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

Question

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 $\mathbb Q(\sqrt 7)$ ?

Please help.

Answer 1

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.