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May 17, 2014 at 13:48 comment added user49214 Yes, the same is true if $n = 3$ and $s\leq 7$. In this case $-K_X$ is nef. On the other hand if $n\geq 4$, $-K_X$ is not nef as soon as $s>1$. I was wondering If it is possibile to give an integer $h$ such that the result hold for $n\geq 4$ and any $s\leq h$.
May 17, 2014 at 13:22 comment added Remke Kloosterman This depends (again) on $n$ and $s$. Consider the case where $n=2$. If $s<9$ then $-K_X$ is ample and the answer is yes. If $s=9$ then you find that $K_X^2=0$. If the 9 points are the base points of a pencil of cubics then you find that $h^0(-k K_x)\geq k+1$ for every $k>0$, and Riemann-Roch gives you that $\chi(-k K_X)=1$.
May 17, 2014 at 11:59 comment added user49214 Do you know if something is known for $D = -K_X = (n+1)H-(n-1)E_1-...-(n-1)E_s$? Thank you very much.
May 17, 2014 at 11:56 comment added user49214 I see your point. I hoped there was a sort of Serre-Grothendieck theroem ensuring that $h^i(X,\mathcal{O}_X(kD)) = 0$ for $i>0$ and $k\gg 0$.
May 17, 2014 at 11:40 history answered Remke Kloosterman CC BY-SA 3.0