Find the degree of $\mathbb{Q}(\zeta_{37} + \zeta_{37}^{10} + \zeta_{37}^{26})$ over $\mathbb{Q}$

Question

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 $\mathbb{Q}(\alpha)$.

Answer 1

Method is due to Gauss, about 30 years before Galois. Note that $1,10,26$ is a subgroup of the multiplicative group in Z/37Z without 0. Method done in detail in Cox, Galois Theory, in modern terms. For this example, see page 50 in Reuschle

Root of $$ x^{12} + x^{11} + 2 x^{10} - 20 x^9 - 13 x^8 - 19 x^7 + 85 x^6 + 51 x^5 + 94 x^4 - 2 x^3 - 13 x^2 - 77 x + 47 $$

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parisize = 4000000, primelimit = 500509
? f = x^{12} + x^{11} + 2* x^{10} - 20* x^9 - 13* x^8 - 19* x^7 + 85* x^6 + 51* x^5 + 94* x^4 - 2* x^3 - 13* x^2 - 77* x + 47 
%1 = x^12 + x^11 + 2*x^10 - 20*x^9 - 13*x^8 - 19*x^7 + 85*x^6 + 51*x^5 + 94*x^4 - 2*x^3 - 13*x^2 - 77*x + 47
? factor(f)
%2 = 
[x^12 + x^11 + 2*x^10 - 20*x^9 - 13*x^8 - 19*x^7 + 85*x^6 + 51*x^5 + 94*x^4 - 2*x^3 - 13*x^2 - 77*x + 47 1]

? poldisc(f)
%3 = 3621909936633833561849479709324950637917
? factor(poldisc(f))
%4 = 
[37 11]

[47 2]

[149 2]

[211 2]

[223 2]

[433 2]

? polgalois(f)
  ***   at top-level: polgalois(f)
  ***                 ^------------
  *** polgalois: sorry, galois of degree higher than 11 is not yet implemented.
  ***   Break loop: type 'break' to go back to GP
break> break

? 

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Answer 2

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 $\mathbb{Q}$.