# The Stacks Project

## Tag: 01TH

This tag has label morphisms-lemma-quasi-finite-at-point-characterize and it points to

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Lemma 25.21.6. 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:
1. The morphism $f$ is quasi-finite at $x$.
2. The point $x$ is isolated in $X_s$.
3. The point $x$ is closed in $X_s$ and there is no point $x' \in X_s$, $x' \not = x$ which specializes to $x$.
4. For any pair of affine opens $\mathop{\rm Spec}(A) = U \subset X$, $\mathop{\rm Spec}(R) = V \subset S$ 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 25.17.10 (and Lemma 25.16.4). 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 $\mathop{\rm Spec}(A) = U \subset X$, $\mathop{\rm Spec}(R) = V \subset S$ 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 25.16.2 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 $\mathop{\rm Spec}(\overline{A})$ 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 $\mathop{\rm Spec}(\overline{A})$ specializing to $\overline{\mathfrak q}$. This implies that $\dim(\overline{A}_{\overline{q}}) = 0$. Hence by Algebra, Definition 9.117.3 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. $\square$

\begin{lemma}
\label{lemma-quasi-finite-at-point-characterize}
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:
\begin{enumerate}
\item The morphism $f$ is quasi-finite at $x$.
\item The point $x$ is isolated in $X_s$.
\item The point $x$ is closed in $X_s$
and there is no point $x' \in X_s$, $x' \not = x$
which specializes to $x$.
\item For any pair of affine opens
$\Spec(A) = U \subset X$, $\Spec(R) = V \subset S$ 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 $\mathfrak q$.
\end{enumerate}
\end{lemma}

\begin{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 $X_s$.

\medskip\noindent
Note that $X_s$ is a Jacobson scheme by
Lemma \ref{lemma-ubiquity-Jacobson-schemes}
(and
Lemma \ref{lemma-base-change-finite-type}).
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 $x$.

\medskip\noindent
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
$\Spec(A) = U \subset X$, $\Spec(R) = V \subset S$ 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 \ref{lemma-locally-finite-type-characterize} 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 $\Spec(\overline{A})$ 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 $\Spec(\overline{A})$
specializing to $\overline{\mathfrak q}$.
This implies that $\dim(\overline{A}_{\overline{q}}) = 0$.
Hence by
Algebra, Definition \ref{algebra-definition-quasi-finite}
we see that $R \to A$ is quasi-finite at $\mathfrak q$, i.e.,
$X \to S$ is quasi-finite at $x$ by definition.

\medskip\noindent
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.
\end{proof}


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