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Let $f:X \to Y$ be a flat morphism, such that each fiber is isomorphic to the affine space $\mathbb{A}^n$. Then is is true that $f$ is a Zariski affine bundle? If not, is it at least an ├Ętale affine bundle?

If not, is there some reasonable natural ipotesis to add to make this true?

My definition of affine bundle is a map $f:X \to Y$ such that there is an open cover ${U_{\alpha}}$ of $Y$ with $f^{-1} U_{\alpha} \cong U_{\alpha} \times \mathbb{A}^n$, and $f$ restricted to $f^{-1}(U_{\alpha})$ corresponds to the first projection.

The question is somewhat related to: Affine bundles over varieties

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Did you check Angelo's answer to mathoverflow.net/questions/58009/non-locally-trivial-an-bundles ? It seems to be very related to your question –  Rami Feb 2 '13 at 23:07
    
If instad of the affine space you would have the protective one, than I think that the answer s positive. If you are interested I can try to write a proof –  Rami Feb 3 '13 at 2:10
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@Rami: In the case of projective spaces, it is only true for étale localisation (not Zariski). Such fibrations are called Severi-Brauer schemes. –  Damian Rössler Feb 3 '13 at 14:27
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...although: what I said in my comment applies if the fibres are geometrically projective spaces. Maybe what you say works if the fibres are actually projective spaces. –  Damian Rössler Feb 3 '13 at 14:34
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@Damian: no, this is false even if fibers are actually projective spaces (over an algebraically closed field). A standard example: if $X\subset\mathbb P^4$ is a smooth cubic, then, projecting it from a line $\ell\subset X$ to $\mathbb P^2$, we obtain a morphism $\tilde X\to\mathbb P^2$ (where $\tilde X$ is the blow-up of $X$ at $\ell$). Over a Zariski open subset of $\mathbb P^2$, all its fibers are smooth plane conics (so they are isomorphic to $\mathbb P^1$). If it were a locally trivial bundle in Zariski topology, $X$ would be rational, which is not the case. –  Serge Lvovski Feb 3 '13 at 16:19

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