Lemma 9.24.1 (Kummer extensions). Let $L/K$ be a Galois extension of fields whose Galois group is $\mathbf{Z}/n\mathbf{Z}$. Assume moreover that the characteristic of $K$ is prime to $n$ and that $K$ contains a primitive $n$th root of $1$. Then $L = K[z]$ with $z^ n \in K$.

Proof. Let $\zeta \in K$ be a primitive $n$th root of $1$. Let $\sigma$ be a generator of $\text{Gal}(L/K)$. Consider $\sigma : L \to L$ as a $K$-linear operator. Note that $\sigma ^ n - 1 = 0$ as a linear operator. Applying linear independence of characters (Lemma 9.13.1), we see that there cannot be a polynomial over $K$ of degree $< n$ annihilating $\sigma$. Hence the minimal polynomial of $\sigma$ as a linear operator is $x^ n - 1$. Since $\zeta$ is a root of $x^ n - 1$ by linear algebra there is a $0 \neq z \in L$ such that $\sigma (z) = \zeta z$. This $z$ satisfies $z^ n \in K$ because $\sigma (z^ n) = (\zeta z)^ n = z^ n$. Moreover, we see that $z, \sigma (z), \ldots , \sigma ^{n - 1}(z) = z, \zeta z, \ldots \zeta ^{n - 1} z$ are pairwise distinct which guarantees that $z$ generates $L$ over $K$. Hence $L = K[z]$ as required. $\square$

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