Field Theory – Finding Intermediate Fields of $\Bbb{Q}(\zeta_7)$

Question

Let $F= \Bbb{Q}(\zeta_7)$ with $\zeta_7 = e^{2\pi/7}$.

a) What is the Galois group of $F$ over $\Bbb{Q}$?

b)Find all intermediate fields between $\Bbb{Q}$ and $F$. (Write each in the form $\Bbb{Q}(\alpha)$ for some specific $\alpha \in F$.)

c) For each intermediate field $E$ above, give the Galois group of $E$ over $\Bbb{Q}$.

a) Since $\Bbb{Q}(\zeta_7)$ is the splitting field for the cyclotomic polynomial $x^6+x^5+…+1$, $Gal(\Bbb{Q}(\zeta_7)/\Bbb{Q}) \cong U_7 \cong \Bbb{Z}^{\times}_7 \cong \Bbb{Z}_6$.

b) $Gal(\Bbb{Q}(\zeta_7)/\Bbb{Q}) \cong \Bbb{Z}_6$

So we have the following diagram for $\Bbb{Z}_6$,

And a corresponding diagram for $\Bbb{Q}(\zeta_7)$ for some $\alpha$ and $\beta$

Since $\Bbb{Q}(\zeta_7)$ has complex numbers, we can look at $\Bbb{Q}(i)$ to see if it has degree 2 or 3 over $\Bbb{Q}$. Since $x^2+1$ is its minimal polynomial, we see that $[\Bbb{Q(i)} : \Bbb{Q}]=2$ and so it must correspond to $\Bbb{Z}_3$, and $\beta=i$.

Sending every number to its conjugate is one possible automorphism. Of course, it has order 2 and so is isomorphic to $\Bbb{Z}_2$. Since it fixes every real number, the corresponding intermediate field must be of the form $\Bbb{Q}(\alpha)$ for some real number $\alpha$, and must have degree 3 over $\Bbb{Q}$.

If we look at $\zeta_7 + \zeta_7^6$, we see that

$\cos(2\pi/7) + i\sin(2\pi/7) + cos(12\pi/7) + i\sin(12\pi/7)$

$= \cos(2\pi/7) + i\sin(2\pi/7) + \cos(-2\pi/7) + i\sin(-2\pi/7)$

$= 2\cos(2\pi/7)$

which is irrational. So $\Bbb{Q}(\zeta_7 + \zeta_7^6)$ is either a proper subfield of $\Bbb{Q}(\alpha)$ or is equal to it. But from the diagram, we know that there is only one other intermediate field, so $\Bbb{Q}(\zeta_7 + \zeta_7^6) = \Bbb{Q}(\alpha)$.

c) The corresponding Galois group of $\Bbb{Q}(i)$ is isomorphic to $\Bbb{Z}_3$ and that of $\Bbb{Q}(\zeta_7 + \zeta_7^6)$ is isomorphic to $\Bbb{Z}_2$.

Do you think this is correct?

Thanks in advance

Answer 1

Here's a trick that often works in this kind of problem.

Take an element $\alpha$ of $F$ --- an element that generates $F$ over the ground field.

Take a subgroup $H$ of the Galois group.

Form $\sum_{\sigma{\rm\ in\ }H}\sigma(\alpha)$. This is guaranteed to be invariant under $H$ and thus to live in some subfield. Unless you're unlucky, it generates the fixed field of $H$.

In particular, this is what @ccorn has done in the other answer.