How to properly verify that $E\times E'$ has no non-trivial effective divisors with Kodaira dimension zero

Question

Let $E$ and $E'$ be elliptic curves over $\mathbb{C}$. I am pretty confident that the only effective divisor $D\subset E\times E'$ with Kodaira dimension zero is the trivial divisor.

How to prove this in a simple manner?

Answer 1

If $D$ has Iitaka dimension zero, then $\dim H^0 ( n E) =1$ for all $n$, because if it were any larger than $\dim H^0(k ne) \geq k+1$.

If we have an automorphism $\sigma$ with $[ n \sigma (D)] = [n D]$ in the divisor class group, we would have two linearly independent sections unless in fact $\sigma(D)= D$.

For $\sigma$ translation by an $n$-torsion point, for any $D$, $\sigma(D)- D$ is $n$-torsion, and so $[n \sigma(D) ] =[nD]$. (To see this, one can use the exponential exact sequence $H^1( A, \mathcal O_X) \to H^1(A, \mathcal O_X^\times) \to H^2(A, \mathbb Z)$. Translation acts trivially on the first and last terms, meaning that $\sigma (\sigma(D)-D) =\sigma(D)-D$ and thus $\sigma^n=1$ implies $\sigma(D)- D$ is $n$-torsion.)

So to have Kodaira dimension $0$, $D$ would have to be invariant under translation by $n$-torsion for all $n$. Thus, it must be either empty or Zariski dense, and therefore it is empty.