Picard number and torsion of Neron-Severi group of abelian varieties over a number field

Question

Let $A$ be an abelian variety over a number field $k$ and let $NS_A$ denote its Neron-Severi scheme. Then the group of $k$-rational points of $NS_A$ is a finitely generated abelian group, i.e. $NS_A(k)=H^0(G_k,NS_A(\bar k)) = \mathbb{Z}^\rho \oplus Torsion$. I'd like to know:

What does a torsion element look like? Can it be ample? Is there an easy example, maybe of an elliptic curve, where the torsion of $NS_A(k)$ is non-trivial?

Is the rank $\rho$ of $NS_A(k)$, i.e. the Picard number, bounded by 1 and the square of the dimension of $A$?

Is the group of $\bar k$-rational points of $NS_A$ also a finitely generated abelian group?

Answer 1

There were quite a few different questions, so forgive me if my answer is somewhat fragmented.

The Néron-Severi group $NS(X)$ (divisors modulo algebraic equivalence) is finitely generated over any field for any non-singular projective variety $X$, this is Severi's theorem of the base (at least for the case of characteristic zero). In what follows however I am assume that the variety is defined over $\mathbb{C}$ for simplicity. Some answers to your other questions:

• In the case of curves, $NS(X)\cong\mathbb{Z}$, with the isomorphism given by the degree map.

• More generally, there is a natural injective morphism $NS(X) \to H^2(X,\mathbb{Z})$. This can then be used to get an upper bound for the Picard number.

• Torsion in $NS(X)$ also naturally lives inside $H_1(X,\mathbb{Z})$, so if this is torsion free then so is $NS(X)$. This is the case free for abelian varieties (over $\mathbb{C}$ say), since $H_1(X,\mathbb{Z})$ can identified with a lattice $\Lambda \subset \mathbb{C}^g$ such that $X\cong\mathbb{C}^g/\Lambda$.

• Finally, torsion divisors can never be ample because they are numerically trivial, and ampleness is preserved under numerical equivalence.

Answer 2

Regarding the bounds for the Picard number $\rho$, the lower bound $\rho \geq 1$ comes from the fact that there exist divisors which are not algebraically equivalent to 0 (alternatively, one can argue that $\operatorname{id}_A$ is a symmetric endomorphism of $A$). The upper bound $\rho \leq (\dim A)^2$ (which holds only in characteristic 0) is Exercise 2.6(5) in Birkenhake-Lange's book Complex abelian varieties.

It is possible to show further that if $A$ is an abelian variety over $\mathbf{C}$ of dimension $\geq 2$, then $\rho = (\dim A)^2$ if and only if $A$ is isogenous to a power of some CM elliptic curve (see Exercise 5.6(10) in the same book).