I need to find the degree of the splitting field of $x^{4}-4$ over $\mathbb{Q}$ and $\mathbb{Z}/5 \mathbb{Z}$.
I know that the roots are $\pm\sqrt{2}, \pm \sqrt{2}i$, so the splitting field is $F(\sqrt{2},i)\cong \mathbb{Q}(\sqrt{2},i)$ and $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}] = 2$. Since $i \notin \mathbb{Q}(\sqrt{2})$ then $[\mathbb{Q}(\sqrt{2},i): \mathbb{Q}(\sqrt{2})] = 2$, because $x^{2}+1$ is irreducible over $\mathbb{Q}(\sqrt{2})$ then by multiplication of degrees $[\mathbb{Q}(\sqrt{2},i):\mathbb{Q}] = 4$, is that correct?
In the other case we have that $x^{4}-4 = (x^{2}-2)(x^{2}+2) = (x^{2}-2)(x^{2}-3)$ over $\mathbb{Z}/5 \mathbb{Z}$, then the roots are $\pm\sqrt{2}, \pm\sqrt{3}$, and the slitting field would be $\mathbb{Z}/5 \mathbb{Z}(\sqrt{2},\sqrt{3})$ which is of degree 4.
I do not know if what I did is right?
Thanks