 Solution

The polynomial $x^2-2$ is an irreducible polynomial over $\mathbb{Q}$, so $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$. This intermediate field is real, so it does not contain $i$. Hence $x^2+1$ has no root in it and adjoining $i$ has degree two. The tower law givesA basis is $1,\sqrt{2},i,i\sqrt{2}$. Square-bracket adjoining notation in this algebraic example gives the same field as round-bracket adjoining notation.