Top degree local cohomology under action by a non-zerodivisor

Let $R$ be a noetherian commutative ring of dimension $n$, and let $M$ be a faithful finite $R$-module. Let $I$ be a proper ideal of $R$, and let $x\in I$ be a non-zerodivisor on $M$.

When does multiplication by $x$ induce an injection $H^n_I(M)\hookrightarrow H^n_I(M)$?

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The local cohomology modules are $I$ torsion, so multiplication by $x$ will actually be the $0$ map. – Hailong Dao Apr 4 '11 at 1:44
Uh-oh! Really? – Harry Gindi Apr 4 '11 at 2:01
Not necessarily the zero map, but definitely not injective. For example, when $M = R$ is a Gorenstein local ring and $I = m$ is the maximal ideal, the top local cohomology is the injective hull of $R/m$, which is $m$-torsion. – Graham Leuschke Apr 4 '11 at 2:38
@Harry: The $I$-torsionness follows from the fact that there is an isomorphism ${\rm H}^i_I(M) = \varinjlim_d {\rm Ext}^i_R(R/I^d, M)$ – Steven Sam Apr 4 '11 at 2:41
Graham was right, let me give a careful answer then. – Hailong Dao Apr 4 '11 at 4:37

1 Answer

Graham was right, the map is not necessarily $0$ as I wrote in the first comment. However, it is true that $H_I^n(M)$ is $I$-torsion, so it will be injective if and only if $H_I^n(M)=0$.

Amusingly, I will observe that the map is actually surjective.

Apply $\Gamma_I(-)$ to the sequence:

$$M \stackrel{x}{\to} M \to M/xM$$

to get $$\to H_I^n(M) \stackrel{x}{\to} H_I^n(M) \to H_I^n(M/xM) \to$$

Now note that $\dim M/xM < \dim M = n$ because $x$ is $M$-regular, so $H_I^n(M/xM) = 0$. (In general, $H_I^n(N) =0$ for $n>\dim N$). So the multiplication by $x$ map is surjective, as claimed (may be this is what you had in mind anyway).

For completeness, the question of when $H_I^n(M) =0$ is rather subtle. It will be true, for example, if $I$ can be generated up to radical by at most $n-1$ elements, because you can calculate local cohomology using the Cech complex on those elements.

Another instance is when $R$ is a complete local domain, and $\dim R/I>0$ (this is known as the Hartshorne-Lichtenbaum vanishing theorem). I do not know an easy equivalent condition off the top of my head.

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Heh, for what it's worth, I already knew it was surjective by that exact argument =). – Harry Gindi Apr 4 '11 at 4:57
@Harry: I am glad you agree! – Hailong Dao Apr 4 '11 at 5:02