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Let $\zeta$ = $\cos(\frac{2\pi}{7} ) + i\sin(\frac{2\pi}{7} )$ , let $\alpha = \zeta +\zeta^{-1} $ note that $\zeta^{-1} =\zeta^6 $

I try to find the minimal polynomial of $\alpha$ over $\mathbb Q$. I only managed to show that the degree of the minimal polynomial is 3. My attempt so far:

$\alpha^3 = \zeta^3+ 3\zeta^{-1}\zeta^2+3\zeta\zeta^{-2}+\zeta^{18} = \zeta^3+\zeta^4+3\alpha $

And I don't know how to continue, Thank you for your help

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  • $\begingroup$ שלום. When you wrote $\zeta^{18}$ did you mean $\zeta^{-3}$? $\endgroup$ – J. W. Tanner Sep 16 '19 at 1:19
  • $\begingroup$ Look up here, or many other places. Like this search list or this. $\endgroup$ – Jyrki Lahtonen Sep 16 '19 at 5:06
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Now add the equation $$ α^4=ζ^4+ζ^{-4}+4(ζ^2+ζ^{-2})+6=ζ^4+ζ^3+4α^2-2 $$ to your consideration to find that $$ α^4-α^3-4α^2+3α+2=0 $$ This has a root at $2=1+1^{-1}$, the remaining factor is $$ α^3 + α^2 - 2α - 1=0 $$


You could of course also start at $$ 0=\frac{ζ^7-1}{ζ-1}=1+ζ+ζ^2+ζ^3+ζ^4+ζ^5+ζ^6=1+α+(α^2-2)+(α^3-3α) $$

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  • $\begingroup$ Thanks !!, how did you know that 2 is a root of the polynomial? $\endgroup$ – משה לוי Sep 15 '19 at 20:18
  • $\begingroup$ Because in that derivation up to that point $ζ=1$ is also a valid root of $ζ^7-1=0$. $\endgroup$ – Lutz Lehmann Sep 15 '19 at 20:20
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Note that $\zeta^7=1$ and $\zeta\ne1$ so $\color{red}{\zeta^6}+\color{purple}{\zeta^5}+\color{blue}{\zeta^4+\zeta^3}+\color{purple}{\zeta^2}+\color{red}{\zeta}+1=0$.

Now $\alpha=\zeta^{-1}+\zeta, \alpha^2=\zeta^{-2}+\zeta^2+2, $ and $\alpha^3=\zeta^{-3}+\zeta^3+3\zeta^{-1}+3\zeta$,

so we have $\color{red}\alpha+\color{purple}{\alpha^2-2}+\color{blue}{\alpha^3-3\alpha}+1=0$ or $\alpha^3+\alpha^2-2\alpha-1=0$.

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  • $\begingroup$ Note that $\zeta^{-1}=\zeta^6, \zeta^{-2}=\zeta^5, $ and $\zeta^{-3}=\zeta^4$ $\endgroup$ – J. W. Tanner Sep 15 '19 at 20:24

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