# Representative in ideal class group coprime to the conductor

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?

• 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? Sep 22 '19 at 19:41
• 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!!! Sep 22 '19 at 20:18
• @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! Sep 22 '19 at 21:10
• @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. Sep 22 '19 at 21:20
• @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 $\mathcal O$. Sep 22 '19 at 21:43

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 $$a \in O_K/C^\times$$, $$I \sim (aO_K \cap O) (J\cap O)$$