Cyclotomic extensions
Euler's totient function (Tag D-JX3YC)
Euler’s totient function \(\phi\colon\ZZ_{\geq 1}\to\ZZ_{\geq 1}\) is \[ \phi(n) \coloneqq \size \theset{ k\in\ZZ \st 1\leq k\leq n,\ \gcd(k ,n ) = 1}. \]
Primitive root of unity (Tag FD-HM4P2)
Let \(F\) be a field and \(n\geq 1\) an integer. An element \(\zeta\in F\) is a primitive \(n\)th root of unity if \(\zeta^n = 1\) and \(\zeta^m\neq 1\) for \(1\leq m<n\), that is, \(\zeta\) has order \(n\) in \(F^\times\).
Computing cyclotomic polynomials $\Phi_n$ (Tag PR-JLDJ6)
Let \(n \geq 1\) and let \(\Phi_n\) be the \(n\)th cyclotomic polynomial. Then \[ x^{n}-1=\prod_{\substack{d \divides n \\ d > 0}} \Phi_{d}(x), \qquad\text{so}\qquad \Phi_n(x) = \qty{x^n-1} \inverseof{\qty{\prod_{\substack{d \divides n \\ 0 < d < n}} \Phi_{d}(x)}}, \] and \[ \Phi_{n}(x)=\prod_{\substack{ d \divides n \\ d > 0} }\left(x^{d}-1\right)^{\mu\left(n/d\right)}, \] where \(\mu\) is the Möbius function: \[ \mu(m) = \begin{cases} 1 & \text{if } m = 1, \\ (-1)^{k} & \text{if } m \text{ is a product of } k \text{ distinct primes}, \\ 0 & \text{if } p^2 \divides m \text{ for some prime } p. \end{cases} \]
The first formula computes \(\Phi_n\) recursively: divide \(x^n - 1\) by the product of the \(\Phi_d\) with \(d \divides n\), \(d < n\), using polynomial long division.
Cyclotomic polynomial $\Phi_p$ for a prime $p$ (Tag FF-MUJDE)
For a prime \(p\), the cyclotomic polynomial \(\Phi_p\) is \[ \Phi_p(x)=\frac{x^p-1}{x-1}=1+x+x^2+\cdots+x^{p-1}. \]
The roots of \(x^p-1\) are the \(p\)th roots of unity, and every one of them other than \(1\) is primitive because \(p\) is prime. So \(x^p-1=(x-1)\Phi_p(x)\), and dividing by \(x-1\) with FF-ED3CD at \(a=1\) gives the sum.
Cyclotomic polynomial $\Phi_{2p}$ for an odd prime $p$ (Tag FF-UACKH)
For an odd prime \(p\), the cyclotomic polynomial \(\Phi_{2p}\) is \[ \Phi_{2p}(x)=\Phi_p(-x)=1-x+x^2-\cdots+x^{p-1}. \]
Since \(p\) is odd, \(-1\) has order \(2\) and \(\zeta\mapsto-\zeta\) is a bijection from the primitive \(p\)th roots of unity to the primitive \(2p\)th roots of unity. Hence \[ \Phi_{2p}(x)=\prod_\zeta(x+\zeta)=(-1)^{p-1}\prod_\zeta(-x-\zeta)=\Phi_p(-x), \] the products running over the primitive \(p\)th roots of unity \(\zeta\), because \(p-1\) is even. Substituting \(-x\) in \(\Phi_p(x)=1+x+\cdots+x^{p-1}\) (FF-MUJDE) gives the sum.
For \(p=2\) the formula fails: \(\Phi_4(x)=x^2+1\), whereas \(\Phi_2(-x)=1-x\).
(Galois group of $\QQ(\zeta_n)/\QQ$)
For a primitive \(n\)th root of unity \(\zeta_n\), \(\QQ(\zeta_n)/\QQ\) is Galois of degree \(\phi(n)\), and \[ \Gal(\QQ(\zeta_n)/\QQ) \cong (\ZZ/n)^\times, \] where the automorphism \(\sigma_a\colon \zeta_n \mapsto \zeta_n^a\) corresponds to \(a\). The group is abelian, so every intermediate field is Galois over \(\QQ\), and the intermediate fields correspond to the subgroups of \((\ZZ/n)^\times\).
