Let $R$ be a Noetherian ring and $I$ an ideal of $R$. If $M$ is a finitely generated $R$-module and $i\neq 0$ is the greatest integer such that $H^i_I(M)$ is nonzero, then $H^i_I(M)$ is not a finitely generated $R$-module.
Can someone show me how to prove it? Thanks a lot!
You can assume that $R$ is local. Since $H_I^i(M)\simeq H_I^i(M/\Gamma_I(M))$ you may also assume that $I$ contains a non-zero divisor on $M$ (why?), say $a$. We have a short exact sequence $$0\to M\stackrel{a}\to M\to M/aM\to 0$$ which gives rise to an exact sequence $$H_I^i(M)\stackrel{a}\to H_I^i(M)\to H_I^i(M/aM)\to 0.$$
Now proceed by induction on $\dim M$.
If $\dim M=1$, then by Grothendieck's Vanishing Theorem we have $i=1$, and moreover $H_I^1(M/aM)=0$. Thus the map $H_I^1(M)\stackrel{a}\to H_I^1(M)$ is surjective, so $H_I^1(M)=aH_I^1(M)$ and by Nakayama $H_I^1(M)$ can't be finitely generated.
If $\dim M>1$, then by Nakayama $H_I^i(M/aM)\ne0$. Since $\dim M/aM\le\dim M-1$ and by the induction hypothesis we get that $H_I^i(M/aM)$ is not finitely generated, so $H_I^i(M)$ is not finitely generated (why?).