# stably birational abelian varieties are isomorphic

Can anybody help me to prove the following result:

Proposition. Let $A$ and $B$ be abelian varieties over a field $k$ of characteristic zero. Assume that $A \times \mathbb{P}_k^n$ and $B \times \mathbb{P}_k^m$ are birational for some $n, m \geq 0$. Then $A$ and $B$ are isomorphic.

Is the result true in positive characteristic?

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$A$ is the albanese variety of $A \times \mathbb{P}^n$ and the albanese variety is a birational invariant (over fields of arbitrary characteristic). –  ulrich Nov 1 '13 at 11:13
In that case, let me ask: why is the albanese a birational invariant? Is there a more direct proof for abelian varieties? –  curlyx Nov 1 '13 at 11:29
The main point in all this is the basic fact that any map from a rational curve to an abelian variety is constant. There are many ways to prove this; you can look up any basic text on abelian varieties. –  ulrich Nov 1 '13 at 12:06
As implicit in Ulrich's comments, this does not seem to be at research level. –  abx Nov 1 '13 at 17:14