The Stacks project

Proof. To show this it suffices to show that being Noetherian is a local property of rings, see Lemma 28.4.3. Any localization of a Noetherian ring is Noetherian, see Algebra, Lemma 10.31.1. By Algebra, Lemma 10.23.2 we see the second property to Definition 28.4.1. $\square$

Proof. A closed immersion is clearly quasi-compact. A composition of quasi-compact morphisms is quasi-compact, see Topology, Lemma 5.12.2. Hence it suffices to show that an open immersion into a locally Noetherian scheme is quasi-compact. Using Schemes, Lemma 26.19.2 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 10.31.5 and Topology, Lemmas 5.9.2 and 5.12.13). $\square$

Proof. By Schemes, Lemma 26.21.6 we have to show that the intersection $U \cap V$ of two affine opens of $X$ is quasi-compact. This follows from Lemma 28.5.3 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.) $\square$

Proof. This is because a Noetherian scheme is a finite union of spectra of Noetherian rings and Algebra, Lemma 10.31.5 and Topology, Lemma 5.9.4. $\square$

Proof. Omitted. Hint: Any quotient, and any localization of a Noetherian ring is Noetherian. For the Noetherian case use again that any subset of a Noetherian space is a Noetherian space (with induced topology). $\square$

Proof. The underlying topological space of a Noetherian scheme is Noetherian (Lemma 28.5.5) and we conclude because a Noetherian topological space has only finitely many irreducible components (Topology, Lemma 5.9.2). $\square$

Proof. Use Lemma 28.5.5 and use that any subset of a Noetherian topological space is quasi-compact (see Topology, Lemmas 5.9.2 and 5.12.13). $\square$

Proof. The second assertion follows from the first (using Schemes, Lemma 26.12.4 and Lemma 28.5.6). 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 26.10.1 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 28.5.6. Denote $V \subset \mathop{\mathrm{Spec}}(R)$ the inverse image of $U_0$ in $\mathop{\mathrm{Spec}}(R)$ by the canonical morphism $\mathop{\mathrm{Spec}}(R) \to X_0$ (see Schemes, Section 26.13.) By construction $V$ is a singleton with unique point corresponding to $x$ (use Schemes, Lemma 26.13.2). By Algebra, Lemma 10.61.1 we see that $\dim (R) = 1$. In other words, we see that $y$ is an immediate specialization of $x$ (see Topology, Definition 5.20.1). 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. $\square$

Proof. Let $x' \leadsto x$ be a specialization in $X$, and let $K/\kappa (x')$ be a finitely generated extension of fields. By Schemes, Lemma 26.13.2 and the discussion following Schemes, Lemma 26.13.3 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 10.119.13. The ring map $\mathcal{O}_{X, x} \to R$ induces the morphism $f : \mathop{\mathrm{Spec}}(R) \to X$, see Schemes, Lemma 26.13.1. This morphism has all the desired properties by construction. $\square$

Proof. 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

1. $t \not\in T_{i - 1} \cup T_{i - 2} \cup \ldots \cup T_0$,

2. there exist a nontrivial specialization $t \leadsto t'$ with $t' \in T_{i - 1}$, and

3. for any nontrivial specialization $t \leadsto t'$ with $t' \in T$ we have $t' \in T_{i - 1} \cup T_{i - 2} \cup \ldots \cup T_0$.

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 10.60.9. 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 $\mathop{\mathrm{Spec}}(\mathcal{O}_{S, t_0})$ (Schemes, Lemma 26.13.2), we see that $i \leq d$. Thus $\bigcup T_ i = T_ d \cup \ldots \cup T_0$ is a finite subset of $T$.

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

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 28.5.4. $\square$

Proof. Recall that $\dim (\mathcal{O}_{S, s}) < \infty $ for any $s \in S$, see Algebra, Proposition 10.60.9. 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 $T_0$. $\square$

Proof. 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 $T'$.

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 28.5.12 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. $\square$