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.



The Galois group and relations among the roots of a polynomial
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.)


