Suppose we have a principally polarized abelian variety X over the complex number field. Given two ample, effective divisors D_1, D_2 such that the global sections of both line bundles are 1 dimensional vector spaces. If D_1 and D_2 have the same first Chern class, then is it true that D_1 and D_2 differ by a translation, i.e., there is x in X such that D_1 + x = D_2 as divisors? Why or why not?
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This is true. For $A_{/\mathbb{C}}$ an abelian variety, $L$ an ample line bundle on $A$, then any line bundle $M \in \operatorname{Pic}^0(A)$  over $\mathbb{C}$, this is equivalent to having first Chern class zero  is of the form $T_x^{*} L \otimes L^{1}$ for some $x \in A$. (e.g. Theorem 1 on p. 77 of Mumford's Abelian Varieties). Applying this theorem with $L = L(D_2)$, $M = L(D_1)  L(D_2)$, we get that $L_1  L_2 = T_x^*(L_2)  L_2$, so $L_1 = T_x^*(L_2)$. So $D_1$ and $x+D_2$ (meaning translation of $D_2$ by $x$!) must be linearly equivalent, but by your assumption $h^0(L(D_1)) = h^0(L(D_2)) = 1$, they are each the unique effective divisors in their linear equivalence classes, so we must have $D_1 = x + D_2$. 

