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Let $K$ be an imaginary quadratic number field. Let $\mathcal O$ be an order in $K$. Can it happen, that there are $h(\mathcal O)>1$ fractional proper $\mathcal O$-ideals, representing the ideal classes of $\mathcal O$, each with norm equal to $1$? Does this always happen?

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(Note: I have edited the answer to cover all cases.)

Let's call this "property 1". I claim that $\mathcal{O}$ has property 1 if and only if $h(\mathcal{O}_K)$ is odd.

First let me prove it when $\mathcal{O} = \mathcal{O}_K$. Then for a fractional ideal to have norm 1, it must be of the form $I/\bar I$, where $\bar I$ is the Galois-conjugate ideal (here I am tacitly using that ideals in $\mathcal{O}_K$ have unique factorization into prime ideals). Since conjugation is inversion in the class group, this means that the ideal class is a square. If all ideal classes are squares then the squaring map is surjective, hence injective, and so the 2-torsion subgroup is trivial. So the group has odd order.

Now consider the general case where $\mathcal{O}$ is an order of index $f$ in $\mathcal{O}_K$. I'll show that $\mathcal{O}$ has property 1 iff $\mathcal{O}_K$ has property 1. If $I$ is a proper fractional $\mathcal{O}$-ideal of norm 1, then $\mathcal{O}_KI$ (the product lattice) has norm 1, since norm is multiplicative. Since the map $\pi_f \colon \mathrm{Pic}(\mathcal{O}) \to \mathrm{Pic}(\mathcal{O}_K)$ sending the class of $I$ to the class of $\mathcal{O}_KI$ is surjective, this shows that if $\mathcal{O}$ has property 1, then so does $\mathcal{O}_K$.

Conversely, suppose $\mathcal{O}_K$ has property 1. Then for any ideal class $[I]$ in $\mathrm{Pic}(\mathcal{O}_K)$, I may choose $I$ to have norm 1 and also to be prime to $f$. Now, one can check that the ideals classes in the pre-image $\pi_f^{-1}([I])$ all have representatives $J$ which are simply index $f$ sublattices in $I$. [Let me not prove this here, but this is exactly what Siegel is doing explicitly in the paper you link to in the comments below. Note, in general not all index $f$ sublattices are proper $\mathcal{O}$-ideals. In fact, the number of such is precisely the size of the kernel of the map $\pi_f$, at least assuming there are no extra units in $\mathcal{O}_K^\times$.] Note that each these $\mathcal{O}$-ideals $J$ has norm 1 as well. So property 1 holds for $\mathcal{O}$.

As for your question of how often this happens, the 2-part of the class group is related to the number of primes dividing the discriminant. If, for example, there are at least 2 odd primes $p, q$ dividing the discriminant of $K$, then the 2-part is non-trivial. Indeed, the unique ideal above $p$ is 2-torsion in the class group of the maximal order, but not principal. But if $K$ has prime discriminant then indeed the class number is odd.

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  • $\begingroup$ On the other hand, in this question we have even class number and at the same time each class is represented by a fractional ideal of norm equal to $1$. How is this possible? $\endgroup$ – Shimrod Mar 30 at 9:27
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    $\begingroup$ Ah, you're right! My proof does not actually show the "only if" in the case of non-maximal orders (because it's conceivable that the ideal in question is not prime to the conductor, and hence my tacit use of unique factorization of ideals is not allowed). I will edit in the correct statement and proof. $\endgroup$ – Ari Shnidman Mar 30 at 16:01
  • $\begingroup$ Why and how is then Siegel computing the character $\left(\frac{f}{N(\mathfrak c)}\right)$? This character depends on norm modulo $f$, but if in each class is a fractional ideal of norm $1$ should not be this always the trivial character? $\endgroup$ – Shimrod Mar 30 at 18:11
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    $\begingroup$ No idea, I don't have time to stare at his paper. Maybe it has to do with the definition of the norm though. In the "proper" definition (the one I was using) the norm of a proper O-ideal is computed relative to O itself (not O_K). Maybe Siegel is computing the norm as the index relative to O_K. $\endgroup$ – Ari Shnidman Mar 30 at 19:02

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