Let, $y = x^2$. Then, $$x^4-4 = y^2 -4 = (y+2)(y-2) = (x^2+2)(x^2-2) = (x-\sqrt{2})(x+\sqrt{2})(x-i\sqrt{2})(x+i\sqrt{2}).$$ Hence, the splitting field of $x^4-4$ is $K = \mathbb{Q}(i\sqrt{2},\sqrt{2})$. From the tower law we can obtain that $[K:\mathbb{Q}] = 2*2 = 4$. So the Galois group corresponding to $K$ is of order $4$, so either the cyclic group of order $4$ or the Klein $4$ group. By inspecting the automorphisms of $K$ that must preserve the roots, we find that there is no element of order $4$; hence, the Galois group corresponding to $K$ is the Klein $4$ group. We have intermediate fields, namely $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(i\sqrt{2}). $
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