Is $\sqrt{5}$ in $\mathbb{Q}(e^{2\pi i /3})$

Question

Is $\sqrt{5} \in \mathbb{Q}(e^{2\pi i /3})$? My initial guess is no, but I am unsure how to prove it, any tips?

Answer 1

First: show that $1, \sqrt{-3}, \sqrt{5}$ are linearly independent over $\Bbb{Q}$. Otherwise, we can suppose that $$a\sqrt{5} + b\sqrt{-3} +c= 0 $$ for some nonzero integers $a,b,c \in \Bbb{Z}$. Note that both $a,b \neq 0$, otherwise you would have that $\sqrt{-3} \in \Bbb{Q}$ or $\sqrt{5} \in \Bbb{Q}$. So $$(a\sqrt{5} + b\sqrt{-3})^2=c^2$$ $$\sqrt{-15} = \frac{c^2-5a^2+3b^2}{2ab} \in \Bbb{Q}$$ a contradiction.

This is quite standard and tedious, but it is necessary. Now, you can conclude, since $$\dim_{\Bbb{Q}} \Bbb{Q}(e^{\frac{2 \pi i}{3}}) = \dim_{\Bbb{Q}} \Bbb{Q}(\sqrt{-3}) = 2$$ and $1, \sqrt{-3} \in \Bbb{Q}(\sqrt{-3})$ are linealry independent (they are a basis).

Answer 2

Using $\omega$ to denote the cube root of 1, it is clear that the only real (or rational) numbers of the form $a+b\omega$ are those with $b=0$. Now we can see that as $a$ has to be rational we cannot obtain the (real) irrational $\sqrt5$ this way.

Edited: The typo changing $\sqrt3$ to $\sqrt5$. Thanks to Ravi Fernando for pointing out.

Answer 3

Let us denote $\omega = e^{\frac {2\pi i}3}$. We have that $\omega^3 = 1$ and thus it is a root of $x^3-1 = (x-1)(x^2+x+1)$, and since $\omega\neq 1$, it is a root of $x^2+x+1$. Since $\omega\notin\mathbb R$, $x^2+x+1$ is irreducible over $\mathbb Q$ and thus minimal polynomial of $\omega$. We conclude that $\mathbb Q(\omega)\cong \mathbb Q[x]/(x^2+x+1)$ so it is $2$-dimensional vector space over $\mathbb Q$ with basis $\{1,\omega\}$.

Assume now that $\sqrt 5\in\mathbb Q(\omega)$. Then we have that $$\sqrt 5 = a + b\omega = a + b\left(-\frac 12 + i\frac{\sqrt 3}2\right) = \left(a - \frac b 2\right) + ib\frac{\sqrt 3}2,\quad a,b\in\mathbb Q$$

but, $\sqrt 5\in\mathbb R$, so $b = 0$, and thus $a = \sqrt 5$. This is contradiction because $a$ is rational while $\sqrt 5$ is not. Hence, $\sqrt 5\notin \mathbb Q(\omega)$