Past exam of the options II course of the University of Oxford / 2006 / B / 2 / b

Let $\zeta =e^{\pi i/4}$. Its odd powers are exactly the four roots, so $\mathbb{Q}(\zeta )$ is the splitting field. The polynomial $x^4+1$ is irreducible: after the invertible substitution $x\mapsto x+1$, it becomes $x^4+4x^3+6x^2+4x+2$, which satisfies Eisenstein criterion at two. HenceEquivalently the field is $\mathbb{Q}(\sqrt{2},i)$, since $\zeta =(1+i)/\sqrt{2}$, ${\zeta }^2=i$ and $\zeta +{\zeta }^{-1}=\sqrt{2}$.