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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 $K$ ?

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    $\begingroup$ What is the $k$ that appears in the sum? Also, if you want $\sigma_0+\ldots+\sigma_{p-1}=(x_1+\ldots+x_{q-1})^p$, then I don't see what role the $i$'s are playing in the sum. I might be confused, but did you want something like $\sum_{0\leq j_0\leq p,\ldots,0\leq j_{q-1}\leq p:\sum j_i = p,\sum ij_i\equiv m\mod p} \frac{p!}{j_0!\cdots j_{q-1}!}x_0^{j_0}\cdots x_{q-1}^{j_{q-1}}$? $\endgroup$
    – Julian Rosen
    Commented Dec 15, 2018 at 22:49
  • $\begingroup$ @JulianRosen: yes, thanks ... I edited ... $\endgroup$
    – user521337
    Commented Dec 16, 2018 at 3:16

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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.

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  • $\begingroup$ Why is $\mathbb C$-span of $\{y_i\}$ the same as the $\mathbb C$-span of $\{x_i\}$ ? the change matrix is not even a square matrix ... $\endgroup$
    – user521337
    Commented Dec 17, 2018 at 5:57
  • $\begingroup$ Sorry, I should have said this better. The matrix taking $x_0,\ldots,x_{q-1}$ to $y_0,\ldots,y_{q-1}$ is square and Vandermonde, so the span of the $x$'s equals the span of $y_0,\ldots,y_{q-1}$. The extra $y$'s ($y_q,\ldots,y_{p-1}$) are by definition in the span of the $x$'s. $\endgroup$
    – Julian Rosen
    Commented Dec 17, 2018 at 17:04
  • $\begingroup$ ok, yeah I got that, thanks ... could you please explain why $y_k^p=\omega_k$ ? $\endgroup$
    – user521337
    Commented Dec 18, 2018 at 3:10
  • $\begingroup$ Using the multinomial theorem, $y_k^p=\sum_{j_i} \frac{p!}{j_0!\cdots j_{q-1}!}x_0^{q_0}\cdots x_{q-1}^{j_{q-1}}\zeta^{k(0j_0+1j_1+\ldots (q-1)j_{q-1})}$. Now if $0j_0+1j_1+\ldots (q-1)j_{q-1}\equiv m\mod p$, then $\zeta^{k(0j_0+1j_1+\ldots (q-1)j_{q-1})}= \zeta^{km}$. So $y_k^p=\sum_m \zeta^{km}\sigma_m$. $\endgroup$
    – Julian Rosen
    Commented Dec 19, 2018 at 13:35

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