Let $f(x) \in \mathbb{Z}[x]$ be a monic irreducible polynomial of degree $n$, and let $\alpha_1, \alpha_2, … , \alpha_n \in \overline{\mathbb{Q}}$ be the $n$ distinct roots of $f(x)$.
Following Bewersdorff's "Galois Theory for Beginners" (and older sources?) I want to define the Galois group of $f(x)$ as follows. Let $I \subset \mathbb{Q}[x_1, x_2, … , x_n]$ be the ideal consisting of those polynomials $g(x_1, x_2, … , x_n)$ in $n$ variables such that $$g(\alpha_1, \alpha_2, … , \alpha_n) = 0$$
I propose to call $I$ the Galois ideal of $f(x)$. If there is a more standard term for this, please let me know.
Now the Galois group of $f(x)$ may be defined as the group $G \subset S_n$ consisting of those permutations $\sigma \in S_n$ such that $$g \in I \Longrightarrow g_\sigma \in I$$
where $g_\sigma$ is the polynomial $g$ transformed by permuting the variables using the permutation $\sigma$.
Certain polynomials will be trivially members of $I$ for any polynomial $f(x)$. Specifically, $f(x_1), f(x_2), … , f(x_n) \in I$, and also the elementary symmetric polynomials minus the coefficients of $f$ will be members, e.g. $x_{1}x_{2} … x_{n} – (-1)^{n} c_0$, where $c_0$ is the constant coefficient of $f$.
For some ("most"?!) polynomials $f(x)$, the Galois ideal is generated only by these trivial members, and for such polynomials the Galois group is the full symmetric group. To the extent that there are non-trivial generators of $I$, the Galois group will be a proper subgroup of $S_n$.
For example, if $f(x) = x^4 – 2$, and $\alpha_1 = \sqrt[4]{2}$, $\alpha_2 = -\sqrt[4]{2}$, $\alpha_3 = i\sqrt[4]{2}$, $\alpha_4 = -i\sqrt[4]{2}$, then the non-trivial generators of $I$ are $$g_1(x_1,x_2,x_3,x_4) = x_1 + x_2$$ $$g_2(x_1,x_2,x_3,x_4) = x_3 + x_4$$ Accordingly, the Galois group is generated only by the permutations $$1234 \rightarrow 2134$$ $$1234 \rightarrow 1243$$ $$1234 \rightarrow 3412$$
My question is, is there an algorithm to find the non-trivial generators of the Galois ideal? It seems that this ideal actually gives more information about the polynomial $f(x)$ than the Galois group, since it is trivial to find the Galois group given the Galois ideal, but not conversely.
Best Answer
The Galois ideal is one of the prime factors of the ideal generated by the trivial elements. Thus, an algorithm for primary decomposition in $\mathbb Q[x_1,...x_n]$, of which there are several, will do the trick.
Proof: The Galois ideal is prime, which is obvious from its definition. It contains the trivial ideal. Since the vanishing set of the trivial ideal has dimension $0$, all prime ideals containing it are maximal, thus all prime ideals are prime factors.
(In terms of "most", if I remember correctly, almost all polynomials with coefficients of bounded height have Galois group the full symmetric group, but I am not an expert on such things.)