[proofplan] The proof constructs a radical tower over $K$ that contains the splitting field $L$. Let $G = \operatorname{Gal}(L/K)$ and let $\{1\} = G_r \trianglelefteq \cdots \trianglelefteq G_0 = G$ be a solubility series with cyclic quotients of prime orders. The Fundamental Theorem of Galois Theory converts this into a tower of intermediate fields $K = F_0 \subset F_1 \subset \cdots \subset F_r = L$ where each $\operatorname{Gal}(F_{i+1}/F_i)$ is cyclic. Setting $N := |G|$, we adjoin a primitive $N$-th root of unity $\zeta_N$ to each level. The key observations are: (1) the cyclotomic step $K \to K(\zeta_N)$ is a radical extension since $\zeta_N$ is a root of $t^N - 1$; (2) over $F_i(\zeta_N)$, each extension to $F_{i+1}(\zeta_N)$ remains cyclic of degree dividing $n_{i+1}$ and the base field contains all necessary roots of unity, so the Converse of Kummer Theory provides a generator that is an $n_{i+1}$-th root of an element in $F_i(\zeta_N)$. The concatenation of these steps is a radical tower from $K$ to $F_r(\zeta_N) = L(\zeta_N) \supset L$. [/proofplan] [step:Extract a tower of intermediate fields from the solubility series via the Fundamental Theorem] Set $G := \operatorname{Gal}(L/K)$. Since $G$ is soluble, there exists a chain of subgroups \begin{align*} \{1\} = G_r \trianglelefteq G_{r-1} \trianglelefteq \cdots \trianglelefteq G_1 \trianglelefteq G_0 = G \end{align*} such that each quotient $G_i / G_{i+1}$ is cyclic of order $n_{i+1} := [G_i : G_{i+1}]$ for $i = 0, 1, \ldots, r-1$. Since $L/K$ is a splitting field of a polynomial over $K$ with $\operatorname{Char}(K) = 0$, the extension $L/K$ is finite, separable, and normal, hence Galois. Each $G_i$ is a subgroup of $G$, and each $G_{i+1}$ is normal in $G_i$ (though not necessarily normal in $G$). By the [Fundamental Theorem of Galois Theory](/theorems/1274), the subgroup $G_i$ corresponds to the intermediate field $F_i := L^{G_i}$, and we obtain a tower \begin{align*} K = L^G = F_0 \subset F_1 \subset \cdots \subset F_r = L^{\{1\}} = L. \end{align*} For each $i = 0, 1, \ldots, r-1$, since $G_{i+1} \trianglelefteq G_i$, the normality criterion of the [Fundamental Theorem of Galois Theory](/theorems/1274) applied to the Galois extension $L/F_i$ with subgroup $G_{i+1}$ of $\operatorname{Gal}(L/F_i) = G_i$ gives that $F_{i+1}/F_i$ is a Galois extension with \begin{align*} \operatorname{Gal}(F_{i+1}/F_i) \cong G_i / G_{i+1}, \end{align*} which is cyclic of order $n_{i+1}$. [guided] The solubility of $G$ provides algebraic structure at the group level; the Fundamental Theorem translates this structure into field-theoretic data. By definition, $G = \operatorname{Gal}(L/K)$ is soluble if there exists a chain of subgroups \begin{align*} \{1\} = G_r \trianglelefteq G_{r-1} \trianglelefteq \cdots \trianglelefteq G_1 \trianglelefteq G_0 = G \end{align*} with each successive quotient $G_i/G_{i+1}$ abelian. Since each $G_i/G_{i+1}$ is a finite abelian group, it is a product of cyclic groups. By