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I am beginner in field extension theory. But I have not learnt Galois theory. I need to solve the following problem:

For which prime numbers $p$ is $$\cos\left(\frac{2 \pi}{p}\right)$$ irrational?

I have spent some time on this question. But I did not make much progress. If someone could provide a full solution, that would be great. But hints will be appreciated as well. Thanks so much.

Robert Soupe
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2 Answers2

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Extanded hints:

  • Undoubtedly you know that the cyclotomic field $L=\Bbb{Q}(\zeta_p)$, with $\zeta_p=e^{2\pi i/p}$ is a degree $p-1$ extension of the rationals.
  • You also probably know that $K=\Bbb{Q}(\cos(2\pi/p))$ is a subfield of $L$, because $\zeta_p+\zeta_p^{-1}=2\cos(2\pi/p)$.
  • The previous equation also shows that $\zeta_p$ is a zero of the polynomial $m(x)=x^2-2\cos(2\pi/p)x+1\in K[x]$.
  • Therefore $[L:K]=$______ , and $[K:\Bbb{Q}]=$_______ (you fill in the blanks).
Jyrki Lahtonen
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We may assume $p>2$ without loss of generality ($\cos\pi$ is rational, of course).

$\cos\left(\frac{2\pi}{p}\right)$ is a root of a Chebyshev polynomial and $\cos\left(\frac{2\pi k}{p}\right)$, for $k=1,2,\ldots,\frac{p-1}{2}$, are algebraic conjugates. If you prove that the previous Chebyshev polynomial is an irreducible polynomial (by proving it is the minimal polynomial for $\cos\left(\frac{2\pi}{p}\right)$, or by relating it with a cyclotomic polynomial), you get that $\cos\left(\frac{2\pi}{p}\right)$ is an algebraic number of degree $\frac{p-1}{2}$ over $\mathbb{Q}$. It follows that $\cos\left(\frac{2\pi}{3}\right)$ belongs to $\mathbb{Q}$ only if $p=2$ or $p=3$.

Jack D'Aurizio
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