## 29.20 Quasi-finite morphisms

A solid treatment of quasi-finite morphisms is the basis of many developments further down the road. It will lead to various versions of Zariski's Main Theorem, behaviour of dimensions of fibres, descent for étale morphisms, etc, etc. Before reading this section it may be a good idea to take a look at the algebra results in Algebra, Section 10.122.

Recall that a finite type ring map $R \to A$ is quasi-finite at a prime $\mathfrak q$ if $\mathfrak q$ defines an isolated point of its fibre, see Algebra, Definition 10.122.3.

reference
Definition 29.20.1. Let $f : X \to S$ be a morphism of schemes.

We say that $f$ is *quasi-finite at a point $x \in X$* if there exist an affine neighbourhood $\mathop{\mathrm{Spec}}(A) = U \subset X$ of $x$ and an affine open $\mathop{\mathrm{Spec}}(R) = V \subset S$ such that $f(U) \subset V$, the ring map $R \to A$ is of finite type, and $R \to A$ is quasi-finite at the prime of $A$ corresponding to $x$ (see above).

We say $f$ is *locally quasi-finite* if $f$ is quasi-finite at every point $x$ of $X$.

We say that $f$ is *quasi-finite* if $f$ is of finite type and every point $x$ is an isolated point of its fibre.

Trivially, a locally quasi-finite morphism is locally of finite type. We will see below that a morphism $f$ which is locally of finite type is quasi-finite at $x$ if and only if $x$ is isolated in its fibre. Moreover, the set of points at which a morphism is quasi-finite is open; we will see this in Section 29.55 on Zariski's Main Theorem.

Lemma 29.20.2. Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ be a point. Set $s = f(x)$. If $\kappa (x)/\kappa (s)$ is an algebraic field extension, then

$x$ is a closed point of its fibre, and

if in addition $s$ is a closed point of $S$, then $x$ is a closed point of $X$.

**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 $k \subset \kappa (\mathfrak q)$ 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.
$\square$

The following lemma is a version of the Hilbert Nullstellensatz.

Lemma 29.20.3. Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ be a point. Set $s = f(x)$. Assume $f$ is locally of finite type. Then $x$ is a closed point of its fibre if and only if $\kappa (s) \subset \kappa (x)$ is a finite field extension.

**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.
$\square$

Lemma 29.20.4. Let $f : X \to S$ be a morphism of schemes which is locally of finite type. Let $g : S' \to S$ be any morphism. Denote $f' : X' \to S'$ the base change. If $x' \in X'$ maps to a point $x \in X$ which is closed in $X_{f(x)}$ then $x'$ is closed in $X'_{f'(x')}$.

**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.
$\square$

Lemma 29.20.5. Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ be a point. Set $s = f(x)$. If $f$ is quasi-finite at $x$, then the residue field extension $\kappa (s) \subset \kappa (x)$ is finite.

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

Lemma 29.20.6. Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ be a point. Set $s = f(x)$. Let $X_ s$ be the fibre of $f$ at $s$. Assume $f$ is locally of finite type. The following are equivalent:

The morphism $f$ is quasi-finite at $x$.

The point $x$ is isolated in $X_ s$.

The point $x$ is closed in $X_ s$ and there is no point $x' \in X_ s$, $x' \not= x$ which specializes to $x$.

For any pair of affine opens $\mathop{\mathrm{Spec}}(A) = U \subset X$, $\mathop{\mathrm{Spec}}(R) = V \subset S$ with $f(U) \subset V$ and $x \in U$ corresponding to $\mathfrak q \subset A$ the ring map $R \to A$ is quasi-finite at $\mathfrak q$.

**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.
$\square$

Lemma 29.20.7. Let $f : X \to S$ be a morphism of schemes. Let $s \in S$. Assume that

$f$ is locally of finite type, and

$f^{-1}(\{ s\} )$ is a finite set.

Then $X_ s$ is a finite discrete topological space, and $f$ is quasi-finite at each point of $X$ lying over $s$.

