non-locally trivial A^n bundles 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?
 A: 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.
A: 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.  
A: 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 $X=X_1\cup X_2$.  
A: 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.
