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Let $R$ be a local ring, $m$ its maximal ideal and $k:= R/m$ its residue field. Suppose that a finite cyclic group $G= \mathbb{Z}/ m \mathbb{Z}$ has a linear nontrivial action on $R$. Let $R^G$ be a ring of invariant elements of $R$ by this action.

Let $E:= E_R(k)$ be an injective hull of $k$

Question 1 Is $G$ naturally acts on $E$?

Question 2 Is $E$ injective hull of $k$ as a $R^G$-module?

Actually I'm thinking the case where $R$ is the local ring of an isolated hypersurface singularity $(f=0)$ over $\mathbb{C}$ and $f$ is a $\mathbb{Z}_m$-eigenfunction.

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I assume that $I := E$? – Karl Schwede Feb 1 '12 at 15:28
Thank you. I corrected it. – tarosano Feb 1 '12 at 17:26
up vote 1 down vote accepted

In the case of a hypersurface of dimension $d$, or any Gorenstein singularity of dimension $d$, $E \cong H^d_{\mathfrak{m}}(R)$ (of course, this isomorphism is up to multiplication by a unit). $G$ should act on $H^d_{\mathfrak{m}}(R)$ directly (you should even be able to do this explicitly via Cech cohomology). This should be enough for question 1 in your setting.

For question 2, I'm pretty sure the answer is no in general. In particular, the socle of $E$ as an $R^G$-module, is probably not 1-dimensional. Note the socle of the injective hull of the residue field is always 1-dimensional (see for example Bruns and Herzog's book).

Let me give an example. Suppose $R = \mathbb{C}[x,y]$, $\mathfrak{m}$ is the origin, and $G = \mathbb{Z}/2$ acts on $R$ by multiplying the variables by ${-1}$. Then $R^G = \mathbb{C}[x^2, xy, y^2]$. The socle (elements killed by the maximal ideal) of $H^2_{\mathfrak{m}}(R)$ as an $R$-module is just the Cech class $[1/(xy)]$. The socle as an $R^G$-module however also includes the elements $[1/(x^2y)], [1/(xy^2)], [1/(x^2y^2)]$ since the relevant maximal ideal of $R^G$ is $(x^2, xy, y^2)$.

The point is that the socle as an $R^G$-module is all elements killed by $x^2, xy$ and $y^2$.

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Karl, I think you want $G$ to act by $-1$, not $i$. – Graham Leuschke Feb 1 '12 at 17:06
Graham, of course you are right. I'll fix it. – Karl Schwede Feb 2 '12 at 1:21
Thank you for the answer. Please take a look at my additional question if possible.… – tarosano Feb 2 '12 at 11:09

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