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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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  • $\begingroup$ It's equivalent to find an element in the dual ideal whose index is coprime to the conductor. This is a rank 2 lattice, can't you just search it by going one by one through the elements or else by generating random ones? $\endgroup$ – Will Sawin Sep 22 at 19:41
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    $\begingroup$ The laziest possible way is to apply Chebotarev to the Hilbert class field to find a prime in the given ideal class. The explicit form of the error term for Chebotarev gives you a bound on the norm of such an ideal (of the form << discriminant^{O(1)}). I couldn’t tell if you wanted a computationally reasonable method, sorry if so!!! $\endgroup$ – alpoge Sep 22 at 20:18
  • $\begingroup$ @WillSawin When you say that dual ideal, do you mean the dual of $\mathfrak{a}$, or do you mean the dual of $\mathcal{O}$? In a monogenic order, which all imaginary quadratic orders are, the dual of the ideal is the order, itself. So, are you saying, `check all the elements of $\mathcal{O}$?' I was hoping for something less bruteforce, haha. @alpoge I was hoping for a computationally reasonable method; thanks for the tip, though! $\endgroup$ – Laarz Sep 22 at 21:10
  • $\begingroup$ @Laarz I mean the dual of $\mathfrak a$. At least in a maximal order, each element in the dual of $\mathfrak a$ defines an ideal with ideal class $\mathfrak a$. The problem with looking for an explicit construction is that you're looking for hay in a haystack. Checking random elements is actually going to be way more computationally efficient than any explicit method. It's easy to see that this takes a number of tries proportional to $\log \log \log $ of the discriminant. $\endgroup$ – Will Sawin Sep 22 at 21:20
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    $\begingroup$ @Laarz The dual of the ideal would be the set of all elements in the fraction field whose product with $\mathfrak a$ lies inside $\mathcal O$, and certainly doesn't equal $\mathcal O$. $\endgroup$ – Will Sawin Sep 22 at 21:43
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There is probably better algorithmically but

  • $C = \{ b\in O_K,bO_K\subset O\}$ is a $O$ and $O_K$ ideal.

Choose some $O_K$-representatives of $O_K/C$ and let $f(a+C) = a O_K \cap O$ whose class doesn't depend on the representative $a$.

$f(O_K/C)$ contains the kernel of $\phi : Cl(O) \to Cl(O_K), \phi(I) =IO_K$.

  • The invertible classes of $O$-ideals are in the image of $f(O_K/C^\times)$ (I don't know how to prove this, it is p.12 of kconrad)

  • If a $O_K$ ideal $J$ is coprime with $C$ then $J = (J \cap O) O_K$.

For an invertible ideal $I \subset O$ there is $b/d\in K$ such that $J=\frac{b}{d} IO_K$ has norm coprime with $N(C)$,

then $I \in (J\cap O)\ker(\phi)$ which means for some $a \in O_K/C^\times$, $I \sim (aO_K \cap O) (J\cap O)$

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