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?

  • 3
    $\begingroup$ For an abelian variety A, the Neron-Severi group is torsion-free and (when tensored with Q) is isomorphic to the set of symmetric elements in the endomorphism algebra. The endomorphism algebras of abelian varieties (as algebras with involution) are well understood (see, for example, Mumford, Abelian Varieties). $\endgroup$
    – mephisto
    May 24, 2011 at 18:03

2 Answers 2


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.

  • 1
    $\begingroup$ $H^1(X,\mathbb{Z})$ is always torsion free. In fact, $NS(X)$ injects into $H^2(X,\mathbb{Z})$ via the exponential sequence. $\endgroup$
    – naf
    May 24, 2011 at 17:45
  • $\begingroup$ Yes I think I need $H_1(X,\mathbb{Z})$ instead, I shall edit my answer accordingly. Although should not numerically trivial divisors also be homologically trivial, so sent to zero in $H^2(X,\mathbb{Z})$? $\endgroup$ May 24, 2011 at 18:02
  • $\begingroup$ @Ulrich. Actually I see now why $NS(X)\to H^2(X,\mathbb{Z})$ is in fact injective. Thanks! $\endgroup$ May 24, 2011 at 19:05
  • $\begingroup$ How about the torsion-freeness of $H_1(X,\mathbb{Z})$ for abelian varieties over number fields? $\endgroup$ May 25, 2011 at 10:13
  • 1
    $\begingroup$ This means that there are injections $F(X) \to F(\overline{X})$, where $\overline{X}$ is the base change of $X$ to $\mathbb{C}$, and $F$ can be the Picard group, Néron-severi group or your favourite cohomology theory. So if there is no torsion in the Néron-severi group over $\mathbb{C}$ then there is certainly none over the ground field. $\endgroup$ May 25, 2011 at 11:47

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).


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