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Say we are given a smooth complex algebraic variety $Y$ which is quasi-projective, and $X$ a second complex manifold together with a holomorphic map $f:X\rightarrow Y$ which is of finite fibers, is it true that $X$ remains quasi-projective? Is the condition of smoothness essential, namely can we do the same for analytic maps from a complex analytic space $X$ to a quasi-projective variety $Y$? I consider a very ample line bundle $L$ on $Y$ that realizes an embedding of $Y$ into some projective space, but I do not see how $f^*L$ should be expected to do the same thing.

Many thanks!

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If you assume that $X \to Y$ is finite, with $X$ a complex analytic space, then $Y$ a quasi-projective variety implies that $X$ is. This is a general version of Riemann's existence theorem, maybe due to Grauert and Remmert when $X$ and $Y$ are normal and to Grothendieck in general (perhaps with additional assumption that $X \to Y$ is unramified). (But I'm not sure if this is the correct attribution, and you should also probably look in the literature for the most general precise statements.)

If you just assume finite fibres, there is not much to say without making more restrictions. E.g. suppose that $X$ is an open subset of $Y$ (in the complex topology). Then the fibres of $X \to Y$ are finite (either singletons or empty), but $X$ won't be quasi-projective unless it is actually Zariski open in $Y$.

Since $X$ is smooth when $Y$ is, this shows that smoothness is not really the issue here.


Here are some speculations:

I don't know how the details would go, but I imagine that you can stratify $Y$ according to the nature of the fibres of the map $X \to Y$ (i.e. consider the strata along which the fibres of $X \to Y$ are of some fixed cardinality, counted with multiplicity). My guess would then be that $X$ is quasi-projective if and only this stratification is an algebraic stratification of $Y$. (The idea is that $X \to Y$ will be finite when restricted to the strata, so Riemann existence would apply. The details could be tricky, though, and my guess might be a little too naive.)

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thanks a lot!. Do you know where I should go to for this general form of Riemann's existence theorem? –  turtle May 23 '12 at 18:48
    
@turtle: Dear turtle, I just typed "Riemann existence theorem Grauert" into google, and got many relevant links. Did you try that? Also, you can look in Grothendieck's Bourbaki seminar on flat descent (I think it is the first of the FGAs), or one of the Grauert--Remmert books on complex analytic geometry. But I would try googling first. Regards, –  Matt E May 24 '12 at 11:52
    
@turtle: Dear turtle, Also, my googling suggests that perhaps my assertion is a little too strong. Maybe if one doesn't assume $X$ and $Y$ are normal, then one has to assume that the map is not just finite, but also unramified. But you should probably do a careful literature search to get the bottom of the most precise general form that is true. Regards, –  Matt E May 24 '12 at 11:56
    
thanks a great deal for your reply! –  turtle May 29 '12 at 18:46
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The answer to your question is yes. If the map $f: X \rightarrow Y$ is quasi-finite, i.e. it has finite fibers, and $Y$ is quasi-projective, then $X$ must be quasi-projective. By Stein factorization, $f$ factors into an open immersion followed by a finite morphism, so it suffices to treat the case when $f$ is finite, since an open subset of a quasi-projective is again quasi-projective. Furthermore, by composing with an embedding of $Y$ into a projective compactification $\bar{Y}$ you may assume that $Y$ is projective.

In this case, where $f: X \rightarrow Y$ is a finite map with $Y$ projective, take an ample line bundle $L$ on $Y$ and show that $f^*L$ is ample on $X$. This is pretty easy, e.g. Seshadri's criterion.

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Dear Parsa, $X$ is assumed to be merely a complex manifold. Any (complex) open subset of $Y$ is a complex manifold if $Y$ is smooth, and most of these are not quasi-projective varieties (because they are typically not Zariski open). Regards, –  Matt E May 29 '12 at 23:00
    
@MattE Yes, I was working in the algebraic category. –  Parsa May 31 '12 at 15:30
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