- For \(n>2\), let \(\zeta_n\) be a primitive \(n\)th root of unity. Prove that \[ [\QQ(\zeta_n+\zeta_n^{-1}):\QQ]=\frac{1}{2}\varphi(n), \] where \(\varphi\) is Euler’s totient function.
Solution
Set \[ K=\QQ(\zeta_n), \qquad F=\QQ(\zeta_n+\zeta_n^{-1}). \]
1. One has \([K:F]\le2\).
Put \[ a=\zeta_n+\zeta_n^{-1}\in F. \] Then \(\zeta_n\) satisfies \[ T^2-aT+1=0 \] over \(F\), because \[ \zeta_n^2-a\zeta_n+1 =\zeta_n^2-(\zeta_n+\zeta_n^{-1})\zeta_n+1=0. \] Since \(K=F(\zeta_n)\), it follows that \([K:F]\le2\).
2. Complex conjugation gives a nontrivial \(F\)-automorphism of \(K\).
Complex conjugation sends \[ \zeta_n\longmapsto\zeta_n^{-1} \] and fixes \[ \zeta_n+\zeta_n^{-1}. \] Thus it fixes \(F\) pointwise. Because \(n>2\), a primitive \(n\)th root of unity is not equal to its inverse: otherwise \(\zeta_n^2=1\), forcing its order to divide \(2\). Hence complex conjugation is nontrivial on \(K\).
3. Therefore \([K:F]=2\).
By 2, the extension \(K/F\) has at least two \(F\)-automorphisms, so it cannot have degree \(1\). Combined with 1, this gives \[ [K:F]=2. \]
4. The cyclotomic extension satisfies\[ [K:\QQ]=\varphi(n). \]
The minimal polynomial of a primitive \(n\)th root of unity over \(\QQ\) is the cyclotomic polynomial \(\Phi_n\), whose degree is \(\varphi(n)\). Hence \[ [\QQ(\zeta_n):\QQ]=\varphi(n). \]
5. Hence\[ [\QQ(\zeta_n+\zeta_n^{-1}):\QQ]=\frac{\varphi(n)}2. \]
By the tower law and <1>3–<1>4, \[ \varphi(n) =[K:\QQ] =[K:F][F:\QQ] =2[F:\QQ]. \] Therefore \[ [F:\QQ]=\frac{\varphi(n)}2. \]