The Stacks project

Proof. The second statement follows from the first by elementary topology. According to Schemes, Lemma 26.18.5 to prove the first statement we may replace $X$ by $X_ s$ and $S$ by $\mathop{\mathrm{Spec}}(\kappa (s))$. Thus we may assume that $S = \mathop{\mathrm{Spec}}(k)$ is the spectrum of a field. In this case, let $\mathop{\mathrm{Spec}}(A) = U \subset X$ be any affine open containing $x$. The point $x$ corresponds to a prime ideal $\mathfrak q \subset A$ such that $\kappa (\mathfrak q)/k$ is an algebraic field extension. By Algebra, Lemma 10.35.9 we see that $\mathfrak q$ is a maximal ideal, i.e., $x \in U$ is a closed point. Since the affine opens form a basis of the topology of $X$ we conclude that $\{ x\} $ is closed. $\square$

Proof. If the extension is finite, then $x$ is a closed point of the fibre by Lemma 29.20.2 above. For the converse, assume that $x$ is a closed point of its fibre. Choose affine opens $\mathop{\mathrm{Spec}}(A) = U \subset X$ and $\mathop{\mathrm{Spec}}(R) = V \subset S$ such that $f(U) \subset V$. By Lemma 29.15.2 the ring map $R \to A$ is of finite type. Let $\mathfrak q \subset A$, resp. $\mathfrak p \subset R$ be the prime ideal corresponding to $x$, resp. $s$. Consider the fibre ring $\overline{A} = A \otimes _ R \kappa (\mathfrak p)$. Let $\overline{\mathfrak q}$ be the prime of $\overline{A}$ corresponding to $\mathfrak q$. The assumption that $x$ is a closed point of its fibre implies that $\overline{\mathfrak q}$ is a maximal ideal of $\overline{A}$. Since $\overline{A}$ is an algebra of finite type over the field $\kappa (\mathfrak p)$ we see by the Hilbert Nullstellensatz, see Algebra, Theorem 10.34.1, that $\kappa (\overline{\mathfrak q})$ is a finite extension of $\kappa (\mathfrak p)$. Since $\kappa (s) = \kappa (\mathfrak p)$ and $\kappa (x) = \kappa (\mathfrak q) = \kappa (\overline{\mathfrak q})$ we win. $\square$

Proof. The residue field $\kappa (x')$ is a quotient of $\kappa (f'(x')) \otimes _{\kappa (f(x))} \kappa (x)$, see Schemes, Lemma 26.17.5. Hence it is a finite extension of $\kappa (f'(x'))$ as $\kappa (x)$ is a finite extension of $\kappa (f(x))$ by Lemma 29.20.3. Thus we see that $x'$ is closed in its fibre by applying that lemma one more time. $\square$

Proof. This is clear from Algebra, Definition 10.122.3. $\square$

Proof. Assume $f$ is quasi-finite at $x$. By assumption there exist opens $U \subset X$, $V \subset S$ such that $f(U) \subset V$, $x \in U$ and $x$ an isolated point of $U_ s$. Hence $\{ x\} \subset U_ s$ is an open subset. Since $U_ s = U \cap X_ s \subset X_ s$ is also open we conclude that $\{ x\} \subset X_ s$ is an open subset also. Thus we conclude that $x$ is an isolated point of $X_ s$.

Note that $X_ s$ is a Jacobson scheme by Lemma 29.16.10 (and Lemma 29.15.4). If $x$ is isolated in $X_ s$, i.e., $\{ x\} \subset X_ s$ is open, then $\{ x\} $ contains a closed point (by the Jacobson property), hence $x$ is closed in $X_ s$. It is clear that there is no point $x' \in X_ s$, distinct from $x$, specializing to $x$.

