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This is a question related to the two dimensional McKay correspondence. Let $R = \mathbb{C}[x,y]$, and $G$ a finite group acting on $R$. Recall that a $G$-equivariant $R$-module is an $R$-module with a $G$ action, such that the multiplication map $R \otimes M \to M$ is $G$-equivariant, i.e., $g *(rm) = (g\cdot r) (g *m)$.

It is well known that the category of such modules, $R\text{-Mod}_G$, is equivalent to the category of modules over the (noncommutative, unital) skew group algebra $R \# G$, which is the algebra on the underlying vector space $R \otimes_{\mathbb{C}} \mathbb{C}[G]$, and multiplication defined by $(r\otimes g)(r' \otimes g') = r(g\cdot r')\otimes gg'$.

I would like to know what the projective objects are in the latter category. I know that Quillen-Suslin implies that finitely generated projective $R$-modules are free, and so I suspect that all the finitely generated projective equivariant $R$-modules are also free. However I would like to know a proof of this in the setting of the skew group ring. I am aware that the group ring $\mathbb{C}[G]$ is semisimple, so one would hope that $R \# G$ also has a similar decomposition, but I can only see this as vector spaces, rather then algebras. Moreover I'm still not sure how I would use this to get projective objects in $R\# G$-Mod.

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  • $\begingroup$ Almost immediately (sigh...) after posting this I found this question which is very related. I would still like to know why the decomposition of $R \# G$ holds as algebras, and I think that the question about projectives is still unclear, as certainly $R\# G$ can't be semisimple (there are $R\# G$ modules which are not projective, right?) $\endgroup$ – DKS Mar 22 at 18:03
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A first observation is that the centre $Z=Z(R\#G)$ is isomorphic to the fixed point ring $R^G$. Next, we can define a ring homomorphism $$ R\# G \to \mathrm{End}_Z(R), \quad r\otimes g \mapsto (t\mapsto rg(t)). $$ Auslander showed that this is actually an isomorphism.

We therefore have an equivalence between the category $\mathrm{proj}(R\# G)$ of finitely generated projective $R\# G$-modules, and the category $\mathrm{add}_Z(R)$ of $Z$-module direct summands of $R^n$ for $n\geq1$.

We also have an equivalence between $\mathrm{proj}(R\# G)$ and the semisimple category $\mathrm{rep}_{\mathbb C}\,G=\mathrm{mod}\,\mathbb C[G]$.

This sends a $G$-representation $V$ to $R\otimes_{\mathbb C}V$, with its natural $R\#G$-module structure. This will be projective since there exists some $W$ such that $V\oplus W$ is a free $\mathbb C[G]$-module. Alternatively it is extension of scalars along the inclusion $\mathbb C[G]\subset R\#G$.

Conversely it sends a projective $R\#G$-module $P$ to the quotient $P/(xP+yP)$. Alternatively this is extension of scalars along the quotient map $R\#G\to\mathbb C[G]$ which sends $x,y$ to zero.

This latter equivalence shows that the isomorphism classes of indecomposable projective $R\#G$-modules are in bijection with the isomorphism classes of irreducible $G$-representations.

Update: The paper by Auslander is 'On the purity of the branch locus' Amer J Math 84 (1962), and note that he takes $R=[[x,y]]$ with $G$ acting linearly.

A good reference is chapter 5 of 'Cohen-Macaulay Representations' by Leuschke and Wiegand.

Note that if we pass to the quotient fields, then this is just Galois theory. Let $L=\mathbb C(x,y)$ and $K=L^G$. Then $L/K$ is a Galois extension with Galois group $G$, and so $L\#G\cong \mathrm{End}_K(L)$.

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  • $\begingroup$ Thank you for your excellent answer. Do you happen to have a reference for the result of Auslander that you quoted? $\endgroup$ – DKS Apr 2 at 12:00
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You need the action to be faithful to have the centre be the fixed point ring. The endomorphism ring and the skew group ring are isomorphic when the group is a small subgroup of GL.

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