degree of extension of Q with primitive roots of unity

Question

Question

Show that $[\mathbb{Q}(\zeta+\zeta^{-1}):\mathbb{Q}]=\phi(n)/2$, where $\zeta$ is a primitive $n$-th root of unity.

Attempt

Let $\sigma: z\mapsto \bar{z}$,$\sigma(\zeta)=\zeta^{-1}$.Since $\mathbb{Q}(\zeta)=\mathbb{Q}(\zeta^{-1})$ we have $\sigma\in Gal(\mathbb{Q}(\zeta)/\mathbb{Q})$.

It is $[\mathbb{Q}(\zeta):\mathbb{Q}]=\phi(n)$, and $[\mathbb{Q}(\zeta):\mathbb{Q}(\zeta)^{<\sigma>}]=|<\sigma>|=2$.

I have o show that $\mathbb{Q}(\zeta)^{<\sigma>}=\mathbb{Q}(\zeta+\zeta^{-1})$

Obviously $\sigma(\zeta+\zeta^{-1})=\zeta+\zeta^{-1}\Rightarrow \mathbb{Q}(\zeta+\zeta^{-1}) \subset\mathbb{Q}(\zeta)^{<\sigma>}$

Is this correct so far? How could I show that $\mathbb{Q}(\zeta+\zeta^{-1}) \supset\mathbb{Q}(\zeta)^{<\sigma>}$?

Answer 1

You need $n\ge 3$. Then $K=\Bbb Q(\zeta+\zeta^{-1})$ embeds in $\Bbb R$ but $L=\Bbb Q(\zeta)$ doesn't. Therefore $L\ne K$. Also $\zeta$ is a zero of $X^2-(\zeta+\zeta^{-1})X+1=0$ so that $|L:K|\le2$. Therefore $|L:K|=2$. But $|L:\Bbb Q|=\phi(n)$, etc.

Of course $\sigma$ acts as complex conjugation on $L$, so its fixed field is $L\cap\Bbb R=K$.