Determine $[ \mathbb{Q} (\sqrt{2}, e^{2\pi i/3}) : \mathbb{Q}]$

Question

I want to determine $[ \mathbb{Q} (\sqrt{2}, e^{2\pi i/3}) : \mathbb{Q}]$. I can see that $[ \mathbb{Q} (\sqrt{2}) : \mathbb{Q}] = 2$ because $2$ is the degree of the polynomial $x^2 -2$, which is used because it has $\sqrt{2}$ as a root. However, I don't know what to do with determine $e^{2\pi i/3}$.

Answer 1

You can verify that $e^{2i\pi/3}$ is a cube root of unity, from which we can deduce that $\displaystyle f(x) = \frac{x^3-1}{x-1} = x^2 + x + 1$ is its minimal polynomial. If $f$ is irreducible over $\mathbb{Q}(\sqrt{2})$—or, equivalently, if $f$ has no roots in this field—then the extension $\mathbb{Q}(\sqrt{2}, e^{2i\pi/3})/\mathbb{Q}(\sqrt{2})$ will be of degree $2$. To see that roots of $f$ do not exist in $\mathbb{Q}(\sqrt{2})$, note that these roots are complex with nonzero imaginary component (look at its discriminant or expand the cube roots of unity according to Euler's theorem).

To finish up, apply the multiplicativity formula for degrees on the tower $\mathbb{Q} \subset \mathbb{Q}(\sqrt{2}) \subset \mathbb{Q}(\sqrt{2}, e^{2i\pi/3})$.