[proofplan] We prove the three parts in sequence. For (1), we use the characteristic hypothesis to show that $t^n - \lambda$ is separable in $L$: the formal derivative $nt^{n-1}$ and $t^n - \lambda$ share no common root since $n$ is invertible in $L$. With $n$ distinct roots $\alpha_1, \ldots, \alpha_n$ in hand, the ratios $\alpha_i \alpha_1^{-1}$ are $n$ distinct roots of $t^n - 1$, so they exhaust the $n$-th roots of unity, and one must be primitive. For (2), we fix a root $\alpha$ of $t^n - \lambda$ and observe that every root has the form $\mu^k \alpha$; since $\mu \in K(\mu)$, the splitting field is $L = K(\mu)(\alpha)$. Each $\sigma \in \operatorname{Gal}(L/K(\mu))$ is determined by $\sigma(\alpha) = \mu^{i(\sigma)}\alpha$, and the map $\sigma \mapsto i(\sigma) \bmod n$ is an injective homomorphism into $\mathbb{Z}/n\mathbb{Z}$, so the Galois group is cyclic of order dividing $n$. For (3), we identify $[L : K(\mu)] = \deg(\operatorname{min}_{K(\mu)}(\alpha))$ and note that $\operatorname{min}_{K(\mu)}(\alpha)$ divides $t^n - \lambda$, so the degree equals $n$ precisely when $t^n - \lambda$ is irreducible over $K(\mu)$. [/proofplan] [step:Show $t^n - \lambda$ has $n$ distinct roots in $L$ using the characteristic hypothesis] Since $L$ is the splitting field of $f(t) := t^n - \lambda$ over $K$, the polynomial $f$ splits completely in $L[t]$. We show its roots are distinct by verifying that $f$ and its formal derivative $f'(t) = nt^{n-1}$ share no common root in $L$. Let $\alpha \in L$ be a root of $f$, so $\alpha^n = \lambda$. Since $\lambda \neq 0$, we have $\alpha \neq 0$, and therefore $f'(\alpha) = n\alpha^{n-1} \neq 0$ provided $n \neq 0$ in $L$. The hypothesis states that either $\operatorname{char} K = 0$ (in which case $n \in \mathbb{N}$ is invertible in $K \subset L$) or $0 < \operatorname{char} K \nmid n$ (in which case $n \not\equiv 0 \pmod{\operatorname{char} K}$, so $n \neq 0$ in $K \subset L$). In both cases, $n$ is a nonzero element of $L$, and since $\alpha \neq 0$, we conclude $f'(\alpha) = n\alpha^{n-1} \neq 0$. Since no root of $f$ is a root of $f'$, the polynomial $f$ has no repeated roots in $L$. Therefore $f$ has exactly $\deg(f) = n$ distinct roots $\alpha_1, \ldots, \alpha_n$ in $L$. [guided] The goal is to show separability of $t^n - \lambda$: that it has $n$ distinct roots in its splitting field. The standard criterion for a polynomial over a field to have a repeated root is that it shares a common root with its formal derivative. The formal derivative of $f(t) = t^n - \lambda$ is $f'(t) = nt^{n-1}$. Suppose $\alpha \in L$ is a common root, i.e., $f(\alpha) = 0$ and $f'(\alpha) = 0$. From $f(\alpha) = 0$ we get $\alpha^n = \lambda \neq 0$ (since $\lambda \in K^\times$), so $\alpha \neq 0$. From $f'(\alpha) = n\alpha^{n-1} = 0$ and $\alpha \neq 0$ (hence $\alpha^{n-1} \neq 0$ since $L$ is a field), we would need $n = 0$ in $L$. This is where the characteristic hypothesis is consumed. If $\operatorname{char} K = 0$, then $K$ (and hence $L \supset K$) has characteristic zero, so $n \in \mathbb{N}$ is nonzero in $L$. If $\operatorname{char} K = p > 0$ with $p \nmid n$, then $n \not\equiv 0 \pmod{p}$, so $n \neq 0$ in $L$. In either case, $n \neq 0$ in $L$, contradicting $n\alpha^{n-1} = 0$. Therefore $f$ and $f'$ share no common root in $L$, which means $f$ has no repeated roots. Since $f$ has degree $n$ and splits completely in $L$ (by definition of the splitting field), it has exactly $n$ distinct roots $\alpha_1, \ldots, \alpha_n$ in $L$. What fails when $\operatorname{char} K$ divides $n$? In that case, $n = 0$ in $K$, so $f'(t) = nt^{n-1} = 0$ identically. Every root of $f$ is automatically a root of $f'$. For example, $t^p - \lambda$ over a field of characteristic $p$ has the form $(t - \alpha)^p$ in the splitting field (if an $\alpha$ with $\alpha^p = \lambda$ exists), so all roots coincide. The characteristic hypothesis is essential for separability. [/guided] [/step] [step:Construct a primitive $n$-th root of unity in $L$ from the ratios of roots] Fix a root $\alpha_1 \in L$ of $t^n - \lambda$. For each $i \in \{1, \ldots, n\}$, define $\omega_i := \alpha_i \alpha_1^{-1} \in L$ (this is well-defined since $\alpha_1 \neq 0$). Then \begin{align*} \omega_i^n = \frac{\alpha_i^n}{\alpha_1^n} = \frac{\lambda}{\lambda} = 1, \end{align*} so each $\omega_i$ is an $n$-th root of unity. The $n$ elements $\omega_1, \ldots, \omega_n$ are pairwise distinct: if $\omega_i = \omega_j$, then $\alpha_i \alpha_1^{-1} = \alpha_j \alpha_1^{-1}$, hence $\alpha_i = \alpha_j$, which forces $i = j$ since the roots are distinct. Since $t^n - 1$ is a polynomial of degree $n$ and has at most $n$ roots in the field $L$, the elements $\omega_1, \ldots, \omega_n$ are exactly all $n$-th roots of unity in $L$. Among any complete set of $n$-th roots of unity there exists a primitive one (i.e., one of order exactly $n$): the $n$-th roots of unity form a subgroup of $L^\times$ of order $n$. Since $L^\times$ is abelian and $n$ is not divisible by $\operatorname{char} L$ (by hypothesis), this subgroup is cyclic of order $n$, and any generator is a primitive $n$-th root of unity. Let $\mu$ denote such a primitive $n$-th root of unity. Since $\mu = \omega_i = \alpha_i \alpha_1^{-1} \in L$ for some $i$, we have $\mu \in L$. This completes Part (1). [guided] The idea is simple: the $n$ roots of $t^n - \lambda$ are distinct (by the previous step), and taking their ratios with a fixed root produces $n$ distinct $n$-th roots of unity — which must include a primitive one. Fix any root $\alpha_1$ and form $\omega_i := \alpha_i / \alpha_1$ for each $i$. The computation $\omega_i^n = \alpha_i^n / \alpha_1^n = \lambda / \lambda = 1$ confirms each $\omega_i$ is an $n$-th root of unity. Distinctness of the $\alpha_i$ forces distinctness of the $\omega_i$ (since $\alpha_1 \neq 0$ and multiplication by $\alpha_1^{-1}$ is injective). We now have $n$ distinct roots of the degree-$n$ polynomial $t^n - 1$, so these are *all* the $n$-th roots of unity in $L$. The set of $n$-th roots of unity forms a finite subgroup of the multiplicative group $L^\times$. A finite subgroup of the multiplicative group of a field is always cyclic (this is a standard result: if the subgroup had two distinct cyclic factors $\mathbb{Z}/a\mathbb{Z} \times \mathbb{Z}/b\mathbb{Z}$ with $a, b > 1$, the polynomial $t^{\operatorname{lcm}(a,b)} - 1$ would have more roots than its degree, contradicting the fact that a polynomial over a field has at most as many roots as its degree). A cyclic group of order $n$ has a generator, and that generator is by definition a primitive $n$-th root of unity. Why do we need $\operatorname{char} L \nmid n$ here? Because the group of $n$-th roots of unity has order $n$ only when $t^n - 1$ has $n$ distinct roots, which requires separability of $t^n - 1$. The derivative of $t^n - 1$ is $nt^{n-1}$; the same argument as in the previous step shows this is nonzero at any root of $t^n - 1$ precisely when $n \neq 0$ in $L$. [/guided] [/step] [step:Show $L = K(\mu)(\alpha)$ by expressing all roots in terms of $\mu$ and a single root $\alpha$] Fix a root $\alpha \in L$ of $t^n - \lambda$. The $n$ elements $\alpha, \mu\alpha, \mu^2\alpha, \ldots, \mu^{n-1}\alpha$ are pairwise distinct (since $\alpha \neq 0$ and $\mu^i \neq \mu^j$ for $0 \le i < j \le n - 1$, as $\mu$ has order $n$). Each is a root of $t^n - \lambda$: \begin{align*} (\mu^k \alpha)^n = \mu^{kn} \alpha^n = 1 \cdot \lambda = \lambda \quad \text{for each } k \in \{0, 1, \ldots, n-1\}. \end{align*} Since $t^n - \lambda$ has degree $n$, these $n$ elements exhaust its roots. Therefore every root of $t^n - \lambda$ lies in $K(\mu)(\alpha)$ (since $\mu \in K(\mu)$ and $\alpha \in K(\mu)(\alpha)$, the products $\mu^k \alpha$ lie in $K(\mu)(\alpha)$). The splitting field $L$ is the smallest extension of $K$ containing all roots of $t^n - \lambda$. Since $K(\mu)(\alpha) \subset L$ (both $\mu$ and $\alpha$ lie in $L$) and all roots of $t^n - \lambda$ lie in $K(\mu)(\alpha)$, the minimality of the splitting field forces $L = K(\mu)(\alpha)$. [/step] [step:Show $L/K(\mu)$ is Galois and construct an injective homomorphism $\operatorname{Gal}(L/K(\mu)) \hookrightarrow \mathbb{Z}/n\mathbb{Z}$] Since $\mu \in K(\mu)$, the previous step shows that $t^n - \lambda$ splits completely over $K(\mu)(\alpha) = L$. As established in the first step, $t^n - \lambda$ is separable. Therefore $L$ is the splitting field of the separable polynomial $t^n - \lambda \in K(\mu)[t]$ over $K(\mu)$, which implies $L/K(\mu)$ is a Galois extension. Every $\sigma \in \operatorname{Gal}(L/K(\mu))$ fixes $K(\mu)$ pointwise — in particular, $\sigma$ fixes the coefficients of $t^n - \lambda \in K(\mu)[t]$ — so $\sigma$ permutes the roots of $t^n - \lambda$. Since $\sigma(\alpha)$ is a root, there exists a unique $i(\sigma) \in \{0, 1, \ldots, n-1\}$ with $\sigma(\alpha) = \mu^{i(\sigma)}\alpha$. Define \begin{align*} \chi \colon \operatorname{Gal}(L/K(\mu)) &\to \mathbb{Z}/n\mathbb{Z} \\ \sigma &\mapsto i(\sigma) \bmod n. \end{align*} **$\chi$ is a group homomorphism.** Let $\sigma, \tau \in \operatorname{Gal}(L/K(\mu))$ with $\sigma(\alpha) = \mu^i \alpha$ and $\tau(\alpha) = \mu^j \alpha$. Then \begin{align*} (\sigma \circ \tau)(\alpha) = \sigma(\tau(\alpha)) = \sigma(\mu^j \alpha) = \sigma(\mu^j) \cdot \sigma(\alpha) = \mu^j \cdot \mu^i \alpha = \mu^{i+j}\alpha, \end{align*} where $\sigma(\mu^j) = \mu^j$ because $\mu \in K(\mu)$ and $\sigma$ fixes $K(\mu)$. Therefore $\chi(\sigma \circ \tau) = (i + j) \bmod n = \chi(\sigma) + \chi(\tau)$. **$\chi$ is injective.** Suppose $\chi(\sigma) = 0$, i.e., $\sigma(\alpha) = \mu^0 \alpha = \alpha$. Since $L = K(\mu)(\alpha)$ by the previous step, every element of $L$ is a $K(\mu)$-polynomial expression in $\alpha$. A $K(\mu)$-automorphism $\sigma$ that fixes $\alpha$ must fix every such expression (since $\sigma$ fixes $K(\mu)$ and fixes $\alpha$, it fixes $\alpha^k$ by multiplicativity for each $k$, hence fixes every $K(\mu)$-linear combination of powers of $\alpha$). Therefore $\sigma = \operatorname{id}_L$, and $\ker(\chi) = \{\operatorname{id}_L\}$. [guided] This step establishes the Galois-theoretic structure of $L/K(\mu)$. There are two ingredients: (i) $L/K(\mu)$ is Galois, and (ii) its Galois group embeds into $\mathbb{Z}/n\mathbb{Z}$. **Why is $L/K(\mu)$ Galois?** A finite extension is Galois if and only if it is the splitting field of a separable polynomial over the base field. The polynomial $t^n - \lambda$ lies in $K[t] \subset K(\mu)[t]$, is separable (shown in the first step), and splits completely over $L$ (by definition of the splitting field). Since $L = K(\mu)(\alpha)$ is generated over $K(\mu)$ by the roots of $t^n - \lambda$, $L$ is the splitting field of $t^n - \lambda$ over $K(\mu)$. **Constructing the homomorphism.