Let $(R, \mathfrak{m})$ be a local ring and $X = Spec(R)$. Let $Y = V(I)$ be a closed subscheme of $X$, defined by an ideal $I \subset R$, and let $P \in X$ (in fact, $P \in Y$) be the closed point. Let $(\hat{X}, \mathcal{O}_{\hat{X}})$ be the formal completion of $X$ along $Y$ ($\hat{X} = Y$ as a topological space, and the sheaf of rings $\mathcal{O}_{\hat{X}}$ is $\varprojlim \mathcal{O}_X/\tilde{I}^n$).

Can we write the cohomology of $\mathcal{O}_{\hat{X}}$ with support in the closed point $P$ in terms of local cohomology on $R$? Specifically, do we have isomorphisms $H^j_P(\hat{X}, \mathcal{O}_{\hat{X}}) \simeq \varprojlim H^j_{\mathfrak{m}}(R/I^n)$?

Without the "support at P" subscript, this is Corollaire 4.1.7 of EGA III, Grothendieck's comparison theorem. So essentially what I am asking is: is there an analogue of the comparison theorem for local cohomology?

One natural strategy for deducing a local comparison theorem from the global one would be to use the long exact sequence relating $H^j_P(\hat{X}, \mathcal{O}_{\hat{X}})$ to $H^j(\hat{X}, \mathcal{O}_{\hat{X}})$ and $H^j(\hat{X} \setminus P, \mathcal{O}_{\hat{X}})$, but this runs into trouble because the inclusion $X \setminus P \hookrightarrow X$ won't be proper, so the global comparison theorem will fail for $X \setminus P$.

SGA 2 would seem to be the natural place to look for results of this type, but its Expose IX (where formal completions are treated) doesn't address local cohomology, as far as I can tell.

  • $\begingroup$ The topological space of the formal scheme $\widehat{X}$ consists of the single point $P$ (so $\widehat{X} - P$ is empty). The functor of global sections on $\widehat{X}$ with supports at $P$ is the same as the one without mention of the supports, so it is an exact functor (even coincides with the identity functor, so to speak). Hence, $H^j_P(\widehat{X},\cdot)$ vanishes for $j > 0$ and is the identity functor for $j = 0$. Is there some motivation to guide the way to an interesting reformulation, or is this idle curiosity? $\endgroup$ – user29720 Feb 27 '13 at 6:10
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    $\begingroup$ @kreck: I think the poster is completing along $I = 0$, so the topological space of $\widehat{X}$ is the same as that of $Spec(R/I)$. @Nick: The obstruction to the isomorphism you want lies in the $\lim^1$ of the projective system $\{ H^{j-1}_m(R/I^n) \}$. I don't see a reason why these should vanish in general. Have you tried looking at homogeneous ideals (so it becomes a question in projective geometry)? $\endgroup$ – anon Feb 28 '13 at 2:33
  • $\begingroup$ @anon: Ah, so I was misreading; thanks for the correction. $\endgroup$ – user29720 Feb 28 '13 at 12:33
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    $\begingroup$ Actually, now it seems to me that this is always true for any noetherian ring $R$: each $H^i_m(R/I^n)$ is an artinian $R$-module, so the ML condition in the comment above is trivially verified (as any projective system of artinian $R$-modules is automatically ML). $\endgroup$ – anon Feb 28 '13 at 15:27
  • $\begingroup$ I agree with anon's comment. Proposition 2.2 in the reference in my answer below has several parts and the Gorenstein assumption may be needed for the proof of other parts of the proposition, but I don't see it used in the proof of the statement you want. $\endgroup$ – Mahdi Majidi-Zolbanin Feb 28 '13 at 19:25

Dear Nick: If your local ring $R$ is Gorenstein, then $H^i_p(\hat{X},\mathcal{O}_{\hat{X}})\cong\varprojlim H^i_{\mathfrak{m}}(R/I^n)$ for all $i$, as you want. This is proved, for instance, in Proposition 2.2, page 334 of A. Ogus' Local cohomological dimension of algebraic varieties, Annals of Mathematics, Vol. 98, No. 2 (1973). He attributes it to Peskine Szpiro. You should also check Proposition A8 and Corollary A9 on pages 362 and 363 of Ogus' paper, as well as Propositions 2.1 and 2.2 on pages 106-107 of Peskine-Szpiro's Dimension projective finie et cohomologie locale.

I would like to add that as anon commented above, the proof given by Ogus is achieved by showing that when $R$ is Gorenstein the suggested (by anon) $\mathrm{lim}^1$ vanishes.

Edit. Following anon's comment, I don't think Gorenstein assumption is needed for this.

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  • $\begingroup$ Excellent. Thank you very much for this. $\endgroup$ – Nick Switala Mar 1 '13 at 1:18

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