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# A question about an application of Moeln'sMolien's formula to find the generators and relations of an invariant ring

In the very beginning of the book "Introduction to Invariants and Moduli" Shigeru Mukai proves Moeln's Molien's formula for the Hilbert series of the invariant ring of a finite group action on $\mathbb C^n$. For example, in the case of the standard action of Quaternions on $\mathbb C^2$ the Hilbert series is $\frac{1-t^{12}}{(1-t^4)^2(1-t^6)}$.

After this Mukai explains that this formula hints us that the ring of invariants can be generated by two elements of order $4$ and one element of order $6$, and indeed such elements can be found: $A=x^4+y^4$, $B=x^2y^2$, and $C=xy(x^4-y^4)$. Then one finds a relation $C^2=A^2B -4B^3$ and this gives a complete description of the ring of invariants.

Moreover the same procedure is shown to work in several other cases (e.g. binary icosahedral group).

My question is a follows: Is there some theorem that say that this heuristics works often? Namely, if we have an action of a finite group $G$ on $\mathbb C^n$, in order to describe the ring of invariants, we first look on the denominator of the Hilbert series (given by Moeln's Molien's formula) and try to associate an invariant polynomial of degree $n$ to each factor $(1-t^n)$ (so that this gives us a full set of generators). Or at least, in practice, is this the first thing that one tries to do?

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# A question about an application of Moeln's formula to find the generators and relations of an invariant ring

In the very beginning of the book "Introduction to Invariants and Moduli" Shigeru Mukai proves Moeln's formula for the Hilbert series of the invariant ring of a finite group action on $\mathbb C^n$. For example, in the case of the standard action of Quaternions on $\mathbb C^2$ the Hilbert series is $\frac{1-t^{12}}{(1-t^4)^2(1-t^6)}$.

After this Mukai explains that this formula hints us that the ring of invariants can be generated by two elements of order $4$ and one element of order $6$, and indeed such elements can be found: $A=x^4+y^4$, $B=x^2y^2$, and $C=xy(x^4-y^4)$. Then one finds a relation $C^2=A^2B -4B^3$ and this gives a complete description of the ring of invariants.

Moreover the same procedure is shown to work in several other cases (e.g. binary icosahedral group).

My question is a follows: Is there some theorem that say that this heuristics works often? Namely, if we have an action of a finite group $G$ on $\mathbb C^n$, in order to describe the ring of invariants, we first look on the denominator of the Hilbert series (given by Moeln's formula) and try to associate an invariant polynomial of degree $n$ to each factor $(1-t^n)$ (so that this gives us a full set of generators). Or at least, in practice, is this the first thing that one tries to do?