Let $A$ be an abelian variety over a field $k$, and $D$ a divisor on $A$. if $a\in A(k)$, I am not sure how to define $D+a$, the translation of $D$?
My guess is that, if $D=\Sigma_{i=1}^mn_iZ_i$ where $Z_i\ne Z_j$ if $i\ne j$, then, we ca define $Z_i+a$ and hence $D+a$ is just $\Sigma_{i=1}^mn_i(Z_i+a)$. Is this true?
An abelian variety is an algebraic group, meaning you have an addition-morphism $$A \times_k A \xrightarrow{+} A.$$ Be restricting this on the second component to a $k$-point $a \in A(k)$, you get a morphism $$A = A \times_k \{a\} \xrightarrow{+a} A,$$ where $\{a\} = \operatorname{Spec}(k(a)) = \operatorname{Spec} k$. This morphism is an isomorphism because it has an inverse $-a$, constructed by composing with the inverse morphism $A \xrightarrow{i} A, a \mapsto -a$. Then $D+a$ is constructed by pushing-forward $D$ along $+a$, or equivalently by pulling-back $D$ along $-a$.