Prove that $\mathbb Q(\cos\tfrac\pi7)\neq\mathbb Q(\cos\tfrac\pi9)$

Question

Let $\tau=2\pi$ be the full angle. (tau)

For any integer $k$ and any angle $\theta$, $\cos(k\theta)$ is a polynomial in $\cos\theta$. In particular, $\cos(2\theta)=2\cos^2\theta-1$, which shows that $\mathbb Q(\cos\tfrac\tau n)\subseteq\mathbb Q(\cos\tfrac\tau{2n})$. Furthermore, if $n$ is odd, then $\cos(\tfrac\tau{2n})=-\cos(\tfrac\tau2-\tfrac\tau{2n})=-\cos(\tfrac{n-1}2\tfrac\tau n)$, which shows that $\mathbb Q(\cos\tfrac\tau{2n})\subseteq\mathbb Q(\cos\tfrac\tau n)$, so the fields are equal: $\mathbb Q(\cos\tfrac\tau n)=\mathbb Q(\cos\tfrac\pi n)$.

$x=2\cos\tfrac\tau9$ has minimal polynomial $x^3-3x+1=0$, and $y=2\cos\tfrac\tau7$ has minimal polynomial $y^3+y^2-2y-1=0$. These can be gotten from cyclotomic polynomials, using $2\cos\theta=e^{i\theta}+e^{-i\theta}$.

I tried writing $y=ax^2+bx+c$ with rational $a,b,c$, but that led to a mess of $3$ cubic equations in $3$ variables; I see no way to tell whether a solution exists:

$$0x^2+0x+0=y^3+y^2-2y-1$$ $$=\Big(9a^3 + 9ab^2 + 9a^2c - 3a^2b + 3ac^2 + 3b^2c + 3a^2 + b^2 + 2ac - 2a\Big)x^2 \\ +\Big(-6a^3 - 3ab^2 - 3a^2c + 27a^2b + 18abc + 3b^3 + 3bc^2 - a^2 + 6ab + 2bc - 2b\Big)x \\ +\Big(a^3 - 9a^2b - 6abc - b^3 + c^3 - 2ab + c^2 - 2c - 1\Big).$$

How else can we show that $\mathbb Q(\cos\tfrac\tau7)\neq\mathbb Q(\cos\tfrac\tau9)$?

I know that the cyclotomic fields $\mathbb Q(e^{i\tau/(2n)})$ are all distinct for $n\in\mathbb N$. (See Theorem 4.1 in Keith Conrad's notes.) But their real subfields $\mathbb Q(\cos\tfrac\tau{2n})$ are not all distinct; for $n=1,2,3$ they are just $\mathbb Q$. Maybe for $n>3$ they are distinct?

Answer 1

According to the previous Theorem (3.4) in those notes,

$$\mathbb Q(e^{i\tau/m})\cap\mathbb Q(e^{i\tau/n})=\mathbb Q(e^{i\tau/\text{GCD}(m,n)}).$$

Intersecting with $\mathbb R$, we get

$$\mathbb Q(\cos\tfrac\tau m)\cap\mathbb Q(\cos\tfrac\tau n)=\mathbb Q\big(\cos\tfrac\tau{\text{GCD}(m,n)}\big).$$

Suppose $\mathbb Q(\cos\tfrac\tau{2m})=\mathbb Q(\cos\tfrac\tau{2n})$ where $m\geq n>3$, and let $g=\text{GCD}(m,n)$. We have $\text{GCD}(2m,2n)=2g$. The field's intersection with itself is itself, so

$$\mathbb Q(\cos\tfrac\tau{2m})=\mathbb Q(\cos\tfrac\tau{2n})=\mathbb Q\big(\cos\tfrac\tau{2g}\big).$$

Taking degrees over $\mathbb Q$, if $g=1$ then

$$\tfrac12\phi(2m)=\tfrac12\phi(2n)=1$$

which is impossible since $\phi(2n)>2$ when $2n>6$. If $g>1$ then

$$\tfrac12\phi(2m)=\tfrac12\phi(2n)=\tfrac12\phi(2g),$$

and since $m$ and $n$ are multiples of $g$, the totients $\phi(2m)$ and $\phi(2n)$ are multiples of $\phi(2g)$. Any extra factor of $2$ in $2n$, compared to $2g$, would contribute a factor of $2$ to the totient, thus making $\phi(2n)>\phi(2g)$. Any extra odd prime factor $p$ in $2n$ would contribute either $p$ or $p-1$ to the totient, again making $\phi(2n)>\phi(2g)$. Therefore, we must have $n=g$, and similarly $m=g$, so $m=n$.

(For the specific case in the OP, take $m=9$ and $n=7$.)

Answer 2

The minimal polynomial of $x=\cos\left(\dfrac{\pi}{9}\right)$ and of $x=\cos\left(\dfrac{\pi}{7}\right)$ are $f(x)=8x^3-6x-1$ and $g(x)=8x^3-4x^2-4x+1$ respectively. It follows that the two irrational numbers are of third degree but distinct and $\mathbb Q$-linear independant so $\mathbb Q(\cos\tfrac\pi7)\neq\mathbb Q(\cos\tfrac\pi9)$. What happen is similar, say, to the cases $\sqrt2$ and $\sqrt3$ in degree two.

NOTE.-A derived problem, equivalent to this answer, could be, a good exercise for beginners:

Prove that for all triple of rational $(a,b,c)$ one has $$a\cos\left(\dfrac{\pi}{9}\right)+b\cos\left(\dfrac{\pi}{9}\right)^2+c\ne \cos\left(\dfrac{\pi}{7}\right)$$

Answer 3

Let $a=2\cos(2\pi/9),b=2\cos(2\pi/7)$ and let $f(x) =x^3-3x+1$,$g(x)=x^3+x^2-2x-1$ be minimal polynomials of $a, b$ respectively and let $K=\mathbb{Q} (a), L=\mathbb{Q} (b) $.

We have $$\text{disc} (f(x)) =\text{disc} (\mathbb{Z} [a])=([O_K:\mathbb{Z} [a]]) ^2\text{disc}(O_K)$$ and then $$81=([O_K:\mathbb{Z} [a]]) ^2\text{disc}(O_K)$$ It follows that $$[O_K:\mathbb{Z} [a]] = 1, 3,\text{ or }9$$ and hence $$\text{disc} (O_K) =1, 9,\text{ or }81$$ But $\text{disc} (O_K) \neq 1$ as $K\neq \mathbb{Q} $ we have $\text{disc} (O_K) =9 \text{ or } 81$.

In same fashion using $$\text{disc} (g(x)) =\text{disc} (\mathbb{Z} [b])=([O_L:\mathbb{Z} [b]]) ^2\text{disc}(O_L)$$ we get $\text {disc} (O_L) =49$. It follows that $K\neq L$ as the discriminants of these fields are not same.