refining the chain (inserting intermediate subgroups corresponding to the cyclic factors of each $G_i/G_{i+1}$), we may assume without loss of generality that each quotient $G_i/G_{i+1}$ is cyclic of some order $n_{i+1} \geq 1$. (We can even arrange for each $n_{i+1}$ to be prime, though this is not necessary for the argument.) Now we invoke the [Fundamental Theorem of Galois Theory](/theorems/1274). We verify its hypotheses: $L/K$ is the splitting field of $f \in K[t]$ over $K$, and since $\operatorname{Char}(K) = 0$, every irreducible factor of $f$ is separable. Therefore $L/K$ is a finite, separable, normal extension, i.e., a finite Galois extension with $\operatorname{Gal}(L/K) = G$. The Fundamental Theorem establishes a bijection between subgroups of $G$ and intermediate fields $K \subset F \subset L$, given by $H \mapsto L^H$. Under this bijection, define $F_i := L^{G_i}$ for each $i$. Then: - $F_0 = L^{G_0} = L^G = K$ (since $L/K$ is Galois). - $F_r = L^{G_r} = L^{\{1\}} = L$ (since the identity fixes everything). - The index-degree formula gives $[F_i : K] = [G : G_i]$, so the tower is strictly increasing: $K = F_0 \subsetneq F_1 \subsetneq \cdots \subsetneq F_r = L$. For the Galois group of each step: $G_{i+1}$ is normal in $G_i$, and $G_i = \operatorname{Gal}(L/F_i)$. The normality criterion from the Fundamental Theorem (applied to the Galois extension $L/F_i$ with the pair of subgroups $G_{i+1} \trianglelefteq G_i$) states that $F_{i+1}/F_i$ is a normal extension, and since $L/F_i$ is separable (as $\operatorname{Char}(K) = 0$), $F_{i+1}/F_i$ is in fact Galois with \begin{align*} \operatorname{Gal}(F_{i+1}/F_i) \cong G_i / G_{i+1}. \end{align*} By construction, $G_i / G_{i+1}$ is cyclic of order $n_{i+1} = [G_i : G_{i+1}] = [F_{i+1} : F_i]$. The situation at this point: we have a tower $K = F_0 \subset F_1 \subset \cdots \subset F_r = L$ in which each step $F_{i+1}/F_i$ is a cyclic Galois extension. To make this into a radical tower, we need each step to be generated by an $n_{i+1}$-th root. The Converse of Kummer Theory provides exactly this conclusion -- but it requires the base field to contain a primitive $n_{i+1}$-th root of unity. The fields $F_i$ may not contain such roots of unity, so the next step adjoins them. [/guided] [/step] [step:Adjoin a primitive $N$-th root of unity and verify the cyclotomic step is radical] Define $N := |G| = [L : K] = n_1 \cdot n_2 \cdots n_r$. Since $\operatorname{Char}(K) = 0$, the polynomial $t^N - 1 \in K[t]$ is separable (its formal derivative $Nt^{N-1}$ is nonzero since $N \neq 0$ in $K$). Let $\zeta_N$ be a primitive $N$-th root of unity in an algebraic closure of $K$. The extension $K(\zeta_N)/K$ is a radical extension: $\zeta_N$ is a root of $t^N - 1 \in K[t]$, so $\zeta_N^N = 1 \in K$. [guided] We need a single root of unity that works for every level of the tower simultaneously. The order of each cyclic quotient $G_i/G_{i+1}$ is $n_{i+1}$, and the Converse of Kummer Theory for the $i$-th step will require a primitive $n_{i+1}$-th root of unity. Since $n_{i+1}$ divides $N = n_1 \cdots n_r$ (which equals $|G|$), a primitive $N$-th root of unity $\zeta_N$ contains all necessary roots of unity: indeed, $\zeta_N^{N/n_{i+1}}$ is a primitive $n_{i+1}$-th root of unity for each $i$. We verify that $\zeta_N$ exists and is separable over $K$. The polynomial $t^N - 1$ has formal derivative $Nt^{N-1}$. Since $\operatorname{Char}(K) = 0$, the integer $N$ is nonzero in $K$, so $\gcd(t^N - 1, Nt^{N-1}) = 1$. Therefore $t^N - 1$ has $N$ distinct roots in an algebraic closure, and a primitive $N$-th root of unity $\zeta_N$ exists among them. The extension $K(\zeta_N)/K$ is radical: $\zeta_N$ satisfies $\zeta_N^N = 1 \in K$, so $\zeta_N$ is an $N$-th root of the element $1 \in K$. This is the first step of the radical tower we are constructing. Why did we choose $N = |G|$ rather than $N = \operatorname{lcm}(n_1, \ldots, n_r)$? Either choice works, but $|G|$ is more directly available and ensures divisibility by each $n_{i+1}$. The specific value of $N$ does not affect the argument. [/guided] [/step] [step:Show that each augmented extension $F_{i+1}(\zeta_N)/F_i(\zeta_N)$ is cyclic of degree dividing $n_{i+1}$] For each $i = 0, 1, \ldots, r-1$, consider the tower $F_i \subset F_i(\zeta_N) \subset F_{i+1}(\zeta_N)$ and $F_i \subset F_{i+1} \subset F_{i+1}(\zeta_N)$. Since $F_{i+1}/F_i$ is Galois (from the first step), $F_{i+1}(\zeta_N)/F_i(\zeta_N)$ is also Galois: indeed, $F_{i+1}$ is the splitting field of some separable polynomial $g \in F_i[t]$, and $g$ remains a polynomial over $F_i(\zeta_N) \supset F_i$, so $F_{i+1}(\zeta_N)$ contains the splitting field of $g$ over $F_i(\zeta_N)$, and the generators of $F_{i+1}(\zeta_N)$ over $F_i(\zeta_N)$ are roots of separable polynomials over $F_i(\zeta_N)$. The restriction map \begin{align*} \operatorname{Gal}(F_{i+1}(\zeta_N)/F_i(\zeta_N)) &\to \operatorname{Gal}(F_{i+1}/F_i) \\ \sigma &\mapsto \sigma|_{F_{i+1}} \end{align*} is a well-defined injective group homomorphism. It is well-defined because each $\sigma \in \operatorname{Gal}(F_{i+1}(\zeta_N)/F_i(\zeta_N))$ fixes $F_i(\zeta_N) \supset F_i$ and maps $F_{i+1}$ to itself (since $F_{i+1}/F_i$ is normal). It is injective because an element of $\operatorname{Gal}(F_{i+1}(\zeta_N)/F_i(\zeta_N))$ is determined by its action on $F_{i+1}$ and on $\zeta_N$, and $\zeta_N \in F_i(\zeta_N)$ is fixed by hypothesis. Therefore $\operatorname{Gal}(F_{i+1}(\zeta_N)/F_i(\zeta_N))$ is isomorphic to a subgroup of $\operatorname{Gal}(F_{i+1}/F_i) \cong \mathbb{Z}/n_{i+1}\mathbb{Z}$. Since every subgroup of a cyclic group is cyclic, $\operatorname{Gal}(F_{i+1}(\zeta_N)/F_i(\zeta_N))$ is cyclic of order dividing $n_{i+1}$. [guided] Adjoining $\zeta_N$ to the tower could, in principle, destroy the cyclic structure of each step. This step verifies that it does not: the Galois group can only shrink (or stay the same), and it remains cyclic. We consider the composite field $F_{i+1}(\zeta_N) = F_{i+1} \cdot F_i(\zeta_N)$, the field generated by $F_{i+1}$ and $F_i(\zeta_N)$ inside a common algebraic closure. **Galois property.