Does Q(Zeta) = Q(Zeta_12), where Zeta is ANY 12th roots of unity & Zeta_12 = cos(2pi/12)+isin(2pi/12).
In fact, does the general case where we have n instead of 12 hold?
I managed to prove that Q(Zeta) is a subset of Q(Zeta_12) - this follows from the fact that Q(Zeta) is the smallest field with Q & Zeta in it & Zeta = (Zeta_12)^k where k = 1,2,...,12. Therefore Q(Zeta) is a subset of Q(Zeta_12).
But how can I prove that Q(Zeta) is a subset of Q(Zeta_12)? ie - how can I show that Zeta_12 is an element of Q(Zeta) ? :-s
Many thanks x
The reason for me asking this question in the first place is that I need to find the Galois Group of Q(Zeta)/Q.
Now this would be incredibly easy if Q(Zeta) = Q(Zeta_12) as, because Zeta_12 is primitive this is just Z12*.
But I don't know if they are equal! :-s
If not, does anyone have any idea how to answer the question?
Many thanks. x
well, this can't be always true. I mean -1 is one of the 4th roots of unity, but but has *imaginary* numbers in it.
That is the first root of unity for in this case 12, is always a primitive one, so it generates all of the other roots, so you will always have
Basically you get equality iff for gcd(a,n)=1
Oh awesome - thanks Gamma! :-D
So how would you deal with this question "Find the Galois Group of Q(Zeta)/Q, where Zeta is ANY 12th root of unity. & list all the fields between Q(Zeta) & Q" ?
Obviously we can use Gal[Q(Zeta_n)/Q] = Z*n because Zeta in our case is not primitive. So how should this question be tackled?
Many thanks again. :-)
You know that the nth root of unity . It is a primitive root of unity iff gcd(a,n)=1. If it is not primitive, basically just divide out by the greatest common divisor so it will be a primitive . Then you have just reduced it to figuring out what is the galois group of the rationals adjoined a primitive nth root of unity.
To answer this question, one must think a bit about the cyclotomic polynomials and their extensions. You will find that the galois group has order the Euler Totient function (the number of things relatively prime to n. This is the degree of the extension and it can be shown that the cyclotomoic polynomials are irreducible and (therefore as Q has characteristic 0) separable. You will find that it turns out the Galois Group is in fact isomorphic to .
The isomorphismn is given by for send a to .