# On a special type of subring of $\mathbb C[x_0,…,x_{q-1}]$

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 $$K$$ ?

• 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 $\sum_{0\leq j_0\leq p,\ldots,0\leq j_{q-1}\leq p:\sum j_i = p,\sum ij_i\equiv m\mod p} \frac{p!}{j_0!\cdots j_{q-1}!}x_0^{j_0}\cdots x_{q-1}^{j_{q-1}}$? – Julian Rosen Dec 15 '18 at 22:49
• @JulianRosen: yes, thanks ... I edited ... – user521337 Dec 16 '18 at 3:16

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.
• 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 ... – user521337 Dec 17 '18 at 5:57
• 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. – Julian Rosen Dec 17 '18 at 17:04
• ok, yeah I got that, thanks ... could you please explain why $y_k^p=\omega_k$ ? – user521337 Dec 18 '18 at 3:10
• 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 $y_k^p=\sum_m \zeta^{km}\sigma_m$. – Julian Rosen Dec 19 '18 at 13:35