[proofplan] The proof proceeds in two stages. First, we establish a single-step Galois closure lemma: given a Galois extension $M/K$ and an element $\gamma \in M$, the splitting field of $\prod_{\sigma \in \operatorname{Gal}(M/K)} (t^n - \sigma(\gamma))$ over $M$ is both radical over $M$ and Galois over $K$. This works because the defining polynomial is invariant under the action of $\operatorname{Gal}(M/K)$, and in characteristic $0$ it is separable. Second, we apply this construction inductively to each step of the given radical tower $K = F_0 \subset F_1 \subset \cdots \subset F_r = L$. At each stage we enlarge the current Galois radical extension to absorb the next radical step and its cyclotomic requirements, producing a tower that is simultaneously radical and Galois over $K$. [/proofplan] [step:Establish a key lemma: adjoin all Galois conjugates of a single radical step to obtain a Galois radical extension] [claim:Galois Closure of a Single Radical Step] Let $K$ be a field of characteristic $0$, let $M/K$ be a finite Galois extension, let $\gamma \in M$, and let $n \geq 1$ be an integer. Define the polynomial \begin{align*} g := \prod_{\sigma \in \operatorname{Gal}(M/K)} (t^n - \sigma(\gamma)) \in M[t]. \end{align*} Let $E$ be the splitting field of $g$ over $M$. Then: (a) $E/K$ is a Galois extension. (b) $E/M$ is a radical extension. More precisely, $E$ is obtained from $M$ by adjoining finitely many $n$-th roots. [/claim] [proof] [step:Show that $g \in K[t]$ and that $g$ is separable over $K$] We first verify that $g$ has coefficients in $K$. Let $\tau \in \operatorname{Gal}(M/K)$. Since $\tau$ permutes $\operatorname{Gal}(M/K)$ by left multiplication (the map $\sigma \mapsto \tau \circ \sigma$ is a bijection $\operatorname{Gal}(M/K) \to \operatorname{Gal}(M/K)$), applying $\tau$ to the coefficients of $g$ yields \begin{align*} \tau(g) &= \prod_{\sigma \in \operatorname{Gal}(M/K)} (t^n - \tau(\sigma(\gamma))) = \prod_{\sigma \in \operatorname{Gal}(M/K)} (t^n - (\tau \circ \sigma)(\gamma)) = \prod_{\rho \in \operatorname{Gal}(M/K)} (t^n - \rho(\gamma)) = g, \end{align*} where we substituted $\rho = \tau \circ \sigma$, which ranges over all of $\operatorname{Gal}(M/K)$ as $\sigma$ does. Since $\tau(g) = g$ for every $\tau \in \operatorname{Gal}(M/K)$ and $M/K$ is Galois, the coefficients of $g$ lie in the fixed field $M^{\operatorname{Gal}(M/K)} = K$. Hence $g \in K[t]$. We next verify separability. Since $K$ has characteristic $0$, the polynomial $t^n - \sigma(\gamma)$ is separable over $\bar{K}$ for each $\sigma$: if $\sigma(\gamma) = 0$, then $t^n - \sigma(\gamma) = t^n$ has the single root $0$ with multiplicity $n$, but this case contributes the factor $t^n$ which we may treat separately; if $\sigma(\gamma) \neq 0$, the roots of $t^n - \sigma(\gamma)$ in $\bar{K}$ are $\zeta^k \alpha_\sigma$ for $k = 0, 1, \ldots, n-1$, where $\alpha_\sigma$ is any fixed $n$-th root of $\sigma(\gamma)$ and $\zeta$ is a primitive $n$-th root of unity. These $n$ roots are distinct because $\zeta$ has exact order $n$ and $\alpha_\sigma \neq 0$. For the full product $g$, two factors $t^n - \sigma_1(\gamma)$ and $t^n - \sigma_2(\gamma)$ with $\sigma_1 \neq \sigma_2$ can share a common root $\alpha$ only if $\alpha^n = \sigma_1(\gamma)$ and $\alpha^n = \sigma_2(\gamma)$, which