Let \(n>2\) and let \(\zeta\) be a primitive \(n\)th root of unity. Prove that \[ [\QQ(\zeta+\zeta^{-1}):\QQ]=\frac{\varphi(n)}2. \]
Solution
Set \[ K=\QQ(\zeta),\qquad F=\QQ(\zeta+\zeta^{-1}). \]
1. One has \([K:F]\le2\).
The element \(\zeta\) satisfies \[ T^2-(\zeta+\zeta^{-1})T+1=0 \] over \(F\). Since \(K=F(\zeta)\), this gives \([K:F]\le2\).
2. Complex conjugation is a nontrivial \(F\)-automorphism of \(K\).
Complex conjugation sends \(\zeta\) to \(\zeta^{-1}\) and fixes \(\zeta+\zeta^{-1}\), so it fixes \(F\) pointwise. Because \(n>2\), a primitive \(n\)th root of unity satisfies \(\zeta\ne\zeta^{-1}\); otherwise \(\zeta^2=1\), contradicting its order. Hence conjugation is nontrivial on \(K\).
3. Therefore \([K:F]=2\).
By 2, \(K/F\) has a nontrivial automorphism, so \(K\ne F\). Combined with 1, this forces \([K:F]=2\).
4. The cyclotomic extension has degree \([K:\QQ]=\varphi(n)\).
The minimal polynomial of a primitive \(n\)th root of unity over \(\QQ\) is the cyclotomic polynomial \(\Phi_n\), whose degree is \(\varphi(n)\).
5. Hence\[ [F:\QQ]=\frac{\varphi(n)}2. \]
The tower law gives \[ \varphi(n)=[K:\QQ]=[K:F][F:\QQ]=2[F:\QQ]. \] Dividing by \(2\) gives the result.