Solvability by Radicals

$F(\zeta_n)$ is the splitting field over $F$ of $x^n - 1$, which is separable in characteristic $0$ (its derivative $nx^{n-1}$ shares no root with $x^n-1$ since $0$ is not a root of $x^n - 1$). Hence $F(\zeta_n)/F$ is Galois. Any $\sigma \in \mathrm{Gal}(F(\zeta_n)/F)$ is determined by $\sigma(\zeta_n) = \zeta_n^{a_\sigma}$ for a unique $a_\sigma \in (\mathbb{Z}/n\mathbb{Z})^\times$ (it must send a primitive root to a primitive root, and all primitive $n$-th roots of unity are powers $\zeta_n^a$ with $\gcd(a,n)=1$). The map $\sigma \mapsto a_\sigma$ is an injective group homomorphism $\mathrm{Gal}(F(\zeta_n)/F) \to (\mathbb{Z}/n\mathbb{Z})^\times$, since $(\sigma\tau)(\zeta_n) = \sigma(\zeta_n^{a_\tau}) = \zeta_n^{a_\sigma a_\tau}$. As $(\mathbb{Z}/n\mathbb{Z})^\times$ is abelian, $\mathrm{Gal}(F(\zeta_n)/F)$, being isomorphic to a subgroup of it, is abelian.

(1) If $\alpha$ is one root of $x^n - a$, the full set of roots is $\{\alpha, \zeta_n\alpha, \zeta_n^2\alpha, \dots, \zeta_n^{n-1}\alpha\}$, since $(\zeta_n^k\alpha)^n = \zeta_n^{kn}\alpha^n = \alpha^n = a$ and these $n$ elements are distinct (as the $\zeta_n^k$ are distinct and $\alpha \ne 0$ — if $a = 0$ the statement is trivial with $K = F$). Since $\zeta_n \in F$ already, adjoining $\alpha$ adjoins all roots: $K = F(\alpha)$. For $\sigma \in \mathrm{Gal}(K/F)$, $\sigma(\alpha)$ is again a root of $x^n - a$, so $\sigma(\alpha) = \zeta_n^{c_\sigma}\alpha$ for some $c_\sigma \in \mathbb{Z}/n\mathbb{Z}$ (well-defined mod $n$ since $\zeta_n^n = 1$). Define $\chi: \mathrm{Gal}(K/F) \to \mathbb{Z}/n\mathbb{Z}$ by $\chi(\sigma) = c_\sigma$. This is a homomorphism: $(\sigma\tau)(\alpha) = \sigma(\zeta_n^{c_\tau}\alpha) = \zeta_n^{c_\tau}\sigma(\alpha) = \zeta_n^{c_\tau + c_\sigma}\alpha$, using that $\sigma$ fixes $F \ni \zeta_n$. It is injective because $\sigma$ is determined by its action on the generator $\alpha$ of $K/F$ ($\sigma(\alpha) = \alpha \Rightarrow \sigma = \mathrm{id}$). So $\mathrm{Gal}(K/F)$ embeds in the cyclic group $\mathbb{Z}/n\mathbb{Z}$, hence is cyclic of order dividing $n$.

(2) Let $\sigma$ generate $G := \mathrm{Gal}(K/F)$, $|G| = n$. Since $\zeta_n \in F \subseteq K$, consider the Lagrange resolvent: for $\beta \in K$ define

By linear independence of characters (Dedekind/Artin), the distinct automorphisms $1, \sigma, \dots, \sigma^{n-1}: K \to K$ are linearly independent over $K$, so the map $\beta \mapsto \sum_i \zeta_n^{-i}\sigma^i(\beta)$ is not the zero map; choose $\beta \in K$ with $\alpha := (\beta,\zeta_n) \ne 0$. Apply $\sigma$ to $\alpha$:

reindexing the sum over $i+1 \bmod n$ (valid since $\sigma^n = \mathrm{id}$). Hence $\sigma(\alpha^n) = (\sigma\alpha)^n = \zeta_n^n \alpha^n = \alpha^n$, so $\alpha^n$ is fixed by the generator $\sigma$ of $G$, hence by all of $G$; by the Fundamental Theorem of Galois Theory (the fixed field of $G$ is $F$), $\alpha^n \in F$. Finally $\sigma^k(\alpha) = \zeta_n^k \alpha$ for $k = 0, \dots, n-1$ are $n$ distinct elements (since $\alpha \ne 0$ and the $\zeta_n^k$ are distinct), so the orbit of $\alpha$ under $G$ has size $n = |G|$, forcing $[F(\alpha):F] \ge n$ (distinct conjugates give distinct roots of the minimal polynomial, whose degree is at least the orbit size) and since $F(\alpha) \subseteq K$ with $[K:F]=n$, we get $F(\alpha) = K$. Thus $K = F(\alpha)$ with $\alpha^n \in F$, as required.

