If \(n>2\) and \(\zeta\) is a primitive \(n\)th root of unity over \(\mathbb{Q}\), then \([\mathbb{Q}(\zeta + \zeta^{-1}): \mathbb{Q}]=\phi(n)/2.\)
Solution
Put \[ L=\QQ(\zeta), \qquad L^+=\QQ(\zeta+\zeta^{-1}). \]
1. One has\[ [L:\QQ]=\phi(n). \]
The minimal polynomial of a primitive \(n\)th root of unity over \(\QQ\) is the \(n\)th cyclotomic polynomial \(\Phi_n(x)\), whose degree is \(\phi(n)\). Hence \[ [\QQ(\zeta):\QQ]=\deg\Phi_n=\phi(n). \]
2. The element \(\zeta\) is algebraic of degree at most \(2\) over \(L^+\).
Let \[ t=\zeta+\zeta^{-1}\in L^+. \] Then \(\zeta\) satisfies \[ x^2-tx+1=0, \] so \[ [L:L^+]\le2. \]
3. The field \(L^+\) is contained in \(\RR\), whereas \(\zeta\notin\RR\).
Since \[ \zeta+\zeta^{-1}=2\cos(2\pi/n), \] the generator of \(L^+\) is real, so \(L^+\subseteq\RR\).
If a primitive \(n\)th root of unity is real, it must be \(1\) or \(-1\), whose orders are \(1\) and \(2\). Since \(n>2\), the primitive root \(\zeta\) is not real. Thus \(\zeta\notin L^+\).
4. Therefore\[ [L:L^+]=2. \]
By 2 the degree is at most \(2\). By 3 the inclusion \(L^+\subsetneq L\) is proper, so the degree is not \(1\).
5. Hence\[ [\QQ(\zeta+\zeta^{-1}):\QQ]=\frac{\phi(n)}2. \]
The tower law and 1, 4 give \[ \phi(n)=[L:\QQ]=[L:L^+][L^+:\QQ]=2[L^+:\QQ]. \] Solving for \([L^+:\QQ]\) gives the asserted degree.