[proofplan] The proof has three stages. First, we invoke the [Radical Extensions Have Galois Radical Towers](/theorems/1284) theorem to obtain a Galois extension $E/K$ that is simultaneously radical and contains $L$. Second, we use the radical tower $K = E_0 \subset E_1 \subset \cdots \subset E_r = E$ to construct a chain of subgroups $G_i = \operatorname{Gal}(E/E_i)$ in $\operatorname{Gal}(E/K)$, and apply the [Fundamental Theorem of Galois Theory](/theorems/1274) (Part 3) to identify each successive quotient $G_i / G_{i+1}$ with $\operatorname{Gal}(E_{i+1}/E_i)$. Since each step of the tower is either cyclotomic or a pure radical adjunction over a field already containing the requisite roots of unity, the [Galois Group of Cyclotomic Extensions](/theorems/1278) and the [Galois Group of Splitting Field of $t^n - \lambda$](/theorems/1281) show that each quotient is abelian, making $\operatorname{Gal}(E/K)$ soluble. Third, the [Fundamental Theorem of Galois Theory](/theorems/1274) identifies $\operatorname{Gal}(L/K)$ as a quotient of $\operatorname{Gal}(E/K)$, and [Persistence of Solubility](/theorems/1285) (Part 2) shows that quotients of soluble groups are soluble. [/proofplan]

[step:Lift the radical extension to a Galois radical extension $E/K$ containing $L$]Since $f$ is soluble by radicals, there exists a radical extension $M/K$ such that $L \subset M$, where $L$ is the splitting field of $f$ over $K$. By the [Radical Extensions Have Galois Radical Towers](/theorems/1284) theorem, applied to the radical extension $M/K$ over the characteristic-$0$ field $K$, there exists an extension $E/M$ such that $E/K$ is both Galois and radical. Since $M/K$ is radical and $E/M$ extends $M$, we have the containment $L \subset M \subset E$, so $L$ is an intermediate field of the Galois extension $E/K$. The radical tower for $E/K$ takes the form \begin{align*} K = E_0 \subset E_1 \subset E_2 \subset \cdots \subset E_r = E, \end{align*} where each step $E_{i+1}/E_i$ is either a cyclotomic extension (adjunction of a root of unity) or a pure radical extension of the form $E_{i+1} = E_i(\alpha_i)$ with $\alpha_i^{n_i} \in E_i$ for some integer $n_i \geq 1$, and the relevant primitive $n_i$-th root of unity already lies in $E_i$.[/step]

[guided]The hypothesis "$f$ is soluble by radicals" means, by definition, that the splitting field $L$ of $f$ over $K$ is contained in some radical extension $M/K$. A radical extension is a field obtained from $K$ by a finite sequence of adjunctions, each adjoining either a root of unity or an $n$-th root of an element already present. The problem is that $M/K$ need not be Galois, and without the Galois property we cannot apply the Galois correspondence to read off the group structure. The [Radical Extensions Have Galois Radical Towers](/theorems/1284) theorem resolves this: it states that for any radical extension $M/K$ over a characteristic-$0$ field, there exists an extension $E \supset M$ such that $E/K$ is simultaneously Galois and radical. We verify the hypotheses of [Theorem 1284](/theorems/1284): - $K$ has characteristic $0$ (given in the theorem statement). - $M/K$ is a radical extension (by the solubility-by-radicals hypothesis). The theorem produces $E \supset M \supset L$ with $E/K$ Galois and $E/K$ radical. The radical tower for $E/K$ can be refined so that each step adjoins either a primitive root of unity (a cyclotomic step) or a pure $n$-th root over a field already containing the relevant roots of unity (a Kummer step). This refinement is possible because in characteristic $0$, one can always insert a cyclotomic step before a Kummer step: to adjoin $\alpha$ with $\alpha^n \in E_i$, first extend $E_i$ to $E_i(\zeta_n)$ (where $\zeta_n$ is a primitive $n$-th root of unity), then adjoin $\alpha$ over $E_i(\zeta_n)$.[/guided]

