 Solution

By the preceding argument take a nonzero integer $m\in \mathfrak{a}$. The ideal property gives $m{\mathcal{O}}_L\subseteq \mathfrak{a}$. If $d=[L:\mathbb{Q}]$, an integral basis identifies the additive group of ${\mathcal{O}}_L$ with ${\mathbb{Z}}^d$, soThe natural projection onto ${\mathcal{O}}_L/\mathfrak{a}$ is surjective. Thus the latter is an abelian group of order at most $|m{|}^d$, provingIn particular the ideal is a full-rank integer lattice; this permits the integral-basis argument in the next part.