The splitting field of $x^4-4$

Question

I have to find the splitting field of $x^4-4$.

I say that $$x^2-4=(x^2-2)(x^2+2)=(x-\sqrt 2)(x+\sqrt 2)(x-i\sqrt 2)(x+i\sqrt 2)$$

then I would say that the splitting field is given by $E=\mathbb Q(i,\sqrt 2)=\mathbb Q(i\sqrt 2)$ but how can I justify this?

Moreover, I know that if a polynomial $p(x)\in\mathbb Q[X]$ is irreducible, then the splitting field is given by $\mathbb Q[X]/(p(x))$, but how can I apply this theorem if $p(x)$ is a product of two irreducible polynomials like here?

Finally, I would say that the degree of the decomposition is two, but I don't really know how to justify. My justification would be that $\mathbb Q(i\sqrt 2)=\{a+i\sqrt 2\mid a,b\in\mathbb Q\}$ but I first have two prove that $E=\mathbb Q(i\sqrt 2)$.

Answer 1

As you have determined, the splitting field must contain both $\alpha$ and $\alpha i$. Given that, it also contains $\frac{\alpha i}{\alpha} = i$. So the splitting field contains $K=\mathbb{Q}[i,\alpha]$. But clearly the polynomial splits in $K$, so that $K$ is minimal with that property and thus is the splitting field.

Answer 2

Knowing that the sppliting field is the smallest field that contains all roots of $f(x) \in \mathbb{K[x]}$.

If you let $\alpha = \sqrt{2}$ then $\alpha $ a root of the polynomial. Moreover we have that if $\xi \neq 1$ is a primitive root of the unity, say $\xi = \cos \frac{2\pi}{4} + i\sin \frac{2\pi}{4} \Rightarrow \xi = i$ then $\alpha, -\alpha, \alpha i, \alpha i $ are the roots of the polynomial.

This being said we have that the sppliting field is $L= Gal(x^4-4,\mathbb{Q}) = \mathbb{Q}[i,\alpha]$.