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This question is a little out of context of the full problem, but I'm basically trying to show that $\mathbb{R} \cap \mathbb{Q}[\omega] \subset \mathbb{Q}[\sqrt{5}]$ where $\omega = e^{\frac{2\pi i}{5}}$, the 5th root of unity. I think I see that $\mathbb{R} \cap \mathbb{Q}[\omega]$ is the set of real elements of $\mathbb{Q}[\omega]$, thus I determined the elements are just sums of rational multiples of $\cos(\frac{2\pi x}{5})$ where $0\leq x \leq 4$. So if I show $\cos(\frac{2\pi x}{5})$ is in $\mathbb{Q}[\sqrt{5}]$ then I'm done.

I'm just not so sure how to do this but was thinking it involved how $\cos(\frac{2\pi x}{5})$ is seen when plotted on the unit circle.

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    $\begingroup$ this page might help. $\endgroup$ – rogerl Nov 29 '17 at 14:32
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    $\begingroup$ Are you allowed to use the exact value? $\cos\dfrac{2\pi}5=\dfrac{\sqrt 5-1}4$.. $\endgroup$ – Bernard Nov 29 '17 at 14:33
  • $\begingroup$ From Dirichlet's unit theorem and the unit group of $\mathbb{Q}(\zeta_n)$ we can compute easily the unit group of its real subfield. $\endgroup$ – reuns Dec 2 '17 at 1:02
  • $\begingroup$ You will only be using $\cos(2\pi x/5)$ when $x$ is an integer, so I don't understand the need to leave the $x$ in there? $\endgroup$ – Jyrki Lahtonen Dec 2 '17 at 7:51
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You can use explicit formulas for angles in a pentagon (they will involve golden ratio). But it's also possible to get through algebraically. Your original field is defined by roots of unity:

$$z^5-1=0$$ Split (with $z\neq 0$): $$(z-1)(z^4+z^3+z^2+z+1)=z^2(z-1)((z^2+z^{-2})+(z+z^{-1})+1)=0$$ We are looking for real parts of the solutions. Take a look at the last term, writing $w=z+z^{-1}$ and noticing $w^2=z^2+z^{-2}+2$. You get:

$$w^2+w-1=0$$ with solutions $$w=\frac{-1\pm \sqrt{5}}{2}$$

We have $w=z+z^{-1}$ and we know that $z$ lie on the unit circle, so $w=2\operatorname{Re} (z)$, and thus $$\operatorname{Re}(z)\in\mathbb{Q}(\sqrt{5}) $$

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$\mathbb{Q}(\sqrt{5})$ is a subfield of $\mathbb{Q}(\omega)$ by Gauss sums. If we consider $$ a = 1+\omega-\omega^2-\omega^3+\omega^4 = 1+2\cos\frac{2\pi}{5}-2\cos\frac{3\pi}{5}$$ we may easily check that $a=\sqrt{5}$ by squaring. On the other hand $$ -\cos\frac{3\pi}{5} = \cos\frac{8\pi}{5} = T_4\left(\cos\frac{2\pi}{5}\right)$$ so $\sqrt{5}\in\mathbb{Q}\left(\cos\frac{2\pi}{5}\right)$, too. By comparing the degrees of the extensions, we get $\mathbb{Q}(\sqrt{5})=\mathbb{Q}\left(\cos\frac{2\pi}{5}\right)$.

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