= Solution
Because $\operatorname{char}K=0$, the finite extension $M/K$ is separable. By the <primitive element theorem>, write $M=K(\theta)$. The <minimal polynomial> of $\theta$ over $K$ has a root in the <algebraically closed field> $L$, so sending $\theta$ to that root defines a $K$-<field embedding> $\iota:M\hookrightarrow L$.
The cyclic group $\langle\sigma\rangle$ acts faithfully on $L$ after replacing $d$ by the order of $\sigma$. The <Artin fixed-field theorem> gives
$$
[L:K]=|\langle\sigma\rangle|
$$
and says that $L/K$ is a finite Galois extension with cyclic Galois group $\langle\sigma\rangle$. The image $E=\iota(M)$ is an <intermediate field>. Every subgroup of a cyclic group is normal, so the <normal subextension criterion> shows that $E/K$ is Galois, and its Galois group is a quotient of $\langle\sigma\rangle$, hence cyclic. Transporting this structure through $\iota$ proves that $M/K$ is Galois with cyclic Galois group. This is the <finite extensions of the fixed field of a finite-order automorphism of an algebraically closed field> theorem.
Algebraic closedness is necessary. Take $L=\mathbb Q$ and $\sigma=\operatorname{id}$, so $K=L^\sigma=\mathbb Q$, and let $M=\mathbb Q(\sqrt[3]{2})$. The extension has degree three but is not normal, since the two nonreal roots of $X^3-2$ do not belong to $M$. It is therefore not Galois.
Back to article page
