Proving a polynomial splits fully in a certain field

Question

I have a polynomial $f(x)=x^6+3$, and define $a$ a root of $f(x)$. Then construct the field $\mathbb{Q}(a)$. I need to prove that $g(x)=x^3+3$ splits completely over $F(a)$. It hints at the fact that $a^2$ is a root of $g(x)$ and $a^3$ is a root of $x^2+3$. However I'm at a complete loss of how to use this information. I can gather that $(x-a^2)(x^2+a^2x+a^4)$ is a factorization of $g(x)$, but I don't think this is the right approach.

Any hints would be appreciated, thanks in advance.

Answer 1

The roots of $g$ are $a^2, \zeta a^2$, and $\zeta^2 a^2$ where $\zeta$ is a third root of unity, so it suffices to show that $\zeta \in \mathbb{Q}(a)$. Recall that $\zeta = \frac{-1 + \sqrt{-3}}{2}$. Since $a^6 = -3$, then $a^3 = \sqrt{-3}$ (this is the second part of your hint), so $$ \zeta = \frac{-1 + \sqrt{-3}}{2} = \frac{-1 + a^3}{2} \in \mathbb{Q}(a) \, . $$

Answer 2

Let $h(x)=x^2+a^2x+a^4$. The discriminant of $h$ is $D=a^4-4a^4=-3a^4$. We have $(a^5)^2=(a^2a^3)^2=a^4(a^3)^2=-3a^4$. Implies that $D$ has a root in $\mathbb{Q}(a)$. This implies that the roots of $h(x)$ are in $\mathbb{Q}(a)$.