[step:Construct the subgroup chain $\{1\} = G_r \trianglelefteq \cdots \trianglelefteq G_0 = \operatorname{Gal}(E/K)$ via the Galois correspondence]Set $G = \operatorname{Gal}(E/K)$ and define, for each $0 \leq i \leq r$, \begin{align*} G_i := \operatorname{Gal}(E/E_i) = \{\sigma \in G : \sigma(a) = a \text{ for all } a \in E_i\}. \end{align*} Since $E_0 = K$ and $E_r = E$, the extreme cases give $G_0 = \operatorname{Gal}(E/K) = G$ and $G_r = \operatorname{Gal}(E/E) = \{1\}$. The inclusions $E_0 \subset E_1 \subset \cdots \subset E_r$ reverse under the Galois correspondence to give \begin{align*} \{1\} = G_r \subset G_{r-1} \subset \cdots \subset G_0 = G. \end{align*} We claim that each $G_{i+1}$ is a normal subgroup of $G_i$. For each $i$, the extension $E_{i+1}/E_i$ is either cyclotomic or a pure radical adjunction $E_{i+1} = E_i(\alpha_i)$ with $\alpha_i^{n_i} \in E_i$ and $\zeta_{n_i} \in E_i$. In both cases, $E_{i+1}$ is the splitting field of a separable polynomial over $E_i$: - **Cyclotomic step:** $E_{i+1} = E_i(\zeta_m)$ is the splitting field of $t^m - 1$ over $E_i$. In characteristic $0$, this polynomial is separable. - **Kummer step:** $E_{i+1} = E_i(\alpha_i)$ where $\alpha_i^{n_i} = \lambda_i \in E_i$ and $\zeta_{n_i} \in E_i$. Then $E_{i+1}$ is the splitting field of $t^{n_i} - \lambda_i$ over $E_i$, since all roots are $\zeta_{n_i}^k \alpha_i$ for $k = 0, 1, \ldots, n_i - 1$ and $\zeta_{n_i} \in E_i \subset E_{i+1}$. In characteristic $0$, $t^{n_i} - \lambda_i$ is separable (its derivative $n_i t^{n_i - 1}$ shares no root with it since $n_i \neq 0$ in $K$). By the [Splitting Fields of Separable Polynomials Are Galois](/theorems/1271) theorem, $E_{i+1}/E_i$ is a Galois extension. Part 2 of the [Fundamental Theorem of Galois Theory](/theorems/1274), applied to the Galois extension $E/E_i$ with intermediate field $E_{i+1}$, states that $G_{i+1} = \operatorname{Gal}(E/E_{i+1})$ is normal in $\operatorname{Gal}(E/E_i) = G_i$ if and only if $E_{i+1}/E_i$ is a normal extension. Since $E_{i+1}/E_i$ is Galois (hence normal), we conclude $G_{i+1} \trianglelefteq G_i$.[/step]

[guided]The Galois correspondence translates the radical tower of fields into a descending chain of subgroups. The map $\Psi$ from Part 1 of the [Fundamental Theorem of Galois Theory](/theorems/1274) sends each intermediate field $E_i$ to the subgroup $G_i = \operatorname{Gal}(E/E_i)$, and the correspondence reverses inclusions: $E_i \subset E_{i+1}$ becomes $G_i \supset G_{i+1}$. The critical question is: why is each $G_{i+1}$ normal in $G_i$? The normality criterion in Part 2 of the [Fundamental Theorem](/theorems/1274) states that, within a Galois extension $E/E_i$, the subgroup $\operatorname{Gal}(E/E_{i+1})$ is normal in $\operatorname{Gal}(E/E_i)$ if and only if $E_{i+1}/E_i$ is a normal extension. We verify the hypotheses of the Fundamental Theorem for the extension $E/E_i$. Since $E/K$ is Galois and $K \subset E_i \subset E$: - $E/E_i$ is finite (since $[E : E_i] \leq [E : K] < \infty$). - $E/E_i$ is separable (separability is inherited from $E/K$ since $K \subset E_i$). - $E/E_i$ is normal ($E$ is a splitting field of some collection of polynomials over $K$, hence also over $E_i \supset K$). Therefore $E/E_i$ is a finite Galois extension and the Fundamental Theorem applies. The normality criterion reduces our task to showing that $E_{i+1}/E_i$ is a normal extension. In a cyclotomic step, $E_{i+1} = E_i(\zeta_m)$ is the splitting field of $t^m - 1$ over $E_i$. The polynomial $t^m - 1$ has formal derivative $mt^{m-1}$, and since $\operatorname{Char} K = 0$, the constant $m$ is nonzero in $K$, so $\gcd(t^m - 1, mt^{m-1}) = 1$. Therefore $t^m - 1$ is separable, and by [Splitting Fields of Separable Polynomials Are Galois](/theorems/1271), $E_{i+1}/E_i$ is Galois. In a Kummer step, $E_{i+1} = E_i(\alpha_i)$ where $\alpha_i^{n_i} = \lambda_i \in E_i$ and $\zeta_{n_i} \in E_i$. The roots of $t^{n_i} - \lambda_i$ are $\zeta_{n_i}^k \alpha_i$ for $k = 0, 1, \ldots, n_i - 1$. Since $\zeta_{n_i} \in E_i \subset E_i(\alpha_i)$, all roots lie in $E_{i+1}$, so $E_{i+1}$ is the splitting field of $t^{n_i} - \lambda_i$ over $E_i$. The polynomial is separable in characteristic $0$ (its derivative $n_i t^{n_i - 1}$ has $0$ as its only root, but $\lambda_i \neq 0$ implies $0$ is not a root of $t^{n_i} - \lambda_i$). By [Splitting Fields of Separable Polynomials Are Galois](/theorems/1271), $E_{i+1}/E_i$ is Galois. In both cases $E_{i+1}/E_i$ is normal, so the Fundamental Theorem gives $G_{i+1} \trianglelefteq G_i$. Note that the normality is only relative: $G_{i+1} \trianglelefteq G_i$, not $G_{i+1} \trianglelefteq G$. This is sufficient for a solubility series, which requires each term to be normal in the preceding one, not in the entire group.[/guided]

