[proofplan] We prove both directions. For the forward direction, we show that if $f$ is solvable by radicals, then all roots of $f$ lie in a radical extension of $k$, which we enlarge to a Galois radical extension; the Galois group of a radical tower is built from cyclic extensions (using the presence of roots of unity in characteristic zero), so it is solvable, and $\operatorname{Gal}(K/k)$ is a quotient of a solvable group, hence solvable. For the converse, we show that a solvable Galois group admits a subnormal series with cyclic factors, each cyclic factor corresponds (after adjoining appropriate roots of unity) to an extension generated by a single radical via Kummer theory, and chaining these radical extensions produces a radical tower containing $K$. [/proofplan] [step:Reduce to showing that radical extensions have solvable Galois groups (forward direction)] Assume $f$ is solvable by radicals over $k$. By definition, there exists a radical tower \begin{align*} k = F_0 \subset F_1 \subset \cdots \subset F_m \end{align*} where each $F_{i+1} = F_i(\alpha_i)$ with $\alpha_i^{n_i} \in F_i$ for some $n_i \ge 1$, and all roots of $f$ lie in $F_m$. In particular, the splitting field $K$ of $f$ over $k$ satisfies $K \subset F_m$. We first enlarge this tower to one that is Galois over $k$. Let $N$ be the normal closure of $F_m$ over $k$. Since $\operatorname{char}(k) = 0$, every algebraic extension of $k$ is separable, so $N/k$ is a finite Galois extension. Moreover, $N$ is still a radical extension of $k$: if $\sigma \in \operatorname{Gal}(N/k)$, then $\sigma(\alpha_i)^{n_i} = \sigma(\alpha_i^{n_i}) \in \sigma(F_i)$, so each conjugate field $\sigma(F_m)$ is again a radical extension, and the compositum of finitely many radical extensions is a radical extension. Thus we may assume WLOG that $F_m/k$ is Galois. Since $K \subset F_m$ and $F_m/k$ is Galois, the [quotetheorem:1274] gives $\operatorname{Gal}(K/k) \cong \operatorname{Gal}(F_m/k) / \operatorname{Gal}(F_m/K)$. A quotient of a solvable group is solvable, so it suffices to show $\operatorname{Gal}(F_m/k)$ is solvable. [guided] Assume $f$ is solvable by radicals over $k$. By definition, there exists a radical tower \begin{align*} k = F_0 \subset F_1 \subset \cdots \subset F_m \end{align*} where each $F_{i+1} = F_i(\alpha_i)$ with $\alpha_i^{n_i} \in F_i$ for some $n_i \ge 1$, and all roots of $f$ lie in $F_m$. In particular, the splitting field $K$ of $f$ over $k$ satisfies $K \subset F_m$. The first reduction is to enlarge this tower to one that is Galois over $k$. Why is this necessary? The Fundamental Theorem of Galois Theory only applies to Galois extensions, and $F_m/k$ need not be Galois (for instance, $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ is not Galois). Let $N$ be the normal closure of $F_m$ over $k$. Since $\operatorname{char}(k) = 0$, every algebraic extension of $k$ is separable, so $N/k$ is a finite Galois extension. Why is $N$ still a radical extension of $k$? If $\sigma \in \operatorname{Gal}(N/k)$, then $\sigma(\alpha_i)^{n_i} = \sigma(\alpha_i^{n_i}) \in \sigma(F_i)$, so each conjugate generator $\sigma(\alpha_i)$ is again a radical over the conjugate field $\sigma(F_i)$. The conjugate field $\sigma(F_m)$ is therefore again a radical extension of $k$, and the compositum of finitely many radical towers (one for each $\sigma \in \operatorname{Gal}(N/k)$, a finite set since $[N:k] < \infty$) is again a radical tower — we concatenate the towers. Thus we may assume WLOG that $F_m/k$ is Galois. The second reduction uses the [quotetheorem:1274]. Since $K \subset F_m$ and $F_m/k$ is Galois, the Fundamental Theorem gives the quotient isomorphism \begin{align*} \operatorname{Gal}(K/k) \cong \operatorname{Gal}(F_m/k) / \operatorname{Gal}(F_m/K). \end{align*} A quotient of a solvable group is solvable: if $G$ is solvable with subnormal series $\{e\} = G_n \trianglelefteq \cdots \trianglelefteq G_0 = G$ having abelian factors, and $\pi: G \twoheadrightarrow Q$ is a surjective homomorphism, then $\{e\} = \pi(G_n) \trianglelefteq \cdots \trianglelefteq \pi(G_0) = Q$ is a subnormal series for $Q$ with abelian factors (since the image of an abelian group under a homomorphism is abelian). Therefore it suffices to show $\operatorname{Gal}(F_m/k)$ is solvable. [/guided] [/step] [step:Adjoin roots of unity and show each layer of the radical tower contributes a cyclic extension] Let $n = n_0 n_1 \cdots n_{m-1}$ be the product of all radical exponents in the tower. Adjoin a primitive $n$-th root of unity $\zeta_n$ to obtain the extended tower \begin{align*} k \subset k(\zeta_n) \subset F_1(\zeta_n) \subset \cdots \subset F_m(\zeta_n). \end{align*} Since $\operatorname{char}(k) = 0$, the polynomial $x^n - 1$ is separable over $k$, so $k(\zeta_n)/k$ is a Galois extension. The Galois group $\operatorname{Gal}(k(\zeta_n)/k)$ embeds into $(\mathbb{Z}/n\mathbb{Z})^\times$ via $\sigma \mapsto j$ where $\sigma(\zeta_n) = \zeta_n^j$. Since $(\mathbb{Z}/n\mathbb{Z})^\times$ is abelian, $\operatorname{Gal}(k(\zeta_n)/k)$ is abelian. Now consider a single layer $F_{i+1}(\zeta_n) / F_i(\zeta_n)$. The field $F_i(\zeta_n)$ contains all $n_i$-th roots of unity (since $n_i \mid n$), and $F_{i+1}(\zeta_n) = F_i(\zeta_n)(\alpha_i)$ where $\alpha_i^{n_i} \in F_i \subset F_i(\zeta_n)$. By [Kummer theory](/theorems/???), the extension $F_i(\zeta_n)(\alpha_i) / F_i(\zeta_n)$ is cyclic of degree dividing $n_i$. Its Galois group is cyclic, hence abelian. [guided] Why do we adjoin roots of unity? The key tool is Kummer theory, which characterises cyclic extensions of degree $n$ over a field containing all $n$-th roots of unity: such extensions are precisely the splitting fields of polynomials $x^n - a$. Without roots of unity, an extension $F(\alpha)/F$ with $\alpha^n \in F$ need not be Galois (for instance, $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ is not Galois since $x^3 - 2$ does not split over $\mathbb{Q}(\sqrt[3]{2})$). After adjoining $\zeta_n$, the field $F_i(\zeta_n)$ contains all $n_i$-th roots of unity. The extension $F_i(\zeta_n)(\alpha_i) / F_i(\zeta_n)$ is then the splitting field of $x^{n_i} - \alpha_i^{n_i}$ over $F_i(\zeta_n)$ (the roots are $\alpha_i, \zeta_{n_i}\alpha_i, \ldots, \zeta_{n_i}^{n_i - 1}\alpha_i$). By Kummer theory, this extension is Galois with cyclic Galois group of order dividing $n_i$, generated by $\sigma: \alpha_i \mapsto \zeta_{n_i}\alpha_i$. The embedding $\operatorname{Gal}(k(\zeta_n)/k) \hookrightarrow (\mathbb{Z}/n\mathbb{Z})^\times$ is defined as follows: each $\sigma \in \operatorname{Gal}(k(\zeta_n)/k)$ permutes the roots of $x^n - 1$ and is determined by $\sigma(\zeta_n) = \zeta_n^j$ for a unique $j \in (\mathbb{Z}/n\mathbb{Z})^\times$. The map $\sigma \mapsto j$ is an injective group homomorphism into the abelian group $(\mathbb{Z}/n\mathbb{Z})^\times$, so $\operatorname{Gal}(k(\zeta_n)/k)$ is abelian. [/guided] [/step] [step:Assemble the subnormal series to conclude $\operatorname{Gal}(F_m(\zeta_n)/k)$ is solvable] The tower $k \subset k(\zeta_n) \subset F_1(\zeta_n) \subset \cdots \subset F_m(\zeta_n)$ produces a chain of subgroups \begin{align*} \{e\} = \operatorname{Gal}(F_m(\zeta_n)/F_m(\zeta_n)) \le \operatorname{Gal}(F_m(\zeta_n)/F_{m-1}(\zeta_n)) \le \cdots \le \operatorname{Gal}(F_m(\zeta_n)/k(\zeta_n)) \le \operatorname{Gal}(F_m(\zeta_n)/k). \end{align*} By the [quotetheorem:1274], each $\operatorname{Gal}(F_m(\zeta_n)/F_i(\zeta_n))$ is a normal subgroup of $\operatorname{Gal}(F_m(\zeta_n)/F_{i-1}(\zeta_n))$ (since $F_i(\zeta_n)/F_{i-1}(\zeta_n)$ is Galois, being a Kummer extension), and the successive quotient is isomorphic to $\operatorname{Gal}(F_i(\zeta_n)/F_{i-1}(\zeta_n))$, which is cyclic. The bottom layer contributes $\operatorname{Gal}(F_m(\zeta_n)/k) / \operatorname{Gal}(F_m(\zeta_n)/k(\zeta_n)) \cong \operatorname{Gal}(k(\zeta_n)/k)$, which is abelian. Every successive quotient in the chain is abelian, so $\operatorname{Gal}(F_m(\zeta_n)/k)$ is solvable. Since $F_m \subset F_m(\zeta_n)$, the restriction map gives $\operatorname{Gal}(F_m/k)$ as a quotient of $\operatorname{Gal}(F_m(\zeta_n)/k)$. A quotient of a solvable group is solvable, so $\operatorname{Gal}(F_m/k)$ is solvable. By the reduction in the first step, $\operatorname{Gal}(K/k)$ is solvable. This completes the forward direction. [guided] We now assemble the pieces. The extended tower $k \subset k(\zeta_n) \subset F_1(\zeta_n) \subset \cdots \subset F_m(\zeta_n)$ has the property that each layer is a Galois extension of the one below (the first layer $k(\zeta_n)/k$ is Galois since it is the splitting field of $x^n - 1$; each subsequent layer $F_i(\zeta_n)/F_{i-1}(\zeta_n)$ is a Kummer extension, hence Galois). By the Fundamental Theorem, normal subgroups of $\operatorname{Gal}(F_m(\zeta_n)/k)$ correspond to intermediate Galois extensions. The chain of fixed fields gives a subnormal series \begin{align*} \{e\} \trianglelefteq G_m \trianglelefteq G_{m-1} \trianglelefteq \cdots \trianglelefteq G_0 \trianglelefteq \operatorname{Gal}(F_m(\zeta_n)/k) \end{align*} where $G_i = \operatorname{Gal}(F_m(\zeta_n)/F_i(\zeta_n))$. Each quotient $G_{i-1}/G_i \cong \operatorname{Gal}(F_i(\zeta_n)/F_{i-1}(\zeta_n))$ is cyclic (by Kummer theory), and $\operatorname{Gal}(F_m(\zeta_n)/k) / G_0 \cong \operatorname{Gal}(k(\zeta_n)/k)$ is abelian. Since every factor in the subnormal series is abelian, the group $\operatorname{Gal}(F_m(\zeta_n)/k)$ is solvable by definition. The final reduction is: $\operatorname{Gal}(F_m/k)$ is a quotient of $\operatorname{Gal}(F_m(\zeta_n)/k)$ (via the restriction homomorphism $\sigma \mapsto \sigma|_{F_m}$, which is surjective since every automorphism of $F_m/k$ extends to $F_m(\zeta_n)/k$). A quotient of a solvable group is solvable. Then $\operatorname{Gal}(K/k)$ is a quotient of $\operatorname{Gal}(F_m/k)$ (by the first step), hence also solvable. [/guided] [/step] [step:Prove the converse: construct a radical tower from a solvable Galois group] Now assume $G = \operatorname{Gal}(K/k)$ is solvable. Since $G$ is solvable, there exists a subnormal series \begin{align*} \{e\} = G_m \trianglelefteq G_{m-1} \trianglelefteq \cdots \trianglelefteq G_1 \trianglelefteq G_0 = G \end{align*} with each quotient $G_i / G_{i+1}$ cyclic of prime order $p_i$. Let $n = p_0 p_1 \cdots p_{m-1}$ and adjoin a primitive $n$-th root of unity $\zeta_n$ to $k$. Define $L = K(\zeta_n)$. The extension $L/k$ is Galois (as the compositum of two Galois extensions $K/k$ and $k(\zeta_n)/k$). Let $H = \operatorname{Gal}(L/K)$. By the [quotetheorem:1274], $H \trianglelefteq \operatorname{Gal}(L/k)$ and $\operatorname{Gal}(L/k) / H \cong \operatorname{Gal}(K/k) = G$. Since $H$ is isomorphic to a subgroup of $\operatorname{Gal}(k(\zeta_n)/k)$, which is abelian, $H$ is abelian. Therefore $\operatorname{Gal}(L/k)$ is solvable (it is an extension of a solvable group $G$ by an abelian group $H$). [guided] For the converse, we start with a solvable Galois group and need to build a radical tower. The key idea is that a cyclic extension of degree $p$ over a field containing $p$-th roots of unity is a Kummer extension, hence generated by a single $p$-th radical. So the strategy is: 1. Refine the solvable group into a subnormal series with cyclic factors of prime order. 2. Adjoin all necessary roots of unity. 