** We must verify that $F_{i+1}(\zeta_N)/F_i(\zeta_N)$ is Galois. Since $F_{i+1}/F_i$ is Galois (established in the first step), $F_{i+1}$ is the splitting field of some separable polynomial $g \in F_i[t]$ over $F_i$. Viewing $g$ as an element of $F_i(\zeta_N)[t]$ (since $F_i \subset F_i(\zeta_N)$), the field $F_{i+1}(\zeta_N)$ contains all roots of $g$ (they lie in $F_{i+1} \subset F_{i+1}(\zeta_N)$) and is generated over $F_i(\zeta_N)$ by these roots together with $\zeta_N$ (which is already in $F_i(\zeta_N)$). Therefore $F_{i+1}(\zeta_N)$ is the splitting field of $g$ over $F_i(\zeta_N)$, and since $g$ is separable, the extension $F_{i+1}(\zeta_N)/F_i(\zeta_N)$ is Galois. **The restriction map.** Define \begin{align*} \rho_i: \operatorname{Gal}(F_{i+1}(\zeta_N)/F_i(\zeta_N)) &\to \operatorname{Gal}(F_{i+1}/F_i) \\ \sigma &\mapsto \sigma|_{F_{i+1}}. \end{align*} We verify that $\rho_i$ is well-defined. Take $\sigma \in \operatorname{Gal}(F_{i+1}(\zeta_N)/F_i(\zeta_N))$. Since $\sigma$ fixes $F_i(\zeta_N) \supset F_i$ pointwise, $\sigma$ is an $F_i$-automorphism. Since $F_{i+1}/F_i$ is normal, $\sigma$ maps $F_{i+1}$ into itself: for any $\alpha \in F_{i+1}$, the minimal polynomial of $\alpha$ over $F_i$ splits in $F_{i+1}$ (by normality), and $\sigma(\alpha)$ is a root of the same polynomial, so $\sigma(\alpha) \in F_{i+1}$. Hence $\sigma|_{F_{i+1}}$ is an $F_i$-automorphism of $F_{i+1}$, i.e., an element of $\operatorname{Gal}(F_{i+1}/F_i)$. We verify that $\rho_i$ is injective. Suppose $\sigma|_{F_{i+1}} = \operatorname{id}_{F_{i+1}}$. Then $\sigma$ fixes $F_{i+1}$ pointwise. It also fixes $\zeta_N$ since $\zeta_N \in F_i(\zeta_N)$ and $\sigma$ fixes $F_i(\zeta_N)$ by definition. Since $F_{i+1}(\zeta_N)$ is generated by $F_{i+1}$ and $\zeta_N$, fixing both forces $\sigma = \operatorname{id}$. Hence $\ker(\rho_i) = \{1\}$. **Conclusion.** The injective homomorphism $\rho_i$ identifies $\operatorname{Gal}(F_{i+1}(\zeta_N)/F_i(\zeta_N))$ with a subgroup of $\operatorname{Gal}(F_{i+1}/F_i) \cong \mathbb{Z}/n_{i+1}\mathbb{Z}$. Every subgroup of a cyclic group is cyclic (a standard fact: if $\mathbb{Z}/m\mathbb{Z}$ has a subgroup of order $d \mid m$, it is generated by the element $m/d$ and is isomorphic to $\mathbb{Z}/d\mathbb{Z}$). Therefore $\operatorname{Gal}(F_{i+1}(\zeta_N)/F_i(\zeta_N))$ is cyclic of some order $d_{i+1}$ dividing $n_{i+1}$. Why might $d_{i+1}$ be strictly smaller than $n_{i+1}$? Adjoining $\zeta_N$ to $F_i$ may partially extend $F_{i+1}$: some of the generators of $F_{i+1}$ over $F_i$ might already lie in $F_i(\zeta_N)$. For instance, if $F_{i+1} = F_i(\sqrt{2})$ and $\zeta_N$ happens to generate an extension containing $\sqrt{2}$, then $F_{i+1}(\zeta_N) = F_i(\zeta_N)$ and the extension has degree $1$. This does not affect the argument -- a cyclic extension