= 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 $\mathbb Z^d$, so
$$
\mathcal O_L/m\mathcal O_L\cong(\mathbb Z/|m|\mathbb Z)^d.
$$
The natural projection onto $\mathcal O_L/\mathfrak a$ is surjective. Thus the latter is an <abelian group> of order at most $|m|^d$, proving
$$
\boxed{\mathcal O_L/\mathfrak a\text{ is finite abelian}}.
$$
In particular the <ideal> is a full-rank integer lattice; this permits the integral-basis argument in the next part.
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