The Stacks project

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. $\square$

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. $\square$

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. $\square$

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. $\square$

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. $\square$

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 $\mathfrak r$. $\square$

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. $\square$

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

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 $\mathfrak r$. $\square$

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. $\square$