curves on abelian surface $E\times E$ It is Exercise 4.16.4 in Kollár, and we are working on $\mathbb{C}$. Let $E$ be a very general smooth elliptic curve and consider $E\times E$. Let $m,n>0$ be coprime integers and define the closed subspace $E_{m,n}\subset X$ be the image of
$$E\rightarrow E\times E,\quad x\mapsto(mx,nx)$$
We know that $E_{1,0},E_{0,1},E_{1,1}$ is a basis of the group of one-cycles on $E\times E$. Hence we can write
$$[E_{m,n}]=a_1[E_{1,0}]+a_2[E_{0,1}]+a_3[E_{1,1}]$$
To compute the coefficients, I need to know the intersection number $E_{m,n}.E_{1,0}$ and $E_{m,n}.E_{1,1}$.
 A: Since the intersection form is bilinear. This reduces the knowing the intersection of any pair out of the 3 curves and there self intersections.
First the self-intersections. Since $E_{1,0}$ and $E_{0,1}$ are factors of the product we have $$E_{1,0}^2 = E_{0,1}^2 = 0.$$
It is well known that the the self intersection of the diagonal of a manifold is equal to the Euler characteristic of the manifold. So we also have $$E_{1,1}^2=0.$$ Or in  AG language, the normal bundle of $E_{1,1}$ is the tangent bundle of $E$, which has degree $0$.
The other intersection numbers are more simple to compute since they all intersection transversally, so we just have to count the intersection points.
$$E_{1,0}.E_{0,1} = 1 .$$
$$E_{1,1}.E_{0,1} = 1 .$$
$$E_{1,1}.E_{1,0} = 1 .$$
By bilinearity we get:
$$ E_{m,n}. E_{1,0} = a_{2} + a_{3}$$
$$ E_{m,n}. E_{1,1} = a_{1} + a_{2}$$
We also need to obtain expressions in terms of $m$ and $n$.
For this we have to compute the number of points in the set $E_{m,n} \cap E_{1,1}$ and $E_{m,n} \cap E_{1,0}$.
Because the tangent space of $E_{m,n}$ is generated by $(m,n)$ which is linearly independant from $(1,0),(1,1), (0,1)$ all the intersection points are transverse and contribute $+1$ to the intersection product.
Added: Counting the intersection points: Note that $E_{m,n} \cap E_{1,0}$ is counting the points $p \in E$ such $np = 0$. This is indeed $n^2$ as you correctly suggested, namely $p = [\frac{k_1}{n}+\frac{k_{2}}{n}i]$ (for the standard lattice), where $k_{i}$ are integers.
So counting the points in the rectangle, we obtain that $$E_{m,n}. E_{1,0} = n^2.$$
Note that $E_{m,n} \cap E_{1,1}$ is counting the points $q \in E$ such $nq = mq$. Lets assume WLOG $m>n$.
Writing $q = [a + bi]$, then $na - ma \in \mathbb{Z}$ and $nb - mb \in \mathbb{Z}$. That is $q = [\frac{k_1}{m-n} +\frac{k_{2}}{m-n}i]$ where $k_{i}$ are integers.
So counting the points in the rectangle again, we obtain that $$E_{m,n}. E_{1,1} = (m-n)^2.$$