forces $\sigma_1(\gamma) = \sigma_2(\gamma)$. When this occurs, the two factors are identical. Let $\gamma_1, \ldots, \gamma_s$ be the distinct elements in the set $\{\sigma(\gamma) : \sigma \in \operatorname{Gal}(M/K)\}$. Then $g = \prod_{j=1}^{s} (t^n - \gamma_j)^{m_j}$ where $m_j$ is the number of $\sigma$ with $\sigma(\gamma) = \gamma_j$. Define $h := \prod_{j=1}^{s} (t^n - \gamma_j) \in K[t]$. The polynomial $h$ is separable: each factor $t^n - \gamma_j$ with $\gamma_j \neq 0$ is separable (characteristic $0$), distinct factors have distinct roots (since $\gamma_j$ are distinct, the roots $\zeta^k \alpha_j$ of one factor satisfy $\alpha_j^n = \gamma_j \neq \gamma_i = \alpha_i^n$), and $g$ and $h$ have the same splitting field over $M$ (since $h$ divides $g$ and they share the same set of roots). [guided] The goal is to show that the polynomial $g$ is defined over $K$ (not just over $M$), and that its splitting field construction is well-behaved in characteristic $0$. **Why $g \in K[t]$:** The key observation is that $\operatorname{Gal}(M/K)$ acts on itself by left multiplication, so applying any $\tau \in \operatorname{Gal}(M/K)$ to $g$ merely permutes the factors of the product. The polynomial $g$ is therefore fixed by every element of the Galois group, and since $M/K$ is Galois, the Galois correspondence gives $M^{\operatorname{Gal}(M/K)} = K$. Each coefficient of $g$ (as a polynomial in $t$) lies in $M$ and is fixed by every $\tau \in \operatorname{Gal}(M/K)$, hence lies in $K$. **Why separability:** In characteristic $0$, every irreducible polynomial is separable. The concern is whether $g$ itself (which is generally reducible) has repeated roots. Two distinct factors $t^n - \gamma_j$ and $t^n - \gamma_i$ (with $\gamma_j \neq \gamma_i$) cannot share a root, since any root $\alpha$ of both would satisfy $\gamma_j = \alpha^n = \gamma_i$, a contradiction. Within a single factor $t^n - \gamma_j$ with $\gamma_j \neq 0$, the $n$ roots $\zeta^k \alpha_j$ are distinct because $\zeta$ is a primitive $n$-th root of unity of exact order $n$. If multiple automorphisms $\sigma$ yield the same value $\sigma(\gamma)$, the product $g$ contains repeated factors, but these do not affect the splitting field. The reduced polynomial $h = \prod_{j=1}^{s} (t^n - \gamma_j)$ is separable, lies in $K[t]$, and has the same splitting field as $g$ over $M$. [/guided] [/step] [step:Show that $E/K$ is Galois] The splitting field $E$ of $g$ over $M$ equals the splitting field of $g$ over $K$, since $M \subset E$ and $g \in K[t]$. (More precisely, $E$ is generated over $K$ by $M$ and the roots of $g$; but $M$ is the splitting field of some separable polynomial $f \in K[t]$, so $E$ is the splitting field over $K$ of the product $f \cdot h$, where $h = \prod_{j=1}^{s}(t^n - \gamma_j) \in K[t]$ is the reduced form of $g$ defined in the previous step.) In characteristic $0$, every polynomial over $K$ is separable (every irreducible factor in $K[t]$ has distinct roots in $\bar{K}$). Hence $f \cdot h$ has separable irreducible factors over $K$, and the splitting field of a separable polynomial over $K$ is a Galois extension of $K$. Therefore $E/K$ is Galois. [guided] We use the characterisation: an extension is Galois over $K$ if and only if it is the splitting field of