Let $K$ be an algebraic number field with ring of integers $\mathcal{O}_K$. A module or generalised integral ideal $\mathfrak{m}$ of $K$ is defined in terms of a finite product over the finite and infinite primes of $K$, i.e.: $$ \mathfrak{m} = \prod_{\mathfrak{p}}\mathfrak{p}^{\nu_{\mathfrak{p}}}, $$ where the product is taken over the finite and infinite primes of $K$, and $\nu_{\mathfrak{p}} \neq 0$ for finitely many primes.
The ray class group of $K$ is defined in terms of the module $\mathfrak{m}$ in the following manner:
Let $J^{\mathfrak{m}}$ denote the group of fractional ideals in $\mathcal{O}_K$ coprime to the module $\mathfrak{m}$.
Let $P^{\mathfrak{m}}$ denote the subgroup of principal fractional ideals in $\mathcal{O}_K$ congruent to $1\ (\textrm{mod}\ \mathfrak{m})$.
Then the ray class group of $K$ with respect to the module $\mathfrak{m}$ is defined a the quotient: $$ \textrm{Cl}^{\mathfrak{m}} := J^{\mathfrak{m}}/P^{\mathfrak{m}}. $$
My question is: Why is the ray class group defined in terms of modules $\mathfrak{m}$ as opposed to just ordinary integral ideals?
What I mean is the following: If $\mathfrak{m}$ is a module and $\mathfrak{m}'$ is $\mathfrak{m}$ except its infinite factors (so that $\mathfrak{m}'$ is an ordinary integral ideal), then we have for any integral ideal $\mathfrak{a} \subset \mathcal{O}_K$: $$ \gcd(\mathfrak{a},\mathfrak{m}) = 1 \quad \iff \quad \gcd(\mathfrak{a},\mathfrak{m}') = 1, $$ and for any principal ideal $\mathfrak{b} \subset \mathcal{O}_K$: $$ \mathfrak{b} \equiv 1\ (\textrm{mod}\ \mathfrak{m}) \quad \iff \quad \mathfrak{b} \equiv 1\ (\textrm{mod}\ \mathfrak{m}'), $$ and so: $$ \textrm{Cl}^{\mathfrak{m}} = \textrm{Cl}^{\mathfrak{m}'}. $$ Thus any ray class group can be expressed in terms of an ordinary integral ideal.
Right?
Of course every ideal is a module, but why consider the infinite places? Is it on account of the class field theoretic existence theorems?
Addendum: I figured it out. It is in general not true that $$ \mathfrak{b} \equiv 1\ (\textrm{mod}\ \mathfrak{m}) \quad \iff \quad \mathfrak{b} \equiv 1\ (\textrm{mod}\ \mathfrak{m}'). $$ Only one implication is true, and so we have a strict inclusion.