The Galois group and relations among the roots of a polynomial

Question

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.

Answer 1

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.)