Why is this divisor on a K3-surface not effective?

Question

Let $D$ be a divisor on a $K3$ surface and set $L:=\mathcal{O}_S(D)$. Riemann Roch theorem : $\chi(L)=\chi(\mathcal{O}_S)+\frac{1}{2}D.(D-K)$ reduces to $h^0(S,L)+h^0(S,L^{-1})=2+\frac{1}{2}D^2+h^1(S,L)$.

Now if $L$ is numerically equivalent to 0 then: 1) either $L$ or $L^{-1}$ has some sections. This follows because $h^0(S,L)+h^0(S,L^{-1})=2+h^1(S,L)$ since $D^2=0$ as $L$ is numerically equivalent to 0.

The next claim is 2) Neither $D$ nor $-D$ is effective and non-trivial, hence $L=\mathcal{O}_S$. But I am not able to understand why. If $D$ is effective and non-trivial, then does it mean $D^2\neq 0$. Any help will be appreciated!

Answer 1

If $D$ is effective and non-trivial, then for any ample divisor $E,$ we have $D.E > 0$ (Nakai-Moishezon Criterion). This contradicts the fact that $D$ is numerically trivial.

Answer 2

If $L$ is numerically equivalent to 0, then $L=\mathcal O_S$, because $H^1(\mathcal O_S)=0$.

If $L^2=0$, then either $L$ or $-L$ is effective, as you say, by Riemann-Roch. It does not mean that $L$ numerically equivalent to 0. An elliptic curve on a K3 gives such an example.