Why the Picard group of a K3 surface is torsion-free Let $X$ be a K3 surface. I want to prove that $Pic(X)\simeq H^1(X,\mathcal{O}^*_X)$ is torsion-free. 
From  D.Huybrechts' lectures on K3 surfaces I read that if $L$ is torsion then the Riemann-Roch formula would imply that $L$ is effective. But then if a section $s$ of $L$ has zeroes then $s^k\in H^0(X,L^k)$ has also zeroes, so no positive power of $L$ can be trivial.
What I am missing is how the Riemann-Roch theorem can imply that if $L$ is torsion then $L$ is effective?

Here are some comments:
You want to show that $h^0(L) >0$, I believe. Riemann-Roch gives $\chi(L) = \chi(0)$. In fact, $(L,D) =0$ for all divisors $D$ on $X$. To see this, note that $L^{\otimes n } = \mathcal{O}_X$ for some $n>0$. Thus, $n(L,D) = (L^{\otimes n},D) = (\mathcal{O}_X,D) =0$.
Now, it should follow from some standard K3 surfaces theory that $h^0(L) >0$. In fact, it equals $\chi(0) + h^1(L) -h^2(L)$. Maybe, you can show that $h^2(L)$ is zero if $L$ is torsion?
