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We know that if $X$ is a smooth connected variety over a field $k$, then any line bundle on $X\times_k\mathbb{A}^1$ is from a line bundle on $X$. This is simply because they have the same Picard group. This has a generalization, in fact, $X\times_k\mathbb{A}^1\to X$ induces an isomorphism of $CH_r(X)\to CH_{r+1}(X\times_k\mathbb{A}^1)$.

But I am thinking about vector bundles on $X\times_k\mathbb{A}^1$, I would not believe that any vector bundle on $X\times_k\mathbb{A}^1$ is from $X$. Can anyone give me a counter example ? I think it might be easy to take $X=\mathbb{P^1}$ where the vector bundles are completely split (into line bundles), while the same should not be true for $X\times_k\mathbb{A}^1$. Can anyone give me an example when I take $X$ to be an Abelian variety? for example an elliptic curve.

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Another way to say this, I suppose, is that you are looking for a non-constant map from $\mathbb{A}^{1}$ to the moduli of vector bundles on $X$. And you can find plenty of curves $X$ for which there is such a map. Of course, this often happens for higher dimensional $X$ as well. –  Chris Brav Feb 12 '12 at 13:07
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up vote 7 down vote accepted

One way to think about $\mathbb{A}^n$-families of bundles on $X$ is in terms of $Ext^1$ groups of bundles on $X$. That is, suppose that $Ext^1(E,F) \simeq k^n$. Then, as Chris Brav mentions in the comments, we can think of this as an $\mathbb{A}^n$-point of the moduli of vector bundles on $X$, which over each point of $\mathbb{A}^n$ restricts to the bundle which is the corresponding extension of $E$ by $F$.

For example, if $X = \mathbb{P}^1$, then $Ext^1(\mathcal{O}(1),\mathcal{O}(-1))$ is $1$-dimensional. We can then think of the corresponding vector bundle on $\mathbb{P}^1 \times \mathbb{A}^1$ as being given by the transition function $g = \begin{pmatrix} z & t \\\ 0 & z^{-1} \end{pmatrix}$, where $t$ is the coordinate on $\mathbb{A}^1$. This gives a family of rank $2$ bundles which are isomorphic to $\mathcal{O} \oplus \mathcal{O}$ when $t \neq 0$, but which is isomorphic to $\mathcal{O}(1) \oplus \mathcal{O}(-1)$ when $t=0$.

You could do something similar on the elliptic curve, where you have a family of rank $2$ bundles generically isomorphic to the unique non-split extension of $\mathcal{O}$ by $\mathcal{O}$, but which specializes to the split extension.

In both of these examples, you can see that these bundles are not pulled back from $X$ because their transition functions cannot be the pull-back of a transition function for a bundle on $X$.

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