# canonical divisor of a surface with nonnegative Kodaira dimension

$S$: a smooth projective surface over $\mathbb{C}$ which has non-negative Kodaira dimension.

$L$: an ample divisor on $S$

Why $K_S.L\ge0?$

I know that :

for some $m\ge1$, there is an effective divisor $E \in |mK_S|$ s.t.$0< E.L=(mK_S).L=m(K_S.L).$

Why is "$=$" possible?

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Since some positive multiple of $K_S$ is effective, by Nakai-Moishezon criterion the case $K_S L=0$ is possible only if $K_S$ is a torsion line bundle, i.e. $E=\mathcal{O}_S$ in your notation.
This implies that $S$ is minimal and $\textrm{kod}(S)=0$. In fact, in this situation one has $12 K_S = \mathcal{O}_S$, hence $K_SL = \frac{1}{12} \mathcal{O}_S L =0$.
In the case $\textrm{kod}(S) \geq 1$, instead, some positive multiple of $K_S$ is an effective, non-trivial line bundle hence if $L$ is ample one has $K_SL >0$.