Let $X$ be an algebraic variety over an alg. closed field with zero char. and let $f:X\to \mathbb{A}^n$ be a smooth surjective morphism, such that all fibers (at closed points) are isomorphic to $\mathbb{A}^m$. Does it follow that $X\cong \mathbb{A}^{n+m}$?

If not, is it true with some additional assumptions? I know that every vector bundle on $\mathbb{A}^n$ is trivial (this is Serre's problem, right?) and that it is even enough to ask it to be locally trivial in etale toplogy. Is every "affine bundle" on $\mathbb{A}^n$ is trivial? I guess it is. The main question is about a general "fibration".

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    $\begingroup$ I suggest you ask this excellent but difficult question on MathOverflow: according to me (but I might be wrong) the specialists here who could answer this question are also active on MathOverflow, but not conversely. Anyway, +1. $\endgroup$ Jul 7, 2013 at 14:09
  • $\begingroup$ @GeorgesElencwajg MO is now part of the stackexchange network, so if you think it is appropriate you can now flag the post to get a moderator to migrate the question. $\endgroup$
    – Ragib Zaman
    Jul 7, 2013 at 14:36
  • $\begingroup$ Dear @Ragib, thanks a lot for the information but at the risk of seeming old-fashioned I prefer not to force the migration and let Kotel make the decision. $\endgroup$ Jul 7, 2013 at 14:49
  • $\begingroup$ @GeorgesElencwajg Not old fashioned at all. I realize now that I should have worded my previous comment differently, as it was as much directed towards KotelKanim as it was to you. My point is simply that instead of having to recreate this question again, OP can just request a moderator to move the question. $\endgroup$
    – Ragib Zaman
    Jul 7, 2013 at 14:53
  • $\begingroup$ Dear Georges and Ragib, I have no objection to Georges suggestion, it sounds like a good idea. I flaged the question for moderator attention. Thank you for your help. $\endgroup$
    – KotelKanim
    Jul 7, 2013 at 14:56

1 Answer 1


I feel like I already answered this question, but it might have been a variant with fibers isomorphic to tori. Let the base $B$ be $\mathbb{A}^2$ with coordinates $s$ and $t$. Begin with $B\times \mathbb{P}^3$, where homogeneous coordinates on $\mathbb{P}^3$ are $[x,y,z,w]$. Let $S$ be the Cartier divisor in $B\times \mathbb{P}^3$ with defining equation $yz-(sx+tw)^2=0$. Let $L$ be the Cartier divisor in $S$ with defining equation $y+z-2(sx+tw)=0$. Let $U$ be the complement of $L$ in $S$. Then $U$ is affine, the morphism $U\to B$ is smooth, and the fiber over every point other than $(0,0)$ is isomorphic to $\mathbb{A}^2$. Of course the fiber over $(0,0)$ is isomorphic to a disjoint union of two copies of $\mathbb{A}^2$. Thus, define $V\subset U$ to be the open subscheme obtained by removing one of these two copies of $\mathbb{A}^2$, i.e., remove the closed subscheme with defining equations $s=t=z=0$. Then $V$ is quasi-affine, and the affine hull is $U$; this follows by Hartog's theorem / the Riemann extension theorem / S2 extension. Therefore $V$ is not isomorphic to an affine space. However, the projection $V\to B$ has all of the requisite properties.

Edit. The older answer I mention above was similar, but a little bit different. That answer was in response to the following similar question, When is a holomorphic submersion with isomorphic fibers locally trivial?.

  • $\begingroup$ Great answer, Jason, and quite counterintuitive result. Amusingly, I too felt like I had asked a similar question and didn't remember exactly the details! $\endgroup$ Jul 8, 2013 at 6:11
  • $\begingroup$ Dear Jason, just to test my understanding of your construction: independently of what the affine hull of $V$ is (I don't know this concept), could we conclude by saying that the complement of a codimension two closed subvariety of an affine variety is never affine? $\endgroup$ Jul 8, 2013 at 9:15
  • $\begingroup$ @GeorgesElencwajg: Dear Georges, By the affine hull of $X$, I think Jason means Spec $\Gamma(X,\mathcal O_X)$. And it is true (in pretty general circumstances) that the complement of a dense affine open is of codimension one, so I think your alternative argument also works. Regards, $\endgroup$
    – Emerton
    Jul 8, 2013 at 12:21
  • $\begingroup$ @Emerton. Dear Matt: thanks a lot for your explanation. And it sure is very reassuring to have an endorsement from you ! $\endgroup$ Jul 8, 2013 at 12:29

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