The Stacks project

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. $\square$