfrak p \subset \mathcal{O}_{X, \xi}$ be the prime corresponding to the generic point of $Y'$, see Schemes, Lemma \ref{schemes-lemma-specialize-points}. By that same lemma the irreducible closed subsets of $X$ in between $Y$ and $Y'$ correspond to primes $\mathfrak q \subset \mathcal{O}_{X, \xi}$ with $\mathfrak p \subset \mathfrak q \subset \mathfrak m_{\xi}$. Hence we see all maximal chains of these are finite and have the same length as $\mathcal{O}_{X, \xi}$ is a catenary ring.
Assume $X$ is Nagata. Let $Z \subset X$ be an integral closed subscheme. Let $z \in Z$. Let $\Spec(A) = U \subset X$ be an affine open containing $z$ such that $A$ is Nagata. Then $Z \cap U \cong \Spec(A/\mathfrak p)$ for some prime $\mathfrak p$, see Schemes, Lemma \ref{schemes-lemma-closed-subspace-scheme} (and Definition \ref{definition-integral}). By Algebra, Definition \ref{algebra-definition-nagata} we see that $A/\mathfrak p$ is Japanese. Hence $Z$ is Japanese by definition. \medskip\noindent Assume every integral closed subscheme of $X$ is Japanese. Let $\Spec(A) = U \subset X$ be any affine open. As $X$ is locally Noetherian we see that $A$ is Noetherian (Lemma \ref{lemma-locally-Noetherian}). 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 $V(\mathfrak p) \subset \Spec(A)$. 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 \ref{schemes-lemma-reduced-closed-subscheme}. In other words, $\Spec(A/\mathfrak p)$ is an affine open of an integral closed subscheme of $X$. This subscheme is Japanese by assumption and by Lemma \ref{lemma-locally-Japanese} we see that $A/\mathfrak p$ is Japanese.