Is $\mathbb{Q}\left( \sqrt[3]{2}, \frac{-1 + i\sqrt{3}}{2}\right):\mathbb{Q}$ a simple extension?

Question

Is the extension $$\mathbb{Q}\left( \sqrt[3]{2}, \frac{-1 + i\sqrt{3}}{2}\right):\mathbb{Q}$$ simple? If so find the minimal polynomial and the basis for the extension.

Answer 1

This extension can be written as $\mathbb{Q}(\sqrt[3]{2}, \zeta_3)$, where $\zeta_3 = e^{2\pi i /3}$. This is a degree 6 extension.

$$p(x) = x^6+3 x^5+6 x^4+3 x^3+9 x+9$$ is the minimal polynomial for $\sqrt[3]{2}+ \zeta_3$. Since this polynomial is degree 6, its roots form number fields of degree 6 over $\mathbb{Q}$ contained in $\mathbb{Q}(\sqrt[3]{2}, \zeta_3)$.

But this actually implies that $\mathbb{Q}(\sqrt[3]{2}, \zeta_3) = \mathbb{Q}(\sqrt[3]{2}+ \zeta_3)$. This gives us our simple extension.

As Rene Schipperus notes, this holds true for any finite extension of a field of characteristic $0$. If we let $\theta = \sqrt[3]{2}+ \zeta_3$ then $1,\theta, \theta^2,\ldots, \theta^5$ should form a basis for the extension.

If we sub in for $\theta$ in the minimal polynomial, we get $$p(\theta) = 63 + 45 \sqrt[3]{2} + 36 (\sqrt[3]{2})^2 + 63\zeta^2 + 45 \sqrt[3]{2}\zeta^2 \\~~~~~~~~~~+ 36 (\sqrt[3]{2})^2\zeta^2 + 63 \zeta + 45 \sqrt[3]{2}\zeta + 36 (\sqrt[3]{2})^2 \zeta$$

Note that we can group terms as follows: $$63(1 + \zeta + \zeta^2)\cdot 45\sqrt[3]{2}(1 + \zeta + \zeta^2)\cdot 36(\sqrt[3]{2})^2(1 + \zeta + \zeta^2)$$

Note that since $1 + \zeta + \zeta^2 = 0$ (Look up cyclotomic polynomials), we have that the above product is $0$. Hence $p(\theta) = 0$.