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.


1 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.

  • $\begingroup$ To get the minimal polynomial of $\sin(2\pi/11)$ over $\Bbb{Q}(\sqrt{11})$ you can similarly work with the automorphisms of $\Bbb{Q}(\zeta_{11},i)$. $\endgroup$ Nov 12, 2018 at 6:48
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    $\begingroup$ 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? :-( $\endgroup$ Nov 12, 2018 at 6:48
  • $\begingroup$ Doesn't your $\sigma (u)$ has a sign typo ... shouldn't $i$ be mapped to $-i$ ? $\endgroup$
    – user
    Nov 12, 2018 at 18:15
  • $\begingroup$ 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. $\endgroup$ Nov 12, 2018 at 18:18

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