a separable polynomial over $K$. Since $M/K$ is Galois, $M$ is the splitting field of some separable polynomial $f \in K[t]$. The field $E$ is generated over $M$ by the roots of $g$, and we showed $g \in K[t]$. Therefore $E$ is generated over $K$ by the roots of $f$ (which give $M$) and the roots of $g$ (which give the rest of $E$). Equivalently, $E$ is the splitting field over $K$ of the product $f \cdot h$, where $h$ is the separable reduction of $g$ constructed in the previous step. In characteristic $0$, every irreducible polynomial over $K$ is separable: if $p \in K[t]$ is irreducible and inseparable, then $p'(t) = 0$, which in characteristic $0$ forces $\deg p = 0$, contradicting irreducibility. Therefore every irreducible factor of $f \cdot h$ is separable, and the splitting field of a polynomial all of whose irreducible factors are separable is a Galois extension. Hence $E/K$ is Galois. [/guided] [/step] [step:Show that $E/M$ is radical] The roots of $g$ in $E$ are the $n$-th roots of $\sigma(\gamma)$ for each $\sigma \in \operatorname{Gal}(M/K)$. For each $\sigma$, pick any root $\alpha_\sigma \in E$ with $\alpha_\sigma^n = \sigma(\gamma)$. Then all roots of $t^n - \sigma(\gamma)$ are of the form $\zeta^k \alpha_\sigma$ where $\zeta$ is a primitive $n$-th root of unity in $E$ (which exists because $t^n - 1$ splits in $E$: its roots are quotients $\alpha_{\sigma_1}/\alpha_{\sigma_2}$ raised to appropriate powers when $\sigma_1(\gamma), \sigma_2(\gamma) \neq 0$; more directly, $E$ contains a splitting field of $t^n - 1$ over $K$ because $t^n - 1$ divides $t^{nm} - \sigma(\gamma)^m$ for appropriate $m$ whenever $\sigma(\gamma) \neq 0$). Enumerate the distinct values $\gamma_1, \ldots, \gamma_s$ of $\{\sigma(\gamma) : \sigma \in \operatorname{Gal}(M/K)\}$ and choose $\alpha_j \in E$ with $\alpha_j^n = \gamma_j$ for each $j$. Then $E = M(\alpha_1, \ldots, \alpha_s)$, and the tower \begin{align*} M \subset M(\alpha_1) \subset M(\alpha_1, \alpha_2) \subset \cdots \subset M(\alpha_1, \ldots, \alpha_s) = E \end{align*} is a radical tower: at each step $M(\alpha_1, \ldots, \alpha_{j-1}) \subset M(\alpha_1, \ldots, \alpha_j)$, the element $\alpha_j$ satisfies $\alpha_j^n = \gamma_j \in M \subset M(\alpha_1, \ldots, \alpha_{j-1})$. [/step] [/proof] [step:Recall the structure of a radical extension and set up the inductive framework] By hypothesis, $L/K$ is a radical extension: there exists a tower of fields \begin{align*} K = F_0 \subset F_1 \subset F_2 \subset \cdots \subset F_r = L \end{align*} where, for each $i = 1, \ldots, r$, we have $F_i = F_{i-1}(\alpha_i)$ with $\alpha_i^{n_i} \in F_{i-1}$ for some integer $n_i \geq 1$. We construct the desired Galois radical extension $E/K$ with $L \subset E$ by induction on the number of steps $r$ in the radical tower. **Base case ($r = 0$).** If $r = 0$, then $L = K$, and we take $E = K$. The extension $K/K$ is both Galois and radical (with an empty tower). **Inductive hypothesis.