Timeline for The Galois group and relations among the roots of a polynomial
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| May 26, 2012 at 15:51 | comment | added | Will Sawin | That is correct. We are. | |
| May 26, 2012 at 15:28 | comment | added | Carsten | And $rGr^{-1}$ is a conjugate subgroup of $G$, right? So there is a 1:1 correspondence between cosets and ideals, and then some kind of onto mapping from that onto the conjugate subgroups. I think we're saying the same thing. | |
| May 26, 2012 at 14:01 | comment | added | Will Sawin | Consider the action of $S_n$ on the Galois ideal. The galois group is defined to be the subgroup that fixes the ideal, that is, the stabilizer. By the orbit-stabilizer theorem, the orbit of the Galois ideal under the $S_n$ action consists of $[S_n:G]$ ideals, each corresponding to a coset of $S_n/G$. Each of those ideals is a Galois ideal, since each one is the same set of equations with a relabeling of $x_1,...,x_n$, and there is no canonical labeling of $x_1,...,x_n$. The Galois group of a Galois ideal is its stabilizer: for the coset $rG$, this is $rGr^{-1}$. | |
| May 26, 2012 at 12:34 | comment | added | Carsten | Thanks for your responses, Will. The only thing I still don't understand is your comment about cosets. Whichever version of the Galois ideal you look at, it is clear from the definition that the corresponding Galois group is indeed a group. And a coset is not a group. I still think the different Galois ideals must correspond to conjugate subgroups of $G \subset S_n$. In your example, as you say, $A_3 \vartriangleleft S_3$, so $A_3$ has no distinct conjugate subgroups. That makes sense. I just don't see where cosets come into play. | |
| May 19, 2012 at 18:50 | comment | added | Will Sawin | We can view this explicitly in $\mathbb Q(\omega)$, where $\omega$ is a primitive third root of unity. Take the equation $f(x)=x^3-2$, then one Galois ideal will have the equations $x_1-\omega x_2, x_2-\omega x_3$ while the other will have $x_3 - \omega x_2$, $x_2-\omega x_1$. Both have the same symmetry group, but they are different. | |
| May 19, 2012 at 18:49 | comment | added | Will Sawin | The Galois ideal is one of the prime factors. The other prime factors are other versions of the Galois ideal. I don't think they quite correspond to conjugate subgroups. Instead, it corresponds to the cosets in $S_n/G$. For instance, an irreducible cubic with discriminant a perfect square should have a trivial ideal that splits into two Galois ideals, each giving the same subgroup, $A_3$, since $A_3$ is a normal subgroup. | |
| May 19, 2012 at 17:38 | comment | added | Carsten | 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. | |
| May 19, 2012 at 2:42 | comment | added | Carsten | OK, I see what you mean now. I will try to look up the decomposition algorithms. | |
| May 19, 2012 at 2:40 | vote | accept | Carsten | ||
| May 19, 2012 at 2:28 | comment | added | Will Sawin | 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. | |
| May 19, 2012 at 2:25 | comment | added | Will Sawin | 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 $f$.) | |
| May 19, 2012 at 2:16 | comment | added | Carsten | 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 $f(x) = x^4 - 2$ ? | |
| May 19, 2012 at 2:01 | history | answered | Will Sawin | CC BY-SA 3.0 |