- #1
PsychonautQQ
- 781
- 10
Homework Statement
Let c be a primitive 3rd root of unity and b be the third real root of four. Now consider the extension Q(c,b):Q. Find the degree of this extension, show that it is Galois, and calculate Gal(Q(c,b):Q) and then use the Galois group to calculate all intermediate fields.
Homework Equations
The Attempt at a Solution
The minimal polynomial of b over Q is x^3-4. The minimal polynomial of c over Q(b) is x^2+x+1. Therefore [Q(c,b):Q]=6.
Char(Q)=0 so the extension is separable. Q(c,b) is a splitting field for x^3-4, thus it is a finite splitting field over Q. Thus the extension Q(c,b):Q is Galois.
I am having a bit of trouble calculating the Galois group. I know there must be a map m such that m(c)=c^2 whilst fixing everything else, and there is a map n such that n(b)=cb whilst fixing everything else. Therefore all the roots of the minimal polynomial of b can be obtained by applying n to b, i.e. n*n(b)=n(cb)=(c^2)b whilst applying n to (c^2)b gives n((c^2)b)=b. The map m will take c to the other root of it's minimal polynomial in Q and then back again, m(c) = c^2 and m(c^2)=(c^2)^2=c^4=c.
So my first (naïve) thought was that this Galois group would be z3 x z2, although I suspected this would be incorrect because there is some overlapping of the maps (since m takes b to a root that is now effected by the map n). I showed that m*n(b+c+c^2) does not equal n*m(b+c+c^2) and thus this group is not abelian. I know the order of the group must be 6 because that is the degree of the extension (which is Galois) and I know that the group is non abelian.
Non abelian groups of order 6... S_3, D_3, maybe some SDP Z3xZ2. Perhaps the group must be S_3 because it contains a 3-cycle (the map m) and a transposition (the map n).
Anyway, that's where I'm at. I believe the Galois group will be S_3. if this turns out to be true, then I will then need to find all subgroups of S_3 and then go through the gritty business of using the subgroups of the Galois group to find the intermediate fields. Subgroups of S_3 will be A_3, which is all elements of S_3 that can be expressed as an even number of transpositions and A_3 will have order 3, thus it will be cyclic of order 3. S_3 also has a subgroup generated by the transposition that is one of the two generators for S_3, in this case the map n. Thus <n> will be a subgroup of S_3 of order 2. So I believe that <n> and A_3 are the only two proper subgroups of S_3.
I'll leave it at this for now and have somebody check my work, for all I know the Galois group isn't even S_3, and if it is I'm not sure if it only has 2 proper subgroups.