** Every $\sigma \in \operatorname{Gal}(L/K(\mu))$ permutes the roots $\{\mu^k \alpha : 0 \le k \le n-1\}$ of $t^n - \lambda$ (because $\sigma$ fixes the coefficients $1, -\lambda \in K \subset K(\mu)$). In particular, $\sigma(\alpha)$ equals $\mu^{i(\sigma)}\alpha$ for a unique $i(\sigma) \in \{0, 1, \ldots, n - 1\}$. The map $\chi(\sigma) = i(\sigma) \bmod n$ is a homomorphism because of the computation \begin{align*} (\sigma \circ \tau)(\alpha) &= \sigma(\mu^j \alpha) = \mu^j \sigma(\alpha) = \mu^j \mu^i \alpha = \mu^{i+j}\alpha. \end{align*} The step $\sigma(\mu^j) = \mu^j$ is valid because $\mu \in K(\mu)$ and $\sigma$ is a $K(\mu)$-automorphism. If $\mu$ were not in $K(\mu)$ — which cannot happen by definition, but consider the analogous argument over $K$ when $\mu \notin K$ — then $\sigma$ might send $\mu$ to a different root of unity, and $\chi$ would fail to be a homomorphism. For injectivity, suppose $\sigma(\alpha) = \alpha$. Since $L = K(\mu)(\alpha)$, every element of $L$ is a $K(\mu)$-linear combination of $1, \alpha, \alpha^2, \ldots, \alpha^{d-1}$ (where $d = [L : K(\mu)]$). The automorphism $\sigma$ fixes each coefficient (in $K(\mu)$) and fixes each power $\alpha^k$ (since $\sigma(\alpha^k) = \sigma(\alpha)^k = \alpha^k$), hence fixes every element of $L$. Therefore $\sigma = \operatorname{id}_L$. [/guided] [/step] [step:Conclude that $\operatorname{Gal}(L/K(\mu))$ is cyclic and $[L : K(\mu)]$ divides $n$] The injective homomorphism $\chi \colon \operatorname{Gal}(L/K(\mu)) \hookrightarrow \mathbb{Z}/n\mathbb{Z}$ identifies $\operatorname{Gal}(L/K(\mu))$ with a subgroup of the cyclic group $\mathbb{Z}/n\mathbb{Z}$. Every subgroup of a cyclic group is cyclic, so $\operatorname{Gal}(L/K(\mu))$ is cyclic. Since $L/K(\mu)$ is Galois, $|\operatorname{Gal}(L/K(\mu))| = [L : K(\mu)]$. By Lagrange's theorem applied to the subgroup $\operatorname{im}(\chi) \subset \mathbb{Z}/n\mathbb{Z}$: \begin{align*} [L : K(\mu)] = |\operatorname{Gal}(L/K(\mu))| = |\operatorname{im}(\chi)| \mid |\mathbb{Z}/n\mathbb{Z}| = n. \end{align*} In particular, $L/K(\mu)$ is a cyclic (hence abelian, hence Galois) extension with $[L : K(\mu)] \mid n$. This completes Part (2). [guided] The embedding $\chi \colon \operatorname{Gal}(L/K(\mu)) \hookrightarrow \mathbb{Z}/n\mathbb{Z}$ has two consequences. **Cyclicity.** The image $\operatorname{im}(\chi)$ is a subgroup of $\mathbb{Z}/n\mathbb{Z}$. Since $\mathbb{Z}/n\mathbb{Z}$ is cyclic of order $n$, every subgroup of $\mathbb{Z}/n\mathbb{Z}$ is cyclic. (Concretely, the subgroups of $\mathbb{Z}/n\mathbb{Z}$ are $\langle d \bmod n \rangle \cong \mathbb{Z}/(n/d)\mathbb{Z}$ for each divisor $d$ of $n$.) Since $\chi$ is injective, $\operatorname{Gal}(L/K(\mu)) \cong \operatorname{im}(\chi)$, which is cyclic. **Degree divides $n$.** By Lagrange's theorem, $|\operatorname{im}(\chi)|$ divides $|\mathbb{Z}/n\mathbb{Z}| = n$. Since $L/K(\mu)$ is Galois, the degree of the extension equals the order of its Galois group: $[L : K(\mu)] = |\operatorname{Gal}(L/K(\mu))| = |\operatorname{im}(\chi)|$. Therefore $[L : K(\mu)]$ divides $n$. Note that the degree can be strictly less than $n$. For instance, if $\lambda$ is already an $n$-th power in $K(\mu)$ — say $\lambda = \beta^n$ for some $\beta \in K(\mu)$ — then $\alpha = \beta$ lies in $K(\mu)$, so $L = K(\mu)$ and $[L : K(\mu)] = 1$. The precise condition distinguishing $[L : K(\mu)] = n$ from $[L : K(\mu)] < n$ is addressed in Part (3). [/guided] [/step] [step:Characterise $[L : K(\mu)] = n$ via