Let $G$ be a finite group with $n$ elements and let $\mathbb{Q}[x\_1,\cdots,x\_n]^G = \mathbb{Q}[g\_1,\cdots,g\_m]$. Then there exist polynomials $s\_j \in \mathbb{Q}[y\_1,\cdots,y\_m]$ for $j=1,\cdots,n$ such that
$e\_j(x\_1,\cdots,x\_n) = s\_j(g\_1(x\_1,\cdots,x\_n), \cdots, g\_m(x\_1,\cdots,x\_n))$ for $j=1,\cdots,n$ where $e\_j$ is the $j$-elementary symmetric polynomial in $x\_k$. Consider the polynomial 
$p(t,y\_1,\cdots,y\_m) = t^n - s\_1(y\_1,\cdots,y\_m) t^{n-1}+\cdots+(-1)^n s\_n(y\_1,\cdots,y\_m)$ which is a polynomial in $R:=\mathbb{Q}[t,y\_1,\cdots,y\_m]$.
Is it true, that this polynomial is irreducible in $R$?



A question about the invariants of a finite group
I think it's irreducible in $R$ as you suggest. Here's a sketch which I think works. If the polynomial factored in a non-trivial way, then because of the $t^n$ term the factors must have degree less than $n$ in $t$ (consider the factorization in $\mathbb{Q}(y\_1,y\_2,\ldots,y\_n)[t]$; note also that we can assume that the factors over $R$ are monic polynomials in $t$). Now specialise via $y\_i\mapsto g\_i$ and we get a non-trivial factorization of the specialised polynomial in $\mathbb{Q}[g\_1,\ldots,g\_m][t]$ and hence in $\mathbb{Q}[x\_1,\ldots,x\_n][t]$ and so in $\mathbb{Q}(x\_1,\ldots,x\_n)[t]$. But we know the complete factorization in this ring, it's just $\prod(t-x\_i)$, so our given factorization must specialise into factors of the form $\prod\_{i\in I}(t-x\_i)$ for some subsets $I$ of $\{1,2,\ldots,n\}$ (with each $I$ not empty or the whole thing), and the constant term of each factor must be in $\mathbb{Q}[g\_1,\ldots,g\_m]$ and hence $G$-invariant. This is a contradiction.


