Solution:Let $\mathbb E$ be the splitting field of $x^4+1$ over $\mathbb Q$.Then $x^4+1$ splits into linear factors in $\mathbb E$.
$$x^4+1=(x^2-i)(x^2+i)=(x-\sqrt i)(x+\sqrt i)(x-\sqrt {-i})(x+\sqrt {-i})$$
$$=(x-e^{i\pi/4})(x+e^{i\pi/4})(x-e^{i3\pi/4})(x+\ e^{i3\pi/4}).$$
Since $\mathbb E$ is the splitting field of $x^4+1$ over $\mathbb Q$ then $\mathbb E$ is the smallest field containing $\mathbb Q$ and the roots of $x^4+1$.
Thus,$$\mathbb E=\mathbb Q(e^{i3\pi/4},-e^{i3\pi/4},e^{i\pi/4},-e^{i3\pi/4}).$$Since $e^{i3\pi/4}$ can be obtained by taking the cube of $e^{i\pi/4}$.Thus,$\mathbb E=\mathbb Q(e^{i\pi/4})=\mathbb Q(\frac{1+i}{\sqrt 2})$.
Since $1\in \mathbb Q$,so there is no need to adjoin 1.Thus,$\mathbb E=\mathbb Q(e^{i\pi/4})=\mathbb Q(\frac{1+i}{\sqrt 2})=\mathbb Q(i,\sqrt 2)=\mathbb Q(\sqrt {-2})$
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Best Answer
This seems to be OK. Another, more general approach, is to look at this and noting that $x^4+1$ is the cyclotomic polynomial $\Phi_8(x)$, hence $x^4+1=(x-\zeta)(x-\zeta^3)(x-\zeta^5)(x-\zeta^7)$, where $\zeta \in \mathbb{C}$ is a primitive root of unity, with $\zeta^8 =1$. The splitting field over $\mathbb{Q}$ of your polynomial is $\mathbb{Q}(\zeta)$, which is indeed $\mathbb{Q}(i,\sqrt{2})$.