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An elementary question of algebraic geometry which was posted on math.stackexchange but received no answer there.

Let $f: X \rightarrow Y$ be a morphism of schemes. Assume that

(i) $f: X \rightarrow Y$ is locally finite, in this sense: $Y$ can be covered by affine open sets $U_i=spec\ A_i$, and $f^{-1}(U_i)$ can be in turn covered by affine open sets $V_{i,j} = spec \ B_{i,j}$, in such a way that each $B_{i,j}$ is finitely generated as an $A_i$-module.

(ii) $f$ is quasi-compact.

Then does it imply that $f$ is a finite morphism?

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Dear Joel, If $f$ is not separated, then I think the answer is no; e.g. if we take the affine line with the doubled origin, then there is a natural morphism from this to the affine line (identifying the two origins) which is locally finite and q.c., but not finite (since not affine, indeed not separated). If $f$ is separated, then the answer is yes, see this answer: math.stackexchange.com/a/387260/221 Best wishes, Matt –  Emerton May 10 '13 at 4:45
    
Thank you Matt. I saw your proof on math.stackexchange, and it is nice and simple: I feel bad to have missed it. Just out of curiosity: is this result somewhere to be found in the literature? I had looked for it in Hartshorne (and I think it is not there) and in EGA (but then it should be there somewhere, and I must have missed it)... –  Joël May 10 '13 at 13:36
    
Dear Joel, I hadn't seen this before, but I wouldn't be surprised if it was hidden somewhere in EGA. (I'm not as familiar with the contents of EGA as I should be.) Best wishes, Matt –  Emerton May 10 '13 at 23:15
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1 Answer

up vote 4 down vote accepted

$\newcommand\Spec{\mathrm{Spec}}$

$ $

Since finiteness is local in the base, we may assume that $Y = \Spec(A)$ is affine.

The assumption then becomes that $f: X \rightarrow \Spec(A)$ is quasi-compact, and that $X$ may be covered by finitely many (by quasi-compactness) open affines $\Spec(B)$ which are finite over $A$.

Suppose that $f$ is an open immersion. We show that $X$ is closed and hence finite in $Y$. Since conditions (i) and (ii) are local on the base, we may assume that $Y = \Spec(A)$ is connected. Then, for each $\Spec(B)$ in our cover of $X$, $\Spec(B)$ is open in $X$ and hence $Y$, and yet finite and thus closed in $Y$. Since $Y$ is connected, it follows that either $\Spec(B) = X = Y$, or $X$ is the empty set.

Suppose that $f$ is separated. Then, by Zariski's main theorem, there exists a factorization $$X \rightarrow \Spec(A') = Y' \rightarrow \Spec(A) = Y$$ where the first map is an open immersion and the second is finite. It follows that for each $B$ there is a map $A \rightarrow A' \rightarrow B$ which makes $B$ finite over $A$. It follows that $B$ is also finite over $A'$. Hence, by the argument above, $X$ is finite over $Y'$, and thus $X \rightarrow Y$ is also finite.

Suppose, however, that $f$ is not separated. Then take $X$ to be the affine line with the origin doubled and $Y$ to be the affine line. Then the map $X \rightarrow Y$ is not finite because it is not even affine.

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I see Matt Emerton answered the question before I finished. Fortunately both answers are in agreement! His argument is a little different, but it does ultimately use the fact that quasi-finite and proper implies finite, which also comes down to Zariski's main theorem in the end. –  Sausage Roll May 10 '13 at 5:25
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Thanks, Sausage. By the Zariski's main theorem, you certainly mean the version given by Grothendieck for quasi-finite morphism. A remark: there is another proof (sketched in SGA1, exposé 1) of quasi-finite + proper implies finite, that does not use Zariski's main theorem but instead a faithfully flat descent argument to reduce to the case where the basis is a complete local ring. So your proof and Matt's may be seen as two completely different proof, which is nice. –  Joël May 10 '13 at 13:47
    
Dear Joel; yes, I meant the EGA version. I didn't know that argument for finiteness in SGA1, I'll take a look. Cheers, Sausage. –  Sausage Roll May 11 '13 at 1:02
    
Dear Sausage, The idea is that over the complete local ring, the quasi-finite scheme becomes the disjoint union of a part whose closed fibre is non-empty, which is then finite by a Nakayama-type argument, and a part lying just over the complement of the closed point, which is lower dimensional. By induction on dim'n that part can be put in something finite over the complement of the closed point, which can then be extended to something finite over the whole base (i.e. including the closed point) by normalization. (Hopefully I haven't gotten this too messed up.) The initial step, where ... –  Emerton May 11 '13 at 11:39
    
... you make the decomposition, is the key one, and is where completeness is used. It comes up in studying torsion in ab. varieties with bad reduction: when you break up the $n$-torsion into a finite part over $\mathbb Z_p$ and a part which lives just over $\mathbb Q_p$ (as occurs from time to time in Mazur, Ribet, etc.) you are using this same SGA 1 techinque. (Although the citations in that case, if there are any (!) , are probably to SGA 7 , where ab. varieties, Neron models, etc. are discussed.) Cheers, Matt –  Emerton May 11 '13 at 11:41
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