Determining of the intermediate fields of the $12$th cyclotomic field Let $\zeta$ be a 12th primitive root over $\mathbb Q$. Determine all intermediate field of $\mathbb Q(\zeta)/\mathbb Q$.
My problem is that this is a task from an old exam where you were not allowed to use some sheets which you prepared at home.
So I would like to know if there are some possibilities in terms of determining the intermediate field without knowing the 12th cyclotomic polynomial (I guess nobody have to memorize all of them..)
So I am looking for something like a "trick" or way how to solve this task. Maybe something which I dont know yes.
Thanks in advance!
 A: The Galois group is $(\mathbb Z/12\mathbb Z)^* \cong V_4$, hence there are $3$ proper intermediate fields of degree $2$, which are made up of square roots.
Obviously the fourth and the third root of unity give rise to the intermediate fields $\mathbb Q(i)$ and $\mathbb Q(i\sqrt{3})$. So the third one is given by $\mathbb Q(\sqrt{3})$.
A: It’s worth while recognizing what the twelfth roots of unity are. You know that the cube roots of unity are, they’re $1$ and $(-1\pm i\sqrt3)/2$. You also know what the fourth roots of unity are, they’re $\pm1$ and $\pm i$. Combine them in an appropriate way and you’ll get the twelfth roots; but looking at the irrationalities I’ve written out, you can see immediately what the intermediate quadratic fields are.
And although you certainly didn’t need the twelfth cyclotomic polynomial, it’s easy to calculate, as $X^4-X^2+1$. You do this by looking first for the sixth polynomial, which is $X^2-X+1$, and noticing that the primitive twelfth roots of unity are the square roots of the primitive sixth roots.
