Following Beauville's book "Complex algebraic Surfaces", in order to prove Castelnuovo's rationality criterion i need to prove one lemma and one proposition.
There is one point of proof of lemma V.8 at page 56 that i can't understand. I'm in the hypothesis that $S$ is a minimal surface with $K^2<0$ (denoting with K the canonical divisor) and i need to find an effective divisor B on $S$ such that $(K.B)<0$. To do this i look at hyperplane sections. Let H be an hyperplane section on $S$, if $(H.K)<0$ i win cause B=H is the divisor i'm looking for.
If $(H.K)=0$, Riemann-Roch inequality applied to the divisor $K+nH$ ($n$ positive) gives
$$h^0(K+nH)\geq \chi(\mathcal{O}_S)-\frac{1}{2}((K+nH)^2-((K+nH).K))-h^0(-nH).$$
My assertion is that for sufficiently large $n$, $h^0(K+nH)=dimH^0(S,\mathcal{O}_S(K+nH))\geq 0$. But this is true only if for sufficiently large $n$, $h^0(-nH)=dimH^0(S,\mathcal{O}_S(-nH))=0$. Is the last equality true?
I have the same problem in the proof of Proposition V.6 at page 57. I'm in the hypotheis that $S$ is a minimal surface with $q(S)=P_2(S)=0$ (then $p_g(S)=0$ and so $\chi(\mathcal{O}_S)=1$). I take a curve $C$ such that $(K.C)<0$, $|K+C|=\emptyset$ and I apply Riemann-Roch inequality at the divisor $K+C$. I obtain $$0=h^0(K+C)\geq 1+\frac{1}{2}(C^2+C.K)-h^0(-C).$$ Again I wonder if $h^0(-C)=dimH^0(S,\mathcal{O}_S(-C))=0$ and how to prove it.