MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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

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.

share|cite|improve this question
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 '12 at 23:01
When you contract curves, the resulting morphism is not finite. – Angelo Feb 29 '12 at 5:46
up vote 12 down vote accepted

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.

share|cite|improve this answer

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.

share|cite|improve this answer
I just want to say wow, that's really interesting. – Karl Schwede Feb 29 '12 at 19:54
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 '12 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 '12 at 9:27

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.