We now prove the main theorem, in two directions.

Proof technique. ($\Leftarrow$) is a direct construction: use Kummer's Lemma (2) repeatedly to realize a solvable series as a tower of radical extensions, after first adjoining enough roots of unity via the cyclotomic lemma. ($\Rightarrow$) is a reduction argument: given a radical tower, adjoin roots of unity to every stage to make each step a genuine Kummer (cyclic) extension, apply Kummer's Lemma (1) to get cyclic Galois quotients at each stage, and use the correspondence between subnormal series of subgroups and subnormal series of fixed fields to build a solvable series for $\mathrm{Gal}(E/F)$ from it, invoking closure properties of solvable groups (subgroups and quotients of solvable groups are solvable).

($\Leftarrow$) Solvable $\Rightarrow$ solvable by radicals. Suppose $G = \mathrm{Gal}(E/F)$ is solvable, so there is a chain

with each $G_{i}/G_{i+1}$ abelian. Every finite abelian group has a composition series with cyclic quotients (refine using cyclic factors from the structure theorem, or simply refine any abelian group's subgroup chain — abelian groups are solvable with cyclic composition factors), so after refining we may assume each quotient $G_i/G_{i+1}$ is cyclic of some order $n_i$. Let $n = |G| = [E:F]$ and let $\zeta_n$ be a primitive $n$-th root of unity (working inside a fixed algebraic closure containing $E$). Set $E' = E(\zeta_n)$ and $F' = F(\zeta_n)$. We claim $F(\zeta_n)$, and more generally $F'$, is a radical extension of $F$: indeed $F' = F(\zeta_n)$ with $\zeta_n^n = 1 \in F$, so this is a single radical step. We must show $E'/F'$ is Galois with solvable — in fact, a refinement showing cyclic-step-buildable — Galois group, and that $E'$ embeds in a radical tower over $F'$; then since $F'/F$ is itself radical, composing gives a radical tower over $F$ containing $E \subseteq E'$. $E/F$ is Galois (splitting field of a separable polynomial, as $\mathrm{char} F = 0$), say $E$ is the splitting field of $f$. Then $E'= E(\zeta_n)$ is the splitting field of $f(x)(x^n-1)$ over $F'$ (it is generated over $F'$ by the roots of $f$ together with $\zeta_n$, all of which already lie in it, and it is the smallest such field), hence $E'/F'$ is Galois. Restriction gives a homomorphism

which is well-defined because $\tau(E) = E$ for $\tau \in \mathrm{Gal}(E'/F')$: indeed $E/F$ is Galois so $E$ is the splitting field of $f$ over $F$, and $\tau$ fixes $F$ (as $F \subseteq F'$) and permutes the roots of $f$ among themselves since $\tau$ is a field automorphism of $E'$ fixing coefficients of $f$; hence $\tau(E) \subseteq E$, and since $\tau$ is injective on the finite-dimensional space $E$ over $F$, $\tau(E) = E$. The map $\rho$ is injective: if $\tau|_E = \mathrm{id}_E$ then, since $\tau$ also fixes $F' \supseteq \zeta_n$ pointwise and $E' = E(\zeta_n)$, $\tau = \mathrm{id}_{E'}$. So $\mathrm{Gal}(E'/F')$ is isomorphic to a subgroup $H$ of $G = \mathrm{Gal}(E/F)$. Subgroups of solvable groups are solvable (restrict the given normal series' intersections with $H$: $H \cap G_i \triangleright H \cap G_{i+1}$, with $(H\cap G_i)/(H \cap G_{i+1})$ embedding via the second isomorphism theorem into $G_i/G_{i+1}$, hence abelian). So $H \cong \mathrm{Gal}(E'/F')$ has a chain

