Let $U$ be an affine open subscheme of an abelian variety $A$ over $\mathbb{C}$. Is $A-U$ an **ample** divisor?


If $\dim A =1$ this is true. 
If $\dim A = 2$, the complement is a divisor $D\_1+\ldots + D\_n$. If all of these are elliptic curves, then the complement is not affine (as it will contain the translate of one of these elliptic curves). So, wlog $D\_1$ is not an elliptic curve. But then $D\_1$ is ample (by Nakai-Moishezon). And this implies that $D\_1 +\ldots +D\_n$ is also ample.


I couldn't figure it out for abelian threefolds.



Is the complement of an affine open in an abelian variety ample?
Welcome new contributor. Yes, that is true. Let $k$ be any field, let $A$ be an Abelian variety over $k$, and let $U\subset A$ be a dense open affine. Denote by $D\subset A$ the complementary divisor with its induced reduced structure. Denote the invertible sheaf of this divisor by $\mathcal{L}:=\mathcal{O}\_A(D).$ Denote by $s$ the global section of $\mathcal{L}$ whose zero locus equals $D.$


Ampleness can be proved after faithfully flat base change over $\text{Spec}\ k$, thus assume that $k$ is algebraically closed. By Lemma II.5.14 of Hartshorne, for every $g\in \mathcal{O}\_A(U)$, there exists an integer $n\_0$ and a section $\widetilde{g}\in \mathcal{L}^{\otimes n\_0}(A)$ such that $\widetilde{g}|\_U$ equals $g\cdot s^{n\_0}|\_U$. Notice that for every integer $r\geq 0$, also $s^r\widetilde{g}|\_U$ equals $g\cdot s^{n\_0+r}|\_U$. Since the $k$-algebra $\mathcal{O}\_A(U)$ is finitely generated, there exists an integer $n\_0>0$ such that the image of $(s|\_U)^{-n}\cdot \mathcal{L}^{\otimes n}(A) \to \mathcal{O}\_A(U)$ generates $\mathcal{O}\_A(U)$ as a $k$-algebra for every $n\geq n\_0$. In particular, the base locus $B\_n$ of the complete linear system of $\mathcal{L}^{\otimes n}$ is disjoint from $U$, and the induced morphism to projective space, $$\phi\_n:A\setminus B\_n \to \mathbb{P}^{d\_n}\_k,$$ restricts as a locally closed immersion on $U$.


The set $A(k)\_{\text{tor}}$ of torsion $k$-points $a$ of $A$ is dense. Thus, denoting by $$\tau\_a:A\to A$$ the morphism of translation by $a$, for every $k$-point $p$ of $A$, the set $\{\tau\_a(p)| a\in A(k)\_{\text{tor}} \}$ is dense in $A$. In particular, not all of these points can lie in the proper, Zariski closed subset $D$. Thus, there exists an integer $m>0$ and an $m$-torsion point $a$ such that $\tau\_a^\*D$ does not contain $p$. In other words, the point $p$ is contained in the open affine subset $\tau\_a^{-1}(U)$. 


Since $a$ is $m$-torsion, $\tau\_a^\*\mathcal{L}^{\otimes n}$ is isomorphic to $\mathcal{L}^{\otimes n}$ for every positive integer $n$ that is divisible by $m$. For such $n$, also the base locus $B\_n$ is disjoint from $\tau\_a^{-1}(U)$, and $\phi\_n$ restricts to a locally closed immersion on $\tau\_a^{-1}(U)$.


The set of translates $\tau\_a^{-1}(U)$ for $a\in A(k)\_{\text{tor}}$ is an open affine covering of $A$. Since $A$ is quasi-compact, there exist finitely many torsion translates of $U$ that cover $A$. For $m>1$ equal to the least common multiple of the orders of those finitely many torsion points, for every positive integer $n$ that is divisible by $m$, the base locus $B\_n$ is disjoint from each of these open translates in this open covering. Thus, the base locus $B\_n$ is empty. Moreover, the $k$-morphism $\phi\_n$ restricts as a locally closed immersion on each of these open translates. 


If $\phi\_n$ had a positive dimensional fiber, that fiber would intersect one of these open translates in a positive dimensional subvariety, and that would contradict that $\phi\_n$ is an immersion on that open translate. Thus, every fiber of $\phi\_n$ is finite. Since $\phi\_n$ is a morphism between proper $k$-schemes that has finite fibers, the morphism $\phi\_n$ is a finite morphism. Since the pullback of an ample invertible sheaf by a finite morphism is ample, the invertible sheaf $\mathcal{L}^{\otimes n} \cong \phi\_n^\*\mathcal{O}(1)$ is ample on $A$. Finally, since $\mathcal{L}^{\otimes n}$ is ample, also $\mathcal{L}$ is ample.


