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10.121. Quasi-finite maps

Consider a ring map $R \to S$ of finite type. A map $\mathop{\rm Spec}(S) \to \mathop{\rm Spec}(R)$ 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 10.121.1. 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:

  1. $\mathfrak q$ is an isolated point of $\mathop{\rm Spec}(S)$,
  2. $S_{\mathfrak q}$ is finite over $k$,
  3. there exists a $g \in S$, $g \not\in \mathfrak q$ such that $D(g) = \{ \mathfrak q \}$,
  4. $\dim_{\mathfrak q} \mathop{\rm Spec}(S) = 0$,
  5. $\mathfrak q$ is a closed point of $\mathop{\rm Spec}(S)$ and $\dim(S_{\mathfrak q}) = 0$, and
  6. 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 $\mathop{\rm Spec}(S)$, i.e., $\{\mathfrak q\}$ is open in $\mathop{\rm Spec}(S)$. Because $\mathop{\rm Spec}(S)$ is a Jacobson space (see Lemmas 10.34.2 and 10.34.4) we see that $\mathfrak q$ is a closed point. Hence $\{\mathfrak q\}$ is open and closed in $\mathop{\rm Spec}(S)$. By Lemmas 10.20.3 and 10.22.3 we may write $S = S_1 \times S_2$ with $\mathfrak q$ corresponding to the only point $\mathop{\rm Spec}(S_1)$. 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 10.114.4. We have proved (1) implies (2).

Suppose $S_{\mathfrak q}$ is finite over $k$. Then $S_{\mathfrak q}$ is Artinian local, see Lemma 10.52.2. So $\mathop{\rm Spec}(S_{\mathfrak q}) = \{\mathfrak qS_{\mathfrak q}\}$ by Lemma 10.52.6. 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 $S$-modules 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 $\mathop{\rm Spec}(S)$ we see that $\mathfrak q$ is an isolated point of $\mathop{\rm Spec}(S)$. We have proved that (3) implies (1). In other words (1), (2) and (3) are equivalent.

Assume $\dim_{\mathfrak q} \mathop{\rm Spec}(S) = 0$. This means that there is some open neighbourhood of $\mathfrak q$ in $\mathop{\rm Spec}(S)$ 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 $D(g) = \mathop{\rm Spec}(S_g)$ is a finite discrete set, see Proposition 10.59.6. 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 10.113.6 gives the implication (5) $\Rightarrow$ (4). The implication (4) $\Rightarrow$ (6) follows from Lemma 10.115.3. The implication (6) $\Rightarrow$ (5) follows from Lemma 10.34.9. 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. $\square$

Lemma 10.121.2. 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 $F = \mathop{\rm Spec}(S \otimes_R \kappa(\mathfrak p))$ be the fibre of $\mathop{\rm Spec}(S) \to \mathop{\rm Spec}(R)$, see Remark 10.16.8. Denote $\overline{\mathfrak q} \in F$ the point corresponding to $\mathfrak q$. The following are equivalent

  1. $\overline{\mathfrak q}$ is an isolated point of $F$,
  2. $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$ is finite over $\kappa(\mathfrak p)$,
  3. there exists a $g \in S$, $g \not \in \mathfrak q$ such that the only prime of $D(g)$ mapping to $\mathfrak p$ is $\mathfrak q$,
  4. $\dim_{\overline{\mathfrak q}}(F) = 0$,
  5. $\overline{\mathfrak q}$ is a closed point of $F$ and $\dim(S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) = 0$, and
  6. 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 10.121.1 for the $\kappa(\mathfrak p)$-algebra $S \otimes_R \kappa(\mathfrak p)$ and the prime $\overline{\mathfrak q}$, hence they are equivalent. $\square$

Definition 10.121.3. Let $R \to S$ be a finite type ring map. Let $\mathfrak q \subset S$ be a prime.