For a prime \(p\), \((\ZZ/p)^\times\) is cyclic of order \(p-1\), so \(\QQ(\zeta_p)\) has exactly one subfield of each degree dividing \(p-1\). For positive integers \(a\neq b\), \(\QQ(\zeta_a) = \QQ(\zeta_b)\) if and only if one of \(a,b\) is odd and the other is twice it.
Radical extensions
($x^n - a$)
For \(a\in\QQ^\times\), the splitting field of \(x^n - a\) over \(\QQ\) is \(\QQ(\zeta_n, a^{1/n})\), and the tower \[ \QQ \subseteq \QQ(\zeta_n) \subseteq \QQ(\zeta_n, a^{1/n}) \] has first step abelian of degree \(\phi(n)\) and second step cyclic of degree dividing \(n\). The Galois group embeds in \(\ZZ/n \semidirect (\ZZ/n)^\times\), with \((\ZZ/n)^\times\) acting on \(\ZZ/n\) by multiplication, and the embedding is an isomorphism if and only if \(x^n - a\) is irreducible over \(\QQ(\zeta_n)\).
For \(x^3 - 2\), the tower \(\QQ \subseteq \QQ(\zeta_3)\subseteq \QQ(\zeta_3, 2^{1/3})\) has degrees \(2\) and \(3\), so the Galois group has order \(6\); it is nonabelian because \(\QQ(2^{1/3})\) is not normal over \(\QQ\), so it is \(S_3\cong\ZZ/3\semidirect(\ZZ/3)^\times\).
Solvability
Solvable group (Tag D-DFIDP)
Let \(G\) be a group with identity \(e\). The group \(G\) is solvable if there is a finite normal series \[ \theset{e} = G_0 \normal G_1 \normal \cdots \normal G_n = G \] such that \(G_{i}/G_{i-1}\) is abelian for every \(1\leq i\leq n\).
Solvable finite group (Tag FD-T7IQV)
A finite group \(G\) is solvable if it has a composition series whose composition factors are all abelian.
Solvability via the derived series (Tag T-QPMGT)
A group \(G\) is solvable if and only if its derived series reaches the trivial group, that is, \(G^{(n)}=\theset{e}\) for some \(n\geq 0\).
$S_n$ is solvable if and only if $n\leq 4$ (Tag T-EN5H4)
Let \(n\geq1\). The symmetric group \(S_n\) is solvable if and only if \(n\leq 4\).
For \(n\leq4\), \(S_n\) is isomorphic to a subgroup of \(S_4\), and subgroups of solvable groups are solvable. The series \(\theset{e}\normal V_4\normal A_4\normal S_4\), with \(V_4=\theset{e,(12)(34),(13)(24),(14)(23)}\), has abelian quotients \(V_4\), \(C_3\), and \(C_2\), so \(S_4\) is solvable.
For \(n\geq5\), \(A_n\) is simple and nonabelian. Its commutator subgroup \([A_n,A_n]\) is normal and nontrivial, so \([A_n,A_n]=A_n\), and the derived series of \(A_n\) is constant at \(A_n\) and never reaches \(\theset e\). Hence \(A_n\) is not solvable, and neither is \(S_n\), which contains it.
(Solvability by radicals)
Over a field \(K\) of characteristic \(0\), a polynomial \(f\in K[x]\) is solvable by radicals if and only if \(\Gal(\SF(f)/K)\) is a solvable group. The symmetric group \(S_n\) is solvable if and only if \(n\leq 4\). For example, \(f = x^5-4x+2\) is irreducible by Eisenstein’s criterion at \(2\) and has exactly three real roots, so complex conjugation gives a transposition in its Galois group and irreducibility gives a \(5\)-cycle; hence its Galois group is \(S_5\), and \(f\) is not solvable by radicals.