Splitting field of $x^4+64$ over $\mathbb{Q}$

Question

I am trying to find the Galois group of the splitting field of $x^4+64$ over $\mathbb{Q}$.

From what I can see so far, $x^4+64$ factors into $(x-(2+2i))(x-(2-2i))(x+(2+2i))(x+(2-2i))$.

How do I get the splitting field from here? It looks to me like it should be $\mathbb{Q}(i)$ or $\mathbb{Q}(1+i)$ or something like that. It's also important to note that $(1+i)^3=-2(1-i)$ so maybe it should be $\mathbb{Q}((1+i), (1+i)^3)$?

WolframAlpha factors the polynomial over $\mathbb{Q}((-1)^\frac{3}{4}\sqrt{2}, \sqrt[4]{-1}\sqrt{2})$. Why is this? Is it the correct splitting field?

Answer 1

Well, it suffices to generate $1\pm i$ as then the field operations yield the other routes. $1\pm i$ have modulus $\sqrt{2}$ so i.e. they equal $\sqrt{2}\cdot\exp(i\theta)$ for suitable real values of $\theta$, so Wolfram's expressions are correct as e.g. $(-1)^{3/4}=\exp(-3\pi i/4)$ (principal value) $=\frac{-1+i}{\sqrt{2}}$ and from that we can recover $-2+2i$ and $2-2i$.

Is it $\Bbb Q(i)$? Can any $x+iy$ with $x,y\in\Bbb Q$ be attained by $1\pm i$ and field operations? Well, yes. $$x+iy=x+\frac{(1+i)-(1-i)}{2}y$$Is an element of the splitting field, and obviously the splitting field is contained in $\Bbb Q(i)$, so there is equality.