  1. If the equivalent conditions of Lemma 10.121.2 are satisfied then we say $R \to S$ is quasi-finite at $\mathfrak q$.
  2. 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 10.121.4. 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 over $k$ of dimension $0$. Hence it is finite over $k$ for example by Lemma 10.114.4. $\square$

Lemma 10.121.5. 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 $\mathop{\rm Spec}(S_{fg}) \to \mathop{\rm Spec}(R_f)$ is homeomorphic to an open subset of the fibre of $\mathop{\rm Spec}(S) \to \mathop{\rm Spec}(R)$. Hence the lemma follows from part (1) of the equivalent conditions of Lemma 10.121.2. $\square$

Lemma 10.121.6. 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 10.121.1. Because $S \otimes_R R' \to S'$ 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. Let $S_i'$ be the image of $S_i \otimes_{\kappa(\mathfrak p)} \kappa(\mathfrak p')$ in $S' \otimes_{R'} \kappa(\mathfrak p')$. Then $S' \otimes_{R'} \kappa(\mathfrak p') =S'_1 \times S'_2$ and $S'_1$ 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 $\mathop{\rm Spec}(S_1')$ in the disjoint union decomposition of the fibre: $\mathop{\rm Spec}(S' \otimes_{R'} \kappa(\mathfrak p')) = \mathop{\rm Spec}(S_1') \amalg \mathop{\rm Spec}(S_1')$. (See Lemma 10.20.2.) Since $S_1'$ is finite over a field, it is Artinian ring, and hence $\mathop{\rm Spec}(S_1')$ is a finite discrete set. (See Proposition 10.59.6.) We conclude $\mathfrak q'$ is isolated in its fibre as desired. $\square$

Lemma 10.121.7. 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 10.121.8. Let $R \to S$ be a ring map of finite type. Let $R \to R'$ be any ring map. Set $S' = R' \otimes_R S$.

  1. The set $\{\mathfrak q' \mid R' \to S' \text{ quasi-finite at }\mathfrak q'\}$ is the inverse image of the corresponding set of $\mathop{\rm Spec}(S)$ under the canonical map $\mathop{\rm Spec}(S') \to \mathop{\rm Spec}(S)$.
  2. If $\mathop{\rm Spec}(R') \to \mathop{\rm Spec}(R)$ is surjective, then $R \to S$ is quasi-finite if and only if $R' \to S'$ is quasi-finite.
  3. 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 10.115.6) and Lemma 10.121.1. 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 10.121.9. Let $A \to B$ and $B \to C$ be finite type ring homomorphisms. 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. Using property (3) of Lemma 10.121.2: By assumption 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. $\square$

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

Lemma 10.121.10. 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 10.24.1. Hence $\mathfrak pS_{\mathfrak p}$ is a locally nilpotent ideal, see Lemma 10.31.2. 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 10.60.3 and 10.114.4) $\kappa(\mathfrak p) \to S_{\mathfrak p}/\mathfrak pS_{\mathfrak p}$ 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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    \section{Quasi-finite maps}
    \label{section-quasi-finite}
    
    \noindent
    Consider a ring map $R \to S$ of finite type.
    A map $\Spec(S) \to \Spec(R)$ 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.
    
    \begin{lemma}
    \label{lemma-isolated-point}
    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:
    \begin{enumerate}
    \item $\mathfrak q$ is an isolated point of $\Spec(S)$,
    \item $S_{\mathfrak q}$ is finite over $k$,
    \item there exists a $g \in S$, $g \not\in \mathfrak q$ such that
    $D(g) = \{ \mathfrak q \}$,
    \item $\dim_{\mathfrak q} \Spec(S) = 0$,
    \item $\mathfrak q$ is a closed point of $\Spec(S)$ and
    $\dim(S_{\mathfrak q}) = 0$, and
    \item the field extension $k \subset \kappa(\mathfrak q)$ is finite
    and $\dim(S_{\mathfrak q}) = 0$.
    \end{enumerate}
    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 $S_{\mathfrak q} = S_g$.
    \end{lemma}
    
    \begin{proof}
    Suppose $\mathfrak q$ is an isolated point of $\Spec(S)$, i.e.,
    $\{\mathfrak q\}$ is open in $\Spec(S)$.
    Because $\Spec(S)$ is a Jacobson space (see
    Lemmas \ref{lemma-finite-type-field-Jacobson} and
    \ref{lemma-jacobson})
    we see that $\mathfrak q$ is a closed point. Hence
    $\{\mathfrak q\}$ is open and closed in $\Spec(S)$.
    By
    Lemmas \ref{lemma-disjoint-decomposition} and
    \ref{lemma-disjoint-implies-product} we may
    write $S = S_1 \times S_2$ with $\mathfrak q$
    corresponding to the only point $\Spec(S_1)$.
    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 \ref{lemma-Noether-normalization}.
    We have proved (1) implies (2).
    