[step:Show each quotient $G_i / G_{i+1}$ is abelian]By Part 3 of the [Fundamental Theorem of Galois Theory](/theorems/1274), applied to the Galois extension $E/E_i$ with the normal subgroup $G_{i+1} \trianglelefteq G_i = \operatorname{Gal}(E/E_i)$, the restriction map induces an isomorphism \begin{align*} G_i / G_{i+1} \xrightarrow{\;\sim\;} \operatorname{Gal}(E_{i+1}/E_i). \end{align*} (Here we use the notation from Part 3: $G_{i+1} = \operatorname{Gal}(E/E_{i+1})$ is normal in $G_i = \operatorname{Gal}(E/E_i)$, and the fixed field $E^{G_{i+1}} = E_{i+1}$, so the quotient $G_i/G_{i+1}$ is isomorphic to $\operatorname{Gal}(E^{G_{i+1}}/E_i) = \operatorname{Gal}(E_{i+1}/E_i)$.) We now show $\operatorname{Gal}(E_{i+1}/E_i)$ is abelian in each case: - **Cyclotomic step:** $E_{i+1} = E_i(\zeta_m)$. By the [Galois Group of Cyclotomic Extensions](/theorems/1278), with $K$ replaced by $E_i$ and $n$ replaced by $m$, the map $\phi \mapsto \bar{j}$ (where $\phi(\zeta_m) = \zeta_m^j$) defines an injective group homomorphism $\operatorname{Gal}(E_{i+1}/E_i) \hookrightarrow (\mathbb{Z}/m\mathbb{Z})^\times$. The hypothesis $\operatorname{Char} E_i \nmid m$ is satisfied since $\operatorname{Char} K = 0$ implies $\operatorname{Char} E_i = 0$, and $0$ does not divide any positive integer. Since $(\mathbb{Z}/m\mathbb{Z})^\times$ is abelian, any subgroup of it is abelian, so $\operatorname{Gal}(E_{i+1}/E_i)$ is abelian. - **Kummer step:** $E_{i+1} = E_i(\alpha_i)$ with $\alpha_i^{n_i} = \lambda_i \in E_i$ and $\zeta_{n_i} \in E_i$. By the [Galois Group of Splitting Field of $t^n - \lambda$](/theorems/1281), with $K$ replaced by $E_i$, $n$ replaced by $n_i$, $\lambda$ replaced by $\lambda_i$, and $\alpha$ replaced by $\alpha_i$, the map $\sigma \mapsto \bar{k}$ (where $\sigma(\alpha_i) = \zeta_{n_i}^k \alpha_i$) defines an injective group homomorphism $\operatorname{Gal}(E_{i+1}/E_i) \hookrightarrow \mathbb{Z}/n_i\mathbb{Z}$. The hypothesis that $E_i$ contains a primitive $n_i$-th root of unity is satisfied by the construction of the tower. Since $\mathbb{Z}/n_i\mathbb{Z}$ is cyclic (and in particular abelian), any subgroup of it is cyclic, so $\operatorname{Gal}(E_{i+1}/E_i)$ is cyclic and hence abelian. In both cases, $G_i / G_{i+1} \cong \operatorname{Gal}(E_{i+1}/E_i)$ is abelian.[/step]

