# minimal polynomial of $\sin (2\pi/7)$ over $\mathbb Q(\sqrt 7)$

How to find the minimal polynomial of $$\sin (2\pi/7)$$ over $$\mathbb Q(\sqrt 7)$$ ? And the minimal polynomial of $$\sin (2\pi/11)$$ over $$\mathbb Q(\sqrt {11})$$ ?

I know that the minimal polynomial of $$\sin (2\pi/7)$$ over $$\mathbb Q$$ is $$\dfrac 1{64} (64x^6-112x^4+56x^2-7)$$ , but what is the minimal polynomial over $$\mathbb Q(\sqrt 7)$$ ?

Everything happens inside the field $$L=\Bbb{Q}(\zeta,i)$$, where $$\zeta=e^{2\pi i/7}$$. The field $$K=\Bbb{Q}(\sqrt{7})$$ is a subfield of $$L$$. This implies that the question can be answered by applying Galois theory. See below.

The extension $$L/\Bbb{Q}$$ is Galois as a compositum of the seventh cyclotomic field and the quadratic extension $$\Bbb{Q}(i)$$. Therefore $$Gal(L/\Bbb{Q})\simeq C_6\times C_2$$. Any automorphism is uniquely determined by where it maps $$\zeta$$ and $$i$$, and all the twelve possible combinations occur.

Because (check out Gauss sums or look at this old answer) $$\zeta+\zeta^2+\zeta^4=\frac{-1+i\sqrt7}2$$ we can write $$\sqrt7=-2i(\zeta+\zeta^2+\zeta^4)-i.$$ As $$\overline{\zeta+\zeta^2+\zeta^4}=\zeta^3+\zeta^5+\zeta^6$$ it follows that $$\sqrt7$$ is stable under the automorphism $$\sigma$$ defined by $$\sigma(i)=-i$$, $$\sigma(\zeta)=\zeta^3$$. Because the restriction of $$\sigma$$ to $$\Bbb{Q}(\zeta)$$ has order six, so does $$\sigma$$. Therefore we can conclude that $$Gal(L/K)=\langle\sigma\rangle\simeq C_6.$$

What this means is that the $$K$$-conjugates of $$u=2\sin(2\pi/7)=-i(\zeta-\zeta^{-1})$$ are $$\sigma(u)=i(\zeta^3-\zeta^{-3})=-2\sin(6\pi/7)$$ and $$\sigma^2(u)=-i(\zeta^9-\zeta^{-9})=2\sin(4\pi/7).$$ As an extra exercise you are invited to verify that $$\sigma^3(u)=u$$.

Therefore the minimal polynomial of $$u$$ over $$K$$ is the cubic $$m(x)=(x-2\sin(2\pi/7))(x+2\sin(6\pi/7))(x-2\sin(4\pi/7)).$$ Also leaving it to you to crunch out those sums of powers of $$\zeta$$ and $$i$$, and to prove that $$m(x)=x^3-\sqrt7 x^2+\sqrt7\in K[x].$$ Getting the minimal polynomial of $$u/2$$ from here is, of course, trivial.

• To get the minimal polynomial of $\sin(2\pi/11)$ over $\Bbb{Q}(\sqrt{11})$ you can similarly work with the automorphisms of $\Bbb{Q}(\zeta_{11},i)$. Nov 12, 2018 at 6:48
• But, I'm somewhat unhappy with this. Somehow I suspect a useful known trig identity can be located... Making those heavy calculations with roots of unity in the end unnecessary? :-( Nov 12, 2018 at 6:48
• Doesn't your $\sigma (u)$ has a sign typo ... shouldn't $i$ be mapped to $-i$ ?
– user
Nov 12, 2018 at 18:15
• Thanks @users. Both $\sigma(u)$ and $\sigma^2(u)$ had the wrong sign in their first forms (but were correct when written as values of sine. Nov 12, 2018 at 18:18