The elementary symmetric functions are a homogeneous system of parameters for the invariant ring of a permutation representation For a permutation representation of order $n$ over a field $F$, I need to show that the elementary symmetric functions $s_1,s_2,\ldots,s_n$ form a homogeneous system of parameters for the ring of polynomial invariants $R$.
I have shown that the elementary symmetric functions are algebraically independent, so it suffices to show that the extension $F[s_1,\ldots,s_n]\subset R$ is finite. I know that for a permutation representation, the ring of polynomial invariants $R$ is generated by $s_n$ and the orbit sums of simple monomials, but I can't figure out how to show that it is finitely generated as an $F[s_1,\ldots,s_n]$-module.
 A: There is a nice Noetherian trick here. Your invariant ring $R$ is a subring of the polynomial ring $A=\mathbb{F}[x_1,x_2,\cdots,x_n]$. Now, the homogeneous symmetric functions $s_i$ form a homogeneous system of parameters (h.s.o.p.) for $A$:
We will first show that $A$ is integral over $\mathbb{F}[s_1,s_2,\cdots s_n]$. It is enough to show this for the $x_i$'s because they generate $A$ as an algebra and because integral elements form a ring. Now each $x_i$ is clearly a root of the following natural polynomial (in $t$)$$\prod_{\sigma\in S_n}(t-\sigma\cdot x_i)$$ where $S_n$ is the symmetric group on $n$ elements. Indeed, because of the identity $e\in S_n$, the product always contains the term $(t-x_i)$. The polynomial is also monic but more importantly its coefficients lie in $\mathbb{F}[s_1,s_2,\cdots s_n]$. They are just the symmetric functions $s_i$. 
Now, the above mean that each $x_i$ is integral over $\mathbb{F}[s_1,s_2,\cdots s_n]$ and hence all of $A$ is. Now $A$ is finitely generated as an algebra and integral over $\mathbb{F}[s_1,s_2,\cdots s_n]$ and this implies it is finitely generated as a $\mathbb{F}[s_1,s_2,\cdots s_n]$-module (see Finiteness of Integral Closure).
Finally the invariant ring $R$ is a submodule of $A=\mathbb{F}[x_1,x_2,\cdots,x_n]$ hence it is also finitely generated (that is because $A$ is a Noetherian module, it is not always true that submodules of finitely generated modules are finitely generated...).
Now all this is actually from two exercises (13.2 and 13.3) in Eisenbud (Commutative Algebra with a view to a kill). However, I would like to mention this geometric way to view the situation that sheds light to the "parameter" part of h.s.o.p.
We can identify $\mathbb{F}[s_1,s_2,\cdots s_n]$ with $\mathbb{F}[y_1,y_2,\cdots y_n]$ (where $y_i$'s are indeterminates) because the $s_i$ are algebraically independent, which means that its Spec is the affine space of dim $n$. Then, the fact that $A$ is finitely generated over $\mathbb{F}[s_1,s_2,\cdots s_n]$ means that the map $$(x_1,x_2,\cdots,x_n)\rightarrow (s_1(x_1,x_2,\cdots,x_n),s_2(\cdots),\cdots,s_n(\cdots))$$ has finite fibers.
This map restricted on the complement of the Braid arrangement, i.e. restricted on $X=\{(x_1,x_2,\cdots,x_n):x_i\neq x_j\}$ is a covering map that results from a group action. Now, since our group $G$ acts as a subgroup of $S_n$, we will have a tower of coverings $$X\rightarrow X/G\rightarrow X/S_n$$ Since the fibers of $X\rightarrow X/S_n$ are finite, so must be the fibers of $X/G\rightarrow X/S_n$. This should imply that $R$ is finitely generated as a $\mathbb{F}[s_1,s_2,\cdots, s_n]$ module.
I would appreciate it if someone made this last part rigorous, as my algebraic geometry is very very limited... Is there significant hindrance with the fact that $X$ is not all of the affine space? Maybe this cannot lead to a different proof but just provide visualization...?
