## 10.122 Quasi-finite maps

Consider a ring map $R \to S$ of finite type. A map $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is quasi-finite at a point if that point is isolated in its fibre. This means that the fibre is zero dimensional at that point. In this section we study the basic properties of this important but technical notion. More advanced material can be found in the next section.

Lemma 10.122.1. Let $k$ be a field. Let $S$ be a finite type $k$ algebra. Let $\mathfrak q$ be a prime of $S$. The following are equivalent:

$\mathfrak q$ is an isolated point of $\mathop{\mathrm{Spec}}(S)$,

$S_{\mathfrak q}$ is finite over $k$,

there exists a $g \in S$, $g \not\in \mathfrak q$ such that $D(g) = \{ \mathfrak q \} $,

$\dim _{\mathfrak q} \mathop{\mathrm{Spec}}(S) = 0$,

$\mathfrak q$ is a closed point of $\mathop{\mathrm{Spec}}(S)$ and $\dim (S_{\mathfrak q}) = 0$, and

the field extension $k \subset \kappa (\mathfrak q)$ is finite and $\dim (S_{\mathfrak q}) = 0$.

In this case $S = S_{\mathfrak q} \times S'$ for some finite type $k$-algebra $S'$. Also, the element $g$ as in (3) has the property $S_{\mathfrak q} = S_ g$.

**Proof.**
Suppose $\mathfrak q$ is an isolated point of $\mathop{\mathrm{Spec}}(S)$, i.e., $\{ \mathfrak q\} $ is open in $\mathop{\mathrm{Spec}}(S)$. Because $\mathop{\mathrm{Spec}}(S)$ is a Jacobson space (see Lemmas 10.35.2 and 10.35.4) we see that $\mathfrak q$ is a closed point. Hence $\{ \mathfrak q\} $ is open and closed in $\mathop{\mathrm{Spec}}(S)$. By Lemmas 10.21.3 and 10.24.3 we may write $S = S_1 \times S_2$ with $\mathfrak q$ corresponding to the only point $\mathop{\mathrm{Spec}}(S_1)$. Hence $S_1 = S_{\mathfrak q}$ is a zero dimensional ring of finite type over $k$. Hence it is finite over $k$ for example by Lemma 10.115.4. We have proved (1) implies (2).

Suppose $S_{\mathfrak q}$ is finite over $k$. Then $S_{\mathfrak q}$ is Artinian local, see Lemma 10.53.2. So $\mathop{\mathrm{Spec}}(S_{\mathfrak q}) = \{ \mathfrak qS_{\mathfrak q}\} $ by Lemma 10.53.6. Consider the exact sequence $0 \to K \to S \to S_{\mathfrak q} \to Q \to 0$. It is clear that $K_{\mathfrak q} = Q_{\mathfrak q} = 0$. Also, $K$ is a finite $S$-module as $S$ is Noetherian and $Q$ is a finite $S$-module since $S_{\mathfrak q}$ is finite over $k$. Hence there exists $g \in S$, $g \not\in \mathfrak q$ such that $K_ g = Q_ g = 0$. Thus $S_{\mathfrak q} = S_ g$ and $D(g) = \{ \mathfrak q \} $. We have proved that (2) implies (3).

Suppose $D(g) = \{ \mathfrak q \} $. Since $D(g)$ is open by construction of the topology on $\mathop{\mathrm{Spec}}(S)$ we see that $\mathfrak q$ is an isolated point of $\mathop{\mathrm{Spec}}(S)$. We have proved that (3) implies (1). In other words (1), (2) and (3) are equivalent.

Assume $\dim _{\mathfrak q} \mathop{\mathrm{Spec}}(S) = 0$. This means that there is some open neighbourhood of $\mathfrak q$ in $\mathop{\mathrm{Spec}}(S)$ which has dimension zero. Then there is an open neighbourhood of the form $D(g)$ which has dimension zero. Since $S_ g$ is Noetherian we conclude that $S_ g$ is Artinian and $D(g) = \mathop{\mathrm{Spec}}(S_ g)$ is a finite discrete set, see Proposition 10.60.7. Thus $\mathfrak q$ is an isolated point of $D(g)$ and, by the equivalence of (1) and (2) above applied to $\mathfrak qS_ g \subset S_ g$, we see that $S_{\mathfrak q} = (S_ g)_{\mathfrak qS_ g}$ is finite over $k$. Hence (4) implies (2). It is clear that (1) implies (4). Thus (1) – (4) are all equivalent.