3. Use Kummer theory to identify each cyclic layer with a radical extension. Why adjoin $\zeta_n$ to the base field? We need roots of unity at each level to apply Kummer theory. Rather than adjoining them level by level, it is cleaner to adjoin all of them at once. The resulting extension $L = K(\zeta_n)$ is Galois over $k$ because it is the compositum of the Galois extensions $K/k$ and $k(\zeta_n)/k$. The group $\operatorname{Gal}(L/k)$ is solvable because it fits into the short exact sequence \begin{align*} 1 \to H \to \operatorname{Gal}(L/k) \to G \to 1 \end{align*} where $H = \operatorname{Gal}(L/K)$ is abelian (as a subgroup of the abelian group $\operatorname{Gal}(k(\zeta_n)/k)$) and $G = \operatorname{Gal}(K/k)$ is solvable by hypothesis. An extension of a solvable group by a solvable (in particular, abelian) group is solvable. [/guided] [/step] [step:Use Kummer theory to realise each cyclic layer as a radical extension] Since $\operatorname{Gal}(L/k)$ is solvable, it admits a subnormal series with cyclic quotients of prime order. This subnormal series corresponds, via the Galois correspondence, to a tower of intermediate fields \begin{align*} k = E_0 \subset E_1 \subset \cdots \subset E_r = L \end{align*} where each $E_{i+1}/E_i$ is a Galois extension with cyclic Galois group of prime order $p_i$. Since $\zeta_n \in L$ and $p_i \mid n$, each $E_i$ contains a primitive $p_i$-th root of unity (because $\zeta_n^{n/p_i}$ is a primitive $p_i$-th root of unity, and $\zeta_n \in E_r$). However, we need $\zeta_{p_i} \in E_i$, not merely in $E_r$. We address this: after reindexing, we may arrange the tower so that the roots-of-unity layers come first (they are abelian extensions of $k$, hence can be placed at the bottom of any solvable tower). Alternatively, we enlarge the base: since $k(\zeta_n)/k$ is abelian, we may replace $k$ with $k(\zeta_n)$ and consider the tower over $k(\zeta_n)$, which corresponds to the subgroup $\operatorname{Gal}(L/k(\zeta_n)) \le \operatorname{Gal}(L/k)$. This subgroup is still solvable (as a subgroup of a solvable group). Its subnormal series gives a tower \begin{align*} k(\zeta_n) = E_0' \subset E_1' \subset \cdots \subset E_s' = L \end{align*} where each $E_{i+1}'/E_i'$ is cyclic of prime order $q_i$ with $q_i \mid n$. Now each $E_i'$ contains $\zeta_n$, hence in particular a primitive $q_i$-th root of unity. By [Kummer theory](/theorems/???), a cyclic extension of prime degree $q$ over a field containing a primitive $q$-th root of unity is generated by a $q$-th root: $E_{i+1}' = E_i'(\beta_i)$ where $\beta_i^{q_i} \in E_i'$. Therefore the tower $k \subset k(\zeta_n) \subset E_1' \subset \cdots \subset L$ is a radical tower. The first step $k \subset k(\zeta_n)$ is itself radical (adjoin $\zeta_n$, which satisfies $\zeta_n^n = 1 \in k$). Since $K \subset L$, all roots of $f$ lie in $L$, and $L$ is contained in a radical tower over $k$. Therefore $f$ is solvable by radicals. This completes the converse. [guided] This is where the algebra meets the field theory. We have a solvable group $\operatorname{Gal}(L/k)$ and need to turn its group-theoretic structure into field-theoretic data (radical extensions). The crucial input is Kummer's theorem: if a field $F$ contains a primitive $p$-th root of unity, then every cyclic extension of $F$ of degree $p$ has the form $F(\sqrt[p]{a})$ for some $a \in F$. This is precisely the translation between "cyclic Galois group" and "radical extension" that makes Galois's criterion work. There is a subtlety: the subnormal series of $\operatorname{Gal}(L/k)$ corresponds to a tower of extensions, but to apply Kummer theory at level $i$, we need the $p_i$-th roots of unity to already be present in $E_i$, not just in $L$. This is why we work over $k(\zeta_n)$ rather than over $k$: by placing all roots of unity in the base, every intermediate field automatically contains them. The tower over $k$ is then: first adjoin $\zeta_n$ (a radical step, since $\zeta_n^n = 1$), then follow the Kummer tower from $k(\zeta_n)$ up to $L$. Each step adjoins a single radical, so the entire tower is a radical extension. Since $K \subset L$, the roots of $f$ lie in this radical tower, and $f$ is solvable by radicals. [/guided] [/step]