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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.

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If you have Maple you can enter interface(verboseproc=3) and then eval(galois) to see how Maple calculates Galois groups. The main galois() function calls various other functions which you can display similarly by entering eval(galois/absres) and so on. Presumably one can do the same with various other systems. –  Neil Strickland May 19 '12 at 15:32
    
Thanks Neil-- I don't have Maple but I did check Pari, and it has a case-by-case algorithm for poly's of degree up to 11, which I found uninspiring. As mentioned, I'm really more interested in the Galois ideal than in the Galois group, and I'm not aware of anything that computes the non-trivial generators of this ideal. I am going to try to look up Eisenbud's paper on primary decomposition in Inventiones 110 if I get a chance. –  Carsten May 19 '12 at 16:35
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1 Answer

up vote 3 down vote accepted

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.)

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OK, forgive me for being dense, but how does a primary decomposition in $\mathbb{Q}[x_1,x_2,...,x_n]$ take into account the particular polynomial $f(x)$? Can you give me an example of how this works for $f(x) = x^4 - 2$ ? –  Carsten May 19 '12 at 2:16
    
It is the primary decomposition of the trivial ideal, which takes into account the particular polynomial. (Note that the trivial ideal is generated by just the elementary symmetrical polynomials minus the coefficients of $f$.) –  Will Sawin May 19 '12 at 2:25
    
So for that $f$ we have the equations $x_1+x_2+x_3+x_4$, $x_1x_2+...+x_3x_4$, $x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_l$, and $x_1x_2x_3x_4=2$. This ideal has three prime factors, one for each coset of $D_4$ in $S_4$. The first factor is generated by $x_1+x_3$, $x_2+x_4$, $x_1^2+x_2^2$, $x_2^2+x_3^2$, $x_3^2+x_4^3$, $x_4^2+x_1^2, etc. The other factors are similar, but permuted. –  Will Sawin May 19 '12 at 2:28
    
OK, I see what you mean now. I will try to look up the decomposition algorithms. –  Carsten May 19 '12 at 2:42
    
I guess I was confused because the trivial ideal is determined as soon as you know $f(x)$, but to find the Galois ideal you have to specify a numbering of the roots $\alpha_1, \alpha_2, ... \alpha_n$. Maybe you're telling me that there are conjugate Galois ideals corresponding to the conjugate subgroups of the Galois group $G \subset S_n$? I'm going to have to chew on this some more. –  Carsten May 19 '12 at 17:38
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