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Let $R$ be a commutative Noetherian local ring, and $S$ be an $R$-algebra. Let $x_1,\dots,x_t$ be elements, in the maximal ideal of $R$, which is a regular sequence on both $R$ and $S$, and let $I$ be the ideal generated by $x_1, \dots, x_t$. Then, for every $n\ge 1$, $x_1^n, \dots, x_t^n$ is also regular on both $R$ and $S$ and we have an isomorphism $S/(x_1^n, \dots, x_t^n)S \cong \text{Ext}^t_R\left (R/(x_1^n, \dots, x_t^n)R, S \right) $.

Since $\text{H}^t_I(S)\cong \varinjlim_n \text{Ext}^t_R\left (R/(x_1^n, \dots, x_t^n)R, S \right) $, so we have a canonical map $\rho: S/(x_1, \dots, x_t)S \to \text{H}^t_I(S)$.

My question is: When can we say $\rho(1+ (x_1, \dots, x_t)S) \neq 0$?

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  • $\begingroup$ Since it's a regular sequence, it should be ok, no? Ie, you are looking at the maps $$S/(x_1, \dots, x_t)S \to S/(x_1^n, \dots, x_t^n)S$$ induced by multiplication by $$x_1^{n-1}\cdots x_t^{n-1}$$ I think. It's a straightforward exercise to see that that map is injective when the $x_i$ form a regular sequence. Probably a good source for this is Hochster's local cohomology notes. dept.math.lsa.umich.edu/~hochster/615W11/loc.pdf $\endgroup$
    – Karl Schwede
    Commented Sep 16 at 0:34
  • $\begingroup$ @KarlSchwede: hmm, for some reason I thought this would have something to do with the "canonical element " conjecture/Theorem... but maybe I was wrong? Could you please let me know if exactly what I want is in Hochster's notes? $\endgroup$
    – uno
    Commented Sep 16 at 3:38
  • $\begingroup$ You want the Koszul description of local cohomology (Section 7 in Hochster's notes). Then you want to argue that the maps in the sequence are injections. That's a straightforward exercise on regular elements: $$(x_1^n, \dots, x_t^n) : (x_1 \cdots x_t) = (x_1^{n-1}, \dots, x_t^{n-1})$$ which can be found in Hochster's notes, Discussion 11.3. $\endgroup$
    – Karl Schwede
    Commented Sep 16 at 14:43
  • $\begingroup$ @KarlSchwede: Thanks a lot ... I think I understand the point now ... so just to reiterate, the canonical map is injective, and $S$ being an algebra plays no role, right? We just needed the sequence to be regular on the module (I think even regular on the ring is not needed?) $\endgroup$
    – uno
    Commented Sep 17 at 12:08
  • $\begingroup$ I believe that's correct. $\endgroup$
    – Karl Schwede
    Commented Sep 18 at 19:39

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