The $ 1 \operatorname{ mod } \mathfrak{m}$ congruence relation in ray $ P^{\mathfrak{m}}$ of the ideal theoretic ray class group

Question

Let $K$ be a number field, $\mathcal{O}_K$ its ring of integers and $S$ a subset of the real places. Let $\mathfrak{m} \subset \mathcal{O}_K$ an ideal. The ideal theoretic ray class group of $\mathfrak{m} $ and $S$ is the quotient group

$$ I^{\mathfrak{m}}/P^{\mathfrak{m}} $$

where $ I^{\mathfrak{m}} $ is the group of fractional ideals co-prime to $ \mathfrak{m} $, and the "ray" $ P^{\mathfrak{m}}$ is the group consisting of all principal ideals in the ring of integers of $K$ having the form $( \alpha)$ where $ \alpha $ is multiplicatively congruent to $ 1 \operatorname{ mod } \mathfrak{m}$ such that $\alpha $ are positive at the places of $S$.

Could somebody explain the original motivation for imposing for elements of the "ray" $ P^{\mathfrak{m}}$ this rather exotic congruence condition for the elements $\alpha $ to be congruent to $ 1 \operatorname{ mod } \mathfrak{m}$?

I suppose that the name "ray" came from the assumption that these in addition should be positive at the real places of $S$. This suggests these "behave" similar like the positive reals $\mathbb{R}_{ \ge 0 }$, which can visualized geometrically as a "ray". I noticed that in this discussion the reason for the name "ray" was discussed, but that doesn't exactly address my issue, since as far as I understood the punch lines from there correctly, the name "ray" based on the positivity assumption for real places.

But I would like to understand the reason for the condition $ 1 \operatorname{ mod } \mathfrak{m}$ for finite places/ primes, which looks very obscure to me and seemingly falls unexpectedly from the sky.

Answer 1

Gauß had studied classes of binary quadratic forms with arbitrary discriminant. Dedekind realized that the class groups of forms with fundamental discriminant are ideal class groups (in the strict sense) of quadratic number fields. In order to find something similar for forms with discriminant $\Delta = df^2$, where $d$ is a fundamental discriminant, it was necessary to put conditions modulo $f$ on the generators of principal ideals.

After Hilbert had realized that unramified abelian extensions were class fields in the sense that the splitting of prime ideals is governed by their order in the class groups, it was only natural to try and incorporate ramified abelian extensions. The simplest such extensions are cyclotomic extensions of the rationals, and their Galois groups are subgroups of $({\mathbb Z}/m{\mathbb Z})^\times$. The splitting of primes $p$ depends on their behavior modulo $m$, and so it is quite natural to look for a generalized ideal class group that realizes such residue class groups. Again, this could have been looked up in Gauss's Disquisitiones: He showed that the class group of primitive quadratic forms with discriminant $\Delta = m^2$ is isomorphic to $({\mathbb Z}/m{\mathbb Z})^\times$.

What I'm trying to say is that this "exotic condition" is a very natural one for everyone familiar with Gauss's theory of binary quadratic forms and Dedekind's explanation of these groups in terms of ideal classes in orders.

Answer 2

In short, because the multiplicative order in $\mathbb{O}_K/\mathfrak{m}$ is important. Consider the classical result that a prime $p$ in $\mathbb{Z}$ has inertial degree its multiplicative order mod $m$ and ramification index $\phi(p^k)$ for $p^k \vert \vert m$, in the extension $\mathbb{Q}(\zeta_m)/\mathbb{Q}$. Thus, to say that $p$ splits completely is exactly saying that $p\equiv 1$ mod $m$. In fact, this is exactly showing that $\mathbb{Q}(\zeta_m)$ is the ray class field for the (strict) ray class group of $m\mathbb{Z}$.

Surely there are other flavours of explanation, but I like this intuition.