# Find the minimal polynomial of $\zeta_9+\zeta^{-1}_9$ over $\Bbb Q$

Knowing the degree of $\Bbb Q(\zeta_9+\zeta^{-1}_9)$ over $\Bbb Q$ is 3, now I want to find the minimal polynomial of $\zeta_9+\zeta^{-1}_9$ over $\Bbb Q$. I tried to use the relation $\zeta_9$ is the root of $x^6＋x^3＋1$ and $x^8＋x^7+x^6+x^5+x^4+x^3+x^2+x+1$. But it does not seems to work. Now I am stuck. Could someone please help？ Thanks so much！

• I have two approaches to this problem that you can read about here; I've found the second one in particular to be most useful on many occasions: math.stackexchange.com/questions/2230280/… May 31, 2017 at 4:42
• Ha! I just noticed you're the one from yesterday when I used the same tool to determine $\sqrt{3}$ is not in that one field. Like I said, quite versatile. Anyways, take care. May 31, 2017 at 4:46

$\zeta_9$ is a root of $p(x)=\frac{x^9-1}{x^3-1}=x^6+x^3+1$ and $$\frac{p(x)}{x^3} = 1+x^{3}+x^{-3} = 1+\left(x+\frac{1}{x}\right)^3- 3\left(x+\frac{1}{x}\right)$$ hence $\zeta_9+\zeta_9^{-1}=2\cos\frac{2\pi}{9}$ is a root of $\color{red}{z^3-3z+1}$.
• This answer, however, is more an example of a root of a primitive symmetric polynomial - if $f(x)$ is irreducible of degree $n$ such that $x^nf(1/x)=f(x)$, then we can use this technique to find a minimal polynomial for $\alpha+1/\alpha$ where $\alpha$ is a root of $f$. @WillJagy Apr 13, 2017 at 19:47