Let $K$ be an imaginary quadratic number field and let $O_K\subset K$ be the ring of algebraic integers in $K$. Let us call a lattice $\Lambda\subset\mathbb C$ normalized if the tori $\mathbb C/\Lambda$ and $\mathbb C/O_K$ have the same area.
The following is true:
A normalized lattice $\Lambda\subset\mathbb C$ is a $O_K$-module (i.e. $\alpha\lambda\in\Lambda$ for all $\alpha\in O_K$, $\lambda\in\Lambda$) if and only if $|\lambda|^2\in\mathbb Z$ for every $\lambda\in\Lambda$.
What is a proof of this statement (if possible direct/elegant/short)? Are there any similar/analogous results?
It is easy to see the $\Rightarrow$ implication: if a normalized lattice $\Lambda\subset\mathbb C$ is a $O_K$ module then for every $\lambda\in\Lambda$ we see that $\lambda O_K\subset\Lambda$, and since $\Lambda$ is normalized, we get (if $\lambda\neq0$) $|\lambda|^2=[\Lambda:\lambda O_K]$, hence $|\lambda|^2\in\mathbb Z$. What is, however, a proof of the $\Leftarrow$ implication? (I can prove it with a random-looking calculation, but a conceptual proof should exist.)
context: The statement is equivalent to the description of the ideal class group of $K$ in terms of quadratic forms (any $\Lambda$ gives us a quadratic form, namely $\lambda\mapsto|\lambda|^2$). I just want to understand this description and its proof geometrically (so everything should be well-known, just not to me).