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I am wondering to what extent the functors "ring of invariants under a group action $G$" and "divided power envelope with respect to a $G$-stable ideal" commute.

To be precise, let $R$ be a commutative ring (with unit) and $G$ a group acting on $R$ by ring automorphisms. I am happy to assume that $R$ is a noetherian adic ring and that $G$ is profinite acting continuously on $R$, but I do not want to assume that $G$ is finite.

Let $S:=R^G$ be the subring of invariants, and suppose given an ideal $I\subseteq R$ which is $G$-stable. Denote by $J:=I\cap S$ the contraction of $J$ to $S$, and let $D(R,I)$ and $D(S,J)$ be the divided power envelopes of $R$ with respect to $I$ and $S$ with respect to $J$. (I should probably also complete these algebras with respect to a $G$-stable ideal of $R$ that is contained in the ideal of definition, so the induced $G$-action on the completed PD algebras is continuous).

Then $G$ acts on $D(R,I)$ and we have $D(S,J)\subseteq D(R,I)^G$.

Is this inclusion an equality?
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  • $\begingroup$ Rather than creating new tags "divided" and "powers" I retagged with existing tags. I hope this question gets an answer. $\endgroup$
    – David White
    Commented Oct 26, 2011 at 4:02
  • $\begingroup$ I mushed "divided" and "powers" together. $\endgroup$
    – S. Carnahan ♦
    Commented Oct 26, 2011 at 4:10
  • $\begingroup$ @S.Carnahan, I thought about doing that, but was unsure if it would really be a good new tag to create. I don't know much about this area, but if this seems like a tag likely to catch on I'm all for creating it. $\endgroup$
    – David White
    Commented Oct 26, 2011 at 4:48
  • $\begingroup$ The following example, which shows the answer to my question is "no", was explained to me by BCnrd: Let $R:=\mathbf{Z}[\zeta_n][X]$ for $n>1$ and $\zeta_n$ a primitive $n$-th root of unity, with $G:=\mu_n$ acting on $R$ via $\zeta\cdot X:=\zeta X$. The PD envelope of $R$ relative to $I:=(x)$ contains the $G$-invariant element $X^n/n!$, whereas the PD envelope of $R^G=\mathbf{Z}[\zeta_n][X^n]$ relative to $I^G=(x^n)$ does not (as can be easily checked by grading everything by degree). $\endgroup$
    – B. Cais
    Commented Oct 26, 2011 at 18:01
  • $\begingroup$ @B.Cais, if BCnrd doesn't want to post that as an answer for some reason, then I suggest you post it as an answer (in CW mode) and accept it. That way this question won't pop back to the front-page as "unanswered" $\endgroup$
    – David White
    Commented Oct 27, 2011 at 22:32

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