irreducibility of $t^n - \lambda$ over $K(\mu)$] Since $L = K(\mu)(\alpha)$, the degree of the extension is \begin{align*} [L : K(\mu)] = [K(\mu)(\alpha) : K(\mu)] = \deg(\operatorname{min}_{K(\mu)}(\alpha)), \end{align*} where $\operatorname{min}_{K(\mu)}(\alpha)$ denotes the minimal polynomial of $\alpha$ over $K(\mu)$. Since $\alpha$ is a root of $t^n - \lambda \in K[t] \subset K(\mu)[t]$, the minimal polynomial $\operatorname{min}_{K(\mu)}(\alpha)$ divides $t^n - \lambda$ in $K(\mu)[t]$. In particular, $\deg(\operatorname{min}_{K(\mu)}(\alpha)) \le n$. **Forward direction.** Assume $[L : K(\mu)] = n$. Then $\deg(\operatorname{min}_{K(\mu)}(\alpha)) = n = \deg(t^n - \lambda)$. Since $\operatorname{min}_{K(\mu)}(\alpha)$ divides $t^n - \lambda$ and both are monic polynomials of the same degree, $\operatorname{min}_{K(\mu)}(\alpha) = t^n - \lambda$. Since minimal polynomials are irreducible, $t^n - \lambda$ is irreducible over $K(\mu)$. **Backward direction.** Assume $t^n - \lambda$ is irreducible over $K(\mu)$. Since $\alpha$ is a root of the irreducible polynomial $t^n - \lambda \in K(\mu)[t]$, this polynomial is (up to scalar multiple, but both are monic) the minimal polynomial of $\alpha$ over $K(\mu)$: $\operatorname{min}_{K(\mu)}(\alpha) = t^n - \lambda$. Therefore $[L : K(\mu)] = \deg(t^n - \lambda) = n$. Combining both directions: \begin{align*} [L : K(\mu)] = n \quad &\iff \quad \deg(\operatorname{min}_{K(\mu)}(\alpha)) = n \\ &\iff \quad t^n - \lambda \text{ is irreducible over } K(\mu). \end{align*} This completes Part (3) and the proof. [guided] The final part connects the degree of the extension to an algebraic property of the polynomial $t^n - \lambda$. The bridge is the minimal polynomial of the generator $\alpha$. Recall the standard fact: for any algebraic element $\alpha$ over a field $F$, the degree $[F(\alpha) : F]$ equals the degree of the minimal polynomial $\operatorname{min}_F(\alpha)$. Moreover, $\operatorname{min}_F(\alpha)$ is the unique monic irreducible polynomial in $F[t]$ having $\alpha$ as a root, and it divides every polynomial in $F[t]$ that vanishes at $\alpha$. From Part (2), $L = K(\mu)(\alpha)$, so $[L : K(\mu)] = \deg(\operatorname{min}_{K(\mu)}(\alpha))$. Since $\alpha^n = \lambda$, the polynomial $t^n - \lambda$ vanishes at $\alpha$, so $\operatorname{min}_{K(\mu)}(\alpha)$ divides $t^n - \lambda$ in $K(\mu)[t]$. This gives the inequality $\deg(\operatorname{min}_{K(\mu)}(\alpha)) \le n$. Equality $\deg(\operatorname{min}_{K(\mu)}(\alpha)) = n$ holds precisely when $\operatorname{min}_{K(\mu)}(\alpha)$ and $t^n - \lambda$ have the same degree. Since $\operatorname{min}_{K(\mu)}(\alpha)$ divides $t^n - \lambda$ and both are monic, equal degree forces equality of the polynomials: $\operatorname{min}_{K(\mu)}(\alpha) = t^n - \lambda$. In particular, $t^n - \lambda$ must be irreducible over $K(\mu)$ (since minimal polynomials are always irreducible). Conversely, if $t^n - \lambda$ is irreducible over $K(\mu)$, then the only monic irreducible divisor of $t^n - \lambda$ in $K(\mu)[t]$ is $t^n - \lambda$ itself. Since $\operatorname{min}_{K(\mu)}(\alpha)$ is a monic irreducible divisor of $t^n - \lambda$, it must equal $t^n - \lambda$, and the degree is $n$. This equivalence is quite natural: irreducibility of $t^n - \lambda$ over $K(\mu)$ means that adjoining $\alpha$ (a root of $t^n - \lambda$) produces an extension of the maximum possible degree $n$. Reducibility means that $\alpha$ satisfies a polynomial relation of degree less than $n$ over $K(\mu)$, so the extension is smaller. [/guided] [/step]