Assume that $x$ is closed in $X_ s$ and that there is no point $x' \in X_ s$, distinct from $x$, specializing to $x$. Consider a pair of affine opens $\mathop{\mathrm{Spec}}(A) = U \subset X$, $\mathop{\mathrm{Spec}}(R) = V \subset S$ with $f(U) \subset V$ and $x \in U$. Let $\mathfrak q \subset A$ correspond to $x$ and $\mathfrak p \subset R$ correspond to $s$. By Lemma 29.15.2 the ring map $R \to A$ is of finite type. Consider the fibre ring $\overline{A} = A \otimes _ R \kappa (\mathfrak p)$. Let $\overline{\mathfrak q}$ be the prime of $\overline{A}$ corresponding to $\mathfrak q$. Since $\mathop{\mathrm{Spec}}(\overline{A})$ is an open subscheme of the fibre $X_ s$ we see that $\overline{q}$ is a maximal ideal of $\overline{A}$ and that there is no point of $\mathop{\mathrm{Spec}}(\overline{A})$ specializing to $\overline{\mathfrak q}$. This implies that $\dim (\overline{A}_{\overline{q}}) = 0$. Hence by Algebra, Definition 10.122.3 we see that $R \to A$ is quasi-finite at $\mathfrak q$, i.e., $X \to S$ is quasi-finite at $x$ by definition.

At this point we have shown conditions (1) – (3) are all equivalent. It is clear that (4) implies (1). And it is also clear that (2) implies (4) since if $x$ is an isolated point of $X_ s$ then it is also an isolated point of $U_ s$ for any open $U$ which contains it. $\square$

Proof. Suppose $T$ is a scheme which (a) is locally of finite type over a field $k$, and (b) has finitely many points. Then Lemma 29.16.10 shows $T$ is a Jacobson scheme. A finite Jacobson space is discrete, see Topology, Lemma 5.18.6. Apply this remark to the fibre $X_ s$ which is locally of finite type over $\mathop{\mathrm{Spec}}(\kappa (s))$ to see the first statement. Finally, apply Lemma 29.20.6 to see the second. $\square$

Proof. It is immediate that (3) implies (2). Lemma 29.20.6 shows that (2) is equivalent to (1). Assume (2) and let $\mathop{\mathrm{Spec}}(k) \to S$ be as in (3). Denote $s \in S$ the image of $\mathop{\mathrm{Spec}}(k) \to S$. Then $X_ k$ is the base change of $X_ s$ via $\mathop{\mathrm{Spec}}(k) \to \mathop{\mathrm{Spec}}(\kappa (s))$. Hence every point of $X_ k$ is closed by Lemma 29.20.4. As $X_ k \to \mathop{\mathrm{Spec}}(k)$ is locally of finite type (by Lemma 29.15.4), we may apply Lemma 29.20.6 to conclude that every point of $X_ k$ is isolated, i.e., $X_ k$ has a discrete underlying topological space. $\square$

Proof. Assume $f$ is quasi-finite. It is quasi-compact by Definition 29.15.1. Let $x \in X$. We see that $f$ is quasi-finite at $x$ by Lemma 29.20.6. Hence $f$ is quasi-compact and locally quasi-finite.

Assume $f$ is quasi-compact and locally quasi-finite. Then $f$ is of finite type. Let $x \in X$ be a point. By Lemma 29.20.6 we see that $x$ is an isolated point of its fibre. The lemma is proved. $\square$

Proof. Assume $f$ is quasi-finite. In particular $f$ is locally of finite type and quasi-compact (since it is of finite type). Let $s \in S$. Since every $x \in X_ s$ is isolated in $X_ s$ we see that $X_ s = \bigcup _{x \in X_ s} \{ x\} $ is an open covering. As $f$ is quasi-compact, the fibre $X_ s$ is quasi-compact. Hence we see that $X_ s$ is finite.

