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?



When Are Quotients Complete Intersections?
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*
<http://www.ams.org/mathscinet-getitem?mr=640951>


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.