of degree $1$ is a radical extension (trivially). [/guided] [/step] [step:Apply the Converse of Kummer Theory to make each augmented step radical] Let $d_{i+1} := [F_{i+1}(\zeta_N) : F_i(\zeta_N)]$ for $i = 0, 1, \ldots, r-1$. By the previous step, $\operatorname{Gal}(F_{i+1}(\zeta_N)/F_i(\zeta_N))$ is cyclic of order $d_{i+1}$, where $d_{i+1}$ divides $n_{i+1}$, which in turn divides $N$. Since $\zeta_N \in F_i(\zeta_N)$ and $d_{i+1}$ divides $N$, the element $\zeta_N^{N/d_{i+1}}$ is a primitive $d_{i+1}$-th root of unity lying in $F_i(\zeta_N)$. The hypotheses of the [Converse of Kummer Theory](/theorems/1283) are therefore satisfied: - $F_i(\zeta_N)$ is a field containing a primitive $d_{i+1}$-th root of unity. - $F_{i+1}(\zeta_N)/F_i(\zeta_N)$ is a cyclic Galois extension of degree $d_{i+1}$. By the [Converse of Kummer Theory](/theorems/1283), there exists $\beta_i \in F_{i+1}(\zeta_N)$ with $\beta_i^{d_{i+1}} \in F_i(\zeta_N)$ and $F_{i+1}(\zeta_N) = F_i(\zeta_N)(\beta_i)$. Therefore $F_{i+1}(\zeta_N)$ is obtained from $F_i(\zeta_N)$ by adjoining a $d_{i+1}$-th root of an element of $F_i(\zeta_N)$: each step in the augmented tower is a radical extension. [guided] This is the heart of the proof: the Converse of Kummer Theory converts each cyclic step into a radical step, provided the base field has the necessary roots of unity. Fix $i \in \{0, 1, \ldots, r-1\}$. From the previous step, the extension $F_{i+1}(\zeta_N)/F_i(\zeta_N)$ is a cyclic Galois extension of degree $d_{i+1}$, where $d_{i+1}$ divides $n_{i+1}$ and hence divides $N = |G|$. The [Converse of Kummer Theory](/theorems/1283) states: if $K'$ is a field containing a primitive $m$-th root of unity and $L'/K'$ is a cyclic Galois extension of degree $m$, then $L' = K'(\beta)$ for some $\beta \in L'$ with $\beta^m \in K'$. We verify the hypotheses with $K' = F_i(\zeta_N)$, $L' = F_{i+1}(\zeta_N)$, and $m = d_{i+1}$: 1. **$F_i(\zeta_N)$ contains a primitive $d_{i+1}$-th root of unity.** The element $\zeta_N$ is a primitive $N$-th root of unity, meaning $\zeta_N^N = 1$ and $\zeta_N^k \neq 1$ for $0 < k < N$. Since $d_{i+1}$ divides $N$, the element $\omega_i := \zeta_N^{N/d_{i+1}}$ satisfies $\omega_i^{d_{i+1}} = \zeta_N^N = 1$. Moreover, if $\omega_i^j = 1$ for some $0 < j < d_{i+1}$, then $\zeta_N^{jN/d_{i+1}} = 1$, which requires $N \mid jN/d_{i+1}$, i.e., $d_{i+1} \mid j$. Since $0 < j < d_{i+1}$, this is impossible. Hence $\omega_i$ is a primitive $d_{i+1}$-th root of unity lying in $F_i(\zeta_N)$. 2. **$F_{i+1}(\zeta_N)/F_i(\zeta_N)$ is a cyclic Galois extension of degree $d_{i+1}$.** This was established in the previous step. By the [Converse of Kummer Theory](/theorems/1283), there exists $\beta_i \in F_{i+1}(\zeta_N)$ such that $\beta_i^{d_{i+1}} \in F_i(\zeta_N)$ and $F_{i+1}(\zeta_N) = F_i(\zeta_N)(\beta_i)$. In concrete terms: $\beta_i$ is a $d_{i+1}$-th root of some element $\lambda_i \in F_i(\zeta_N)$, so the extension $F_i(\zeta_N) \to F_{i+1}(\zeta_N)$ is obtained by adjoining a radical. This is precisely what it means for each step of the augmented tower to be a radical extension. [/guided] [/step] [step:Assemble the radical tower $K \to K(\zeta_N) \to F_1(\zeta_N) \to \cdots \to L(\zeta_N) \supset L$] We now concatenate the extensions constructed above into a single radical tower from $K$ to a field containing $L$. **The cyclotomic step.