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I hope this question is not too simple:

Let $F,E,B$ complex algebraic varieties such that there exists a fiber bundle $$F\to E\to B$$ where all morphisms are assumed to be algebraic.

Question:

If $F$ and $B$ are projective (resp. quasi-projective) is then total space $E$ also projective (resp. quasi-projective)

I think both statements are wrong but I can't construct counterexamples.

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  • $\begingroup$ What's your definition of a fiber bundle? $\endgroup$
    – Angelo
    Commented Apr 13, 2013 at 8:49
  • $\begingroup$ I would say locally trivial in the Zariski topology! $\endgroup$
    – Oliver Straser
    Commented Apr 13, 2013 at 9:18
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    $\begingroup$ Locally trivial meaning locally a product? This is an extremely restrictive notion; usually one uses the étale topology. Also, if you want an analogue of the topological notion of fiber bundle even having isomorphic fibers may be too strong. For example, a smooth projective morphism is topologically a fiber bundle, but does not have isomorphic fibers in general. $\endgroup$
    – Angelo
    Commented Apr 13, 2013 at 9:42
  • $\begingroup$ I think Hartshorne Proposition 7.10. of Chapter II is related your question $\endgroup$
    – Xiaobo Zhuang
    Commented Apr 13, 2013 at 9:59
  • $\begingroup$ Ok, i am sorry but maybe my question was to naive. First i really thought about the obvious (naive) definition of a fiber bundle in the Zariski topology, of course if there are nice results in the form of: We have an algebraic morphism $E\to B$ which is a fiber bundle in the etale topology. If $F$ and $B$ are projective then is $ E$ (for maybe a very restrictive class of varieties $B$ and $F$) then so is $E$ I would be glad to hear about it. @ZhuangXiaobo thank you for the reference! – O. Straser 3 hours ago $\endgroup$
    – Oliver Straser
    Commented Apr 13, 2013 at 14:32

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