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Let $f: X \to Y$ be a morphism of varieties such that its fibres are isomorphic to $\mathbb{A}^n$. Since the definition of a vector bundle stipulates that $f$ be locally the projection $U \times \mathbb{A}^n \to U$, it is likely that there exist morphisms that are not locally of that form, but I can't come up with an example.

So the question is: what is an example of a morphism with fibres $\mathbb{A}^n$ that is not locally trivial? not locally isotrivial?

UPDATE: what if one assumes vector space structure on the fibres?

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Most of the non-locally trivial vector bundles I've seen show up are not locally trivial for 'another' reason: the rank is not constant. One gets them as kernels of maps between bundles: for example, the map $(x,v)\in\mathbb C\times\mathbb X\to(x,xv)\in\mathbb C\times\mathbb X$, viewed as an endomorphism of the trivial line bundle on $\mathbb C$, has a non-locally trivial kernel. – Mariano Suárez-Alvarez Mar 9 '11 at 23:12
(The $\mathbb X$s should be $\mathbb C$s...) – Mariano Suárez-Alvarez Mar 9 '11 at 23:13
@Mariano, this is a little confusing, since no matter how you look at it, I'm pretty sure that a vector bundle should have locally constant rank. The example you give corresponds to an injective map of locally free sheaves which is not an inclusion of vector bundles, precisely because the kernel is not a vector bundle. (But it's a nice example for understanding that confusing notion!) – Dave Anderson Mar 10 '11 at 5:10
In the definition of vector bundle it's not only required that the total space locally looks like $U\times \mathbb{A}^n$, but also that the resulting transition functions be linear. The object you are interested in in you question -I think- would be called an $\mathbb{A}^n$-bundle, where $\mathbb{A}^n$ is thought of as a variety, not a vectoor space. [hence my edit in the title; feel free to restore the previous version if you prefer!] – Qfwfq Mar 10 '11 at 19:36
In the absence of local triviality, I'm not sure it makes sense to ask about vector space structure on fibers. As @unknown points out, the definition of a vector bundle requires that the gluing maps be linear isomorphisms on fibers, but this makes sense only when you have a local trivialization. Without this requirement, you can of course put any vector space structure you like on each ${\Bbb A}^n$ fiber. – Dave Anderson Mar 10 '11 at 21:34
up vote 16 down vote accepted

In Jack's example the fiber is not scheme-theoretically $\mathbb A^1$. You can get a counterexample by taking $Y$ to be a nodal curve, $Y'$ its the normalization, with one of the two points in the inverse image of the node removed, and $X = Y' \times \mathbb A^1$.

If we assume that the map is smooth, this becomes quite subtle. It is false in positive characteristic. Let $k$ be a field of characteristic $p > 0$. Take $Y = \mathbb A^1 = \mathop{\rm Spec}k[t]$, $Y' = \mathop{\rm Spec} k[t,x]/(x^p - t)$. Of course $Y' \simeq \mathbb A^1$, but the natural map $Y' \to Y$ is an inseparable homemorphism. Now embed $Y'$ in $\mathbb P^1 \times Y$ over $Y$, and take $X$ to be the difference $(\mathbb P^1 \times Y) \smallsetminus Y'$.

On the other hand, it is not so hard to show that in characteristic 0 the answer is positive for $n = 1$ (if $Y$ is reduced), and I believe it is known to be true $n = 2$. The general case seems estremely hard.

I am afraid that Sasha'a argument does not work; if the fiber does not have a vector spaces structure, there is not reason that choosing points gives you a trivialization.

[Edit] The question has been updated with "what if one assumes vector space structure on the fibres?"?

Well, $\mathbb A^n$ can always be given a vector space structure. In my first example, the fibers are canonically isomorphic to $\mathbb A^1$, so they have a natural vector space structure.

However, if the map $X \to Y$ is smooth, and the vector space stucture is allowed to "vary algebraically" that is, if the zero section $Y \to X$ is a regular function, the addition gives a regular function $X \times_Y X \to X$, and scalar multiplication gives a regular function $\mathbb A^1 \times X \to X$, then $X$ is in fact a vector bundle. The proof uses some machinery: one uses smoothness to construct bases locally in the étale topology, showing that $X$ is étale locally trivial over $Y$, and descent theory to show that in fact $X$ is Zariski locally trivial.

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I think such map is locally trivial if and only if it is smooth. The "only if" part is clear. So, assume $f$ is smooth. Take any point $y_0 \in Y$ and $n+1$ affine-linearly independent points in the fiber over $y_0$. Then choose local sections through this points (this is where you need smoothness), denote them by $x_0,\dots,x_n$. Let $U$ be the set of $y \in Y$ over which $x_i$ are affine-lnearly independent. Then we have a map $f^{-1}(U) \to A^n$, taking a point $x$ to $(t_0,t_1,\dots,t_n) \in \{ \sum t_i = 1 \} \subset A^{n+1}$ such that $x = \sum t_ix_i$. This gives a local trivialization.

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Does your definition of varieties allow them to be disconnected? If so, let $Y$ be any variety, $P\in Y$ a closed point, $U$ the complement of $P$, $X_1$ a vector bundle over $U$, $X_2$ a vector bundle over $P$, and $X=X_1\cup X_2$.

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I meant $X$ and $Y$ to be irreducible – Dima Sustretov Mar 10 '11 at 11:01

EDIT: See the comments for why this isn't a good example. Angelo gives a similar example that actually works.

Suppose that $Y=C$ is a cuspidal curve, and let $\tilde C\to C$ be the normalization. Put $X = \mathbb{A}^1\times \tilde C$, and let $X\to Y$ be the obvious map. The fibers are all $\mathbb{A}^1$'s, but over the singularity in $Y$ the map cannot locally be the projection.

If the map $f$ is a submersion and $Y$ is smooth then I think things should work out, at least in characteristic zero.

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But here the fiber over the origin is equal to two copies of ${\mathbb A}^1$, not one. – Steven Landsburg Mar 10 '11 at 7:25
More precisely, the fiber over the origin is a nonreduced scheme of multiplicity 2, whose reduction is A^1. (He said cuspidal, not nodal.) – David Speyer Mar 10 '11 at 13:01
Ah, yes of course. This is a bad example. – Jack Huizenga Mar 10 '11 at 14:11

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