[guided]Part 3 of the [Fundamental Theorem of Galois Theory](/theorems/1274) provides the bridge between the subgroup chain and the Galois groups of the individual steps. We verify its hypotheses for each $i$: - $E/E_i$ is a finite Galois extension (established in the previous step). - $G_{i+1} \trianglelefteq G_i$ (established in the previous step). - The fixed field $E^{G_{i+1}} = E_{i+1}$ (by Part 1 of the Fundamental Theorem, since $G_{i+1} = \operatorname{Gal}(E/E_{i+1})$ and $\Phi \circ \Psi = \operatorname{id}$). Part 3 then gives the isomorphism $G_i / G_{i+1} \cong \operatorname{Gal}(E_{i+1}/E_i)$. Why does the Galois group of a cyclotomic step embed into $(\mathbb{Z}/m\mathbb{Z})^\times$? [Theorem 1278](/theorems/1278) constructs the embedding as follows: any $\phi \in \operatorname{Gal}(E_i(\zeta_m)/E_i)$ is determined by $\phi(\zeta_m)$, which must be another primitive $m$-th root of unity (since $\phi$ permutes the roots of $t^m - 1$ and preserves the order of $\zeta_m$). Writing $\phi(\zeta_m) = \zeta_m^j$ with $\gcd(j, m) = 1$, the map $\phi \mapsto \bar{j} \in (\mathbb{Z}/m\mathbb{Z})^\times$ is an injective homomorphism. The group $(\mathbb{Z}/m\mathbb{Z})^\times$ is abelian (it is a subgroup of the commutative ring $\mathbb{Z}/m\mathbb{Z}$ under multiplication), so $\operatorname{Gal}(E_{i+1}/E_i)$ is abelian. Why does the Galois group of a Kummer step embed into $\mathbb{Z}/n_i\mathbb{Z}$? [Theorem 1281](/theorems/1281) constructs the embedding: any $\sigma \in \operatorname{Gal}(E_i(\alpha_i)/E_i)$ must send $\alpha_i$ to another root of $t^{n_i} - \lambda_i$, i.e., $\sigma(\alpha_i) = \zeta_{n_i}^k \alpha_i$ for some $k \in \{0, 1, \ldots, n_i - 1\}$. The map $\sigma \mapsto \bar{k}$ is well-defined because $\zeta_{n_i}$ has exact order $n_i$, and it is a homomorphism because \begin{align*} (\sigma \circ \tau)(\alpha_i) = \sigma(\zeta_{n_i}^j \alpha_i) = \zeta_{n_i}^j \sigma(\alpha_i) = \zeta_{n_i}^j \zeta_{n_i}^k \alpha_i = \zeta_{n_i}^{j+k} \alpha_i, \end{align*} where we used $\sigma(\zeta_{n_i}) = \zeta_{n_i}$ since $\zeta_{n_i} \in E_i$ and $\sigma$ fixes $E_i$. Injectivity holds because $\sigma(\alpha_i) = \alpha_i$ forces $\sigma = \operatorname{id}$ (since $\alpha_i$ generates $E_{i+1}$ over $E_i$). The group $\mathbb{Z}/n_i\mathbb{Z}$ is cyclic, so any subgroup is cyclic and hence abelian. The fact that the Kummer Galois group is not just abelian but cyclic is stronger than needed here. However, the cyclic structure plays a crucial role in the converse direction (the [Soluble Implies Radical](/theorems/1288) theorem), where one must construct radical extensions from cyclic quotients using the [Converse of Kummer Theory](/theorems/1283).[/guided]

[step:Conclude that $\operatorname{Gal}(E/K)$ is soluble] The chain \begin{align*} \{1\} = G_r \trianglelefteq G_{r-1} \trianglelefteq \cdots \trianglelefteq G_0 = G = \operatorname{Gal}(E/K) \end{align*} satisfies: each $G_{i+1}$ is normal in $G_i$, and each quotient $G_i / G_{i+1}$ is abelian. This is precisely the definition of a solubility series for $G$. Therefore $\operatorname{Gal}(E/K)$ is a soluble group. [/step]