    \medskip\noindent
    Suppose $S_{\mathfrak q}$ is finite over $k$.
    Then $S_{\mathfrak q}$ is Artinian local, see
    Lemma \ref{lemma-finite-dimensional-algebra}. So
    $\Spec(S_{\mathfrak q}) = \{\mathfrak qS_{\mathfrak q}\}$ by
    Lemma \ref{lemma-artinian-finite-length}.
    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 $S$-modules 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).
    
    \medskip\noindent
    Suppose $D(g) =  \{ \mathfrak q \}$. Since $D(g)$ is open by
    construction of the topology on $\Spec(S)$ we see that
    $\mathfrak q$ is an isolated point of $\Spec(S)$.
    We have proved that (3) implies (1).
    In other words (1), (2) and (3) are equivalent.
    
    \medskip\noindent
    Assume $\dim_{\mathfrak q} \Spec(S) = 0$. This means that
    there is some open neighbourhood of $\mathfrak q$ in $\Spec(S)$
    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
    $D(g) = \Spec(S_g)$ is a finite discrete set, see
    Proposition \ref{proposition-dimension-zero-ring}.
    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.
    
    \medskip\noindent
    Lemma \ref{lemma-dimension-closed-point-finite-type-field}
    gives the implication (5) $\Rightarrow$ (4).
    The implication (4) $\Rightarrow$ (6) follows from
    Lemma \ref{lemma-dimension-at-a-point-finite-type-field}.
    The implication (6) $\Rightarrow$ (5) follows from
    Lemma \ref{lemma-finite-residue-extension-closed}.
    At this point we know (1) -- (6) are equivalent.
    
    \medskip\noindent
    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.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-isolated-point-fibre}
    \begin{slogan}
    Equivalent conditions for isolated points in fibres
    \end{slogan}
    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 $F = \Spec(S \otimes_R \kappa(\mathfrak p))$
    be the fibre of $\Spec(S) \to \Spec(R)$, see
    Remark \ref{remark-fundamental-diagram}.
    Denote $\overline{\mathfrak q} \in F$ the point corresponding to
    $\mathfrak q$. The following are equivalent
    \begin{enumerate}
    \item $\overline{\mathfrak q}$ is an isolated point of $F$,
    \item $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$ is finite over
    $\kappa(\mathfrak p)$,
    \item there exists a $g \in S$, $g \not \in \mathfrak q$ such that
    the only prime of $D(g)$ mapping to $\mathfrak p$ is $\mathfrak q$,
    \item $\dim_{\overline{\mathfrak q}}(F) = 0$,
    \item $\overline{\mathfrak q}$ is a closed point of $F$ and
    $\dim(S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) = 0$, and
    \item the field extension $\kappa(\mathfrak p) \subset \kappa(\mathfrak q)$
    is finite and $\dim(S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}) = 0$.
    \end{enumerate}
    \end{lemma}
    
    \begin{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 \ref{lemma-isolated-point}
    for the $\kappa(\mathfrak p)$-algebra $S \otimes_R \kappa(\mathfrak p)$
    and the prime $\overline{\mathfrak q}$, hence they are equivalent.
    \end{proof}
    
    \begin{definition}
    \label{definition-quasi-finite}
    Let $R \to S$ be a finite type ring map.
    Let $\mathfrak q \subset S$ be a prime.
    \begin{enumerate}
    \item If the equivalent conditions of Lemma \ref{lemma-isolated-point-fibre}
    are satisfied then we say $R \to S$ is {\it quasi-finite at $\mathfrak q$}.
    \item We say a ring map $A \to B$ is {\it quasi-finite}
    if it is of finite type and quasi-finite at all primes of $B$.
    \end{enumerate}
    \end{definition}
    
