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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！

non-abelian group of order 9
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  • 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: https://math.stackexchange.com/questions/2230280/find-the-degree-of-extension-bbb-q%CE%B6-9-%CE%B6%E2%88%921-9-over-bbb-q/2230302#2230302 – Kaj Hansen May 31 '17 at 04:42
  • 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. – Kaj Hansen May 31 '17 at 04:46

1 Answers1

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$\zeta_9$ 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 $\color{red}{z^3-3z+1}$.

Jack D'Aurizio
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  • Neat trick! Very nice. – Thomas Andrews Apr 13 '17 at 13:47
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    see page 174 in https://books.google.com/books?id=wt7lgfeYqMQC&pg=PR1&lpg=PR1&dq=reuschle++tafeln+complexer+primzahlen&source=bl&ots=VGZFPrfUBn&sig=MlQ667PqXaQ9rAvLWkG3_F1rwsk&hl=en&sa=X&ved=0ahUKEwiIwtSvm9TQAhUJ-2MKHXJIA_kQ6AEIODAE#v=onepage&q=reuschle%20%20tafeln%20complexer%20primzahlen&f=false method of Gauss. The primes up to 100 are done pages 1-171 – Will Jagy Apr 13 '17 at 17:56
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    @ThomasAndrews there is a modern treatment of Gauss on cyclotomy in Galois Theory by David A. Cox, but just three or four examples. Reuschle (1875) is the only item I have found with many examples. There was also a discussion in Theory of Numbers by G. B. Mathews, I think that was about 1900. Few examples, bu that is where I found the reference to Reuschle – Will Jagy Apr 13 '17 at 18:06
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    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 – Thomas Andrews Apr 13 '17 at 19:47