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This sounds very classical, but it would be great if someone can point out a reference to me.

Let $G \leq \mathrm{SL}(2, \mathbb{C})$ be a finite group with its natural action on $\mathbb{C}[x, y]$, and let $R$ be the ring of invariants. Denote the maximal ideal of $R$ by $\mathfrak{m}$. How is the quotient $\mathbb{C}[x, y]/\mathfrak{m}$ decomposed into irreducible representations of $G$?

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    $\begingroup$ By Chevalley-Shephard-Todd, $G$ is generated by pseudo-reflections (elements having a codimension $1$ fixed locus) if and only if $\mathbb{C}[x,y]/\mathfrak{m}$ is isomorphic to a copy of the group ring $\mathbb{C}[G]$. In general it seems that $\mathbb{C}[x,y]/\mathfrak{m}$ is isomorphic to a direct sum of $\mathbb{C}[G]$ and some other representation. $\endgroup$
    – Jason Starr
    Commented Feb 16, 2017 at 9:13
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    $\begingroup$ I realize now that a nontrivial subgroup of $\textbf{SL}(2,\mathbb{C})$ cannot be generated by pseudo-reflections, so always $\mathbb{C}[x,y]/\mathfrak{m}$ is strictly bigger than $\mathbb{C}[G]$. For a cyclic group subgroup $G$, $\mathbb{C}[x,y]/\mathfrak{m}$ is isomorphic to $\mathbb{C}[G]$ direct sum with $\mathbb{C}[G]/\mathbb{C}$, the coinvariants. $\endgroup$
    – Jason Starr
    Commented Feb 16, 2017 at 9:19
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    $\begingroup$ There is one more small observation. The trivial character has multiplicity one in $\mathbb{C}[x,y]/\mathfrak{m}$. Indeed, $\mathfrak{m}$ is generated as a $\mathbb{C}[x,y]$-ideal by the invariant elements of positive degree. So every trivial $G$-subrepresentation of $\langle x,y\rangle$ is contained in $\mathfrak{m}$. $\endgroup$
    – Jason Starr
    Commented Feb 16, 2017 at 12:33
  • $\begingroup$ @JasonStarr: Thank you very much, Jason! This is good enough. $\endgroup$
    – Xudong
    Commented Feb 16, 2017 at 13:40
  • $\begingroup$ @JasonStarr: I guess another way of phrasing the statement is that the cardinality of a minimal generating set of any non-free maximal Cohen-Macaulay module over the invariant ring is twice the rank of the module. $\endgroup$
    – Xudong
    Commented Feb 16, 2017 at 15:46

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