curves on abelian surface $E\times E$, part II

Question

It is Chapter II Exercise 4.16.4 in Kollár, and we are working on $\mathbb{C}$. Let $E$ be a very general smooth elliptic curve and consider the abelian surface $X=E\times E$. Let $m,n$ be coprime integers and define the closed subspace $E_{m,n}\subset X$ be the image of $$E\rightarrow E\times E,\quad x\mapsto(mx,nx)$$ These $E_{m,n}$ are all curves. According to my previous question, $E_{1,0},E_{0,1},E_{1,1}$ is a basis of the group of one-cycles on $X$ and $E_{m,n}^2=0$.

Show that every irreducible curve $C$ with $C.C=0$ is a translate of some $E_{m,n}$.

Answer 1

There is an extra condition of $E$ in Kollar's book. Namely, the elliptic curve $E$ is very general in the moduli. Otherwise, for example, if $E=\mathbb C/\mathbb Z+\omega_3\mathbb Z$, where $\omega_3=\exp(2\pi i/3)$. Then the curve given by $E\to E\times E$, $x\to (x,\omega_3 x)$ contributing an extra Picard factor and is not a translate of $E_{m,n}$. The same assumption is needed in your previous question as well.

The extra assumption should be understood as the elliptic curve $E$ is not defined over a quadratic number field. In particular $$Hom(E, E)=\mathbb Z\tag{*}\label{*}.$$

Therefore

$$Hom(E, E\times E)=Hom(E, E)\times Hom(E, E)\cong \mathbb Z\times \mathbb Z.$$

The image is just your $E_{m,n}$. So it suffices to show

Claim. Any irredubible curve $C$ in $E\times E$ with $C\cdot C=0$ implies $C\cong E$.

Proof. First by adjunction formula, $2g_a(C)-2=C\cdot C+C\cdot K=0$, where $g_a$ is the arithmetic genus. So $g_a(C)=1$, and by the assumption that $C$ is irreducible, $C$ is either a nodal rational curve or an elliptic curve.

If $C$ is singular, then composite with the normalization $\mathbb P^1=\tilde{C}\to C\to E\times E$ has to be constant, contradicting the assumption that $C$ is a curve in $E\times E$. So $C\cong E'$ must be a smooth elliptic curve.

Supposedly, $E'$ is isogenous to $E$, but the condition \eqref{*} guarantees that $E'\cong E$ is an isomorphism. See the proof in this post.