Coherent modules and cohomology group

Question

Let $k$ be a field, $S=k[T_0,\ldots,T_r]$, $\mathbb P=\mathbb P_k^r=\operatorname{Proj}S$ and $O$ the structure sheaf of $\mathbb P$. Let $\mathscr F$ be a coherent $O$-module. Why do the is an integer $n_0$ such that for every $i>0,n\geq n_0$ we have that $$H^i(\mathbb P(\mathscr F(N)))=0?$$

Answer 1

This is exactly Serre vanishing. For an example of a reference, Ravi Vakil proves it as Theorem 18.1.4 here: http://math.stanford.edu/~vakil/216blog/FOAGjun1113public.pdf.

The outline is to use the fact that $\mathscr{O}(1)$ is ample to find a surjection $\mathscr{O}(m)^n\rightarrow \mathscr{F}$ and apply the long exact sequence in cohomology to $0\rightarrow \mathscr{G}\rightarrow \mathscr{O}(m)^n\rightarrow\mathscr{F}\rightarrow 0$, where $\mathscr{G}$ is the kernel.

It isn't true that the $r^{th}$ cohomology of $\mathscr{O}(m)^n$ is trivial, but you can twist the exact sequence by $\mathscr{O}(N)$ for $N>m$ to find $H^r(\mathscr{F}(N))=0$ for $N>m$. So $H^r(\mathscr{F}(N))=0$ for $N>>0$ for all $\mathscr{F}$.

Then, we look at what the long exact sequence tells us about $H^{r-1}(\mathscr{F}(N))$. We find that, if $N$ is large enough so that $H^r(\mathscr{G}(N))$ also vanishes, then $H^{r-1}(\mathscr{F}(N))=0$. Continue the process.

The reason why we don't also get that $H^{0}(\mathscr{F}(N))=0$ for $N>>0$ is because $H^{0}(\mathscr{O}(m+N))\neq 0$ for $N$ large (while the higher cohomology vanishes).

Anyways, I think I've basically written what's in the notes, but less well than Vakil, so it's better to look there.