Galois Theory Quadratic Subfield

Question

Let $ζ_7=e^{i2\pi/7}$ be a 7th root of unity. The field $\Bbb{Q}(ζ_7)$ contains a quadratic subfield that can be expressed in the form of $\Bbb{Q}(\sqrt{D})$ where D is an integer. What is D? I understand that there is a field extension of order 6 and therefore there will be a quadratic subfield, but how do we find out what D is?

Answer 1

Not every field extension of degree $ 6 $ has a quadratic subfield. For example, $ \mathbf Q(\sqrt{1 + \sqrt[3]{2}}) $ has no subfield that is quadratic over $ \mathbf Q $. It is, however, true that every Galois extension of degree $ 6 $ has a unique quadratic subfield. It has been pointed out in another answer how to find this subfield of $ \mathbf Q(\zeta_7)/\mathbf Q $.

Answer 2

Let $\displaystyle\alpha = \sum_{n=0}^{p-1} \left(\frac{n}{p}\right) \zeta_p^n$ be the Gauss sum, then $\alpha= \sqrt{-p}$ if $p \equiv 3 \bmod 4$ and $\mathbb{Q}(\alpha)= \mathbb{Q}(\sqrt{-p})$ is the quadratic subfield of $\mathbb{Q}(\zeta_p)$.

$\text{Gal}(\mathbb{Q}(\zeta_p)/\mathbb{Q}) \simeq \mathbb{Z}_p^\times$. Let $H$ be its subgroup of order $\frac{p-1}{2}$. It corresponds to the quadratic residues in $\mathbb{Z}_p^\times$. Then $\mathbb{Q}(\alpha)$ is the fixed field of $H$.

Answer 3

The Galois group is generated by $\zeta\mapsto\zeta^3$, since $3$ generates the cyclic group $(\Bbb Z/7\Bbb Z)^\times$. In this group $3^2=2$, so that $2$ generates the unique subgroup of order $3$, and it might be that the sum of the three transforms of $\zeta$ under this group is a generator of the fixed field. That is, you might try seeing what the minimal polyknomial of $\zeta+\zeta^2+\zeta^4$ is.