TITLE: Divisors in an abelian surface
QUESTION [10 upvotes]: How to compute the Néron-Severi group of the abelian surface $Y = \mathbb{C}/\mathbb{Z}[i] \times \mathbb{C}/\mathbb{Z}[i]$. More generally, are there any result that compute the Néron-Severi group of product of curves?
Suppose that surface $Y$ has a divisor class $[F]$ such that $[F]^{2} > 0$. Why is $[F]$ or $-[F]$ ample?
Thanks.

REPLY [5 votes]: In general, for two curves $C$ and $C'$ you have that $\mbox{Pic}(C\times C')\simeq\mbox{Pic}(C)\times\mbox{Pic}(C')\times\mbox{Hom}(JC,JC')$, where $JC$ denotes the Jacobian of $C$, and I'm sure there must be a similar situation with the Néron-Severi group for the product of curves. Now, if $(A,\lambda)$ is a polarized abelian variety, you have that $\mbox{NS}(A)\simeq\mbox{End}_\lambda(A)$, that is, the endomorphisms of $A$ that are symmetric (invariant under the Rosati involution defined by $\lambda$). In the case of the product of elliptic curves, it is easy to see that we have an isomorphism
$$\mbox{End}_{\lambda_1}(E_1)\times\mbox{End}_{\lambda_2}(E_2)\times\mbox{Hom}(E_1,E_2)\simeq\mathbb{Z}^2\times\mbox{Hom}(E_1,E_2)\to\mbox{End}_{\lambda_1\otimes\lambda_2}(E_1\times E_2),$$
given by
$$(\alpha,\beta,f)\mapsto\left(\begin{array}{cc}\alpha &\lambda_2^{-1}\hat f\lambda_1\\f&\beta\end{array}\right)$$
where $\lambda_1\otimes\lambda_2$ denotes the product polarization and $\lambda_i$ corresponds to the polarization on $E_i$ given by the divisor of a single point. This then induces an isomorphism $\mathbb{Z}^2\times\mbox{Hom}(E_1,E_2)\to\mbox{NS}(A)$ given by
$$(a,b,f)\mapsto (a-\deg f)[\{0\}\times E_2]+(b-1)[E_1\times\{0\}]+[\Gamma_{-f}],$$
where $\Gamma_{-f}$ is the divisor corresponding to the graph of $-f$. This answer is independent of the ground field. 
To answer your second question, at least for abelian varieties over $\mathbb{C}$, it is interesting to note that if $(A,\Theta)$ is a polarized abelian variety of dimension $n$, then a divisor $D$ is nef if and only if $(D^{k}\cdot\Theta^{n-k})\geq 0$ for all $k$ (see Lemma 1.1 in http://arxiv.org/pdf/alg-geom/9712019v1.pdf), and it is ample if it is a strict inequality. So for dimension 2, we have that if $(F^2)>0$, then either $(F\cdot\Theta)>0$ (and in this case $F$ is ample) or $(F\cdot\Theta)\leq 0$ and so $-F$ is at least nef. if $(F\cdot\Theta)=0$, then by the same Lemma, $F$ is linearly equivalent to an effective divisor, and since the intersection of an ample divisor and an effective divisor is positive (if the effective divisor is not numerically trivial), we must have that $F\equiv 0$, and so $(F^2)=0$, a contradiction.
{"set_name": "stack_exchange", "score": 10, "question_id": 407028}