Let \(K=\mathbb Q(2^{1/n},\zeta_n)\), the splitting field of \(x^n-2\) over \(\mathbb Q\). Show that \[ \operatorname{Gal}(K/\mathbb Q)\hookrightarrow (\mathbb Z/n\mathbb Z)\rtimes(\mathbb Z/n\mathbb Z)^\times \] via \(\sigma\mapsto(a,b)\), where \[ \sigma(2^{1/n})=\zeta_n^a2^{1/n},\qquad \sigma(\zeta_n)=\zeta_n^b. \] For \(n=3,4,6\), show that this embedding is an isomorphism onto the dihedral group of order \(2n\).
The formula \([K:\mathbb Q]=n\varphi(n)\) is false in general; the intersection \(\mathbb Q(2^{1/n})\cap\mathbb Q(\zeta_n)\) can be nontrivial.
Solution
Write \(\alpha=2^{1/n}\) and \(G=\operatorname{Gal}(K/\mathbb Q)\).
1. Each \(\sigma\in G\) determines a unique pair\[ (a,b)\in \mathbb Z/n\mathbb Z\times(\mathbb Z/n\mathbb Z)^\times \]by\[ \sigma(\alpha)=\zeta_n^a\alpha,\qquad \sigma(\zeta_n)=\zeta_n^b. \]
The conjugates of \(\alpha\) are the roots \(\zeta_n^a\alpha\) of \(x^n-2\), and an automorphism sends the primitive \(n\)th root \(\zeta_n\) to another primitive \(n\)th root. Since \(K=\mathbb Q(\alpha,\zeta_n)\), the pair determines \(\sigma\) uniquely.
2. This gives an injective homomorphism into the affine group.
If \(\sigma_i\) corresponds to \((a_i,b_i)\), then \[ (\sigma_1\sigma_2)(\alpha)=\zeta_n^{a_1+b_1a_2}\alpha, \qquad (\sigma_1\sigma_2)(\zeta_n)=\zeta_n^{b_1b_2}. \] Hence composition corresponds to \[ (a_1,b_1)(a_2,b_2)=(a_1+b_1a_2,b_1b_2), \] the semidirect-product law. Injectivity follows from 1.
3. For \(n=3,4,6\), the embedding is onto and \(G\cong D_n\).
For these three values, \(\varphi(n)=2\) and \[ (\mathbb Z/n\mathbb Z)^\times=\{\pm1\}. \] Also \(\mathbb Q(\alpha)\subset\mathbb R\), while \(\mathbb Q(\zeta_n)\) is the imaginary quadratic field \(\mathbb Q(\sqrt{-3})\) for \(n=3,6\) and \(\mathbb Q(i)\) for \(n=4\). Hence \[ \mathbb Q(\alpha)\cap\mathbb Q(\zeta_n)=\mathbb Q. \] Eisenstein at \(2\) gives \([\mathbb Q(\alpha):\mathbb Q]=n\), so \[ [K:\mathbb Q]=2n. \] The target affine group also has order \(2n\), hence the injection is an isomorphism. Since the nontrivial unit \(-1\) acts on \(\mathbb Z/n\mathbb Z\) by inversion, \[ G\cong (\mathbb Z/n\mathbb Z)\rtimes\{\pm1\}=D_n. \]
4. The degree formula \([K:\mathbb Q]=n\varphi(n)\) cannot be used for arbitrary \(n\).
For example, when \(n=8\), \[ \sqrt2=\alpha^4\in\mathbb Q(\alpha), \qquad \sqrt2=\zeta_8+\zeta_8^{-1}\in\mathbb Q(\zeta_8). \] Thus the two subfields have nontrivial intersection, so their degrees do not multiply.