Find $a$ such that $\mathbb{Q}\left(\zeta_{n}\right)\cap \mathbb{R}=\mathbb{Q}(a)$

Question

Let $L:=\mathbb{Q}\left(\zeta_{n}\right)$, where $\zeta_{n}=e^{2\pi i/n}$ for some prime $n\geq 3$. Let $K:=L \cap \mathbb{R}$.

Find $a\in L$ such that $K=\mathbb{Q}(a)$.

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What is known:

The degree $[L:K]$ equals $2$.

I can take $a=\cos(2\pi/n)$, as for $b=i\sin(2\pi/n)$, one has the equality $L=\mathbb{Q}(a,b)$:

• "$\subseteq$" is clear, as $\zeta_{n}=a+b$;
• "$\supseteq$" follows from the equalities $\frac{\zeta_{n}+\zeta_{n}^{-1}}{2}=a$ and $\frac{\zeta_{n}-\zeta_{n}^{-1}}{2}=b$.

Now $p-1=[L:\mathbb{Q}]=[L:\mathbb{Q}(a)]\cdot [\mathbb{Q}(a):\mathbb{Q}]$, where the second factor equals $\frac{p-1}{2}$. Hence $[L:\mathbb{Q}(a)]=2$.

As $\mathbb{Q}(a)\subseteq K$ (again by the argument that $\frac{\zeta_{n}+\zeta_{n}^{-1}}{2}=a$) and $[L:\mathbb{Q}(a)]=2=[L:K]$, we must have $\mathbb{Q}(a)=K$.

The problem is that I have no proof for the fact that $[\mathbb{Q}(a):\mathbb{Q}]=\frac{p-1}{2}$. I have found online that the degree of the minimal polynomial of $a=\cos(2\pi/n)$ over $\mathbb{Q}$ equals $\frac{p-1}{2}$.

There is no need to prove it using this. I just wanted to argue why I know that this $a$ works.

Answer 1

It is easier to show that $[L:\mathbb{Q}(a)]=2$ directly. Since $L$ is not contained in $\mathbb{R}$ while $\mathbb{Q}(a)$ is , we have $L\neq\mathbb{Q}(a)$ and $[L:\mathbb{Q}(a)]\geq 2$. Now $\zeta_n^2-2a\zeta_n+1=0$, so $[L:\mathbb{Q}(a)]\leq 2$. Hence $[L:\mathbb{Q}(a)]=2$, as expected.

Since $\mathbb{Q}(a)\subset K\subset L$ and that you know that $[L:K]=2$, we deduce using multiplicativity of degrees that $[K:\mathbb{Q}(a)]=1$, that is $K=\mathbb{Q}(a)$.