# Minimal polynomial- fields

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

• שלום. When you wrote $\zeta^{18}$ did you mean $\zeta^{-3}$? – J. W. Tanner Sep 16 '19 at 1:19
• Look up here, or many other places. Like this search list or this. – Jyrki Lahtonen Sep 16 '19 at 5:06

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 $$0=\frac{ζ^7-1}{ζ-1}=1+ζ+ζ^2+ζ^3+ζ^4+ζ^5+ζ^6=1+α+(α^2-2)+(α^3-3α)$$

• Thanks !!, how did you know that 2 is a root of the polynomial? – משה לוי Sep 15 '19 at 20:18
• Because in that derivation up to that point $ζ=1$ is also a valid root of $ζ^7-1=0$. – Lutz Lehmann Sep 15 '19 at 20:20

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 $$\alpha^3+\alpha^2-2\alpha-1=0$$.

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