Lemma 10.114.6 gives the implication (5) $\Rightarrow $ (4). The implication (4) $\Rightarrow $ (6) follows from Lemma 10.116.3. The implication (6) $\Rightarrow $ (5) follows from Lemma 10.35.9. At this point we know (1) – (6) are equivalent.

The two statements at the end of the lemma we saw during the course of the proof of the equivalence of (1), (2) and (3) above.
$\square$

slogan
Lemma 10.122.2. Let $R \to S$ be a ring map of finite type. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p \subset R$. Let $F = \mathop{\mathrm{Spec}}(S \otimes _ R \kappa (\mathfrak p))$ be the fibre of $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$, see Remark 10.17.8. Denote $\overline{\mathfrak q} \in F$ the point corresponding to $\mathfrak q$. The following are equivalent

$\overline{\mathfrak q}$ is an isolated point of $F$,

$S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$ is finite over $\kappa (\mathfrak p)$,

there exists a $g \in S$, $g \not\in \mathfrak q$ such that the only prime of $D(g)$ mapping to $\mathfrak p$ is $\mathfrak q$,

$\dim _{\overline{\mathfrak q}}(F) = 0$,

$\overline{\mathfrak q}$ is a closed point of $F$ and $\dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) = 0$, and

the field extension $\kappa (\mathfrak p) \subset \kappa (\mathfrak q)$ is finite and $\dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) = 0$.

**Proof.**
Note that $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q} = (S \otimes _ R \kappa (\mathfrak p))_{\overline{\mathfrak q}}$. Moreover $S \otimes _ R \kappa (\mathfrak p)$ is of finite type over $\kappa (\mathfrak p)$. The conditions correspond exactly to the conditions of Lemma 10.122.1 for the $\kappa (\mathfrak p)$-algebra $S \otimes _ R \kappa (\mathfrak p)$ and the prime $\overline{\mathfrak q}$, hence they are equivalent.
$\square$

Definition 10.122.3. Let $R \to S$ be a finite type ring map. Let $\mathfrak q \subset S$ be a prime.

If the equivalent conditions of Lemma 10.122.2 are satisfied then we say $R \to S$ is *quasi-finite at $\mathfrak q$*.

We say a ring map $A \to B$ is *quasi-finite* if it is of finite type and quasi-finite at all primes of $B$.

Lemma 10.122.4. Let $R \to S$ be a finite type ring map. Then $R \to S$ is quasi-finite if and only if for all primes $\mathfrak p \subset R$ the fibre $S \otimes _ R \kappa (\mathfrak p)$ is finite over $\kappa (\mathfrak p)$.

**Proof.**
If the fibres are finite then the map is clearly quasi-finite. For the converse, note that $S \otimes _ R \kappa (\mathfrak p)$ is a $\kappa (\mathfrak p)$-algebra of finite type and of dimension $0$. Hence it is finite over $\kappa (\mathfrak p)$ for example by Lemma 10.115.4.
$\square$

Lemma 10.122.5. Let $R \to S$ be a finite type ring map. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p \subset R$. Let $f \in R$, $f \not\in \mathfrak p$ and $g \in S$, $g \not\in \mathfrak q$. Then $R \to S$ is quasi-finite at $\mathfrak q$ if and only if $R_ f \to S_{fg}$ is quasi-finite at $\mathfrak qS_{fg}$.

