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Suppose I have a quasiprojective variety $X$ and a finite surjective map $$f: X \rightarrow Y$$ to a scheme $Y$. Is it true that $Y$ is quasiprojective as well? It seems like the answer could be no, but I don't know enough examples of non-projective schemes.

flag	asked Feb 28 2012 at 22:49 anon 145●1●1●3

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No. One classic example of a non-projective 3-fold involves contracting a bunch of rational curves in a projective 3-fold. This gives a finite morphism $X\to Y$, with $X$ projective. Since $Y$ is complete but not projective, it is not quasi-projective. I'll see if I can dig it up. – Will Sawin Feb 28 2012 at 23:01

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When you contract curves, the resulting morphism is not finite. – Angelo Feb 29 2012 at 5:46

Karl Schwede · Accepted Answer · 2012-02-28 23:10:24Z UTC

This isn't true in general.

For example, see Section 6 of Conducteur, Descente et Pincement by D. Ferrand. There Ferrand gives an example of a non-normal proper variety $Y$ whose normalization is projective.

If I recall correctly, many examples of proper non-projective schemes have finite maps from projective ones.

Olivier Benoist · Answer 2 · 2012-02-29 17:42:59Z UTC

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It is true if $Y$ is normal. Let me explain why.

I will say that a variety has the Chevalley-Kleiman (CK) property if every finite subset is contained in an affine open. By Corollary 48 of Kollar's "Quotients by finite equivalence relations" arXiv:0812.3608, if $f:X\to Y$ is finite and surjective, then if $X$ has the CK property, $Y$ has it too.

Now, it is clear that a quasi-projective variety has the CK property, and a normal variety with the CK property is quasi-projective by Corollary 2 of "Quasi-projectivity of normal varieties" arXiv:1112.0975.

Putting this together, we see that if $f:X\to Y$ is a finite surjective map of varieties with $X$ quasi-projective and $Y$ normal, then $Y$ is quasi-projective.

answered Feb 29 2012 at 17:42

Olivier Benoist
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I just want to say wow, that's really interesting. – Karl Schwede Feb 29 2012 at 19:54

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I agree. Although there is more elementary way to see this by a nice and easy construction in EGA2, 6.5: To simplify things I assume that all schemes are of finite type over some noetherian ring $R$. In EGA2 it is explained that for a finite morphism of schemes $f: X \to Y$ you can construct a group homomorphism $N_{X/Y}: {\rm Pic}(X) \to {\rm Pic}(Y)$ if $f$ is flat of if $Y$ is normal. Moreover in EGA2, 5.6, it is then shown that if $L$ is an ample line bundle on $X$, then $N_{X/Y}(L)$ is an ample line bundle on $Y$. This shows that $Y$ is quasi-projective if $f$ is flat or $Y$ is normal. – Torsten Wedhorn Mar 1 2012 at 5:18

You are right : this is much simpler. Moreover, it constructs an explicit ample line bundle on $Y$ (I don't think it is possible to get one with the arguments I gave). On the other hand, Kollar's result says something even if $Y$ is not normal. For example, if $Y$ were only known to be an algebraic space, then it is automatically a scheme. – Olivier Benoist Mar 1 2012 at 9:27

Finite map from quasi-projective variety

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