Let $K$ be an imaginary quadratic number field, and $\mathcal{O}_K$ the ring of integers. Let $\mathcal{O}$ be an order. Call the $\textit{conductor}$ $f = [\mathcal{O}_K:\mathcal{O}]$.
Given some $\mathcal{O}$-ideal $\mathfrak{a}$, call the $\textit{norm}$ of $\mathfrak{a}$ to be $N\mathfrak{a} = [\mathcal{O}:\mathfrak{a}]$. We say $\mathfrak{a}$ is $\textit{proper}$ when the set $\{\alpha \in K : \alpha\mathfrak{a} \subseteq \mathfrak{a}\}$ is equal to $\mathcal{O}$. Prove that if $f$ and $N\mathfrak{a}$ are relatively prime, then $\mathfrak{a}$ is proper. Give a counterexample to the converse.
Can anybody help with this? I know that $\mathfrak{a}$ is proper if and only if it is invertible, but beyond that I'm pretty stuck.
Best Answer
For the first part, suppose $\beta\in K$ with $\beta\mathfrak{a}\subset \mathfrak{a}$. Then $\beta\in\mathcal{O}_K$, and thus $$\beta\mathcal{O} = \beta(\mathfrak{a}+f\mathcal{O}) = \beta\mathfrak{a} + \beta f\mathcal{O} \subset \mathfrak{a} + f\mathcal{O}_K. $$ But $f\mathcal{O}_K\subset\mathcal{O}$, so that $\beta\mathcal{O}\subset\mathcal{O}$ and thus $\beta\in\mathcal{O}$ so that $\mathfrak{a}$ is proper.
One possible counterexample is that principal ideals are always proper in this situation. For if $\mathfrak{a} = (a)$ is principal in $\mathcal{O}$, then $$\mathcal{O}\subset S = \{\beta\in K\mid\beta\mathfrak{a}\subset\mathfrak{a}\} = \{\beta\in\mathcal{O}_K\mid \beta a = \alpha a,\ \alpha\in\mathcal{O}\}\subset\mathcal{O}_K.$$ But $\beta a = \alpha a$ implies $\beta = \alpha$, so that $\beta\in\mathcal{O}$ and thus $S = \mathcal{O}$.
It is also true that any ideal in the maximal order $\mathcal{O}_K$ is proper.
Edit in response to comment: it's also worth showing that $\gcd(N(\mathfrak{a}),f)=1$ if and only if $\mathfrak{a}+f\mathcal{O} = \mathcal{O}$. Consider the map $\mathcal{O}/\mathfrak{a}\to \mathcal{O}/\mathfrak{a}$ given by multiplication by $f$. Then $\mathfrak{a}+f\mathcal{O} = \mathcal{O}$ if and only if this map is surjective; since the domain and codomain are finite, this is the same as the map being an isomorphism. But multiplication by $f$ in this group is an isomorphism if and only if $f$ is relatively prime to the order of $\mathcal{O}/\mathfrak{a}$, which is $N(\mathfrak{a})$.