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.