# Is the order of a ray class group bounded by the class number?

Let $$K$$ be an algebraic number field with ring of integers $$\mathcal{O}_K$$, and $$\mathfrak{m}$$ a module of $$K$$.

Let $$J$$ be the group of fractional ideals in $$\mathcal{O}_K$$.

Let $$P$$ be the subgroup of fractional ideals in $$\mathcal{O}_K$$.

Let $$J^{\mathfrak{m}}$$ be the group of fractional ideals coprime to $$\mathfrak{m}$$.

Let $$P^{\mathfrak{m}}$$ be the subgroup of fractional ideals congruent to $$1\ (\textrm{mod}\ \mathfrak{p})$$ for every prime in the factorisation of $$\mathfrak{m}$$.

The class group of $$K$$ is defined as the quotient $$J_K/P_K,$$ and the class number of $$K$$ is the order of its class group.

The ray class group of $$K$$ with respect to the module $$\mathfrak{m}$$ is defined as the quotient $$J_{K}^{\mathfrak{m}}/P_K^{\mathfrak{m}}.$$

My question is: Is the order of the ray class group always bounded by the class number. Or in other words: For any module $$\mathfrak{m}$$, do we always have

$$\lvert J_{K}^{\mathfrak{m}}/P_K^{\mathfrak{m}} \rvert \leq \lvert J_K /P_K\rvert\ ?$$

I know this seems like a very basic question, but I don't think it's as simple as it seems.

It is rather the other direction: $$\lvert J_{K}^{\mathfrak{m}}/P_K^{\mathfrak{m}} \rvert \geq \lvert J_K /P_K\rvert$$.

To see this, note that there is a group homomorphism from $$J_{K}^{\mathfrak{m}}$$ to $$J_K /P_K$$, sending any ideal to its ideal class.

This homomorphism is surjective, by strong approximation theorem, and its kernel contains $$P_K^{\mathfrak{m}}$$.

Therefore it induces a surjective homomorphism from $$J_{K}^{\mathfrak{m}}/P_K^{\mathfrak{m}}$$ to $$J_K /P_K$$.

Note that this classical language is a bit "outdated". The modern language uses adeles (or ideles).

Restating the above, the ray class group $$J_{K}^{\mathfrak{m}}/P_K^{\mathfrak{m}}$$ is nothing but the quotient $$\Bbb A_K^\times / K^\times U_\mathfrak{m}$$, where $$U_\mathfrak{m}$$ is the open subgroup of $$\widehat{\mathcal O_K}^\times \times K_\infty^\times$$ of conductor $$\mathfrak m$$.

Among all the subgroups $$U_\mathfrak m$$, the largest one is the case $$\mathfrak m = 1$$, which is simply $$U_1 = \widehat{\mathcal O_K}^\times \times K_\infty^\times$$. The corresponding ray class group $$J_K^1/P_K^1$$ is nothing but the usual class group $$J_K/P_K$$.

From this point of view, it is apparent that $$J_K/P_K$$ is a quotient of $$J_{K}^{\mathfrak{m}}/P_K^{\mathfrak{m}}$$ for any $$\mathfrak m$$.

For details of the idele version, see e.g. the corresponding wiki page.

By definition, for a given algebraic number field $$K$$, the ray class group relative to the modulus $$\frak M$$ surjects onto the class group. But the appeal to CFT - which translates these groups in terms of Galois groups - gives much more precise information. For an excellent account of the main results and terminologies (in "modulii" as well in "idelic" terms), I recommend D. Garbannati, "CFT summarized", Rocky Mountain J. of Math. 11, 2 (1981).

The class group of $$K$$ is isomorphic to the Galois group over $$K$$ of the maximal abelian unramified extension of $$K$$, whereas the ray class group relative to $$\frak M$$ is isomorphic to the Galois group over $$K$$ of the maximal abelian extension of $$K$$ which is unramified outside $$\frak M$$ (NB: ramification at archimedean primes must be precisely defined, but this is just a question of conventions). The class group is always finite, whereas the ray class group can be infinite. A much studied case is when the places in $$\frak M$$ are just the places defined by the prime ideals of $$K$$ dividing a given rational prime $$p$$ (assume $$p$$ odd to get rid of the archimedean primes). Then CFT asserts that the maximal pro-$$p$$-quotient of the ray class group is isomorphic to $$T \times \mathbf Z_p ^{1+c+d}$$, where $$T$$ is finite, $$\mathbf Z_p$$ denotes the $$p$$-adic integers, $$c$$ is the number of complex places of $$K$$ and $$d$$ is conjecturally null (Leopoldt's conjecture). In a perfect world, the $$p$$-class group would be a quotient of $$T$$, but no. The relations between them are governed by a complicated combination of isomorphism and duality usually baptized "reflection" ("Spiegelung" in German).