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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

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  • $\begingroup$ There are two questions I need answers for. The minimal polynomial of $\xi_7$ over $\mathbb{Q}(\beta)$ and the minimal polynomial of $\beta$ over $\mathbb{Q}$. $\endgroup$ – user319128 Mar 23 '16 at 10:11
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    $\begingroup$ This question I posted might be helpful math.stackexchange.com/questions/1693198/… $\endgroup$ – user281593 Mar 23 '16 at 10:40
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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.

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  • $\begingroup$ and what is the argument for saying that $\sum_{k=0}^{p-1} x^k$ is irreducible over $\mathbb{Q}$ ? $\endgroup$ – reuns Mar 23 '16 at 10:37
  • $\begingroup$ @user1952009 Eisenstein's criterion after substituting $x=t+1$. $\endgroup$ – Jyrki Lahtonen Mar 23 '16 at 10:44
  • $\begingroup$ It is clear that the minimal polynomial of $\zeta_7$ over $\Bbb Q(\beta)$ is $(x-\zeta_7)(x - \overline{\zeta_7}) = x^2 - \beta x + 1$, since $\zeta_7 \not\in \Bbb R$. Also (@user1952009: ) it is a (somewhat involved) theorem that the cyclotomic polynomials are all irreducible over $\Bbb Q$-this is the special case $n = 7$, see here: lehigh.edu/~shw2/c-poly/several_proofs.pdf $\endgroup$ – David Wheeler Mar 23 '16 at 10:55
  • $\begingroup$ @JyrkiLahtonen : thank you. and the proof of the Eisenstein's criterion is even simpler than finding a shift of $\sum_{k=0}^{p-1} x^k$ to which we can apply it : en.wikipedia.org/wiki/… $\endgroup$ – reuns Mar 23 '16 at 11:00
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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 $x^2-\beta x+1$.

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