Working in an order $\mathcal{O}$ in an imaginary quadratic field $K = \mathbb{Q}(\sqrt{d})$ and given an invertible ideal $\mathfrak{a}\subseteq \mathcal{O}$, I would like to produce another integral ideal $\mathfrak{b}$ in the same equivalence class of the ideal class group, i.e. $\mathfrak{b}=\alpha\mathfrak{a}$ for some $\alpha\in K$, such that the norm of $\mathfrak{b}$ is coprime to the conductor of $\mathcal{O}$.

We know that this is possible by, for instance, Cox's *Primes of the Form $x^2 +ny^2$* Corollary 7.17. However, the proof of this fact is existential, using the isomorphism to the form class group, and does not give an explicit method of finding such a representative.

I remember reading a way to do this long ago, and maybe it's a consequence of Artin-Whaples Lemma? But I have been unable to get anywhere myself.

Any and all help is appreciated.

Edit: In this paper, ON SINGULAR MODULI FOR ARBITRARY DISCRIMINANTS, on page 15, the author seems to say that, if we start with a $\mathfrak{p}$-primary ideal $\mathfrak{a}$ that is locally principally generated by some integral $\alpha$ whose norm is supported only at $\mathfrak{p}$ over the conductor (which I am not sure how one would find such an element), then $\mathfrak{a}\sim \tilde{\mathfrak{a}}:= \mathcal{O}_p\cap\bigcap \alpha\mathcal{O}_\mathfrak{q}$.

Why must such an $\alpha$ exist? Further, what principal ideal makes these in the same ideal class? It feels like it should be $\alpha^{-1}$; am I being silly?

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