= Solution
The automorphisms $1,\sigma,\ldots,\sigma^{n-1}$ are distinct and hence linearly independent as maps $L\to L$ by the <linear independence of distinct field embeddings>. Therefore the operator
$$
T=\sum_{j=0}^{n-1}\zeta^{-j}\sigma^j
$$
is not the zero map. Choose $\beta\in L$ for which $\alpha=T(\beta)\ne0$. Reindexing the sum and using $\sigma^n=1$ gives
$$
\sigma(\alpha)
=\sum_{j=0}^{n-1}\zeta^{-j}\sigma^{j+1}(\beta)
=\zeta\alpha.
$$
This is the <Lagrange resolvent eigenvector for a cyclic field automorphism>.
Let $F=L^\sigma$ be the <fixed field>. Since $\sigma(\alpha^n)=\alpha^n$, the element $\alpha$ is a root of
$$
X^n-\alpha^n\in F[X].
$$
Its orbit under $\langle\sigma\rangle$ is
$$
\alpha,\zeta\alpha,\ldots,\zeta^{n-1}\alpha,
$$
and these $n$ elements are distinct because $\alpha\ne0$ and $\zeta$ is a <primitive root of unity>. Hence the <minimal polynomial> has at least $n$ distinct roots. It also divides the displayed degree-$n$ polynomial, so
$$
\boxed{m_{\alpha,F}(X)=X^n-\alpha^n}.
$$
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