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 $H^n\_I(M)\hookrightarrow H^n\_I(M)$?



Top degree local cohomology under action by a non-zerodivisor
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. 