** Assume that for any radical extension of $K$ with a tower of $r - 1$ steps, there exists a Galois radical extension of $K$ containing it. [guided] The inductive strategy is to peel off one radical step at a time. The extension $K \subset F_1 \subset \cdots \subset F_{r-1}$ is itself a radical tower with $r - 1$ steps. By the inductive hypothesis, there exists a field $M$ with $F_{r-1} \subset M$, $M/K$ Galois, and $M/K$ radical. We then face the problem of extending $M$ to absorb the final step $F_{r-1} \subset F_r = F_{r-1}(\alpha_r)$ where $\alpha_r^{n_r} \in F_{r-1}$. This is where the Galois closure lemma from the first step applies: the element $\gamma := \alpha_r^{n_r}$ lies in $F_{r-1} \subset M$, and we adjoin all Galois conjugates of the $n_r$-th root of $\gamma$ to obtain a Galois radical extension over $K$ containing $L$. [/guided] [/step] [step:Apply the inductive hypothesis to the first $r-1$ steps] Consider the sub-tower $K = F_0 \subset F_1 \subset \cdots \subset F_{r-1}$. This is a radical extension of $K$ with $r - 1$ steps. By the inductive hypothesis, there exists a field $M$ such that: (i) $F_{r-1} \subset M$, (ii) $M/K$ is a Galois extension, (iii) $M/K$ is a radical extension. Since $M/K$ is radical, there is a radical tower \begin{align*} K = M_0 \subset M_1 \subset \cdots \subset M_s = M \end{align*} where $M_j = M_{j-1}(\beta_j)$ with $\beta_j^{m_j} \in M_{j-1}$ for each $j = 1, \ldots, s$. [/step] [step:Set $\gamma := \alpha_r^{n_r} \in M$ and apply the Galois closure lemma] The final step of the original radical tower is $F_{r-1} \subset F_r = F_{r-1}(\alpha_r)$ where $\alpha_r^{n_r} \in F_{r-1}$. Set $\gamma := \alpha_r^{n_r}$. Since $F_{r-1} \subset M$, we have $\gamma \in M$. Set $n := n_r$. Apply the Galois Closure of a Single Radical Step (established above) to the Galois extension $M/K$, the element $\gamma \in M$, and the integer $n$. Let $E$ be the splitting field over $M$ of \begin{align*} g := \prod_{\sigma \in \operatorname{Gal}(M/K)} (t^n - \sigma(\gamma)) \in K[t]. \end{align*} By the lemma: (a) $E/K$ is a Galois extension. (b) $E/M$ is a radical extension: there exist elements $\alpha_{\sigma} \in E$ with $\alpha_{\sigma}^n = \sigma(\gamma) \in M$ for each $\sigma \in \operatorname{Gal}(M/K)$, and $E = M(\alpha_\sigma : \sigma \in \operatorname{Gal}(M/K))$. [guided] This is the core of the construction. The element $\gamma = \alpha_r^{n_r}$ lies in $F_{r-1} \subset M$. We need to adjoin an $n$-th root of $\gamma$ to $M$ (to recover $\alpha_r$ and hence $L = F_{r-1}(\alpha_r) \subset F_{r-1}(\alpha_r) \subset M(\alpha_r)$), but merely adjoining one root would not produce a Galois extension over $K$. The solution is to adjoin not just one $n$-th root of $\gamma$, but $n$-th roots of every Galois conjugate $\sigma(\gamma)$ for $\sigma \in \operatorname{Gal}(M/K)$. The defining polynomial $g = \prod_\sigma (t^n - \sigma(\gamma))$ lies in $K[t]$ by the Galois-invariance argument, and its splitting field over $K$ is therefore Galois over $K$. Why do we need all conjugates? If we only adjoined an $n$-th root of $\gamma$ itself, the resulting extension $M(\alpha_r)/K$ would typically not be normal: for an automorphism $\tau$ of $\bar{K}$ fixing $K$, the image $\tau(\alpha_r)$ satisfies $\tau(\alpha_r)^n = \tau(\gamma) = \tau(\sigma(\gamma))$ for some $\sigma$ (since $\gamma \in M$ and $\tau|_M$ is some element of $\operatorname{Gal}(M/K)$). So the Galois conjugates of $\alpha_r$ over $K$ are $n$-th roots of the various $\sigma(\gamma)$, and for the extension to be normal, all such roots must be present. [/guided] [/step] [step:Verify that $L \subset