Let \(L\) be the splitting field of \(x^{15}+2\) over \(\QQ\). Prove that \[ \operatorname{Gal}(L/\QQ) \cong \ZZ/15\ZZ\semidirect (\ZZ/15\ZZ)^\times \cong C_{15}\semidirect(C_4\times C_2), \] and identify the right factor as a Sylow \(2\)-subgroup.
Solution
Let \(\alpha\) be a root of \(x^{15}+2\), so \(\alpha^{15}=-2\), and let \(\zeta=\zeta_{15}\) be a primitive \(15\)th root of unity.
1. The splitting field is\[ L=\QQ(\alpha,\zeta). \]
The roots of \(x^{15}+2\) are \[ \alpha,\zeta\alpha,\dots,\zeta^{14}\alpha. \] Hence adjoining \(\alpha\) and \(\zeta\) gives every root, and any splitting field contains both \(\alpha\) and the quotient \((\zeta\alpha)/\alpha=\zeta\).
2. One has\[ [\QQ(\alpha):\QQ]=15, \qquad [\QQ(\zeta):\QQ]=\varphi(15)=8, \]and therefore\[ \QQ(\alpha)\cap\QQ(\zeta)=\QQ. \]
The polynomial \(x^{15}+2\) is Eisenstein at \(2\), so it is irreducible over \(\QQ\) and \([\QQ(\alpha):\QQ]=15\). The cyclotomic degree is \(\varphi(15)=8\).
The degree of the intersection divides both extension degrees, hence divides \(\gcd(15,8)=1\). Therefore the intersection is \(\QQ\).
3. Hence\[ [L:\QQ]=120. \]
Since the two finite extensions in 2 have relatively prime degrees, they are linearly disjoint over \(\QQ\). Thus \[ [L:\QQ] =[\QQ(\alpha):\QQ]\,[\QQ(\zeta):\QQ] =15\cdot8=120. \]
4. The subgroup\[ N=\operatorname{Gal}(L/\QQ(\zeta)) \]is cyclic of order \(15\).
By 3, \[ [L:\QQ(\zeta)]=15. \] Because \(\zeta\in\QQ(\zeta)\), every map \[ \sigma_a:\alpha\mapsto\zeta^a\alpha, \qquad \zeta\mapsto\zeta, \qquad a\in\ZZ/15\ZZ, \] is an automorphism of \(L\) fixing \(\QQ(\zeta)\). These give \(15\) distinct automorphisms, so \[ N\cong\ZZ/15\ZZ. \] Since \(N\) is the kernel of restriction \[ \operatorname{Gal}(L/\QQ)\to\operatorname{Gal}(\QQ(\zeta)/\QQ), \] it is normal.
5. The cyclotomic Galois group\[ H\cong(\ZZ/15\ZZ)^\times \]embeds in \(\operatorname{Gal}(L/\QQ)\) as the subgroup fixing \(\alpha\).
Because \(\QQ(\alpha)\cap\QQ(\zeta)=\QQ\), every automorphism of \(\QQ(\zeta)/\QQ\) extends uniquely to the compositum while acting trivially on \(\QQ(\alpha)\). Thus for each \[ b\in(\ZZ/15\ZZ)^\times \] there is an automorphism \[ \tau_b:\alpha\mapsto\alpha, \qquad \zeta\mapsto\zeta^b. \] These form a subgroup \[ H\cong\operatorname{Gal}(\QQ(\zeta)/\QQ) \cong(\ZZ/15\ZZ)^\times. \] Since \[ (\ZZ/15\ZZ)^\times\cong C_4\times C_2, \] this subgroup has order \(8\) and is therefore a Sylow \(2\)-subgroup of the group of order \(120\).
6. Conjugation gives the natural multiplicative action\[ \tau_b\sigma_a\tau_b^{-1}=\sigma_{ba}. \]
It suffices to evaluate both sides on \(\alpha\) and \(\zeta\). Since \(\tau_b^{-1}\) fixes \(\alpha\), \[ \tau_b\sigma_a\tau_b^{-1}(\alpha) = \tau_b(\zeta^a\alpha) = \zeta^{ba}\alpha = \sigma_{ba}(\alpha), \] and both automorphisms fix \(\zeta\) after the full conjugation.
7. Therefore\[ \operatorname{Gal}(L/\QQ) \cong C_{15}\semidirect(C_4\times C_2). \]
The subgroups \(N\) and \(H\) have orders \(15\) and \(8\), have trivial intersection, and \[ |NH|=15\cdot8=120=|\operatorname{Gal}(L/\QQ)|. \] Thus the full group is their internal semidirect product. By 6 the action of the complement \(H\cong(\ZZ/15\ZZ)^\times\) on \(N\cong\ZZ/15\ZZ\) is multiplication modulo \(15\).