# Understanding fixed fields (Galois Theory)

1. Dec 2, 2016

### PsychonautQQ

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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 eachother, 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.

2. Dec 7, 2016