# Question on Galois Group

jeffreydk
I am trying to show that if z, z2, z3, ..., zn=1 are n distinct roots of xn-1 in some extension field of Q (the rationals), then GalQQ(z) (the galois group of Q(z) over Q) is abelian. Would I be wrong to say that since the galois group we're talking about here only involves an extension field with one of the roots, namely z, then the only map we could have in the group would be the identity map and therefore it is abelian? Something feels wrong about this but I'm not sure how else there would be other automorphisms in the group.

## Answers and Replies

Staff Emeritus
Gold Member
To say that GalQ Q(z) is abelian requires an imlicit assumption: that GalQ Q(z) actually exists. For that to happen. Q(z) must actually be a Galois extension. Also, if Q(z) is a Galois extension of Q, then #GalQ Q(z) = [Q(z) : Q] -- so if GalQ Q(z) consists only of the identity, then z must actually be a rational number... which is only true if n < 3.

The problem secretly told you that Q(z) is a Galois extension of Q, and you know that automorphisms of Q(z) map roots of z^n - 1 to roots of z^n - 1. You made an unwarranted assumption that led you to conclude that the only automorphism of Q(z) is trivial -- what was it?

If you're still confused, maybe it's worth first working out the Galois group of the entire splitting field of x^n - 1, rather than just of Q(z).

jeffreydk
Yes that makes sense that its a Galois extension, because we're told that each of the roots are distinct (and there are n of them) so it is a splitting field and has characteristic 0, thus it's Galois. Although your argument makes perfect sense, saying that |GalQQ(z)|=[Q(z):Q]=1 if the group is trivial, I am having trouble noticing where my unwarranted assumption was, because to me, if the field is only extending z and not the rest of the roots, there are no combinations of isomorphisms that can be made other than some f:z-->z. I'll try working out the group for the entire splitting field.

Staff Emeritus