**Proof.**
The fibre of $\mathop{\mathrm{Spec}}(S_{fg}) \to \mathop{\mathrm{Spec}}(R_ f)$ is homeomorphic to an open subset of the fibre of $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$. Hence the lemma follows from part (1) of the equivalent conditions of Lemma 10.122.2.
$\square$

Lemma 10.122.6. Let

\[ \xymatrix{ S \ar[r] & S' & & \mathfrak q \ar@{-}[r] & \mathfrak q' \\ R \ar[u] \ar[r] & R' \ar[u] & & \mathfrak p \ar@{-}[r] \ar@{-}[u] & \mathfrak p' \ar@{-}[u] } \]

be a commutative diagram of rings with primes as indicated. Assume $R \to S$ of finite type, and $S \otimes _ R R' \to S'$ surjective. If $R \to S$ is quasi-finite at $\mathfrak q$, then $R' \to S'$ is quasi-finite at $\mathfrak q'$.

**Proof.**
Write $S \otimes _ R \kappa (\mathfrak p) = S_1 \times S_2$ with $S_1$ finite over $\kappa (\mathfrak p)$ and such that $\mathfrak q$ corresponds to a point of $S_1$ as in Lemma 10.122.1. This product decomposition induces a corresponding product decomposition for any $S \otimes _ R \kappa (\mathfrak p)$-algebra. In particular, we obtain $S' \otimes _{R'} \kappa (\mathfrak p') = S'_1 \times S'_2$. Because $S \otimes _ R R' \to S'$ is surjective the canonical map $(S \otimes _ R \kappa (\mathfrak p)) \otimes _{\kappa (\mathfrak p)} \kappa (\mathfrak p') \to S' \otimes _{R'} \kappa (\mathfrak p')$ is surjective and hence $S_ i \otimes _{\kappa (\mathfrak p)} \kappa (\mathfrak p') \to S'_ i$ is surjective. It follows that $S'_1$ is finite over $\kappa (\mathfrak p')$. The map $S' \otimes _{R'} \kappa (\mathfrak p') \to \kappa (\mathfrak q')$ factors through $S_1'$ (i.e. it annihilates the factor $S_2'$) because the map $S \otimes _ R \kappa (\mathfrak p) \to \kappa (\mathfrak q)$ factors through $S_1$ (i.e. it annihilates the factor $S_2$). Thus $\mathfrak q'$ corresponds to a point of $\mathop{\mathrm{Spec}}(S_1')$ in the disjoint union decomposition of the fibre: $\mathop{\mathrm{Spec}}(S' \otimes _{R'} \kappa (\mathfrak p')) = \mathop{\mathrm{Spec}}(S_1') \amalg \mathop{\mathrm{Spec}}(S_2')$, see Lemma 10.21.2. Since $S_1'$ is finite over a field, it is Artinian ring, and hence $\mathop{\mathrm{Spec}}(S_1')$ is a finite discrete set. (See Proposition 10.60.7.) We conclude $\mathfrak q'$ is isolated in its fibre as desired.
$\square$

Lemma 10.122.7. A composition of quasi-finite ring maps is quasi-finite.

**Proof.**
Suppose $A \to B$ and $B \to C$ are quasi-finite ring maps. By Lemma 10.6.2 we see that $A \to C$ is of finite type. Let $\mathfrak r \subset C$ be a prime of $C$ lying over $\mathfrak q \subset B$ and $\mathfrak p \subset A$. Since $A \to B$ and $B \to C$ are quasi-finite at $\mathfrak q$ and $\mathfrak r$ respectively, then there exist $b \in B$ and $c \in C$ such that $\mathfrak q$ is the only prime of $D(b)$ which maps to $\mathfrak p$ and similarly $\mathfrak r$ is the only prime of $D(c)$ which maps to $\mathfrak q$. If $c' \in C$ is the image of $b \in B$, then $\mathfrak r$ is the only prime of $D(cc')$ which maps to $\mathfrak p$. Therefore $A \to C$ is quasi-finite at $\mathfrak r$.
$\square$

Lemma 10.122.8. Let $R \to S$ be a ring map of finite type. Let $R \to R'$ be any ring map. Set $S' = R' \otimes _ R S$.

The set $\{ \mathfrak q' \mid R' \to S' \text{ quasi-finite at }\mathfrak q'\} $ is the inverse image of the corresponding set of $\mathop{\mathrm{Spec}}(S)$ under the canonical map $\mathop{\mathrm{Spec}}(S') \to \mathop{\mathrm{Spec}}(S)$.

