I want to find the splitting field in$\mathbb{C}$ of $x^4-4$ over $\mathbb{Q}$.
I solved for the zeros, which is $i\sqrt2, -i\sqrt2, \sqrt2, -\sqrt2$, so the splitting field, say $E$, is just $\mathbb{Q}(i\sqrt2, \sqrt2)$, or should it just be $\mathbb{Q}(i, \sqrt2)$?
If $E=\mathbb{Q}(i\sqrt2, \sqrt2)$, then I get the order of $\mathbb{Q}(i\sqrt2, \sqrt2)$ over $\mathbb{Q}$ is $$[\mathbb{Q}(i\sqrt2, \sqrt2):\mathbb{Q}]=4$$
Please give me some input. I am still trying to grasp the definition/ methods of dealing with splitting field.
Those fields are the same. You can check this by noting that $i\sqrt{2} \in {\mathbb Q}(i,\sqrt{2})$ and, conversely, that $i \in {\mathbb Q}(i\sqrt{2},\sqrt{2})$.
And, indeed, $[{\mathbb Q}(i,\sqrt{2}) : {\mathbb Q}] = 4$. This follows, for instance, from the fact that $x^4 - 4$ is irreducible over ${\mathbb Q}$. You could also reason that since $[{\mathbb Q}(i) : {\mathbb Q}] = 2$ and $[{\mathbb Q}(\sqrt{2}) : {\mathbb Q}] = 2$ and ${\mathbb Q}(i) \neq {\mathbb Q}(\sqrt{2})$, you must have that $[{\mathbb Q}(i,\sqrt{2}) : {\mathbb Q}] = 4$.