The following result is a classical one:
Let $A$ be an abelian variety over a field $k$ and let $D$ be a divisor on $A$, the following conditions are equivalent:
i) The divisor $2D$ gives a finite morphism to $\mathbb{P}(H^0(A, \mathcal{O}_A(2D))$
ii) The subset of $b\in A(\overline{k})$ such that $t_b^*D =D$ is finite
iii) The subset of $b\in A(\overline{k})$ such that $t_b^*D$ is rationally equivalent to $D$ is finite.
One proof of $i) \Rightarrow ii)$ goes along those lines: we get a morphism $\phi_{2D}: A\to \mathbb{P}^r$ and for any $b$ such that $t_b^*D\sim D$ (and a fortiori $t_b^*2D\sim 2D$) we get an autmorphism $a(b)\in PGL(r+1)$ such that $\phi_{t_b^*2D}=\phi_{2D}\circ t_{-b}=a(b)\circ \phi_{2D}$ this gives a (set theoretic) map $K\to PGL(r+1)$ as $K$ is the set of closed points a subvariety of $A$, which is a subgroup this is a map from an abelian variety to a affine group and is thus constant and therefore there for the connected componets of $K$ are finite and $K$ itself is also.
The problem is that this description relies of course on the fact that the map $K\to PGL(r+1)$ is algebraic, I'm pretty much convinced that it should be so, but I'd like to give a rigorous proof of that fact and I'm stuck. Is there a neat argument for this?