Intermediate fields of $\QQ(2^{1/4},\zeta_8)$

Let \[ \alpha=2^{1/4}. \] Since \(\alpha^2=\sqrt2\) and \(\zeta_8=(1+i)/\sqrt2\), we have \[ L=\QQ(\alpha,i), \] the splitting field of \(x^4-2\). Hence \[ [L:\QQ]=8 \] and \[ G=\operatorname{Gal}(L/\QQ)\cong D_4. \] Take generators \[ r(\alpha)=i\alpha,\quad r(i)=i, \qquad s(\alpha)=\alpha,\quad s(i)=-i, \] so \[ r^4=s^2=1, \qquad srs=r^{-1}. \]

By Galois correspondence, the three subgroups of order \(4\) give the three quadratic intermediate fields: \[ \begin{array}{c|c} \text{subgroup} & \text{fixed field}\\ \hline \langle r\rangle & \QQ(i)\\ \langle r^2,s\rangle & \QQ(\sqrt2)\\ \langle r^2,rs\rangle & \QQ(\sqrt{-2}). \end{array} \]

The five subgroups of order \(2\) give the five quartic intermediate fields: \[ \begin{array}{c|c} \text{subgroup} & \text{fixed field}\\ \hline \langle r^2\rangle & \QQ(\sqrt2,i)=\QQ(\zeta_8)\\ \langle s\rangle & \QQ(\alpha)\\ \langle r^2s\rangle & \QQ(i\alpha)\\ \langle rs\rangle & \QQ((1+i)\alpha)\\ \langle r^3s\rangle & \QQ((1-i)\alpha). \end{array} \]

Together with the endpoints \(\QQ\) and \(L\), these are all intermediate fields, because the displayed subgroups exhaust the subgroup lattice of \(D_4\).