I'm currently working through
Don Bernard Zagier - Zetafunktionen und quadratische Körper (1981)
(German for: Zeta functions and quadratic fields).
$K = \mathbb{Q}(\sqrt{d})$ is a quadratic number field and $\mathcal{O}_K$ it's ring of integers. Zagier (page 88) defines an Ideal of $\mathcal{O}_K$ as a subgroup $\mathfrak{a} \subset \mathcal{O}_K$ with $\mathcal{O}_K \cdot \mathfrak{a} = \mathfrak{a}$. That means: $$ \lambda \in \mathcal{O}_K, \alpha \in \mathfrak{a} \implies \lambda\alpha \in \mathfrak{a}\hspace{45px}(1) $$ Later on (page 89) he defines a Fractional Ideal as finitely a generated subgroup $\mathfrak{a} \subset K$ of $K$ (instead of $\mathcal{O}_K$) for which (1) holds.
Now I'm a little bit confused. Does he really mean (1) or does he mean:
$$ \lambda \in K \,\,(\text{instead of } \mathcal{O}_K), \alpha \in \mathfrak{a} \implies \lambda\alpha \in \mathfrak{a} $$
Also why do we need the extra "finitely generated"?
EDIT:
I now do know, Zagier really meant (1) - otherwise the Fractional Ideals would just be Ideals of $K$ (since $K$ is a field there only exist two such Ideals $\{0\}$ and $K$ itself).
Comments suggest, that the "finitely generated" is equivalent the the statement, that for every Fractional Ideal $\mathfrak{a}$ there there exists $n \in \mathbb{N}$, such that $n\mathfrak{a}$ is an Ideal of $\mathcal{O}_K$.
Now I need to understand why the equivalence holds. Any help is welcome!