Calculating a minimal polynomial over $\mathbb{Q}(\sqrt[3]{2})$

Question

Let $L_1=\mathbb{Q}(\omega\sqrt[3]{2})$ where $\omega=e^\frac{2\pi i}{3}$ and $L_2=\mathbb{Q}(\sqrt[3]{2})$.

I want to calculate $[L_1L_2:L_2]$, that it is the degree of the minimal polynomial over $L_2$ with root $\omega\sqrt[3]{2}$.

I think it must be 2, but all my efforts to prove this has been failed!

I think you are asking what the minimal polynomial of $\omega \sqrt[3]{2}$ is over field $\ell_2 = \mathbb{Q}(\sqrt[3]{2})$. — hardmath, Oct 15 '14 at 19:04
See also http://math.stackexchange.com/questions/349011/find-the-galois-group-of-x3-2-over-mathbbq?rq=1. — Dietrich Burde, Oct 15 '14 at 19:07

score 1 · Answer 1 · answered Oct 15 '14 at 19:02

1

Calculate the minimal polynomial of $\omega$ over the rationals, and see how this is related to the polynomial you seek.

answered Oct 15 '14 at 19:02

quid

42,135

Aaron Maroja · Answer 2 · 2014-10-15T23:33:33.050

Show that $p_{\omega,\mathbb{Q}} = X^2+X+1$ is irreducible over $\mathbb{Q}$ and that $\omega$ is a root of $p_{\omega,\mathbb{Q}}$. Let $\alpha = \sqrt[3]{2}$.

Also keep in mind that if $[\mathbb{Q}[\omega]:\mathbb{Q}] = 2$, $[\mathbb{Q}[\alpha]:\mathbb{Q}] = 3$ where $GDC(2,3) = 1$ and $[\mathbb{Q}[\omega,\alpha]:\mathbb{Q} [\alpha]] = r$ , $[\mathbb{Q}[\omega,\alpha]:\mathbb{Q} [\omega]] = s$. What can you tell about $r$ and $s$? And $p_{\omega,\mathbb{Q}}$? Draw a diagram.

Calculating a minimal polynomial over $\mathbb{Q}(\sqrt[3]{2})$

2 Answers2