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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 ètale 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 ? 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
@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
...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
@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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