[proofplan] The proof divides into three parts matching the three conclusions. For Part (1), we fix a root $\alpha \in L$ of $t^n - \lambda$ and observe that every root of $t^n - \lambda$ has the form $\zeta^k \alpha$ for $0 \le k \le n-1$; since $\zeta \in K$ by hypothesis, all roots lie in $K(\alpha)$, so $L = K(\alpha)$. For Part (2), we construct an injective group homomorphism from $\operatorname{Gal}(L/K)$ into $\mathbb{Z}/n\mathbb{Z}$ by sending each $\sigma$ to the unique $i$ with $\sigma(\alpha) = \zeta^i \alpha$. Injectivity follows from the fact that $\alpha$ generates $L$ over $K$, the homomorphism property follows from the computation $(\sigma \circ \tau)(\alpha) = \zeta^{i+j}\alpha$, and the embedding into $\mathbb{Z}/n\mathbb{Z}$ forces $\operatorname{Gal}(L/K)$ to be cyclic. Since $L/K$ is separable (the roots of $t^n - \lambda$ are distinct) and normal (all roots lie in $L$), the extension is Galois, and $[L:K] = |\operatorname{Gal}(L/K)|$ divides $n$. For Part (3), we use $L = K(\alpha)$ to identify $[L:K] = \deg(\operatorname{min}_K(\alpha))$, which equals $n$ precisely when $t^n - \lambda$ is itself the minimal polynomial of $\alpha$. [/proofplan] [step:Identify all roots of $t^n - \lambda$ and show $L = K(\alpha)$] Let $\alpha \in L$ be a root of $t^n - \lambda$, so that $\alpha^n = \lambda$. Since $\zeta \in K$ is a primitive $n$-th root of unity, the $n$ elements $\alpha, \zeta\alpha, \zeta^2\alpha, \ldots, \zeta^{n-1}\alpha$ are pairwise distinct (as $\alpha \neq 0$ because $\lambda \in K^\times$, and $\zeta^i \neq \zeta^j$ for $0 \le i < j \le n-1$). Each is a root of $t^n - \lambda$: \begin{align*} (\zeta^k \alpha)^n = \zeta^{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 $n$ and admits $n$ distinct roots, these are all of its roots. In particular, $t^n - \lambda$ is separable over $K$. Since $\zeta \in K$, each root $\zeta^k \alpha = \zeta^k \cdot \alpha$ is a $K$-multiple of $\alpha$ and therefore lies in $K(\alpha)$. Consequently, $t^n - \lambda$ splits completely over $K(\alpha)$. Since $L$ is defined as the splitting field of $t^n - \lambda$ over $K$ and $K(\alpha)$ is a subfield of $L$ over which $t^n - \lambda$ splits, the minimality of the splitting field forces $L = K(\alpha)$. [guided] We must show that adjoining a single root $\alpha$ of $t^n - \lambda$ already produces the entire splitting field. This would fail if the other roots required additional generators, but the presence of $\zeta$ in $K$ prevents this. Fix a root $\alpha \in L$ of $t^n - \lambda$. Since $\lambda \in K^\times$, we have $\alpha^n = \lambda \neq 0$, so $\alpha \neq 0$. We first enumerate all roots of $t^n - \lambda$. For each $k \in \{0, 1, \ldots, n-1\}$, the element $\zeta^k \alpha$ satisfies \begin{align*} (\zeta^k \alpha)^n = (\zeta^k)^n \cdot \alpha^n = (\zeta^n)^k \cdot \lambda = 1^k \cdot \lambda = \lambda, \end{align*} so $\zeta^k \alpha$ is a root of $t^n - \lambda$. These $n$ roots are pairwise distinct: if $\zeta^i \alpha = \zeta^j \alpha$ for $0 \le i < j \le n-1$, then $\zeta^{j-i} = 1$ (dividing by $\alpha \neq 0$), contradicting the fact that $\zeta$ has order $n$ and $0 < j - i < n$. Since $t^n - \lambda$ has degree $n$, these $n$ elements exhaust its roots. In particular, $t^n - \lambda$ has no repeated roots, so it is separable. Now, since $\zeta \in K$ by hypothesis, the product $\zeta^k \cdot \alpha$ is a $K$-scalar multiple of $\alpha$ and thus lies in $K(\alpha)$ for every $k$. Therefore every root of $t^n - \lambda$ belongs to $K(\alpha)$, which means $t^n - \lambda$ splits completely over $K(\alpha)$. The splitting field $L$ is, by definition, the smallest subfield of an algebraic closure containing $K$ and all roots of $t^n - \lambda$. Since $K(\alpha) \subset L$ (as $\alpha \in L$) and all roots lie in $K(\alpha)$, no proper subfield of $K(\alpha)$ containing $K$ can contain all roots. Hence $L = K(\alpha)$. Why does this argument depend on $\zeta \in K$? If $\zeta \notin K$, the roots $\zeta^k \alpha$ would not lie in $K(\alpha)$ in general, and the splitting field could be strictly larger than $K(\alpha)$. For example, the splitting field of $t^3 - 2$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt[3]{2}, \zeta_3)$, which has degree 6 over $\mathbb{Q}$, not 3. [/guided] [/step] [step:Show $L/K$ is Galois and construct an injective homomorphism $\operatorname{Gal}(L/K) \hookrightarrow \mathbb{Z}/n\mathbb{Z}$] Since $L$ is the splitting field of the separable polynomial $t^n - \lambda \in K[t]$ (separability was established in the previous step), the extension $L/K$ is Galois. Every $\sigma \in \operatorname{Gal}(L/K)$ permutes the roots of $t^n - \lambda$ and hence maps $\alpha$ to some root $\zeta^{i(\sigma)}\alpha$ for a unique $i(\sigma) \in \{0, 1, \ldots, n-1\}$. Define the map \begin{align*} \chi: \operatorname{Gal}(L/K) &\to \mathbb{Z}/n\mathbb{Z} \\ \sigma &\mapsto i(\sigma) \bmod n, \end{align*} where $i(\sigma)$ is the unique element of $\{0, 1, \ldots, n-1\}$ satisfying $\sigma(\alpha) = \zeta^{i(\sigma)}\alpha$. **$\chi$ is a group homomorphism.** Let $\sigma, \tau \in \operatorname{Gal}(L/K)$ with $\sigma(\alpha) = \zeta^i \alpha$ and $\tau(\alpha) = \zeta^j \alpha$. Then \begin{align*} (\sigma \circ \tau)(\alpha) = \sigma(\tau(\alpha)) = \sigma(\zeta^j \alpha) = \zeta^j \sigma(\alpha) = \zeta^j \cdot \zeta^i \alpha = \zeta^{i+j}\alpha, \end{align*} where the third equality uses $\sigma(\zeta^j) = \zeta^j$ since $\zeta \in K$ and $\sigma$ fixes $K$. Therefore $\chi(\sigma \circ \tau) = i + j \bmod n = \chi(\sigma) + \chi(\tau)$. **$\chi$ is injective.** Suppose $\chi(\sigma) = 0$, i.e., $\sigma(\alpha) = \zeta^0 \alpha = \alpha$. Since $L = K(\alpha)$ by Part (1), every element of $L$ is a $K$-polynomial expression in $\alpha$. A $K$-automorphism $\sigma$ that fixes $\alpha$ must fix every such polynomial expression, so $\sigma = \operatorname{id}_L$. Hence $\ker(\chi) = \{\operatorname{id}_L\}$. [guided] The extension $L/K$ is Galois because $L$ is the splitting field of a separable polynomial over $K$. Recall that separability was verified in the previous step: the $n$ roots $\zeta^k \alpha$ are pairwise distinct, so $t^n - \lambda$ has no repeated roots. We now construct a concrete group homomorphism from $\operatorname{Gal}(L/K)$ into $\mathbb{Z}/n\mathbb{Z}$. Since every $K$-automorphism $\sigma$ of $L$ must permute the roots of $t^n - \lambda$ (as $\sigma$ fixes the coefficients of $t^n - \lambda \in K[t]$), we have $\sigma(\alpha) \in \{\alpha, \zeta\alpha, \ldots, \zeta^{n-1}\alpha\}$. Hence there exists a unique $i(\sigma) \in \{0, 1, \ldots, n-1\}$ with $\sigma(\alpha) = \zeta^{i(\sigma)}\alpha$. Define $\chi(\sigma) = i(\sigma) \bmod n$. We verify that $\chi$ is a group homomorphism. Take $\sigma, \tau \in \operatorname{Gal}(L/K)$ with $\sigma(\alpha) = \zeta^i \alpha$ and $\tau(\alpha) = \zeta^j \alpha$. Computing $(\sigma \circ \tau)(\alpha)$: \begin{align*} (\sigma \circ \tau)(\alpha) &= \sigma(\tau(\alpha)) = \sigma(\zeta^j \alpha). \end{align*} Since $\zeta \in K$ and $\sigma$ is a $K$-automorphism, $\sigma$ fixes $\zeta^j$. Therefore \begin{align*} \sigma(\zeta^j \alpha) = \zeta^j \cdot \sigma(\alpha) = \zeta^j \cdot \zeta^i \alpha = \zeta^{i+j}\alpha. \end{align*} Hence $\chi(\sigma \circ \tau) = (i + j) \bmod n = \chi(\sigma) + \chi(\tau)$ in $\mathbb{Z}/n\mathbb{Z}$. Why is the step $\sigma(\zeta^j \alpha) = \zeta^j \sigma(\alpha)$ valid? Because $\sigma$ is a field homomorphism, $\sigma(\zeta^j \alpha) = \sigma(\zeta^j)\sigma(\alpha)$. The crucial point is that $\zeta^j \in K$ and $\sigma$ fixes $K$ pointwise, so $\sigma(\zeta^j) = \zeta^j$. If $\zeta$ were not in $K$, $\sigma$ could move $\zeta$ to a different root of unity, and the map $\chi$ would not be a homomorphism. For injectivity, suppose $\chi(\sigma) = 0$, meaning $\sigma(\alpha) = \alpha$. Since $L = K(\alpha)$, every element of $L$ can be written as $\sum_{k=0}^{d-1} a_k \alpha^k$ with $a_k \in K$ and $d = [L:K]$. A $K$-automorphism $\sigma$ that fixes $\alpha$ fixes each $\alpha^k$ (by multiplicativity) and each $a_k$ (since $a_k \in K$), hence fixes $\sum a_k \alpha^k$. Therefore $\sigma = \operatorname{id}_L$, so $\ker(\chi) = \{\operatorname{id}_L\}$. Note that injectivity depends on $L = K(\alpha)$: we need $\alpha$ to generate the entire extension. If there were elements of $L$ not expressible in terms of $\alpha$, fixing $\alpha$ would not force $\sigma$ to be the identity. [/guided] [/step] [step:Conclude that $\operatorname{Gal}(L/K)$ is cyclic and $[L:K]$ divides $n$] The injective homomorphism $\chi: \operatorname{Gal}(L/K) \hookrightarrow \mathbb{Z}/n\mathbb{Z}$ identifies $\operatorname{Gal}(L/K)$ with a subgroup of the cyclic group $\mathbb{Z}/n\mathbb{Z}$. Every subgroup of a cyclic group is cyclic, so $\operatorname{Gal}(L/K)$ is cyclic. Since $L/K$ is Galois, $|\operatorname{Gal}(L/K)| = [L:K]$. The order of $\operatorname{Gal}(L/K)$, viewed as a subgroup of $\mathbb{Z}/n\mathbb{Z}$ via $\chi$, divides $|\mathbb{Z}/n\mathbb{Z}| = n$ by Lagrange's theorem. Therefore $[L:K]$ divides $n$. This completes Part (2). [guided] The previous step produced an injective group homomorphism $\chi: \operatorname{Gal}(L/K) \hookrightarrow \mathbb{Z}/n\mathbb{Z}$. This embedding has two immediate consequences. First, $\operatorname{Gal}(L/K) \cong \operatorname{im}(\chi)$, which is a subgroup of $\mathbb{Z}/n\mathbb{Z}$. The group $\mathbb{Z}/n\mathbb{Z}$ is cyclic of order $n$, and every subgroup of a cyclic group is cyclic. (Concretely, the subgroups of $\mathbb{Z}/n\mathbb{Z}$ are exactly the groups $\langle d \bmod n \rangle$ for divisors $d$ of $n$, each of which is cyclic of order $n/d$.) Therefore $\operatorname{Gal}(L/K)$ is cyclic. Second, by Lagrange's theorem applied to the subgroup $\operatorname{im}(\chi) \subset \mathbb{Z}/n\mathbb{Z}$, the order $|\operatorname{im}(\chi)|$ divides $n$. Since $\chi$ is injective, $|\operatorname{Gal}(L/K)| = |\operatorname{im}(\chi)|$, and