Understanding fixed fields (Galois Theory)

[PsychonautQQ]

Homework Statement

Let's look at Q(a,c):Q where a is the third real root of 2 and c is a primitive cube root of unity. Then this extension is Galois and it's Galois group is isomorphic to $D_3=S_3$. The proper subgroups of the Galois group are thus <(12)>,<(23)>,<(23)>,<(123)>. Let (12) switch the roots a and ab, (13) switch the roots a and ab^2, (23) switch the roots b and b^2 and (123) be the 3 cycle that will move all roots.

I'm having trouble understanding the intermediate fields that these groups correspond to.

Homework Equations

The Attempt at a Solution

So <(12)> will switch the roots a and ab, thus a+ab must be an element of the fixed field, as must a*ab. is the entire fixed field of the subgroup <(12)> Q(a+ab) and a*ab is an element of Q(a+ab) somehow? But then again, won't b^2 be fixed by this transposition of (12)? I'm picture the complex plane where a is on the positive x-axis and b and b^2 are complex numbers with positive and negative imaginary components respectively on the left side of the y axis.

And then what would the subgroup <(23)> fix? This is the transpotion that corresponds to complex conjugation, so surely b+b^2 would be fixed, but b+b^2=-1 so this is part of Q. This transposition must also fix a, so is the fixed field <(23)> simply Q(a)?

And then I'm having the same problem with <(13)> as I am with <(12)>. I mean surely <(13)> must fix a+ab^2 as it sends them to each other, but would it fix ab also? My intuition tells me that the fixed field of <(13)> is just Q(a+ab^2) but I'm not exactly sure why.

Then I think <(123)> would fix Q(a+ab+ab^2).

I'd appreciate any feedback and clarification on what's going on here. I'd prefer to understand this in terms of permutations in $S_3$, but I wouldn't mind understanding it in terms of elements of $D_3$ also.

 

[mighty2000]

Hi there,

The intermediate fields corresponding to the subgroups of the Galois group can be understood in terms of fixed fields. Let's start with <(12)>. This subgroup corresponds to the transposition (12), which switches the roots a and ab. As you correctly mentioned, the fixed field of this subgroup is Q(a+ab), since this is the field of elements that are fixed by the transposition. This means that any element in this field will remain unchanged when we apply the transposition (12). So, as you said, a+ab must be in this field, as well as a*ab. However, b^2 is not fixed by this transposition, since it is sent to b which is not equal to itself. Therefore, b^2 is not in the fixed field of <(12)>.

Moving on to <(23)>, this subgroup corresponds to the transposition (23), which switches the roots b and b^2. The fixed field of this subgroup is Q(a), since a is the only root that is fixed by this transposition. As you mentioned, b+b^2 is also fixed by this transposition, but this element is already in Q(a) since it is just -1. So the fixed field of <(23)> is simply Q(a).

For <(13)>, this subgroup corresponds to the transposition (13), which switches the roots a and ab^2. The fixed field of this subgroup is Q(a+ab^2), since this is the field of elements that are fixed by the transposition (13). This means that a+ab^2 must be in this field, as well as a*ab^2. However, ab is not fixed by this transposition, since it is sent to b which is not equal to itself. Therefore, ab is not in the fixed field of <(13)>.

Finally, for <(123)>, this subgroup corresponds to the 3-cycle (123), which moves all of the roots. The fixed field of this subgroup is Q(a+ab+ab^2), since this is the field of elements that are fixed by the 3-cycle (123). This means that any element in this field will remain unchanged when we apply the 3-cycle. So, as you mentioned, a+ab+ab^2 must be in this field. However, a*ab and a*ab^2 are not fixed by