Is $K := \mathbb{Q}(\cos (2\pi / 11))$ a Galois extension over $\mathbb{Q}$?

Question

I believe that it is because $\cos(2\pi / 11) = (\zeta + \zeta^{-1})/2$ where $\zeta = e^{2\pi i/11}$ is a primitive $11$-th root of unity, and so $K$ is a subfield of $\mathbb{Q}(\zeta)$ with corresponding subgroup $H$ of $\mbox{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$. Since the latter is isomorphic to $\mathbb{Z}/10$, $H$ is in particular normal and so $K$ is Galois over $\mathbb{Q}$. Is this argument sufficient? It seems that this argument would also work in the case that $11$ is replaced with a composite number - is that true?

Answer 1

Yes! If $Gal(L/F)$ is abelian, then any intermediate extension $K/F$ corresponds to a normal subgroup and hence is also Galois. In particular, cyclotomic extensions are abelian.