I have this exercise:
Let $S$ a surface equal to the cartesian product of two projective irreducible smooth curves of genus greater or equal than 1. So $S=E \times F$. I want to describe the kodaira's dimension $K$ depending on the genus of the curves.
First let $p$ and $q$ the projections from my surface to the curve $E$ and $F$ respectively.
here what i've done. $K_S$ is nef.
1) if $g(E)=g(F)=1$ i can define $K_S=p^{*}K_E + q^{*}K_F$ where the star mean the pull-back application.
So i get $K_{S}=0$ because $E,F$ are equal to complex torus. So $K=0$.
2) if $g(E)\ge 2$ and $g(F)\ge 2$ i get $K_S=(-2+2g(F))E+(-2+2g(E))F$ with $E^2=F^2=0$ and $EF=1$. This implies that $K_S^{2}>0$ so $K=2$.
3) The problem is here. if $g(E)=1$ $g(F)\ge2$ i get $K_S=(2g(F)-2)E$. i have tried to compute $h^0(mK_S)$ with $m\ge 1$. In this case i find only that $h^i(mK_S)=0$ due to Kawamata vanishing theorem in fact $h^i(mK_S)=h^i(K_S+(m-1)K_S)=_{KVT}=0$ for $i=1,2.$ and $m\ge2$. So using RR i get $h^0(mK_S)=\chi(O_S)+\frac{1}{2} (m^2K_S^2+ K_S^2) $ but $K_S^2=(....)E^2=0$ so finally $h^0(mK_S)=\chi(O_S)=\frac{1}{12}(K_S^2+e(S))$ using noether's formula. Now i don't know how to proceed. Any suggestions?
The last part is simpler than the way you are attempting it.
If $g(E)=1$, then $K_S=q^*K_F$. So for any $m$ we have $h^0(mK_S)=h^0(mK_F)$ by the projection formula. Now $F$ has Kodaira dimension 1, so $h^0(mK_F)$ grows like a linear function of $m$, hence so too does $h^0(mK_S)$. So $S$ has Kodaira dimension 1 also.