Describe the Galois group and the intermediate fields of the cyclotomic extension \(\mathbb Q(\zeta_{12})/\mathbb Q\).
Solution
1. The extension \(\mathbb Q(\zeta_{12})/\mathbb Q\) is Galois of degree \(4\), with\[ \operatorname{Gal}(\mathbb Q(\zeta_{12})/\mathbb Q) \cong(\mathbb Z/12\mathbb Z)^\times =\{1,5,7,11\} \cong C_2\times C_2. \]
The \(12\)th cyclotomic polynomial has degree \[ \varphi(12)=4, \] so \[ [\mathbb Q(\zeta_{12}):\mathbb Q]=4. \] Cyclotomic extensions are Galois, and every automorphism is determined by \[ \sigma_a(\zeta_{12})=\zeta_{12}^a, \qquad a\in(\mathbb Z/12\mathbb Z)^\times. \] The units modulo \(12\) are \(1,5,7,11\), and each nonidentity element has order \(2\), so the group is the Klein four group.
2. The field \(\mathbb Q(\zeta_{12})\) equals\[ \mathbb Q(i,\sqrt3). \]
We have \[ i=\zeta_{12}^3 \] and \[ \sqrt3=\zeta_{12}+\zeta_{12}^{-1}=2\cos(\pi/6). \] Hence \(\mathbb Q(i,\sqrt3)\subseteq\mathbb Q(\zeta_{12})\). Conversely, \[ \zeta_{12}=\cos(\pi/6)+i\sin(\pi/6)=\frac{\sqrt3+i}{2}, \] so \(\zeta_{12}\in\mathbb Q(i,\sqrt3)\).
3. The subgroup\[ H_i=\{1,5\}\le(\mathbb Z/12\mathbb Z)^\times \]fixes exactly \(\mathbb Q(i)\).
Since \(i=\zeta_{12}^3\), \[ \sigma_a(i)=i \iff \zeta_{12}^{3a}=\zeta_{12}^3 \iff 3(a-1)\equiv0\pmod{12} \iff a\equiv1\pmod4. \] Among \(1,5,7,11\), this gives \(a=1,5\). Thus the stabilizer of \(i\) is \(H_i\). Its fixed field has degree \[ [\mathbb Q(\zeta_{12})^{H_i}:\mathbb Q]=\frac4{|H_i|}=2, \] and contains the quadratic field \(\mathbb Q(i)\), so equality holds.
4. The subgroup\[ H_{\sqrt3}=\{1,11\} \]fixes exactly \(\mathbb Q(\sqrt3)\).
The automorphism \(\sigma_{11}\) is complex conjugation because \(11\equiv-1\pmod{12}\). Hence it fixes \[ \zeta_{12}+\zeta_{12}^{-1}=\sqrt3. \] Thus \(\mathbb Q(\sqrt3)\) lies in the fixed field of \(H_{\sqrt3}\). Both fields have degree \(2\) over \(\mathbb Q\), so they are equal.
5. The subgroup\[ H_{\sqrt{-3}}=\{1,7\} \]fixes exactly \(\mathbb Q(\sqrt{-3})=\mathbb Q(\zeta_3)\).
Since \[ \zeta_3=\zeta_{12}^4, \] we have \[ \sigma_a(\zeta_3)=\zeta_3 \iff 4(a-1)\equiv0\pmod{12} \iff a\equiv1\pmod3. \] Among the units modulo \(12\), this gives \(a=1,7\). Hence the fixed field of \(\{1,7\}\) contains \(\mathbb Q(\zeta_3)=\mathbb Q(\sqrt{-3})\); both have degree \(2\) over \(\mathbb Q\), so they are equal.
6. Therefore the complete list of intermediate fields is\[ \boxed{ \mathbb Q, \quad \mathbb Q(i), \quad \mathbb Q(\sqrt3), \quad \mathbb Q(\sqrt{-3}), \quad \mathbb Q(\zeta_{12})}. \]
By the fundamental theorem of Galois theory, intermediate fields correspond bijectively to subgroups of the Klein four group. That group has exactly five subgroups: the whole group, the trivial subgroup, and its three distinct subgroups of order \(2\). The corresponding three quadratic fixed fields were identified in <1>3–<1>5.