with cyclic quotients $H_i/H_{i+1}$ of order dividing $n$ (each is a subquotient of a cyclic group of order $n_i \mid n$... more precisely each $H_i/H_{i+1}$, being a subquotient of an abelian group appearing in the original series for $G$, is a subgroup of a cyclic group of order dividing $n$, hence cyclic of order dividing $n$). By the Fundamental Theorem of Galois Theory applied to $E'/F'$, let $L_i$ be the fixed field of $H_i$; this gives a tower

with $\mathrm{Gal}(L_{i+1}/L_i) \cong H_i/H_{i+1}$ (using that $H_{i+1} \triangleleft H_i$, so $L_{i+1}/L_i$ is Galois with group $H_i/H_{i+1}$), cyclic of order $n_i \mid n$. Since $L_i \supseteq F' \ni \zeta_n$, and $\zeta_n$ has order $n$ hence $\zeta_{n_i} := \zeta_n^{n/n_i}$ is a primitive $n_i$-th root of unity lying in $L_i$, Kummer's Lemma (2) applies at each stage: $L_{i+1} = L_i(\alpha_i)$ with $\alpha_i^{n_i} \in L_i$. Thus $F' = L_0 \subseteq L_1 \subseteq \cdots \subseteq L_s = E'$ is a radical tower over $F'$. Composing with the radical step $F \subseteq F' = F(\zeta_n)$ gives a radical tower

over $F$, and $E \subseteq E'$. Hence $f$ is solvable by radicals.

($\Rightarrow$) Solvable by radicals $\Rightarrow$ solvable. Suppose $E \subseteq K$ for some radical extension $K/F$:

Let $n = n_1 n_2 \cdots n_m$ and let $\zeta_n$ be a primitive $n$-th root of unity. Define $F_i' = F_i(\zeta_n)$ for $i = 0, \dots, m$, so we get an enlarged tower

Each $F_i'/F$ still forms a radical extension: $F_0' = F(\zeta_n)$ with $\zeta_n^n = 1 \in F$; and $F_i' = F_i(\zeta_n) = F_{i-1}(\alpha_i,\zeta_n) = F_{i-1}'(\alpha_i)$ with $\alpha_i^{n_i}\in F_{i-1} \subseteq F_{i-1}'$, so each step $F_{i-1}' \subseteq F_i'$ is again a radical step with the same exponent $n_i$, and now — crucially — $F_{i-1}'$ contains a primitive $n_i$-th root of unity $\zeta_{n_i} = \zeta_n^{n/n_i}$ since $n_i \mid n$. We claim $K'/F$ is Galois. First $F_0'/F$ is Galois with abelian group, by the cyclotomic Lemma. Next, $F_i'/F_{i-1}'$ need not individually be Galois (the other roots $\zeta_{n_i}^k \alpha_i$ might not lie in $F_{i-1}'$ yet if we only adjoined one root), but we remedy this by further enlarging: replace $F_i'$ by the splitting field $\widetilde{F_i}$ over $F$ of $\prod_{j \le i}(x^{n_j} - a_j)$ where $a_j = \alpha_j^{n_j} \in F_{j-1} \subseteq \widetilde{F_{j-1}}$, together with $\zeta_n$, adjoined successively. Concretely, set

