Prove that for all integers $n$, $\cos (2\pi/n)$ is an algebraic number.
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The comment by CropDuster here: http://mathhelpforum.com/advanced-algebra/176372-prove-cos-2pi-n-algebraic-n-nonzero-integer.html may be of interest. – Daniel Pietrobon Dec 13 '12 at 10:07
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@DanielPietrobon: Thank you. – Metta World Peace Dec 13 '12 at 10:16
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3You may look up http://en.wikipedia.org/wiki/Chebyshev_polynomials. – Joel Cohen Dec 13 '12 at 10:27
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If you believe that sums and scalar multiples of numbers algebraic over $\mathbb{Q}$ are algebraic, then we have $1 + x + \ldots + x^{n-1}$ has roots $\cos(2\pi/n) \pm i \sin(2\pi /n)$, hence half their sum.
uncookedfalcon
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We have $\zeta_n = e^{2\pi i/n}$ being an algebraic integer for all $n$ simply because it is a root of $x^n -1$. Now notice that $\zeta_n + \zeta_n^{-1} =2\cos(\frac{2\pi i}{n})$ and hence $\frac{1}{2}(\zeta_n + \zeta_n^{-1})$ is contained in the finite algebraic extension $\Bbb{Q}(\zeta_n)$ from which it follows that it itself is algebraic over $\Bbb{Q}$. This comes from the equivalence of the following facts:
Let $E/F$ be a field extension and $\alpha \in E$. TFAE:
- $\alpha$ is algebraic over $F$
- $\dim_F F[\alpha] < \infty$
- $F[\alpha]$ is a field
- $F[\alpha] = F(\alpha)$.
Lastly I should add that $\frac{1}{2}(\zeta_n + \zeta_n^{-1})$ is in fact an algebraic integer, meaning to say that its minimal polynomial has coefficients in $\Bbb{Z}$.
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just re-write the following $\zeta_n + \zeta_n^{-1} =2\cos(\frac{2\pi }{n})$ – MUH Oct 24 '17 at 08:36