Consider the polynomial $g(x) = x^6 + 2$. Compute the degree of the splitting field $M$ of $g(x)$ over $\mathbb{Q}$.
I've been struggling with this question now for a bit and I'm not really sure what the answer is. Earlier in the question we calculated the splitting field of $f(x) = x^6 + 3$, which I found to be $\mathbb Q(\sqrt[6]-3)$, since we can make the sixth root of unity $\zeta = \frac{1 + \sqrt -3}{2}$ out of $\sqrt[6]-3$.
It would then make sense to me that the splitting field for $g(x)$ is $\mathbb Q(\sqrt[6]-2,\zeta)$, since $g(x)$ certainly splits in this field. However, the hint for this question is that "You may assume that the polynomial $x^2 + 3$ has no root in the extension of $\mathbb Q$ generated by $g(x)$" which invalidates my guess since $\sqrt-3 = 2\zeta-1\in \mathbb Q(\sqrt[6]-2,\zeta)$.
The only other thing I could think of would be if I could somehow be smart with some type of simple extension. However, even if I could find a potential candidate (which I can't using products of $\zeta$ and $\sqrt[6] -2$), using some software I can see the Galois group is of size $12$, which would mean $g(x)$ would have to be a degree $12$ polynomial which clearly isn't true.
Where is the flaw in my logic?
Thanks!