# Question on Inverse Galois Theory

Let $G$ be a finite group, $n=|G|$. Let $\rho:G\rightarrow GL(n,\mathbb{C})$ be the regular representation. Let $G \le H \le S_n$ be another group. Then we have

$\mathbb{Q}[x_1,\cdots,x_n]^H \le \mathbb{Q}[x_1,\cdots,x_n]^G$ (*)

Let $\mathbb{Q}[x_1,\cdots,x_n]^H = \mathbb{Q}[h_1,\cdots,h_M]$ and $\mathbb{Q}[x_1,\cdots,x_n]^G = \mathbb{Q}[g_1,\cdots,g_m]$. Because of (*) there exists $\forall i = 1,\cdots,M$ polynomials $p_i$ with rational coefficients, such that

$h_i= p_i(g_1,\cdots,g_m)$ $\forall i=1,\cdots,M$ (**)

Let $s_1,\cdots,s_r$ (this might be empty) be the algebraic relations, which are fullfilled by the $g_i$, that is $s_j(g_1,\cdots,g_m) = 0$ $\forall j=1 \cdots r$. Let $a_1,\cdots,a_m \in \mathbb{Q}$ such that $s_j(a_1,\cdots,a_m) = 0$ for all $j=1,\cdots,r$. Then by Hilberts Nullstellensatz there exist $\alpha_1,\cdots,\alpha_n \in \mathbb{C}$ such that $g_i(\alpha_1,\cdots,\alpha_n) = a_i$ for all $i=1,\cdots,m$. Then we also have that

$h_i(\alpha_1,\cdots,\alpha_n) = p_i(g_1(\alpha_1,\cdots,\alpha_n),\cdots,g_m(\alpha_1,\cdots,\alpha_n)) = p_i(a_1,\cdots,a_m)$

is a rational number. Especially for $H=S_n$ we get that $h_i= e_i =$ elementary symmetric poylnomial in $x_1,\cdots,x_n$ and the polynomial

$f(x) = (x-\alpha_1)\cdots (x-\alpha_n) = x^n-e_1x^{n-1}+e_2x^{n-2} \cdots \pm e_n$

has rational coefficients. Let $K := \mathbb{Q}(\alpha_1,\cdots,\alpha_n)$. Is it true, that if $f(x)$ is separable, then

$Gal(K/\mathbb{Q}) \le G$?

Motivation for this question are small order groups which I computed manually. For example: $G=C_3=<a>$. Then $\rho(a) =$ companion matrix of $x^3-1$. One can compute using the Reynolds operator, that $k[x,y,z]^G = k[x+y+z,xy+yz+xz,xyz, x^2y+z^2x+y^2z]$. If we let

$\alpha+\beta+\gamma = a$

$\alpha\beta+\alpha\gamma+\beta\gamma = b$

$\alpha\beta\gamma = c$

$\alpha^2\beta+\gamma^2\alpha+\beta^2\alpha = d$

with $a,b,c,d\in \mathbb{Q}$ such that they fullfill the only (computed with Singular) relation: $a^3c + b^3-6abc-abd+9c^2+3cd+d^2 = 0$ Then since this last equation is quadratic in $d$ its discriminant $\Delta = b^2a^2-4ca^3-4b^3-27c^2+18abc$ must be a square in $\mathbb{Q}$. But $\Delta = \Delta_f$ where $\Delta_f$ is the discriminant of $f(x) = (x-\alpha)(x-\beta)(x-\gamma) = x^3-ax^2+bx-c$, hence $\Delta_f$ is also a square in $\mathbb{Q}$. If $f(x)$ is irreducible, then $Gal(K/\mathbb{Q}) = C_3$. If $f(x)$ is reducible, then [ link here] (https://math.stackexchange.com/questions/1596412/does-a-cubic-polynomial-split-in-linear-factors-over-the-rational-numbers) $f$ splits over $\mathbb{Q}$ hence has trivial Galois group. In every case if $f(x)$ is separable, then its Galois group is a subgroup of $C_3$. (The point here is that sucht things as $(x-1)(x^2+1)$ with Galois group $S_2$ can not happen.)