# The Stacks Project

## Tag 02NH

Lemma 28.19.10. Let $f : X \to S$ be a morphism of schemes. The following are equivalent:

1. $f$ is quasi-finite, and
2. $f$ 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 28.19.7. Hence it is quasi-finite by Lemma 28.19.9. $\square$

The code snippet corresponding to this tag is a part of the file morphisms.tex and is located in lines 3394–3402 (see updates for more information).

\begin{lemma}
\label{lemma-quasi-finite}
Let $f : X \to S$ be a morphism of schemes.
The following are equivalent:
\begin{enumerate}
\item $f$ is quasi-finite, and
\item $f$ is locally of finite type, quasi-compact, and has finite fibres.
\end{enumerate}
\end{lemma}

\begin{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.

\medskip\noindent
Conversely, assume $f$ is locally of finite type, quasi-compact
and has finite fibres. Then it is locally quasi-finite by
Lemma \ref{lemma-finite-fibre}. Hence it is quasi-finite by
Lemma \ref{lemma-quasi-finite-locally-quasi-compact}.
\end{proof}

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