28.5 Noetherian schemes
Recall that a ring R is Noetherian if it satisfies the ascending chain condition of ideals. Equivalently every ideal of R is finitely generated.
Definition 28.5.1. Let X be a scheme.
We say X is locally Noetherian if every x \in X has an affine open neighbourhood \mathop{\mathrm{Spec}}(R) = U \subset X such that the ring R is Noetherian.
We say X is Noetherian if X is locally Noetherian and quasi-compact.
Here is the standard result characterizing locally Noetherian schemes.
Lemma 28.5.2. Let X be a scheme. The following are equivalent:
The scheme X is locally Noetherian.
For every affine open U \subset X the ring \mathcal{O}_ X(U) is Noetherian.
There exists an affine open covering X = \bigcup U_ i such that each \mathcal{O}_ X(U_ i) is Noetherian.
There exists an open covering X = \bigcup X_ j such that each open subscheme X_ j is locally Noetherian.
Moreover, if X is locally Noetherian then every open subscheme is locally Noetherian.
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
Lemma 28.5.3. Any immersion Z \to X with X locally Noetherian is quasi-compact.
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
Lemma 28.5.4. A locally Noetherian scheme is quasi-separated.
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
Lemma 28.5.5. A (locally) Noetherian scheme has a (locally) Noetherian underlying topological space, see Topology, Definition 5.9.1.
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
Lemma 28.5.6. Any locally closed subscheme of a (locally) Noetherian scheme is (locally) Noetherian.
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
Lemma 28.5.7. A Noetherian scheme has a finite number of irreducible components.
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
Lemma 28.5.8. Any morphism of schemes f : X \to Y with X Noetherian is quasi-compact.
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
Here is a fun lemma. It says that every locally Noetherian scheme has plenty of closed points (at least one in every closed subset).
Lemma 28.5.9. Any nonempty locally Noetherian scheme has a closed point. Any nonempty closed subset of a locally Noetherian scheme has a closed point. Equivalently, any point of a locally Noetherian scheme specializes to a closed point.
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
Lemma 28.5.10. Let X be a locally Noetherian scheme. Let x' \leadsto x be a specialization of points of X. Then
there exists a discrete valuation ring R and a morphism f : \mathop{\mathrm{Spec}}(R) \to X such that the generic point \eta of \mathop{\mathrm{Spec}}(R) maps to x' and the special point maps to x, and
given a finitely generated field extension K/\kappa (x') we may arrange it so that the extension \kappa (\eta )/\kappa (x') induced by f is isomorphic to the given one.
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
Lemma 28.5.11. Let S be a Noetherian scheme. Let T \subset S be an infinite subset. Then there exists an infinite subset T' \subset T such that there are no nontrivial specializations among the points T'.
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
t \not\in T_{i - 1} \cup T_{i - 2} \cup \ldots \cup T_0,
there exist a nontrivial specialization t \leadsto t' with t' \in T_{i - 1}, and
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
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 28.5.4.
\square
Lemma 28.5.12. Let S be a Noetherian scheme. Let T \subset S be a subset. Let T_0 \subset T be the set of t \in T such that there is no nontrivial specialization t' \leadsto t with t' \in T'. Then (a) there are no specializations among the points of T_0, (b) every point of T is a specialization of a point of T_0, and (c) the closures of T and T_0 are the same.
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
Lemma 28.5.13. Let S be a Noetherian scheme. Let T \subset S be an infinite dense subset. Then there exist a countable subset E \subset T which is dense in S.
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
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