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Let $S_{n}$ denote the permutation group on $n$ letters and $G\subset S_{n}$ a transitive subgroup. The inclusion of $G$ in $S_{n}$ defines an action of $G$ on $\mathbb{C}^{n}$. By finding a generating set of invariant polynomials and relations among these polynomials we may realize the quotient space $\mathbb{C}^{n}/G$ as an algebraic variety.

I have noticed that for $n=2,3,4$ such quotients are always complete intersections. That is, the difference between the order of a minimal generating set for the invariant polynomials, and the order of a minimal generating set for the relations, is equal to $n=\mathrm{dim}(\mathbb{C}^{n}/G)$.

I would like to know if this fact persists for all $n$. If so, what is the property of the group action which makes the quotient a complete intersection? If not what is a counterexample?

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up vote 13 down vote accepted

This paper of Kac and Watanabe may be of interest:

Kac, Victor; Watanabe, Keiichi, Finite linear groups whose ring of invariants is a complete intersection

From the review:

"The following theorems are proved: Let $k$ be a field and $G$ be a finite subgroup of $\text{GL}(n,k)$. The group $G$ acts naturally on the polynomial ring $S=k[x_1,\cdots,x_n]$. Let $R=S^G$.

Theorem A: If $R$ is a complete intersection then $G$ is generated by the set $\lbrace g\in G\colon\text{rank}(g-I)\leq 2\rbrace$.

Theorem B: Let $k={\bf C}$. If $R$ has $m$ generators such that their ideal of relations is generated by $m-n+s$ elements, then $G$ is generated by the set $\lbrace g\in G\colon\text{rank}(g-I)\leq s+2\rbrace$.

Theorem C: If $R$ is a complete intersection then each stabilizer $G_x$, $x\in k^n$, is generated by $\lbrace g\in G_x\colon\text{rank}(g-I)\leq 2\rbrace$. "

In this case, the cyclic group of $S_4$ generated by the long cycle acting on ${\bf C}^4$ should be a counterexample. I should probably calculate the ring of invariants, but I'll leave it at that for now.

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