since $L/K$ is Galois, $[L:K] = |\operatorname{Gal}(L/K)|$. Combining these equalities, $[L:K]$ divides $n$. Note what this does not say: it does not assert that $[L:K] = n$. The degree can be any divisor of $n$. For example, $t^4 - 1$ over $\mathbb{Q}(i)$ has $\alpha = 1$ as a root, so $L = \mathbb{Q}(i)$ and $[L:K] = 1$, which divides 4 but does not equal 4. More interestingly, $t^4 - 4$ over $\mathbb{Q}(i)$ has $\alpha = \sqrt{2}$ as a root, and $[\mathbb{Q}(i, \sqrt{2}) : \mathbb{Q}(i)] = 2$, which divides 4 but is not 4. The precise condition for $[L:K] = n$ is the content of Part (3). [/guided] [/step] [step:Characterise when $[L:K] = n$ via irreducibility of $t^n - \lambda$] Since $L = K(\alpha)$ by Part (1), the degree of the extension is $[L:K] = [K(\alpha):K] = \deg(\operatorname{min}_K(\alpha))$, where $\operatorname{min}_K(\alpha)$ is the minimal polynomial of $\alpha$ over $K$. Since $\alpha$ is a root of $t^n - \lambda \in K[t]$, the minimal polynomial $\operatorname{min}_K(\alpha)$ divides $t^n - \lambda$ in $K[t]$. Therefore $\deg(\operatorname{min}_K(\alpha)) \le n$, with equality if and only if $t^n - \lambda$ is itself irreducible over $K$ (since $\operatorname{min}_K(\alpha)$ is irreducible and divides $t^n - \lambda$, having the same degree forces $t^n - \lambda = c \cdot \operatorname{min}_K(\alpha)$ for some $c \in K^\times$; as both are monic, $c = 1$). Consequently: \begin{align*} [L:K] = n \quad &\iff \quad \deg(\operatorname{min}_K(\alpha)) = n \\ &\iff \quad t^n - \lambda \text{ is irreducible over } K. \end{align*} This completes Part (3) and the proof. [guided] We now connect the degree $[L:K]$ to the irreducibility of $t^n - \lambda$. The bridge is the minimal polynomial of $\alpha$ over $K$. Recall that for any algebraic element $\alpha$ over a field $K$, the degree $[K(\alpha):K]$ equals the degree of the minimal polynomial $\operatorname{min}_K(\alpha)$. Part (1) gives $L = K(\alpha)$, so \begin{align*} [L:K] = [K(\alpha):K] = \deg(\operatorname{min}_K(\alpha)). \end{align*} Since $\alpha$ is a root of $t^n - \lambda \in K[t]$, the minimal polynomial $\operatorname{min}_K(\alpha)$ divides $t^n - \lambda$ in $K[t]$ (because $\operatorname{min}_K(\alpha)$ is the monic polynomial of least degree in $K[t]$ having $\alpha$ as a root, and it divides every polynomial in $K[t]$ that vanishes at $\alpha$). In particular, $\deg(\operatorname{min}_K(\alpha)) \le n$. **Forward direction:** Assume $[L:K] = n$. Then $\deg(\operatorname{min}_K(\alpha)) = n = \deg(t^n - \lambda)$. Since $\operatorname{min}_K(\alpha)$ divides $t^n - \lambda$ and both are monic polynomials of the same degree, $\operatorname{min}_K(\alpha) = t^n - \lambda$. Since minimal polynomials are irreducible over $K$, the polynomial $t^n - \lambda$ is irreducible over $K$. **Backward direction:** Assume $t^n - \lambda$ is irreducible over $K$. Since $\alpha$ is a root of $t^n - \lambda$ and $t^n - \lambda$ is irreducible, $t^n - \lambda$ is (up to scaling by a unit, but both are monic) the minimal polynomial of $\alpha$ over $K$. Hence $\operatorname{min}_K(\alpha) = t^n - \lambda$, giving $[L:K] = \deg(t^n - \lambda) = n$. Thus $[L:K] = n$ if and only if $t^n - \lambda$ is irreducible over $K$, which completes Part (3) and the entire proof. [/guided] [/step]