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. $\square$

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