**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.
$\square$

slogan
Lemma 29.20.8. Let $f : X \to S$ be a morphism of schemes. Assume $f$ is locally of finite type. Then the following are equivalent

$f$ is locally quasi-finite,

for every $s \in S$ the fibre $X_ s$ is a discrete topological space, and

for every morphism $\mathop{\mathrm{Spec}}(k) \to S$ where $k$ is a field the base change $X_ k$ has an underlying discrete topological space.

**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.
$\square$

Lemma 29.20.9. Let $f : X \to S$ be a morphism of schemes. Then $f$ is quasi-finite if and only if $f$ is locally quasi-finite and quasi-compact.

**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.
$\square$

Lemma 29.20.10. Let $f : X \to S$ be a morphism of schemes. The following are equivalent:

$f$ is quasi-finite, and

$f$ is locally of finite type, quasi-compact, and has finite fibres.

**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.
$\square$

Recall that a ring map $R \to A$ is quasi-finite if it is of finite type and quasi-finite at *all* primes of $A$, see Algebra, Definition 10.122.3.

Lemma 29.20.11. Let $f : X \to S$ be a morphism of schemes. The following are equivalent

The morphism $f$ is locally quasi-finite.

For every pair of affine opens $U \subset X$, $V \subset S$ with $f(U) \subset V$ the ring map $\mathcal{O}_ S(V) \to \mathcal{O}_ X(U)$ is quasi-finite.

There exists an open covering $S = \bigcup _{j \in J} V_ j$ and open coverings $f^{-1}(V_ j) = \bigcup _{i \in I_ j} U_ i$ such that each of the morphisms $U_ i \to V_ j$, $j\in J, i\in I_ j$ is locally quasi-finite.

There exists an affine open covering $S = \bigcup _{j \in J} V_ j$ and affine open coverings $f^{-1}(V_ j) = \bigcup _{i \in I_ j} U_ i$ such that the ring map $\mathcal{O}_ S(V_ j) \to \mathcal{O}_ X(U_ i)$ is quasi-finite, for all $j\in J, i\in I_ j$.

Moreover, if $f$ is locally quasi-finite then for any open subschemes $U \subset X$, $V \subset S$ with $f(U) \subset V$ the restriction $f|_ U : U \to V$ is locally quasi-finite.

**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.
$\square$

Lemma 29.20.12. The composition of two morphisms which are locally quasi-finite is locally quasi-finite. The same is true for quasi-finite morphisms.

**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.
$\square$

We will see later (Lemma 29.55.2) that the set $U$ of the following lemma is open.

slogan
Lemma 29.20.13. Let $f : X \to S$ be a morphism of schemes. Let $g : S' \to S$ be a morphism of schemes. Denote $f' : X' \to S'$ the base change of $f$ by $g$ and denote $g' : X' \to X$ the projection. Assume $X$ is locally of finite type over $S$.

Let $U \subset X$ (resp. $U' \subset X'$) be the set of points where $f$ (resp. $f'$) is quasi-finite. Then $U' = U \times _ S S' = (g')^{-1}(U)$.

The base change of a locally quasi-finite morphism is locally quasi-finite.

The base change of a quasi-finite morphism is quasi-finite.

**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.
$\square$

Lemma 29.20.14. Let $f : X \to S$ be a morphism of schemes of finite type. Let $s \in S$. There are at most finitely many points of $X$ lying over $s$ at which $f$ is quasi-finite.

**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.
$\square$

Lemma 29.20.15. Let $f : X \to Y$ be a morphism of schemes. If $f$ is locally of finite type and a monomorphism, then $f$ is separated and locally quasi-finite.

**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.
$\square$

Lemma 29.20.16. Any immersion is locally quasi-finite.

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

Lemma 29.20.17. Let $X \to Y$ be a morphism of schemes over a base scheme $S$. Let $x \in X$. If $X \to S$ is quasi-finite at $x$, then $X \to Y$ is quasi-finite at $x$. If $X$ is locally quasi-finite over $S$, then $X \to Y$ is locally quasi-finite.

**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 $R' = B$.
$\square$

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