The Stacks project

29.16 Points of finite type and Jacobson schemes

Let $S$ be a scheme. A finite type point $s$ of $S$ is a point such that the morphism $\mathop{\mathrm{Spec}}(\kappa (s)) \to S$ is of finite type. The reason for studying this is that finite type points can replace closed points in a certain sense and in certain situations. There are always enough of them for example. Moreover, a scheme is Jacobson if and only if all finite type points are closed points.

Lemma 29.16.1. Let $S$ be a scheme. Let $k$ be a field. Let $f : \mathop{\mathrm{Spec}}(k) \to S$ be a morphism. The following are equivalent:

  1. The morphism $f$ is of finite type.

  2. The morphism $f$ is locally of finite type.

  3. There exists an affine open $U = \mathop{\mathrm{Spec}}(R)$ of $S$ such that $f$ corresponds to a finite ring map $R \to k$.

  4. There exists an affine open $U = \mathop{\mathrm{Spec}}(R)$ of $S$ such that the image of $f$ consists of a closed point $u$ in $U$ and the field extension $\kappa (u) \subset k$ is finite.

Proof. The equivalence of (1) and (2) is obvious as $\mathop{\mathrm{Spec}}(k)$ is a singleton and hence any morphism from it is quasi-compact.

Suppose $f$ is locally of finite type. Choose any affine open $\mathop{\mathrm{Spec}}(R) = U \subset S$ such that the image of $f$ is contained in $U$, and the ring map $R \to k$ is of finite type. Let $\mathfrak p \subset R$ be the kernel. Then $R/\mathfrak p \subset k$ is of finite type. By Algebra, Lemma 10.34.2 there exist a $\overline{f} \in R/\mathfrak p$ such that $(R/\mathfrak p)_{\overline{f}}$ is a field and $(R/\mathfrak p)_{\overline{f}} \to k$ is a finite field extension. If $f \in R$ is a lift of $\overline{f}$, then we see that $k$ is a finite $R_ f$-module. Thus (2) $\Rightarrow $ (3).

Suppose that $\mathop{\mathrm{Spec}}(R) = U \subset S$ is an affine open such that $f$ corresponds to a finite ring map $R \to k$. Then $f$ is locally of finite type by Lemma 29.15.2. Thus (3) $\Rightarrow $ (2).

Suppose $R \to k$ is finite. The image of $R \to k$ is a field over which $k$ is finite by Algebra, Lemma 10.36.18. Hence the kernel of $R \to k$ is a maximal ideal. Thus (3) $\Rightarrow $ (4).

The implication (4) $\Rightarrow $ (3) is immediate. $\square$

Lemma 29.16.2. Let $S$ be a scheme. Let $A$ be an Artinian local ring with residue field $\kappa $. Let $f : \mathop{\mathrm{Spec}}(A) \to S$ be a morphism of schemes. Then $f$ is of finite type if and only if the composition $\mathop{\mathrm{Spec}}(\kappa ) \to \mathop{\mathrm{Spec}}(A) \to S$ is of finite type.

Proof. Since the morphism $\mathop{\mathrm{Spec}}(\kappa ) \to \mathop{\mathrm{Spec}}(A)$ is of finite type it is clear that if $f$ is of finite type so is the composition $\mathop{\mathrm{Spec}}(\kappa ) \to S$ (see Lemma 29.15.3). For the converse, note that $\mathop{\mathrm{Spec}}(A) \to S$ maps into some affine open $U = \mathop{\mathrm{Spec}}(B)$ of $S$ as $\mathop{\mathrm{Spec}}(A)$ has only one point. To finish apply Algebra, Lemma 10.54.4 to $B \to A$. $\square$

Recall that given a point $s$ of a scheme $S$ there is a canonical morphism $\mathop{\mathrm{Spec}}(\kappa (s)) \to S$, see Schemes, Section 26.13.

Definition 29.16.3. Let $S$ be a scheme. Let us say that a point $s$ of $S$ is a finite type point if the canonical morphism $\mathop{\mathrm{Spec}}(\kappa (s)) \to S$ is of finite type. We denote $S_{\text{ft-pts}}$ the set of finite type points of $S$.

We can describe the set of finite type points as follows.

Lemma 29.16.4. Let $S$ be a scheme. We have

\[ S_{\text{ft-pts}} = \bigcup \nolimits _{U \subset S\text{ open}} U_0 \]

where $U_0$ is the set of closed points of $U$. Here we may let $U$ range over all opens or over all affine opens of $S$.

