= Solution
If $a=0$, the <splitting field> is $K$ and its <Galois group> is trivial. Otherwise choose a root $\alpha$ and a primitive $n$th root of unity $\zeta\in K$. Every root is $\zeta^j\alpha$, so $L=K(\alpha)$. Each automorphism is determined by $\sigma(\alpha)/\alpha\in\mu_n$. This is an injective group homomorphism, since the <roots of unity> are fixed by every automorphism. A subgroup of the <cyclic group> $\mu_n$ is cyclic. Thus \b[$\operatorname{Gal}(L/K)$ is cyclic], with order dividing $n$.
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