Splitting field for $x^6-4$

Question

I'm trying to find the splitting field for $x^6-4$ over $\mathbb{F}_7$. I have that $x^6-4=(x^3+2)(x^3-2)=(x^3-5)(x^3-2)$, and both $(x^3-5)$ and $(x^3-2)$ are irreducible over $\mathbb{F}_7$ since neither one has a root in $\mathbb{F}_7$. Moreover, $\mathbb{F}_7$ has $3$ cube roots of $1$: $1,2$, and $4$. So once we have one root in a splitting field, say $r$, the others are $2r$ and $4r$. If we let $k=\mathbb{F}_7[x]/(x^3-5)$ and let $r$ denote the image of $x$ in $k$, then $x^3-5=(x-r)(x-2r)(x-4r)$, and so $k$ is a splitting field for $x^3-5$ over $\mathbb{F}_7$. My question is where do I go from here? I could do the same thing to get a splitting field for $x^3-2$, but I'm not sure how to ultimately get a splitting field for $x^6-4$.

Answer 1

You did almost all the work yourself already. If $\alpha$ is a cube root of $5$ in $k$, then $3 \alpha$ is a cube root of $2$. So $k$ is already a splitting field of $x^6 - 4$.