Determine the splitting field and its degree over $\mathbb Q$ for $x^6 - 4.$

Question

I have calculated its degree and it is also $6,$ am I correct?$x^6 - 4 = (x^3 - 2)(x^3 + 2)$

I am just suspecting my answer as I know by example $(3)$ on pg. $537$ in Dummit and Foote that the spliting field of $x^3 - 2$ only is also $6.$

Answer 1

The degree of this extension will be $6$ due to the following reason: as you pointed out the splitting field is $F=\mathbb{Q}(\zeta_3,\sqrt[3]{2})$. Now because clearly $X^3-2\in \mathbb{Q}[X]$ is the min. poly. of $\sqrt[3]{2}$ and $X^2+X+1\in \mathbb{Q}[X]$ the one of $\zeta_3$ we have $$[\mathbb{Q}(\zeta_3):\mathbb{Q}]=2,\quad [\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$$ and hence in particular $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3\mid [F:\mathbb{Q}]$ and $[\mathbb{Q}(\zeta_3):\mathbb{Q}]=2\mid [F:\mathbb{Q}]$. Thus, $6\mid [F:\mathbb{Q}]$ because $\gcd(2,3)=1$. On the other hand we know $$[F:\mathbb{Q}]=[F:\mathbb{Q}(\zeta_3)]\cdot [\mathbb{Q}(\zeta_3):\mathbb{Q}]\le 3\cdot 2=6$$ proving the claim.