[proofplan] We show that $K$ is the splitting field of the separable polynomial $t^n - a$ over $k$, which makes $K/k$ Galois. Every $k$-automorphism of $K$ must send $\alpha$ to another $n$-th root of $a$, and since $k$ contains all $n$-th roots of unity, these roots are $\zeta^j \alpha$ for $j = 0, \ldots, n-1$. We define a map from $\operatorname{Gal}(K/k)$ to $\mathbb{Z}/n\mathbb{Z}$ by tracking the exponent $j$, verify it is an injective group homomorphism, and conclude that $\operatorname{Gal}(K/k)$ is cyclic of order dividing $n$. [/proofplan] [step:Show that $K$ is the splitting field of $t^n - a$ over $k$] Let $\zeta \in k$ be a primitive $n$-th root of unity, which exists by hypothesis. The roots of $t^n - a$ in an algebraic closure of $k$ are $\zeta^j \alpha$ for $j = 0, 1, \ldots, n-1$. Since $\zeta \in k$ and $\alpha \in K = k(\alpha)$, every root $\zeta^j \alpha$ lies in $K$. Therefore $t^n - a$ splits completely over $K$, and $K = k(\alpha)$ is generated over $k$ by one root, so $K$ is the splitting field of $t^n - a$ over $k$. Since $\operatorname{char} k = 0$, the polynomial $t^n - a$ is separable: its formal derivative is $n t^{n-1}$, which shares no common root with $t^n - a$ (as $n \neq 0$ in $k$ and $a \neq 0$ since $\alpha^n = a$ with $\alpha$ algebraic). By the [Splitting Fields of Separable Polynomials Are Galois](/theorems/1271) theorem, $K/k$ is a Galois extension. [guided] We need to show that $K$ is the splitting field of the polynomial $t^n - a$ and that this extension is Galois. The key observation is that $k$ already contains all the $n$-th roots of unity. Let $\zeta \in k$ be a primitive $n$-th root of unity. The roots of $t^n - a$ in an algebraic closure are precisely $\zeta^j \alpha$ for $j = 0, 1, \ldots, n-1$: each satisfies $(\zeta^j \alpha)^n = \zeta^{jn} \alpha^n = 1 \cdot a = a$, and there are $n$ distinct such roots (since $\zeta$ is primitive), which accounts for all roots of a degree-$n$ polynomial. Since $\zeta \in k \subset K$ and $\alpha \in K$, every product $\zeta^j \alpha$ lies in $K$. So $t^n - a$ splits completely over $K$. Moreover, $K = k(\alpha)$ is generated by a single root of $t^n - a$, and all other roots are $k$-multiples of $\alpha$, so no smaller field between $k$ and $K$ can contain all roots. Therefore $K$ is the splitting field of $t^n - a$ over $k$. Is $t^n - a$ separable? In characteristic zero, $n \neq 0$ in $k$, so the formal derivative $nt^{n-1}$ is not the zero polynomial. The only common root of $t^n - a$ and $nt^{n-1}$ would be $t = 0$, but $0^n - a = -a \neq 0$ (if $a = 0$ then $K = k$ and the result is trivial). So $t^n - a$ has no repeated roots, hence is separable. By the [Splitting Fields of Separable Polynomials Are Galois](/theorems/1271) theorem, the splitting field of a separable polynomial over $k$ is a Galois extension. Therefore $K/k$ is Galois. [/guided] [/step] [step:Define the map $\chi: \operatorname{Gal}(K/k) \to \mathbb{Z}/n\mathbb{Z}$ by tracking the action on $\alpha$] Let $\sigma \in \operatorname{Gal}(K/k)$. Since $\sigma$ is a $k$-automorphism and $\alpha^n = a \in k$, we have $\sigma(\alpha)^n = \sigma(\alpha^n) = \sigma(a) = a$. So $\sigma(\alpha)$ is an $n$-th root of $a$, hence $\sigma(\alpha) = \zeta^{j(\sigma)} \alpha$ for a unique $j(\sigma) \in \{0, 1, \ldots, n-1\}$. Define the map \begin{align*} \chi: \operatorname{Gal}(K/k) &\to \mathbb{Z}/n\mathbb{Z} \\ \sigma &\mapsto \overline{j(\sigma)}. \end{align*} Since $K = k(\alpha)$, every element of $K$ is a $k$-polynomial in $\alpha$, so $\sigma$ is completely determined by $\sigma(\alpha)$, i.e., by $j(\sigma)$. [guided] Every $\sigma \in \operatorname{Gal}(K/k)$ fixes $k$ pointwise. Since $\alpha$ satisfies $\alpha^n = a \in k$, applying $\sigma$ gives $\sigma(\alpha)^n = a$. So $\sigma(\alpha)$ must be one of the $n$-th roots of $a$, which we identified as $\{\zeta^j \alpha : 0 \le j \le n-1\}$. This assigns to $\sigma$ a unique integer $j(\sigma) \in \{0, 1, \ldots, n-1\}$ with $\sigma(\alpha) = \zeta^{j(\sigma)} \alpha$. Why is $\sigma$ determined by $j(\sigma)$? Because $K = k(\alpha)$, so every element of $K$ is a $k$-linear combination of $1, \alpha, \alpha^2, \ldots, \alpha^{d-1}$ where $d = [K:k]$. A $k$-automorphism that fixes $k$ is completely determined by where it sends $\alpha$. Two automorphisms with the same $j$-value agree on $\alpha$ and hence on all of $K$. We define $\chi(\sigma) = \overline{j(\sigma)} \in \mathbb{Z}/n\mathbb{Z}$. [/guided] [/step] [step:Verify that $\chi$ is an injective group homomorphism] Let $\sigma, \tau \in \operatorname{Gal}(K/k)$. Then \begin{align*} (\sigma \circ \tau)(\alpha) = \sigma(\zeta^{j(\tau)} \alpha) = \zeta^{j(\tau)} \sigma(\alpha) = \zeta^{j(\tau)} \zeta^{j(\sigma)} \alpha = \zeta^{j(\sigma) + j(\tau)} \alpha, \end{align*} where the second equality uses $\zeta^{j(\tau)} \in k$ so $\sigma(\zeta^{j(\tau)}) = \zeta^{j(\tau)}$. Therefore $j(\sigma \circ \tau) \equiv j(\sigma) + j(\tau) \pmod{n}$, which gives $\chi(\sigma \circ \tau) = \chi(\sigma) + \chi(\tau)$. So $\chi$ is a group homomorphism. For injectivity: if $\chi(\sigma) = \overline{0}$, then $j(\sigma) \equiv 0 \pmod{n}$, so $\sigma(\alpha) = \zeta^0 \alpha = \alpha$. Since $\sigma$ fixes $\alpha$ and fixes $k$, and $K = k(\alpha)$, $\sigma$ is the identity. Thus $\ker \chi = \{e\}$, and $\chi$ is injective. [guided] We verify the homomorphism property. Let $\sigma, \tau \in \operatorname{Gal}(K/k)$ with $\sigma(\alpha) = \zeta^{j(\sigma)}\alpha$ and $\tau(\alpha) = \zeta^{j(\tau)}\alpha$. Computing the composition: \begin{align*} (\sigma \circ \tau)(\alpha) = \sigma(\tau(\alpha)) = \sigma(\zeta^{j(\tau)} \alpha). \end{align*} The crucial point is that $\zeta^{j(\tau)} \in k$, so $\sigma$ fixes it: $\sigma(\zeta^{j(\tau)}) = \zeta^{j(\tau)}$. Therefore \begin{align*} (\sigma \circ \tau)(\alpha) = \zeta^{j(\tau)} \cdot \sigma(\alpha) = \zeta^{j(\tau)} \cdot \zeta^{j(\sigma)} \alpha = \zeta^{j(\sigma) + j(\tau)} \alpha. \end{align*} Reading off the exponent: $j(\sigma \circ \tau) \equiv j(\sigma) + j(\tau) \pmod{n}$, so $\chi(\sigma \circ \tau) = \chi(\sigma) + \chi(\tau)$ in $\mathbb{Z}/n\mathbb{Z}$. For injectivity, suppose $\chi(\sigma) = \overline{0}$. Then $j(\sigma) \equiv 0 \pmod{n}$, meaning $j(\sigma) = 0$ (since $0 \le j(\sigma) \le n-1$). So $\sigma(\alpha) = \alpha$, and since $\sigma$ also fixes $k$ and $K = k(\alpha)$, $\sigma$ must be the identity automorphism. The kernel is trivial, so $\chi$ is injective. [/guided] [/step] [step:Conclude that $\operatorname{Gal}(K/k)$ is cyclic of order dividing $n$] Since $\chi: \operatorname{Gal}(K/k) \hookrightarrow \mathbb{Z}/n\mathbb{Z}$ is an injective group homomorphism, $\operatorname{Gal}(K/k)$ is isomorphic to a subgroup of the cyclic group $\mathbb{Z}/n\mathbb{Z}$. Every subgroup of a cyclic group is cyclic, so $\operatorname{Gal}(K/k)$ is cyclic. Its order satisfies $|\operatorname{Gal}(K/k)| \mid n$, since the order of any subgroup of $\mathbb{Z}/n\mathbb{Z}$ divides $n$. Since $K/k$ is Galois, $|\operatorname{Gal}(K/k)| = [K:k]$ by the [Fundamental Theorem of Galois Theory](/theorems/1274). Therefore $\operatorname{Gal}(K/k)$ is cyclic of order $[K:k]$, and $[K:k]$ divides $n$. [/step]