Compute all intermediate fields of \(\mathbb Q(2^{1/4},\zeta_8)\) over \(\mathbb Q\).
Solution
Let \(\alpha=2^{1/4}\) and \[ K=\mathbb Q(\alpha,\zeta_8). \] Since \(\alpha^2=\sqrt2\) and \(\zeta_8=(1+i)/\sqrt2\), \[ i=\alpha^2\zeta_8-1, \] so \(K=\mathbb Q(\alpha,i)\), the splitting field of \(x^4-2\).
1. The Galois group is \(D_4\) of order \(8\).
Eisenstein gives \([\mathbb Q(\alpha):\mathbb Q]=4\), and \(i\notin\mathbb Q(\alpha)\subset\mathbb R\), hence \([K:\mathbb Q]=8\). Define \[ \sigma(\alpha)=i\alpha,\quad \sigma(i)=i, \qquad \tau(\alpha)=\alpha,\quad \tau(i)=-i. \] Then \(\sigma^4=\tau^2=1\) and \(\tau\sigma\tau=\sigma^{-1}\), so \[ G=\operatorname{Gal}(K/\mathbb Q)=\langle\sigma,\tau\rangle\cong D_4. \]
2. The three quadratic intermediate fields are\[ \mathbb Q(i),\qquad \mathbb Q(\sqrt2),\qquad \mathbb Q(i\sqrt2). \]
They are the fixed fields of the three order-\(4\) subgroups \[ \langle\sigma\rangle,\qquad \langle\sigma^2,\tau\rangle,\qquad \langle\sigma^2,\sigma\tau\rangle, \] respectively. Each displayed generator is fixed by the corresponding subgroup and generates a quadratic extension, so the fixed field is exactly the one shown.
3. The five quartic intermediate fields are\[ \mathbb Q(\zeta_8),\quad \mathbb Q(\alpha),\quad \mathbb Q(i\alpha),\quad \mathbb Q((1+i)\alpha),\quad \mathbb Q((1-i)\alpha). \]
The five order-\(2\) subgroups of \(D_4\) are \[ \langle\sigma^2\rangle, \langle\tau\rangle, \langle\sigma^2\tau\rangle, \langle\sigma\tau\rangle, \langle\sigma^3\tau\rangle. \] Their respective fixed fields are the five fields displayed above. For example, \(\sigma^2\) fixes \(i\) and \(\alpha^2=\sqrt2\), hence fixes \(\mathbb Q(i,\sqrt2)=\mathbb Q(\zeta_8)\); and \(\sigma\tau\) fixes \((1+i)\alpha\). Each fixed field has degree \(4\), so the indicated degree-\(4\) subfield is the full fixed field.
Together with \(\mathbb Q\) and \(K\), these are all intermediate fields, because they account for all ten subgroups of \(D_4\).