Let me be more precise than the title. (This will be my last attempt to do something with abelian varieties. Sorry for all the basic questions. The answers have been great!)

Let $A$ be a simple abelian variety over a field $k$. Let $g\geq 2$ be the dimension of $A$.

Does there exist an integer $n\geq 1$ such that $A^n = A\times_k A\ldots\times_k A$ contains an abelian variety of dimension less than $g$?

It suffices to prove that $A^n$ contains a curve of genus strictly smaller than $ g$ for some $n\geq 1$.

I'm afraid that this is not true. In fact, if $B\subset A^n$, then $B$ is isogenous to $A^m$ probably. Therefore, $\dim B =mg$. I'm just asking to be sure.

  • 1
    $\begingroup$ The answer to your question is no. Poincaré's complete reducibility theorem. As you surmise, any $B$ in $A^n$ is isogenous to $A^m, m \le n$. $\endgroup$ Aug 1, 2012 at 15:38

2 Answers 2


No (I suppose that $k$ is algebraically closed). This is because Poincaré's complete reducibility theorem contains a unicity statement for the intervening factors (up to isogeny). See Mumford, Abelian varieties, p. 173-174.


In fact, it is no for completely elementary reasons. If $A$ is simple and $B\subset A^n$ is an abelian variety with $\dim B < g$, then $Hom(B,A^n)=Hom(B,A)^n$ is necessarily zero. So $B=0$.


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