Since $\widetilde F_{i-1}$ contains a primitive $n_i$-th root of unity, adjoining one root $\alpha_i$ of $x^{n_i}-a_i$ already adjoins all of them (as in Kummer's Lemma (1): all roots are $\zeta_{n_i}^k\alpha_i$), so in fact $\widetilde F_i = \widetilde F_{i-1}(\alpha_i)$ and this tower is radical over $F$, with $\widetilde F_m \supseteq F_m' = K' \supseteq K \supseteq E$ (each $F_i \subseteq \widetilde F_i$ by induction, since $\alpha_i \in F_i$ and $F_{i-1}\subseteq \widetilde F_{i-1}$). Rename $\widetilde F_i =: N_i$. By construction $N_0/F$ is Galois with abelian group, and $N_i/N_{i-1}$ is a splitting field over $N_{i-1}$ of the separable polynomial $x^{n_i} - a_i$ (separable since $n_i \ne 0$ in characteristic $0$ and roots of unity of order $n_i$ are already present, so $x^{n_i}-a_i$ has $n_i$ distinct roots), hence Galois; by Kummer's Lemma (1), $\mathrm{Gal}(N_i/N_{i-1})$ is cyclic. We now show $N_m/F$ is Galois by induction: $N_0/F$ is Galois. Suppose $N_{i-1}/F$ is Galois. Then $N_{i-1}$ is the splitting field over $F$ of some separable polynomial $g_{i-1}$, and $N_i$ is generated over $F$ by the roots of $g_{i-1}$ together with the roots of $x^{n_i}-a_i$ (as $N_i = N_{i-1}(\text{roots of }x^{n_i}-a_i)$ and $N_{i-1} = F(\text{roots of } g_{i-1})$); hence $N_i$ is the splitting field over $F$ of $g_{i-1}\cdot(x^{n_i}-a_i)$, a separable polynomial (product of separable polynomials in characteristic $0$; distinct irreducible factors are automatically coprime hence contribute no repeated roots collectively beyond what each contributes individually — more directly, in characteristic 0 every polynomial is separable up to repeated irreducible factors, and we may replace $g_{i-1}(x^{n_i}-a_i)$ by its radical/squarefree part without changing the splitting field). So $N_i/F$ is Galois. By induction, $N_m/F$ is Galois. Let $\Gamma = \mathrm{Gal}(N_m/F)$. Restriction gives a normal subgroup chain: let $\Gamma_i = \mathrm{Gal}(N_m/N_i)$ for $i = 0, \dots, m$ (with $N_m = N_m$, so $\Gamma_m = \{e\}$). Since each $N_i/F$ is Galois for the specific tower built (each $N_i$ is itself a splitting field over $F$ as shown above, by the same inductive argument truncated at stage $i$), each $\Gamma_i$ is normal in $\Gamma$ (subgroups fixing a field that is Galois over $F$ are normal, by the Fundamental Theorem of Galois Theory), and

the latter isomorphism because $N_{i-1}/F$ Galois and $\Gamma_i \triangleleft \Gamma_{i-1}$ correspond under the correspondence to $N_i/N_{i-1}$ Galois with that quotient group (this is the standard "tower" consequence of the Fundamental Theorem: if $N_{i-1} \subseteq N_i$ are both Galois over $F$, then $\mathrm{Gal}(N_i/N_{i-1}) \cong \Gamma_{i-1}/\Gamma_i$). So we obtain a chain

where $\Gamma_0/\Gamma_1 \cong \mathrm{Gal}(N_1/N_0)$ — wait, we index so that $\Gamma_{i-1}/\Gamma_i \cong \mathrm{Gal}(N_i/N_{i-1})$, cyclic for $i \ge 1$, and we still must handle $N_0/F$: but $N_0 = F(\zeta_n)$ is itself Galois over $F$ with abelian (not necessarily cyclic) group; refine the trivial subgroup chain of this abelian group's corresponding piece to cyclic quotients as before. Concretely, $\Gamma/\Gamma_0 \cong \mathrm{Gal}(N_0/F)$ is abelian, hence solvable with a further refinement into cyclic quotients; splice this refinement in front of the chain $\Gamma_0 \triangleright \cdots \triangleright \Gamma_m = \{e\}$. This exhibits $\Gamma$ as solvable (a chain of normal subgroups with each successive quotient cyclic, in particular abelian). Finally, $\mathrm{Gal}(E/F)$ is a quotient of $\Gamma = \mathrm{Gal}(N_m/F)$: since $E/F$ is Galois and $E \subseteq N_m$, restriction $\Gamma \to \mathrm{Gal}(E/F)$, $\tau \mapsto \tau|_E$, is a well-defined surjective homomorphism (well-defined because $E/F$ Galois means $E$ is stable under all automorphisms of $N_m/F$ fixing $F$ — same argument as before showing $\tau(E)=E$ — and surjective by the Fundamental Theorem of Galois Theory, since every automorphism of $E/F$ extends to $N_m$ using that $N_m/F$ is a normal, in fact Galois, extension containing $E$). Quotients of solvable groups are solvable (images of a normal chain with abelian quotients remain a normal chain with abelian quotients, by the third isomorphism theorem applied termwise). Hence $\mathrm{Gal}(E/F)$ is solvable. $\blacksquare$