Solution
= Solution
The equality of residue-field sizes says that the residue degree $f(Q/P)$ is one. Since $Q=\alpha O_L$, the principal <ideal> factorization of $\alpha$ has valuation one at $Q$ and zero at every other prime. Taking the relative <ideal> norm gives
$$
N_{L/K}(\alpha)O_K=N_{L/K}(\alpha O_L)
=N_{L/K}(Q)=P^{f(Q/P)}=P.
$$
\b[Thus $P=N_{L/K}(\alpha)O_K$.]