E$ and that $E/K$ is a radical extension] **$L \subset E$:** The identity automorphism $\operatorname{id} \in \operatorname{Gal}(M/K)$ contributes the factor $t^n - \gamma$ to $g$. Since $g$ splits completely in $E$, there exists $\alpha \in E$ with $\alpha^n = \gamma = \alpha_r^{n_r}$. In particular, $\alpha$ is an $n_r$-th root of $\alpha_r^{n_r}$ in $E$. Since $\alpha_r$ is also an $n_r$-th root of $\alpha_r^{n_r}$ and both lie in $\bar{K}$, we have $\alpha_r = \zeta^k \alpha$ for some $n_r$-th root of unity $\zeta^k$. The roots of unity of order $n_r$ lie in $E$ (since $E$ contains all roots of $t^{n_r} - \gamma$, and the quotient of any two such roots is an $n_r$-th root of unity). Therefore $\alpha_r \in E$, and since $F_{r-1} \subset M \subset E$, we get $L = F_{r-1}(\alpha_r) \subset E$. **$E/K$ is radical:** We have the tower \begin{align*} K = M_0 \subset M_1 \subset \cdots \subset M_s = M \subset M(\alpha_{\sigma_1}) \subset M(\alpha_{\sigma_1}, \alpha_{\sigma_2}) \subset \cdots \subset E, \end{align*} where: - The first segment $K = M_0 \subset \cdots \subset M_s = M$ is the radical tower for $M/K$ (which exists by the inductive hypothesis, condition (iii)). - The second segment adjoins the elements $\alpha_{\sigma_j}$ one at a time. At each step, $\alpha_{\sigma_j}^n = \sigma_j(\gamma) \in M \subset M(\alpha_{\sigma_1}, \ldots, \alpha_{\sigma_{j-1}})$, so each step is a radical step with exponent $n$. The concatenation is a radical tower from $K$ to $E$. Hence $E/K$ is a radical extension. [guided] We must verify two things: that $E$ is large enough (it contains $L$), and that the resulting extension is radical over $K$. **Containment $L \subset E$:** The original element $\alpha_r$ satisfies $\alpha_r^{n_r} = \gamma$, and $E$ contains all roots of $t^{n_r} - \gamma$ (since $\operatorname{id} \in \operatorname{Gal}(M/K)$ contributes the factor $t^{n_r} - \gamma$ to $g$, and $g$ splits in $E$). Therefore $\alpha_r$ is one of the roots of $t^{n_r} - \gamma$ in $E$, giving $\alpha_r \in E$. Combined with $F_{r-1} \subset M \subset E$, we get $L = F_{r-1}(\alpha_r) \subset E$. **Radical tower for $E/K$:** The extension $M/K$ is radical by the inductive hypothesis, providing a radical tower from $K$ to $M$. The extension $E/M$ is radical by Part (b) of the lemma, providing a radical tower from $M$ to $E$. Concatenating these two radical towers yields a radical tower from $K$ to $E$. The concatenation is valid because each step in the second tower adjoins an element $\alpha_{\sigma_j}$ satisfying $\alpha_{\sigma_j}^n \in M$, and $M$ is contained in every field in the second segment of the tower. This completes the induction: $E/K$ is a Galois extension (by Part (a) of the lemma), $E/K$ is a radical extension (by the concatenated tower), and $L \subset E$. Hence $E$ is the desired field. [/guided] [/step] [step:Conclude the induction] Combining the results of the preceding steps: starting from the radical extension $L/K$ with radical tower $K = F_0 \subset F_1 \subset \cdots \subset F_r = L$, we have produced (by induction on $r$) a field $E$ with $L \subset E$ such that $E/K$ is both Galois and radical. The extension $E/L$ satisfies $E \supset L$, and $E/K$ is Galois with $E/K$ radical, which is precisely the conclusion of the theorem. [/step]