Let \(n\ge3\), let \(\alpha=2^{1/n}\), and let \(\zeta_n\) be a primitive \(n\)th root of unity.
Show that the splitting field of \(x^n-2\) over \(\QQ\) is \[ L=\QQ(\alpha,\zeta_n), \] and that \[ \Gal(L/\QQ)\hookrightarrow (\ZZ/n\ZZ)\rtimes(\ZZ/n\ZZ)^\times \] via \[ \sigma(\alpha)=\zeta_n^a\alpha, \qquad \sigma(\zeta_n)=\zeta_n^b. \]
Prove that \(\Gal(L/\QQ)\) is dihedral of order \(2n\) exactly for \[ n=3,4,6. \]
Solution
1. The affine embedding.The roots of \(x^n-2\) are\[ \zeta_n^j\alpha, \qquad j\in\ZZ/n\ZZ, \]so the splitting field is \(L=\QQ(\alpha,\zeta_n)\).
Every \(\sigma\in\Gal(L/\QQ)\) permutes these roots, hence \[ \sigma(\alpha)=\zeta_n^a\alpha \] for a unique \(a\in\ZZ/n\ZZ\). Also \[ \sigma(\zeta_n)=\zeta_n^b \] for a unique \(b\in(\ZZ/n\ZZ)^\times\). Since \(\alpha\) and \(\zeta_n\) generate \(L\), the pair \((a,b)\) determines \(\sigma\), giving the claimed injection. Composition is the affine law \[ (a,b)(c,d)=(a+bc,bd). \]
2. The cases \(n=3,4,6\) are dihedral.For \(n=3,4,6\), the cyclotomic field has degree \(2\). In each case\[ \QQ(\alpha)\cap\QQ(\zeta_n)=\QQ: \]for \(n=3,6\) the cyclotomic field is imaginary quadratic while \(\QQ(\alpha)\subset\RR\), and for \(n=4\) it is \(\QQ(i)\) while \(\QQ(\alpha)\) is real. Thus\[ [L:\QQ]=2n. \]There is a rotation\[ r:\alpha\mapsto\zeta_n\alpha,\qquad \zeta_n\mapsto\zeta_n \]of order \(n\), and complex conjugation\[ s:\alpha\mapsto\alpha,\qquad \zeta_n\mapsto\zeta_n^{-1} \]has order \(2\) and satisfies\[ srs=r^{-1}. \]Hence\[ \Gal(L/\QQ)\cong D_{2n}. \]
3. A dihedral group forces a very small cyclotomic quotient.Restriction gives a surjection\[ \Gal(L/\QQ)\twoheadrightarrow \Gal(\QQ(\zeta_n)/\QQ) \cong(\ZZ/n\ZZ)^\times. \]If \(\Gal(L/\QQ)\cong D_{2n}\), then every abelian quotient factors through the abelianization of \(D_{2n}\), which is \(C_2\) for odd \(n\) and \(C_2\times C_2\) for even \(n\). Therefore\[ \varphi(n)\le4 \]and \((\ZZ/n\ZZ)^\times\) must have exponent at most \(2\).
For \(n\ge3\), the cases with \(\varphi(n)=2\) are exactly \(n=3,4,6\). The only additional cases with \(\varphi(n)\le4\) are \[ n=5,8,10,12. \] For \(n=5,10\), the unit group is cyclic of order \(4\), so it cannot be an abelian quotient of a dihedral group. It remains to exclude \(8\) and \(12\).
4. The case \(n=8\) is not dihedral.Here\[ \QQ(\alpha)\cap\QQ(\zeta_8)=\QQ(\sqrt2), \]because both fields contain \(\sqrt2=\alpha^4=\zeta_8+\zeta_8^{-1}\), while the intersection is real and the maximal real subfield of \(\QQ(\zeta_8)\) is exactly \(\QQ(\sqrt2)\). Hence\[ [L:\QQ]=\frac{8\cdot4}{2}=16. \]
Restriction to \(\QQ(\zeta_8)\) is surjective. Its kernel has order \[ 16/4=4 \] and consists of the even translations \[ (0,1),(2,1),(4,1),(6,1). \] Take the cyclotomic automorphism \(\zeta_8\mapsto\zeta_8^5\). Any lift to \(L\) must send \(\alpha\) to \(\zeta_8^a\alpha\) with \(a\) odd, because \[ \alpha^4=\sqrt2=\zeta_8+\zeta_8^{-1} \] is sent to \(-\sqrt2\). Composing with an even translation, we may choose a lift \[ r=(1,5). \] Let complex conjugation be \[ s=(0,7). \] Using the affine multiplication law, \[ r^8=s^2=1, \qquad |r|=8, \qquad srs=r^3. \] They generate a subgroup of order \(16\), hence all of \(\Gal(L/\QQ)\).
If this group were \(D_{16}\), its cyclic subgroup of order \(8\) would be the unique rotation subgroup, and every element outside it would conjugate a generator to its inverse. But here \[ srs=r^3\ne r^{-1}=r^7. \] Therefore the group is not dihedral.
5. The case \(n=12\) is not dihedral.Suppose for contradiction that\[ \Gal(L/\QQ)\cong D_{24}. \]Restriction to \(\QQ(\zeta_{12})\) has quotient\[ (\ZZ/12\ZZ)^\times\cong C_2\times C_2, \]so its kernel would have order \(6\).
An automorphism fixing \(\zeta_{12}\) acts by a translation \[ \alpha\mapsto\zeta_{12}^a\alpha, \] so the kernel is a subgroup of the cyclic translation group \(C_{12}\). Its unique subgroup of order \(6\) is generated by translation by \(2\), which fixes \[ \alpha^6=\sqrt2. \] Hence \(\sqrt2\) would belong to the fixed field of the kernel, namely \(\QQ(\zeta_{12})\).
But \[ \QQ(\zeta_{12})=\QQ(i,\sqrt3) \] has quadratic subfields only \[ \QQ(i),\qquad \QQ(\sqrt3),\qquad \QQ(\sqrt{-3}), \] so it does not contain \(\sqrt2\). Contradiction.
Thus the only dihedral cases are \[ \boxed{n=3,4,6}. \]