Find all fields $\mathbb{Q} \subset E \subset \mathbb{Q}(\zeta_{40})$ with $[E:\mathbb{Q}]=2$ I need to find all fields $E$ such that $\mathbb{Q} \subset E \subset \mathbb{Q}(\zeta_{40})$ and $[E:\mathbb{Q}]=2$, where $\zeta_{40}$ is a primitive root of unity of order $40$. I understand that we have a galois extension, and $Gal(\mathbb{Q}(\zeta_{40})/Q) = (\mathbb{Z}/40 \mathbb{Z})^{\times} = \mathbb{Z_2} \times \mathbb{Z_2} \times \mathbb{Z_4}$. Now, $[E:\mathbb{Q}] = 2$ is equivalent to $[\mathbb{Q}(\zeta_{40}):E]=8$, which is equivalent to $Gal(\mathbb{Q}(\zeta_{40}) / E) \subset Gal(\mathbb{Q}(\zeta_{40}) / \mathbb{Q})$ being a subset of order $8$. Now, I think $\mathbb{Z_2} \times \mathbb{Z_2} \times \mathbb{Z_4}$ has $7$ such subgroups. My question is, how to find the $E$'s that match to each of those subgroups? Say, if I have two automorphisms $\sigma, \tau$ in the galois group, how do I find the fixed-point field of $<\sigma, \tau>$? 
 A: It is simpler to determine the quadratic subfields directly. Since $(5,8)=1$, the cyclotomic field $K_{40}=\mathbf Q(\zeta_{40})$ is the compositum of $K_5=\mathbf Q(\zeta_5)$ and $K_8=\mathbf Q(\zeta_8)$.
It is classically known that $K_5$ is cyclic of degree $4$ over $\mathbf Q$, hence has a unique quadratic subfield which is $\mathbf Q(\zeta_5 + \zeta_5 ^{-1})=\mathbf Q (cos(2\pi/5))=\mathbf Q(\sqrt 5)$ (the last equality comes from a discriminant computation, see e.g. Marcus' "Number Fields", chap. 2, ex. 8). As for $K_8$, it contains (hence is the compositum of) the imaginary $K_4=\mathbf Q(\zeta_4)=\mathbf Q(\sqrt -1)$ and the real $\mathbf Q(\zeta_8 + \zeta_8 ^{-1})=\mathbf Q (cos(\pi/4))=\mathbf Q(\sqrt 2)$. Thus the tri-quadratic extension $\mathbf Q (\sqrt 5, \sqrt 2, \sqrt -1)/\mathbf Q$ is the maximal multi-quadratic subextension of $K_{40}/\mathbf Q$, and the 7 quadratic subextensions themselves will be $\mathbf Q (\sqrt 5), \mathbf Q(\sqrt 2), \mathbf Q(\sqrt 10), \mathbf Q(\sqrt -1), \mathbf Q (\sqrt {-5}), \mathbf Q(\sqrt {-2}), \mathbf Q(\sqrt {-10})$.
