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Let $n>4$, and $(h,n) = 1$. How to show that $[\mathbb{Q}(\tan 2 \pi h/n):\mathbb{Q}]= \phi(n)$ or $\phi(n)/2$ or $\phi(n)/4$ respectively if $\gcd(n,8)<4$ or $\gcd(n,8)=4$ or $\gcd(n,8)>4$.

(In fact, this is a question from J. McCarthy's Algebraic Extensions of Fields, Ch. 2. We know from a previous question that $[\mathbb{Q}(\cos 2\pi h/n):\mathbb{Q}]=\phi(n)/2$ if $n>2$ and $\gcd(n,h)=1$; and also that if $n>4,$ $[\mathbb{Q}(\sin 2\pi h/n):\mathbb{Q}]=\phi(n), \phi(n)/4$ or $\phi(n)/2$, respectively if $\gcd(n,8)<4$, $\gcd(n,8)=4$ or $\gcd(n,8)>4$.)

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@YACP thanks for the reminder! –  Belgi Nov 23 '12 at 18:00

1 Answer 1

This is Theorem 3.11 from the book Irrational numbers of Ivan Niven.

I will try a slightly different approach. Set $K=\mathbb{Q}(\frac{\zeta-\zeta^{-1}}{\zeta+\zeta^{-1}})\subset\mathbb{Q}(\zeta)$, where $\zeta=\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n}$. We have the following:

1. If $4\not\mid n$, then $K=\mathbb{Q}(\zeta)$. If $4\mid n$, then $K$ is the fixed field of the automorphism $\zeta\to-\zeta$ (of $\mathbb{Q}(\zeta)$).

Proof. Take $\zeta\to\zeta^i$ an automorphism of $\mathbb{Q}(\zeta)$ and see when it fixes $\frac{\zeta-\zeta^{-1}}{\zeta+\zeta^{-1}}$.

1 shows that the degree of $\mathbb{Q}(\zeta)$ over $K$ is $1$ if $4\not\mid n$, respectively $2$ when $4\mid n$.

2. $i\in\mathbb{Q}(\zeta)$ if and only if $4\mid n$.

Proof. Well known.

3. $i\in K$ if and only if $8\mid n$.

Proof. $i\in K$ implies $i\in\mathbb{Q}(\zeta)$, so $4\mid n$. In this case we get that $i=\zeta^{\frac{n}{4}}$. Moreover, $K$ is the fixed field of the automorphism $\zeta\to-\zeta$. Therefore $i\in K$ if and only if $\zeta^{\frac{n}{4}}=(-\zeta)^{\frac{n}{4}}$ if and only if $8\mid n.$

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