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Let $\omega$ be a primitive 7th root of 1 over $\Bbb Q$ .Let $\alpha= \omega+\omega^6$. Find the minimum polynomial of $\alpha$ over $\Bbb Q$.

What I have so far is;

$\omega^7=1$

$\alpha=\omega+\omega^6$

$\alpha - \omega = \omega^6$

$\alpha=1/\omega + \omega$

But I don't see how this is going to help as as I still don't have a root and can't figure our how to get the minimal polynomial when there is two variables.

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HINT:

We have $$\omega^6+\omega^5+\omega^4+\omega^3+\omega^2+\omega+1=0$$

Dividing either sides by $\omega^3,$ $$\omega^3+\frac1{\omega^3}+\omega^2+\frac1{\omega^2}+\omega+\frac1\omega+1=0$$

$$\left(\omega+\frac1{\omega}\right)^3-3\cdot\omega\cdot\frac1{\omega}\left(\omega+\frac1{\omega}\right)+\left(\omega+\frac1{\omega}\right)^2-2\cdot\omega\cdot\frac1{\omega}+\omega+\frac1\omega+1=0$$

Replace $\displaystyle\omega+\frac1{\omega}$ with $\alpha$

I leave for you as an exercise to show that this of the smallest degree

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  • $\begingroup$ why ω^6+ω^5+ω^4+ω^3+ω^2+ω+1=0 is this true?? $\endgroup$ – Padraic Apr 16 '14 at 15:03
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    $\begingroup$ @Padraic, What is a primitive 7th root? $\endgroup$ – lab bhattacharjee Apr 16 '14 at 15:10
  • $\begingroup$ Of course, thank you, sorry I had forgotten that fact. $\endgroup$ – Padraic Apr 16 '14 at 15:14
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You could try to compute the powers of $\alpha$ consecutively: $$ \alpha^2 =\omega^2+2\omega^7+\omega^{12}=\omega^2+2+\omega^5$$ $$ \alpha^3 =\omega^3+3\omega^8+3\omega^{13}+\omega^{18}=\omega^3+3\omega+3\omega^6+\omega^4$$ etc. and see if you can get some cancelling among $1,\alpha,\ldots,\alpha^6$ (or possibly earlier?). From the above, expanding $(\alpha^2-2)^2=(\omega^2+\omega^5)^2$ seems to be helpful as well ...

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