Proof. Immediate from Lemma 29.16.1. $\square$

Lemma 29.16.5. Let $f : T \to S$ be a morphism of schemes. If $f$ is locally of finite type, then $f(T_{\text{ft-pts}}) \subset S_{\text{ft-pts}}$.

Proof. If $T$ is the spectrum of a field this is Lemma 29.16.1. In general it follows since the composition of morphisms locally of finite type is locally of finite type (Lemma 29.15.3). $\square$

Lemma 29.16.6. Let $f : T \to S$ be a morphism of schemes. If $f$ is locally of finite type and surjective, then $f(T_{\text{ft-pts}}) = S_{\text{ft-pts}}$.

Proof. We have $f(T_{\text{ft-pts}}) \subset S_{\text{ft-pts}}$ by Lemma 29.16.5. Let $s \in S$ be a finite type point. As $f$ is surjective the scheme $T_ s = \mathop{\mathrm{Spec}}(\kappa (s)) \times _ S T$ is nonempty, therefore has a finite type point $t \in T_ s$ by Lemma 29.16.4. Now $T_ s \to T$ is a morphism of finite type as a base change of $s \to S$ (Lemma 29.15.4). Hence the image of $t$ in $T$ is a finite type point by Lemma 29.16.5 which maps to $s$ by construction. $\square$

Lemma 29.16.7. Let $S$ be a scheme. For any locally closed subset $T \subset S$ we have

\[ T \not= \emptyset \Rightarrow T \cap S_{\text{ft-pts}} \not= \emptyset . \]

In particular, for any closed subset $T \subset S$ we see that $T \cap S_{\text{ft-pts}}$ is dense in $T$.

Proof. Note that $T$ carries a scheme structure (see Schemes, Lemma 26.12.4) such that $T \to S$ is a locally closed immersion. Any locally closed immersion is locally of finite type, see Lemma 29.15.5. Hence by Lemma 29.16.5 we see $T_{\text{ft-pts}} \subset S_{\text{ft-pts}}$. Finally, any nonempty affine open of $T$ has at least one closed point which is a finite type point of $T$ by Lemma 29.16.4. $\square$

It follows that most of the material from Topology, Section 5.18 goes through with the set of closed points replaced by the set of points of finite type. In fact, if $S$ is Jacobson then we recover the closed points as the finite type points.

Lemma 29.16.8. Let $S$ be a scheme. The following are equivalent:

  1. the scheme $S$ is Jacobson,

  2. $S_{\text{ft-pts}}$ is the set of closed points of $S$,

  3. for all $T \to S$ locally of finite type closed points map to closed points, and

  4. for all $T \to S$ locally of finite type closed points $t \in T$ map to closed points $s \in S$ with $\kappa (s) \subset \kappa (t)$ finite.

Proof. We have trivially (4) $\Rightarrow $ (3) $\Rightarrow $ (2). Lemma 29.16.7 shows that (2) implies (1). Hence it suffices to show that (1) implies (4). Suppose that $T \to S$ is locally of finite type. Choose $t \in T$ closed and let $s \in S$ be the image. Choose affine open neighbourhoods $\mathop{\mathrm{Spec}}(R) = U \subset S$ of $s$ and $\mathop{\mathrm{Spec}}(A) = V \subset T$ of $t$ with $V$ mapping into $U$. The induced ring map $R \to A$ is of finite type (see Lemma 29.15.2) and $R$ is Jacobson by Properties, Lemma 28.6.3. Thus the result follows from Algebra, Proposition 10.35.19. $\square$

Lemma 29.16.9. Let $S$ be a Jacobson scheme. Any scheme locally of finite type over $S$ is Jacobson.

Proof. This is clear from Algebra, Proposition 10.35.19 (and Properties, Lemma 28.6.3 and Lemma 29.15.2). $\square$

Lemma 29.16.10. The following types of schemes are Jacobson.

  1. Any scheme locally of finite type over a field.

  2. Any scheme locally of finite type over $\mathbf{Z}$.

  3. Any scheme locally of finite type over a $1$-dimensional Noetherian domain with infinitely many primes.

  4. A scheme of the form $\mathop{\mathrm{Spec}}(R) \setminus \{ \mathfrak m\} $ where $(R, \mathfrak m)$ is a Noetherian local ring. Also any scheme locally of finite type over it.

Proof. We will use Lemma 29.16.9 without mention. The spectrum of a field is clearly Jacobson. The spectrum of $\mathbf{Z}$ is Jacobson, see Algebra, Lemma 10.35.6. For (3) see Algebra, Lemma 10.61.4. For (4) see Properties, Lemma 28.6.4. $\square$


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