Field extensions of $\mathbb{Q}$, $\mathbb{Q}(\xi_7)$ and $\mathbb{Q}(\xi_7+\xi_7^{-1})$

Question

Let $\xi_7$ denote the complex number $e^{2\pi i/7}$ and let $\beta = \xi_7+\xi_7^{-1}$, consider the field extensions $\mathbb{Q} \subset \mathbb{Q}(\beta) \subset \mathbb{Q}(\xi_7) $.

Determine the minimal polynomials of $\xi_7$ over $\mathbb{Q}$, $\mathbb{Q}(\beta)$ and the minimal polynomial of $\beta$ over $\mathbb{Q}$.

For the first question, we know that $x^7-1 = 0 \implies (x-1)(x^6+x^5+x^4+x^3+x^2+x+1) =0 \implies \xi_7$ is a root of $x^6+x^5+x^4+x^3+x^2+x+1$, which is the minimal polynomial, given that this polynomial is irreducible in $\mathbb{Q}$.

However, I'm finding the other questions quite difficult to answer.

Thanks

Answer 1

Hints:

1) The equation known to you implies the more symmetric equation $$ \xi_7^3+\xi_7^2+\xi_7+1+\xi_7^{-1}+\xi_7^{-2}+\xi_7^3=0. $$ Can you write the l.h.s. a linear combination of the powers $(\xi_7+\xi_7^{-1})^k$ with $k=1,2,3$? That gives you the minimal polynomial of $\beta$ over $\Bbb{Q}$.

2) What happens when you multiply both sides of $$ \xi_7+\xi_7^{-1}=\beta $$ by $\xi_7$? You get a polynomial equation in $\xi_7$ with coefficients in...

3) Calculate the extension degrees to prove that you have found the minimal polynomials.

Answer 2

We can just compute the powers of $\zeta_7+\zeta_7^{-1}$. We have $$ \begin{align*} \zeta_7 + \zeta_7^{-1} \;&=\; -1 - \zeta_7^2 - \zeta_7^3 - \zeta_7^4 - \zeta_7^5 \\ (\zeta_7 + \zeta_7^{-1})^2 \;&=\; 2 + \zeta_7^2 + \zeta_7^5 \\ (\zeta_7 + \zeta_7^{-1})^3 \;&=\; -3 - 3\zeta_7^2 - 2\zeta_7^3 - 2\zeta_7^4 - 3\zeta_7^5 \end{align*} $$ In particular we have $$(\zeta_7+\zeta_7^{-1})^3 + (\zeta_7+\zeta_7^{-1})^2 - 2(\zeta_7+\zeta_7^{-1}) - 1 = 0,$$ and no linear combination of smaller powers is zero. Hence the minimal polynomial for $\zeta_7+\zeta_7^{-1}$ is $x^3 + x^2 - 2x - 1$.

For the other question, note that $[\mathbb{Q}(\zeta_7):\mathbb{Q}(\beta)]=2$. It is easy to see that the minimal polynomial is $x^2-\beta x+1$.