If $\mathop{\mathrm{Spec}}(R') \to \mathop{\mathrm{Spec}}(R)$ is surjective, then $R \to S$ is quasi-finite if and only if $R' \to S'$ is quasi-finite.

Any base change of a quasi-finite ring map is quasi-finite.

**Proof.**
Let $\mathfrak p' \subset R'$ be a prime lying over $\mathfrak p \subset R$. Then the fibre ring $S' \otimes _{R'} \kappa (\mathfrak p')$ is the base change of the fibre ring $S \otimes _ R \kappa (\mathfrak p)$ by the field extension $\kappa (\mathfrak p) \to \kappa (\mathfrak p')$. Hence the first assertion follows from the invariance of dimension under field extension (Lemma 10.116.6) and Lemma 10.122.1. The stability of quasi-finite maps under base change follows from this and the stability of finite type property under base change. The second assertion follows since the assumption implies that given a prime $\mathfrak q \subset S$ we can find a prime $\mathfrak q' \subset S'$ lying over it.
$\square$

Lemma 10.122.9. Let $A \to B$ and $B \to C$ be ring homomorphisms such that $A \to C$ is of finite type. Let $\mathfrak r$ be a prime of $C$ lying over $\mathfrak q \subset B$ and $\mathfrak p \subset A$. If $A \to C$ is quasi-finite at $\mathfrak r$, then $B \to C$ is quasi-finite at $\mathfrak r$.

**Proof.**
Observe that $B \to C$ is of finite type (Lemma 10.6.2) so that the statement makes sense. Let us use characterization (3) of Lemma 10.122.2. If $A \to C$ is quasi-finite at $\mathfrak r$, then there exists some $c \in C$ such that

\[ \{ \mathfrak r' \subset C \text{ lying over }\mathfrak p\} \cap D(c) = \{ \mathfrak {r}\} . \]

Since the primes $\mathfrak r' \subset C$ lying over $\mathfrak q$ form a subset of the primes $\mathfrak r' \subset C$ lying over $\mathfrak p$ we conclude $B \to C$ is quasi-finite at $\mathfrak r$.
$\square$

The following lemma is not quite about quasi-finite ring maps, but it does not seem to fit anywhere else so well.

Lemma 10.122.10. Let $R \to S$ be a ring map of finite type. Let $\mathfrak p \subset R$ be a minimal prime. Assume that there are at most finitely many primes of $S$ lying over $\mathfrak p$. Then there exists a $g \in R$, $g \not\in \mathfrak p$ such that the ring map $R_ g \to S_ g$ is finite.

**Proof.**
Let $x_1, \ldots , x_ n$ be generators of $S$ over $R$. Since $\mathfrak p$ is a minimal prime we have that $\mathfrak pR_{\mathfrak p}$ is a locally nilpotent ideal, see Lemma 10.25.1. Hence $\mathfrak pS_{\mathfrak p}$ is a locally nilpotent ideal, see Lemma 10.32.3. By assumption the finite type $\kappa (\mathfrak p)$-algebra $S_{\mathfrak p}/\mathfrak pS_{\mathfrak p}$ has finitely many primes. Hence (for example by Lemmas 10.61.3 and 10.115.4) $\kappa (\mathfrak p) \to S_{\mathfrak p}/\mathfrak pS_{\mathfrak p}$ is a finite ring map. Thus we may find monic polynomials $P_ i \in R_{\mathfrak p}[X]$ such that $P_ i(x_ i)$ maps to zero in $S_{\mathfrak p}/\mathfrak pS_{\mathfrak p}$. By what we said above there exist $e_ i \geq 1$ such that $P(x_ i)^{e_ i} = 0$ in $S_{\mathfrak p}$. Let $g_1 \in R$, $g_1 \not\in \mathfrak p$ be an element such that $P_ i$ has coefficients in $R[1/g_1]$ for all $i$. Next, let $g_2 \in R$, $g_2 \not\in \mathfrak p$ be an element such that $P(x_ i)^{e_ i} = 0$ in $S_{g_1g_2}$. Setting $g = g_1g_2$ we win.
$\square$

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