[step:Deduce that $\operatorname{Gal}(L/K)$ is soluble as a quotient of $\operatorname{Gal}(E/K)$]Since $E/K$ is Galois and $L$ is an intermediate field with $K \subset L \subset E$, Part 3 of the [Fundamental Theorem of Galois Theory](/theorems/1274) applies. The extension $L/K$ is itself Galois: $L$ is the splitting field of the separable polynomial $f \in K[t]$ (separability holds because $\operatorname{Char} K = 0$, so the [Characteristic Zero Separability](/theorems/1262) theorem gives that every irreducible polynomial over $K$ is separable), and the [Splitting Fields of Separable Polynomials Are Galois](/theorems/1271) theorem gives that $L/K$ is Galois. By Part 2 of the [Fundamental Theorem](/theorems/1274), the normality of $L/K$ implies $\operatorname{Gal}(E/L) \trianglelefteq \operatorname{Gal}(E/K)$. By Part 3, the restriction map induces an isomorphism \begin{align*} \operatorname{Gal}(E/K) \,/\, \operatorname{Gal}(E/L) \;\xrightarrow{\;\sim\;}\; \operatorname{Gal}(L/K). \end{align*} Therefore $\operatorname{Gal}(L/K)$ is isomorphic to a quotient of the soluble group $\operatorname{Gal}(E/K)$. By Part 2 of [Persistence of Solubility](/theorems/1285), every quotient of a soluble group is soluble. Hence $\operatorname{Gal}(L/K)$ is soluble.[/step]

[guided]At this stage we have shown that $\operatorname{Gal}(E/K)$ is soluble, but the theorem asks about $\operatorname{Gal}(L/K)$, where $L$ is the splitting field of $f$ and $L \subset E$. The passage from $E$ to $L$ is a quotient operation. We verify the hypotheses needed to apply the Fundamental Theorem to the pair $E/K$ with intermediate field $L$: 1. $E/K$ is a finite Galois extension (established in the first step). 2. $L$ is an intermediate field: $K \subset L \subset E$ (since $L \subset M \subset E$). 3. $L/K$ is a normal extension: $L$ is by definition the splitting field of $f$ over $K$. We verify separability of $f$: since $\operatorname{Char} K = 0$, the [Characteristic Zero Separability](/theorems/1262) theorem states that every irreducible polynomial in $K[t]$ is separable. Each irreducible factor of $f$ is therefore separable, so $f$ itself is separable (its roots in a splitting field are all distinct). By [Splitting Fields of Separable Polynomials Are Galois](/theorems/1271), $L/K$ is Galois, and in particular normal. Part 2 of the [Fundamental Theorem](/theorems/1274) now gives: since $L/K$ is normal, the corresponding subgroup $\operatorname{Gal}(E/L)$ is normal in $\operatorname{Gal}(E/K)$. Part 3 gives: the restriction homomorphism $\operatorname{Gal}(E/K) \to \operatorname{Gal}(L/K)$, $\sigma \mapsto \sigma|_L$, is surjective with kernel $\operatorname{Gal}(E/L)$. By the first isomorphism theorem, $\operatorname{Gal}(L/K) \cong \operatorname{Gal}(E/K) / \operatorname{Gal}(E/L)$. Finally, we apply [Persistence of Solubility](/theorems/1285) (Part 2). This theorem states: if $G$ is a soluble group and $N \trianglelefteq G$, then $G/N$ is soluble. We take $G = \operatorname{Gal}(E/K)$ (shown soluble in the previous step) and $N = \operatorname{Gal}(E/L) \trianglelefteq G$. The conclusion is that $G/N \cong \operatorname{Gal}(L/K)$ is soluble. This is the essential mechanism of the theorem: the radical tower provides solubility of the "big" Galois group $\operatorname{Gal}(E/K)$, and the splitting field $L$ sits inside $E$ as a quotient object. The persistence theorem ensures that solubility, being defined by the existence of an abelian series, passes to quotients — each abelian quotient $G_i / G_{i+1}$ in the series for $G$ projects to an abelian quotient (or collapses) in the induced series for $G/N$.[/guided]