MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
The answer to your question is no. Poincaré's complete reducibility theorem. As you surmise, any $B$ in $A^n$ is isogenous to $A^m, m \le n$. – Felipe Voloch Aug 1 '12 at 15:38
up vote 6 down vote accepted

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.

share|cite|improve this answer

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 $B=0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.