Show that the $i$th local cohomology functor is zero for $i > 0$

Question

Let $I$ be an ideal of a Noetherian ring $R$, and let $M$ be a module over $R$. Let $\Gamma_I(M)$ be the set of all elements $m$ of $M$ for which $I^n m = 0$ for some $n \geq 1$. Then $\Gamma_I(-)$ is a left exact functor. Let $H_I^i(-)$ denote the $i$th right derived functor of $\Gamma_I(-)$. This is the local cohomology functor.

I'm trying to show that if $\overline{M}$ is the quotient module $M/\Gamma_I(M) = M/H_I^0(M)$, then $H_I^i(\overline{M}) \cong H_I^i(M)$ for $i > 0$. I considered the long exact sequence $$\cdots \rightarrow H_I^0(\overline{M}) \rightarrow H_I^1(\Gamma_I(M)) \rightarrow H_I^1(M) \rightarrow H_I^1(\overline{M}) \rightarrow H_I^2(\Gamma_I(M)) \rightarrow \cdots $$ and tried to show that $H^i(\Gamma_I(M)) = 0$ for all $i > 0$. That should do it. This is equivalent to saying that if I take an injective resolution of $\Gamma_I(M)$ $$0 \rightarrow \Gamma_I(M) \rightarrow E^0 \rightarrow E^1 \rightarrow \cdots$$ and apply the functor $\Gamma_I(-)$: $$0 \rightarrow \Gamma_I(E^0) \rightarrow \Gamma_I(E^1) \rightarrow \cdots$$ then this sequence remains exact at any $\Gamma_I(E^i), i > 0$. Is this obvious? I would appreciate a hint, not a full solution.

Answer 1

This is Corollary 2.1.7 of the book Local Cohomology by Brodmann-Sharp. In parts (i) and (ii) they have $H_I^i(\Gamma_I(M))=0, \forall i \gt 0.$ Now use your long exact sequence in the question.

Answer 2

Hints.

1. If $E$ is an injective module, then $\Gamma_I(E)$ is also injective.

2. If $M$ is an $I$-torsion module, then there is an injective resolution of $M$ whose all terms are $I$-torsion.