Extending a field by a 16th primitive root of unity

[PsychonautQQ]

Homework Statement

let c be a primitive 16th root of unity. How many subfields M<Q(c) are there such that [M:Q] = 2

Homework Equations

The Attempt at a Solution

I think the only subfield M of Q(c) such that [M:Q] = 2 is Q(c^8). Then M = {a+b(c^8) such that a,b are elements of Q}. I'm thinking about the other powers of c and trying to think if any other would generate an extension field over Q with a degree of 2. Any number that's relatively prime to 16 would be another primitive 16th root of unity, so we can throw out all odd numbers. Q(c^2) and Q(c^14) would both be degree 8, Q(c^4) and Q(c^12) would both be degree four, Q(c^6) would be of degree 8 and Q(c^10) would also be of degree 8. So it's only [Q(c^8):Q]= 2 correct?

 

[andrewkirk]

I think the only subfield M of Q(c) such that [M:Q] = 2 is Q(c^8).

Think again about that '8' exponent. Note that $c^8=-1$, which is in $\mathbb Q$, so $\mathbb Q(c^8)=\mathbb Q$ (ie the extension is trivial), whence $[\mathbb Q(c^8):\mathbb Q]=1$, not 2.

Q(c^4) and Q(c^12) would both be degree four

Why? I think you are confusing the degree of the extension $[\mathbb Q(c^k):\mathbb Q]$, which is the dimension of the vector space, with the order of the element $c^k$ in the multiplicative group $\{c^j\ :\ j\in\{0,1,...,15\}\}$.