Let $p,q$ be odd primes. Consider the polynomial ring $\mathbb C[x\_0,...,x\_{q-1}]$. For $m=0,1,...,p-1$, let


$$\sigma\_m=\sum\_{0\le j\_0\le p;...;0\le j\_{q-1}\le p; j\_1+...+j\_{q-1}=p; 1.j\_1+...+(q-1)j\_{q-1}\equiv m (\mod p)} \dfrac {p!}{j\_0!...j\_{q-1}!} x\_{0}^{j\_0}...x\_{q-1}^{j\_{q-1}}$$. 


Notice that $\sigma\_0+\sigma\_1+...+\sigma\_{p-1}=(x\_1+...+x\_{q-1})^p$ . 


Let $K$ be the fraction field of $\mathbb C[\sigma\_0,...,\sigma\_{p-1}]$.


For which $p,q$, is it true that $\mathbb C(x\_0,...,x\_{q-1})$ is a finite Galois extension of $K$ ?



On a special type of subring of $\mathbb C[x_0,...,x_{q-1}]$
Since $K$ is generated by $p$ elements, $\mathbb{C}(x\_0,\ldots,x\_{q-1})$ cannot be an algebraic extension of $K$ if $q>p$. I claim that $\mathbb{C}(x\_0,\ldots,x\_{q-1})$ is a finite Galois extension of $K$ whenever $p\geq q$.


Define $\zeta=e^{2\pi i/p}\in\mathbb{C}$, and for $0\leq k\leq p-1$, define
$$
\omega\_k:=\sum\_{i=0}^{p-1} \zeta^{ik}\sigma\_i, y\_k:=\sum\_{i=0}^{q-1} \zeta^{ik}x\_i.
$$
The $\mathbb{C}$-span of $\{\omega\_i\}$ equals the $\mathbb{C}$-span of $\{\sigma\_i\}$, and the $\mathbb{C}$-span of $\{y\_i\}$ equals the $\mathbb{C}$-span of $\{x\_i\}$ (the change of coordinates matrices are Vandermonde, hence invertible). So it suffices to show that
$$
\mathbb{C}(y\_0,\ldots,y\_{p-1})\big/\mathbb{C}(\omega\_0,\ldots,\omega\_{p-1})
$$
is a finite Galois extension. Finally, $y\_k^p = \omega\_k$, so the extension comes from adjoining $p$-th roots. This implies the extension is finite Galois, since $\mathbb{C}$ contains $p$-th roots of unity.


