TITLE: Find the degree of $\mathbb{Q}(\zeta_{37} + \zeta_{37}^{10} + \zeta_{37}^{26})$ over $\mathbb{Q}$
QUESTION [7 upvotes]: Apologies if this is a repeat question. First off, I notice that $1 + 10 + 26 = 37$. Hmmmmm! What could the significance of this possibly be? I have nothing more intelligent to add, except that I would try to see if I could find the degree of $\mathbb{Q}(\zeta_{37})$ over $\mathbb{Q}$ adjoin this $\alpha$ first, then use the tower argument. To that end, maybe I could try to see how many automorphisms in Gal$(\mathbb{Q}(\zeta_{37})/ \mathbb{Q})$ fix $\mathbb{Q}(\alpha)$.

REPLY [3 votes]: This is a supplement of Will Jagy's answer.
$\{1,10,26\}$ is a subgroup of $(\mathbb{Z}/37\mathbb{Z})^{\times}$ implies that the automorphism $\sigma$ on $\mathbb{Q}(\zeta_{37})$ which sends $\zeta_{37}\mapsto \zeta_{37}^{10}$ fixes $\alpha=\zeta_{37} + \zeta_{37}^{10} + \zeta_{37}^{26}$. It is easy to see that no automorphisms other than $1,\sigma,\sigma^2$ fix $\alpha$. Thus $\mathbb{Q}(\alpha)$, being the fixed field of a subgroup of order $3$, has degree 12 over $\mathbb{Q}$.