= Solution
Let $\sigma(\alpha)=\xi_d\alpha$ with $\alpha\ne0$. Then
$$
\sigma^j(\alpha)=\xi_d^j\alpha
\qquad(0\le j<d),
$$
and these $d$ elements are distinct because $\xi_d$ is a <primitive root of unity>. Thus the stabilizer of $\alpha$ in the cyclic <Galois group> is trivial. The orbit-stabilizer form of the <Fundamental theorem of Galois theory> gives
$$
[K(\alpha):K]=d=[L:K],
$$
so
$$
\boxed{L=K(\alpha)}.
$$
Moreover,
$$
\sigma(\alpha^d)=\xi_d^d\alpha^d=\alpha^d.
$$
The fixed field of the whole Galois group is $K$, and therefore
$$
\boxed{\alpha^d\in K}.
$$
This is the basic eigenvector mechanism behind a cyclic <Kummer extension>.
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