Find Galois group of $L:\mathbb{Q}(\sqrt{-5})$ where $L=\mathbb{Q}(\zeta_{20})$

Question

I would like to find the Galois group of the field extension $L:\mathbb{Q}(\sqrt{-5})$, where $L=\mathbb{Q}(\zeta_{20})$ and $\zeta_{20}$ is a primitive 20th root of unity (in $\mathbb{C}$).

We know that the Galois group must be of order four, as $L:\mathbb{Q}$ is Galois and $\mathbb{Q}\subset \mathbb{Q}(\sqrt{-5})\subset L$, so $L:\mathbb{Q}(\sqrt{-5})$ is Galois, and $[L:\mathbb{Q}(\sqrt{-5})]=\frac{[L:\mathbb{Q}]}{[\mathbb{Q}(\sqrt{-5}):\mathbb{Q})]}=\frac{8}{2}=4$, and if $H=\textrm{Gal}(L:\mathbb{Q}(\sqrt{-5}))$, then $|H|=[L:\mathbb{Q}(\sqrt{-5})]$, so it must be isomorphic to either $\mathbb{Z}_4$ or $\mathbb{Z}_2^2$. (Note: $[L:\mathbb{Q}]=\phi(20)=8$, where $\phi$ denotes the Euler phi-function)

I have thus far managed to show that $L=\mathbb{Q}(\zeta_{20})=\mathbb{Q}(\zeta_{4},\zeta_5)=\mathbb{Q}(i)(\zeta_{5})$, and hence that $L:\mathbb{Q}(i)$ is a cyclic field extension of degree four, $\textit{i.e.}\ \textrm{Gal}(L:\mathbb{Q}(i))\cong \mathbb{Z}_4$. I have also noticed that as we may let $\zeta_5=e^{\frac{2}{5}\pi i}=\frac{\sqrt{5}-1}{4}+\frac{i}{4}\sqrt{2\sqrt{5}+10}$, we can express $L$ as: $L=\mathbb{Q}(i,\sqrt{2\sqrt{5}+10})$.

My initial idea was to let $K=\mathbb{Q}(i,\sqrt{-5})$, so that $\mathbb{Q}(\sqrt{-5})\subset K\subset L$, and note that $\textrm{Gal}(L:K)\cong \mathbb{Z}_2$ and $\textrm{Gal}(K:\mathbb{Q}(\sqrt{-5}))\cong \mathbb{Z}_2$, and thus somehow conclude that $\textrm{Gal}(L:\mathbb{Q}(\sqrt{-5}))\cong \mathbb{Z}_2^2$, which I suspect to be the right conclusion. But this reasoning is of course not only erroneous but unsubstantiated.

I subsequently seemed to become unable to come up with better ideas and thus decided to consult the community.

All help or input would, as always, be highly appreciated.

Answer 1

Let $\zeta = \zeta_{20}$. The automorphism $\sigma(\zeta) = \zeta^3$ generates a subgroup of $G = \textrm{Gal}(L/\mathbb{Q})$ isomorphic to $\mathbb{Z}/4\mathbb{Z}$. Then the Galois group of $L/L^{\sigma}$ is isomorphic to $\mathbb{Z}/4\mathbb{Z}$ by the Galois correspondence. We show that $L^{\sigma} = \mathbb{Q}(\sqrt{-5})$. Let $$z = \zeta + \zeta^3 + \zeta^7 + \zeta^9$$ so that $z$ is fixed by $\sigma$. One checks that $$z + \overline{z} = 2\cos(\pi/10) + 2\cos(3\pi/10) + 2\cos(7\pi/10) + 2\cos(9\pi/10) = 0$$ and \begin{equation*} \begin{aligned} z\overline{z} &= 4 + 2\cos(\pi/5) + \cos(2\pi/5) + 2\cos(3\pi/5) + \cos(4\pi/5) \\ &= 4 + 2\cos(\pi/5) + 2\cos(3\pi/5) \\ &= 5. \end{aligned} \end{equation*} From this we conclude that $L^{\sigma} = \mathbb{Q}(z) = \mathbb{Q}(\sqrt{-5})$, thus the Galois group of $L/\mathbb{Q}(\sqrt{-5})$ is $\mathbb{Z}/4\mathbb{Z}$.

Answer 2

Let $L=\mathbf Q(\zeta_{20})$ and $K=\mathbf Q (\sqrt {-5})$. Here is a "Galois only" argument showing that $Gal( L/K)\cong C_4$. Because $<\zeta_{20}>=<\zeta_4>\times <\zeta_5>, L=\mathbf Q(i,\zeta_5)$ = the compositum of the two linearly disjoint extensions $\mathbf Q(i)$ and $\mathbf Q(\zeta_5)$ over $\mathbf Q$, so $Gal(L/\mathbf Q)\cong C_2\times C_4$, with $Gal(L/\mathbf Q(\zeta_5))\cong C_2$ and $Gal(L/\mathbf Q(i)\cong C_4$. Since the maximal qotient of $C_2\times C_4$ killed by $2$ is $C_2\times C_2$, it follows that $L$ contains exactly $3$ quadratic subfields. What are they ? Because $5\equiv 1$ mod $4$, the quadratic subfield of $\mathbf Q(\zeta_5)$ is $\mathbf Q(\sqrt 5)$ (*), hence our 3 quadratic subfields are $\mathbf Q (i), \mathbf Q(\sqrt 5)$ and $K$. Because $K$ and $\mathbf Q(\zeta_5)$ are linearly disjoint over $\mathbf Q, Gal(L/K)\cong C_4$, just as the case was for $Gal(L/\mathbf Q(i))$. Whereas $Gal(L/\mathbf Q(\sqrt 5))\cong C_2\times C_2$ because $L$ is the compositum of $\mathbf Q(\zeta_5)$ and $\mathbf Q(i,\sqrt 5)$ over $\mathbf Q(\sqrt 5)$.

(*) This is a classical result coming from the calculation of discriminants. Given an odd prime $p$, put $p^*=(-1)^{(p-1)/2}p$ ; then the unique quadratic subfield of $\mathbf Q(\zeta_{p})$ is $\mathbf Q(\sqrt {p^*})$. The statement of your OP can be directly generalized as follows : let $L=\mathbf Q(\zeta_{4p})=\mathbf Q(i,\zeta_p)$, with Galois group $C_2\times C_{p-1}$ ; the $3$ quadratic subfields of $L$ are $\mathbf Q (i), \mathbf Q(\sqrt {p^*})$ and $\mathbf Q(\sqrt {-p^*})$, with $Gal(L/\mathbf Q(i))\cong Gal(L/\mathbf Q(\sqrt {-p^*}))\cong C_{p-1}$ and $Gal(L/\mathbf Q(\sqrt {p^*}))\cong C_2\times C_{(p-1)/2}$ ./.