I thought it were obvious that on *any* simple abelian variety, an effective divisor is ample. What is wrong with the following argument?

Let $A$ be the simple abelian variety and $D \subset A$ an effective divisor. Consider the stabilizer 
$$
Z_D: \hspace{3cm} \{x \in A \mid x+D = D\}.
$$
It is an algebraic subgroup of $A$, and hence it is finite. It is a standard fact that the finiteness of $Z_D$ is equivalent to $D$ being ample (the reverse being trivial). 

Reference for the latter fact: see, for example, Lemma 8.5.6 in the abelian varieties chapter of the book *Heights in diophantine geometry* by Bombieri and Gubler. In virtually any proof of the projectivity of abelian varieties (defined initially as *complete* group varieties), this fact is shown and used implicity, if not stated explicitly. (E.g., see J. S. Milne's 1986 article on abelian varieties in *Arithmetic geometry*. Also, I think it should appear, explicitly, in Milne's course notes on abelian varieties, although I did not check this.)

**Application.** A surjective morphism $f: A \to Y$ from a simple abelian variety onto a positive-dimensional projective variety $Y$ is finite.

*Proof.* Choose $H \subset Y$ an ample divisor. The divisor $f^*H$ is effective on $A$, hence it is ample. This is equivalent to $f$ being finite.