Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Does anybody know a reference for a proof of:

Let $f: X \rightarrow Y$ be a quasi-finite proper morphism of varieties. Then $f$ is finite.

Is there one in Hartshorne? I could not find it.


share|cite|improve this question
Your question is only the "second thing" in the title. – user18119 Jan 29 '13 at 15:47
An overkill proof via ZMT: By the Grothendieck version of Zariski's main theorem, $f$ is the composition of an open embedding $g: X\rightarrow Z$ and a finite morphism $h:Z\rightarrow Y$. Then, $g$ is the composition of the graph morphism $X \rightarrow X\times_Y Z$ and the projection $X\times_Y Z \rightarrow Z.$ The first is a base-change of the morphism $Z \rightarrow Z\times_Y Z$, which is a closed embedding and thus proper, and the second is a base change of $f$, and thus proper. Therefore, $g$ is proper and thus an isomorphism (as the image must be both closed and open.) – Nehsb Jan 29 '13 at 17:29
@QiL, oops, found a reference for the first thing while typing and forgot to change the title, thanks. – Joachim Jan 29 '13 at 17:36
@Nehsb, why don't you post this as an answer? – Joachim Jan 29 '13 at 17:37
up vote 4 down vote accepted

The proof is in ÉGA IV 3. It is in two steps.

Step I : reduce to the case of $\mathrm Y = \mathrm{Spec}(\mathrm A)$ where $\mathrm A$ is a complete noetherian local ring.

Step II : quasi-finite $\mathfrak m_{\mathrm A}$-seperated modules are finite for such an $\mathrm A$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.