** The extension $K \to K(\zeta_N)$ is radical: $\zeta_N^N = 1 \in K$. **The Kummer steps.** For each $i = 0, 1, \ldots, r-1$, the extension $F_i(\zeta_N) \to F_{i+1}(\zeta_N)$ is radical: $F_{i+1}(\zeta_N) = F_i(\zeta_N)(\beta_i)$ with $\beta_i^{d_{i+1}} \in F_i(\zeta_N)$. The full tower is: \begin{align*} K \subset K(\zeta_N) = F_0(\zeta_N) \subset F_1(\zeta_N) \subset F_2(\zeta_N) \subset \cdots \subset F_r(\zeta_N) = L(\zeta_N). \end{align*} Each successive extension is obtained by adjoining a single radical: the first step adjoins $\zeta_N$ (an $N$-th root of $1$), and the $(i+1)$-th step adjoins $\beta_i$ (a $d_{i+1}$-th root of an element in the preceding field). This is a radical tower from $K$ to $L(\zeta_N)$. Since $L \subset L(\zeta_N)$, every root of $f$ lies in $L \subset L(\zeta_N)$, and $L(\zeta_N)$ sits atop a radical tower starting at $K$. By definition, $f$ is soluble by radicals. [guided] We assemble the pieces. The radical tower has $r + 1$ steps: **Step 0 (cyclotomic).** $K \to K(\zeta_N)$: adjoining $\zeta_N$, which satisfies $\zeta_N^N = 1 \in K$. This is a radical extension with exponent $N$. **Steps 1 through $r$ (Kummer).** For $i = 0, 1, \ldots, r-1$: $F_i(\zeta_N) \to F_{i+1}(\zeta_N) = F_i(\zeta_N)(\beta_i)$, where $\beta_i^{d_{i+1}} = \lambda_i \in F_i(\zeta_N)$. Each is a radical extension with exponent $d_{i+1}$. We verify that the initial identification $K(\zeta_N) = F_0(\zeta_N)$ is correct: $F_0 = K$ by definition, so $F_0(\zeta_N) = K(\zeta_N)$. We also verify the terminal identification: $F_r = L$ by definition, so $F_r(\zeta_N) = L(\zeta_N)$. The full tower is therefore: \begin{align*} K \subset K(\zeta_N) \subset F_1(\zeta_N) \subset F_2(\zeta_N) \subset \cdots \subset L(\zeta_N). \end{align*} This is a radical tower: each step is obtained by adjoining a single element whose appropriate power lies in the preceding field. Since $L$ is the splitting field of $f$ over $K$, all roots of $f$ lie in $L$. Since $L \subset L(\zeta_N)$ and $L(\zeta_N)$ sits at the top of a radical tower over $K$, every root of $f$ lies in a field obtained from $K$ by a sequence of radical extensions. By definition, $f$ is soluble by radicals. Note that the radical tower terminates at $L(\zeta_N)$, which is potentially larger than $L$ itself. The definition of "soluble by radicals" requires only that the roots of $f$ lie in some field at the top of a radical tower over $K$ -- the tower is permitted to overshoot $L$. The extra elements (namely $\zeta_N$ and its powers) are the price paid for ensuring that Kummer theory applies at every step. [/guided] [/step]