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.

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    $\begingroup$ This is actually the way projectivity of abelian varieties is typically proven, cf. Mumford's Abelian varieties, p. 59 (p. 62 in the first edition). See also Theorem 10.9 of Bhatt's notes. $\endgroup$ Oct 31, 2018 at 22:41

1 Answer 1


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.

  • $\begingroup$ Thank you for this detailed explanation. $\endgroup$
    – Sjoerd
    Nov 1, 2018 at 15:58

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