    \begin{lemma}
    \label{lemma-quasi-finite}
    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 $\kappa(\mathfrak p)$.
    \end{lemma}
    
    \begin{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 over
    $k$ of dimension $0$. Hence it is finite over $k$ for example
    by Lemma \ref{lemma-Noether-normalization}.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-quasi-finite-local}
    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 $\mathfrak qS_{fg}$.
    \end{lemma}
    
    \begin{proof}
    The fibre of $\Spec(S_{fg}) \to \Spec(R_f)$ is homeomorphic
    to an open subset of the fibre of $\Spec(S) \to \Spec(R)$.
    Hence the lemma follows from part (1) of the equivalent conditions of
    Lemma \ref{lemma-isolated-point-fibre}.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-four-rings}
    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 $\mathfrak q'$.
    \end{lemma}
    
    \begin{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 \ref{lemma-isolated-point}.
    Because $S \otimes_R R' \to S'$ 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. Let $S_i'$ be the image of $S_i \otimes_{\kappa(\mathfrak p)}
    \kappa(\mathfrak p')$ in $S' \otimes_{R'} \kappa(\mathfrak p')$.
    Then $S' \otimes_{R'} \kappa(\mathfrak p') =S'_1 \times
    S'_2$ and $S'_1$ 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
    $\Spec(S_1')$ in the disjoint union decomposition
    of the fibre: $\Spec(S' \otimes_{R'} \kappa(\mathfrak p'))
    = \Spec(S_1') \amalg \Spec(S_1')$. (See
    Lemma \ref{lemma-spec-product}.)
    Since $S_1'$ is finite over a field, it is Artinian ring,
    and hence $\Spec(S_1')$ is a finite discrete set.
    (See Proposition \ref{proposition-dimension-zero-ring}.)
    We conclude $\mathfrak q'$ is isolated in its fibre as
    desired.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-quasi-finite-composition}
    A composition of quasi-finite ring maps is quasi-finite.
    \end{lemma}
    
    \begin{proof}
    Suppose $A \to B$ and $B \to C$ are quasi-finite ring maps. By
    Lemma \ref{lemma-compose-finite-type}
    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 $\mathfrak r$.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-quasi-finite-base-change}
    Let $R \to S$ be a ring map of finite type.
    Let $R \to R'$ be any ring map. Set $S' = R' \otimes_R S$.
    \begin{enumerate}
    \item The set
    $\{\mathfrak q' \mid R' \to S' \text{ quasi-finite at }\mathfrak q'\}$
    is the inverse image of the corresponding set of $\Spec(S)$
    under the canonical map $\Spec(S') \to \Spec(S)$.
    \item If $\Spec(R') \to \Spec(R)$ is surjective,
    then $R \to S$ is quasi-finite if and only if $R' \to S'$ is quasi-finite.
    \item Any base change of a quasi-finite ring map is quasi-finite.
    \end{enumerate}
    \end{lemma}
    
    \begin{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 \ref{lemma-dimension-at-a-point-preserved-field-extension})
    and Lemma \ref{lemma-isolated-point}.
    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.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-quasi-finite-permanence}
    Let $A \to B$ and $B \to C$ be finite type ring homomorphisms.
    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 $\mathfrak r$.
    \end{lemma}
    
    \begin{proof}
    Using property (3) of Lemma \ref{lemma-isolated-point-fibre}:
    By assumption 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.
    \end{proof}
    
    \noindent
    The following lemma is not quite about quasi-finite ring maps, but
    it does not seem to fit anywhere else so well.
    
    \begin{lemma}
    \label{lemma-generically-finite}
    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.
    \end{lemma}
    
    \begin{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 \ref{lemma-minimal-prime-reduced-ring}.
    Hence $\mathfrak pS_{\mathfrak p}$ is a locally nilpotent ideal, see
    Lemma \ref{lemma-locally-nilpotent}.
    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 \ref{lemma-finite-type-algebra-finite-nr-primes} and
    \ref{lemma-Noether-normalization})
    $\kappa(\mathfrak p) \to S_{\mathfrak p}/\mathfrak pS_{\mathfrak p}$
    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.
    \end{proof}

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