For a Galois theory course, I need to prove that $\mathbb{Q}(\zeta_{2^{n+2}} + (\zeta_{2^{n+2}})^{-1})$ is a cyclic field extension of $\mathbb{Q}$ of degree $2^n$. Constructing the minimal polynomial does not seem the way to go, so what would be the best approach?
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See here for the first factor in the degree theorem. – Dietrich Burde May 16 '19 at 16:56
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You can do it with minimal polynomials: We have $x=\zeta_{2^{n+2}}+(\zeta_{2^{n+2}})^{-1}=2\cos(\pi/2^{n+1})$, so $T_{2^n}(x/2)=\cos(2^n\cdot\pi/2^{n+1})=0$, where $T_m$ is the Chebyshev polynomial of the first kind, etc.
However, it is easier if you just use the fact the cyclotomic extension $\mathbb{Q}(\zeta_{2^{n+2}})/\mathbb{Q}$ has degree $\phi(2^{n+2})=2^{n+1}$.
Edit: The automorphisms are $\cos(\pi/2^{n+1})\mapsto\cos(k\pi/2^{n+1})$ where $k$ is odd. Recall $(\mathbb{Z}/2^{n+2}\mathbb{Z})^*=C_2\times C_{2^n}$, the $C_{2^n}$ being generated by $5$, so it suffices to show $5^{2^{n-1}}\not\equiv\pm 1\pmod{2^{n+2}}$, which you can find a proof here

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The field degree being a power of 2 does not imply the Galois group is cyclic. The OP needs to be clearer about what the question actually is: just find the degree, just determine the structure of the Galois group, or both? – KCd May 16 '19 at 21:25
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It is to be determined that said field extension is AND cyclic AND of order $2^n$. – MCDC May 16 '19 at 22:14