= Solution
For a nonzero <ideal> $I\subset O_K$, its norm is
$$
N(I)=|O_K/I|.
$$
For $d=[K:\mathbb Q]$ and embeddings $\sigma_1,\ldots,\sigma_d$, define
$$
\operatorname{disc}(\alpha_1,\ldots,\alpha_d)
=\det(\sigma_i(\alpha_j))^2.
$$
The field discriminant is the discriminant of any <integral> <basis> of $O_K$. If $\alpha_1,\ldots,\alpha_d$ is a $\mathbb Z$-basis of $I$, then
$$
N(I)^2=\frac{|\operatorname{disc}(\alpha_1,\ldots,\alpha_d)|}{|D_K|}.
$$
Multiplication by $\alpha$ on an <integral> <basis> has integer <matrix> $M_\alpha$. Its image lattice is $\alpha O_K$, so
$$
N(\alpha O_K)=|\det M_\alpha|.
$$
The <determinant> of this $\mathbb Q$-linear map is the field norm $N_{K/\mathbb Q}(\alpha)$, proving
$$
\boxed{N(\alpha O_K)=|N_{K/\mathbb Q}(\alpha)|.}
$$
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