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

There exists a finite morphism $A\to \mathbf{P}^g_k$. Here's the question.

Does there exist a finite morphism $A\to \mathbf{P}^g_k$ of degree two?

Can we say something about the minimal degree of a finite morphism $A\to \mathbf{P}^g_k$?

share|cite|improve this question
A better analogue of the degree 2 map $E\to\mathbb{P}^1$ for an elliptic curve is the degree 2 map $A\to A/\pm1$ from $A$ to the associated Kummer variety. (Of course, you need to blow up some points if you want a smooth quotient, and then then map is only rational, it's not a morphism.) – Joe Silverman Aug 1 '12 at 14:05
up vote 9 down vote accepted

For a very general, principally polarized Abelian variety $(A,\Theta)$ of dimension $g$ over $\mathbb{C}$, every Cartier divisor $D$ on $A$ is numerically equivalent to $m\Theta$ for some integer $m$. In particular, the intersection number $D^g$ is $m^g \Theta^g$. So the minimal degree of an effective, nonzero divisor is $g!$, not $2$.

share|cite|improve this answer
Your argument works for any abelian variety of dimension at least $3$: by Riemann-Roch, $g!$ divides $D^g$ for any ample divisor $D$. One can also eliminate the $g=2$ case using the fact that any double cover of $\mathbb{P}^2$ with trivial canonical bundle is a $K3$ surface. – ulrich Aug 1 '12 at 13:33
@ulrich -- I agree. – Jason Starr Aug 1 '12 at 14:35
How to prove that any $\mathbb{P}^n$ is finitely covered by an Abelian variety of the same dimension? Maybe it is elementary, but I can not see it. – Tong Jun 20 at 22:39

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.