Conversely, assume $f$ is locally of finite type, quasi-compact and has finite fibres. Then it is locally quasi-finite by Lemma 29.20.7. Hence it is quasi-finite by Lemma 29.20.9. $\square$

Proof. For a ring map $R \to A$ let us define $P(R \to A)$ to mean “$R \to A$ is quasi-finite” (see remark above lemma). We claim that $P$ is a local property of ring maps. We check conditions (a), (b) and (c) of Definition 29.14.1. In the proof of Lemma 29.15.2 we have seen that (a), (b) and (c) hold for the property of being “of finite type”. Note that, for a finite type ring map $R \to A$, the property $R \to A$ is quasi-finite at $\mathfrak q$ depends only on the local ring $A_{\mathfrak q}$ as an algebra over $R_{\mathfrak p}$ where $\mathfrak p = R \cap \mathfrak q$ (usual abuse of notation). Using these remarks (a), (b) and (c) of Definition 29.14.1 follow immediately. For example, suppose $R \to A$ is a ring map such that all of the ring maps $R \to A_{a_ i}$ are quasi-finite for $a_1, \ldots , a_ n \in A$ generating the unit ideal. We conclude that $R \to A$ is of finite type. Also, for any prime $\mathfrak q \subset A$ the local ring $A_{\mathfrak q}$ is isomorphic as an $R$-algebra to the local ring $(A_{a_ i})_{\mathfrak q_ i}$ for some $i$ and some $\mathfrak q_ i \subset A_{a_ i}$. Hence we conclude that $R \to A$ is quasi-finite at $\mathfrak q$.

We conclude that Lemma 29.14.3 applies with $P$ as in the previous paragraph. Hence it suffices to prove that $f$ is locally quasi-finite is equivalent to $f$ is locally of type $P$. Since $P(R \to A)$ is “$R \to A$ is quasi-finite” which means $R \to A$ is quasi-finite at every prime of $A$, this follows from Lemma 29.20.6. $\square$

Proof. In the proof of Lemma 29.20.11 we saw that $P = $“quasi-finite” is a local property of ring maps, and that a morphism of schemes is locally quasi-finite if and only if it is locally of type $P$ as in Definition 29.14.2. Hence the first statement of the lemma follows from Lemma 29.14.5 combined with the fact that being quasi-finite is a property of ring maps that is stable under composition, see Algebra, Lemma 10.122.7. By the above, Lemma 29.20.9 and the fact that compositions of quasi-compact morphisms are quasi-compact, see Schemes, Lemma 26.19.4 we see that the composition of quasi-finite morphisms is quasi-finite. $\square$

Proof. The first and second assertion follow from the corresponding algebra result, see Algebra, Lemma 10.122.8 (combined with the fact that $f'$ is also locally of finite type by Lemma 29.15.4). By the above, Lemma 29.20.9 and the fact that a base change of a quasi-compact morphism is quasi-compact, see Schemes, Lemma 26.19.3 we see that the base change of a quasi-finite morphism is quasi-finite. $\square$

Proof. The fibre $X_ s$ is a scheme of finite type over a field, hence Noetherian (Lemma 29.15.6). Hence the topology on $X_ s$ is Noetherian (Properties, Lemma 28.5.5) and can have at most a finite number of isolated points (by elementary topology). Thus our lemma follows from Lemma 29.20.6. $\square$

Proof. A monomorphism is separated by Schemes, Lemma 26.23.3. A monomorphism is injective, hence we get $f$ is quasi-finite at every $x \in X$ for example by Lemma 29.20.6. $\square$

Proof. This is true because an open immersion is a local isomorphism and a closed immersion is clearly quasi-finite. $\square$

Proof. Via Lemma 29.20.11 this translates into the following algebra fact: Given ring maps $A \to B \to C$ such that $A \to C$ is quasi-finite, then $B \to C$ is quasi-finite. This follows from Algebra, Lemma 10.122.6 with $R = A$, $S = S' = C$ and $R' = B$. $\square$

Proof. Let $x \in X$ with images $y \in Y$ and $s \in S$. Since $g \circ f$ is locally quasi-finite by Lemma 29.20.5 the extension $\kappa (x)/\kappa (s)$ is finite. Hence $\kappa (y)/\kappa (s)$ is finite. Hence $y$ is a closed point of $Y_ s$ by Lemma 29.20.2. Since $f$ is surjective, we see that every point of $Y$ is closed in its fibre over $S$. Thus by Lemma 29.20.6 we conclude that $g$ is quasi-finite at every point. $\square$