Let $K$ be an algebraic number field, $\mathcal{O}_K$ its ring of integers, $\mathfrak{m}$ an integral ideal of $\mathcal{O}_K$. Let $J$ be the set of all fractional ideals, $P$ the set of principal fractional ideals, $P^+$ the set of principal fractional ideals $(\alpha)$ such that $\alpha$ is totally real, $J_\mathfrak{m}$ the group of fractional ideals that are coprime to $\mathfrak{m}$, and $P_\mathfrak{m}^+$ the set of principal fractional ideals $(\alpha)$ such that $\alpha \equiv 1 \mod{\mathfrak{m}}$ and $\alpha$ is totally real. Further, let $(\mathcal{O}_K/\mathfrak{m})^\times$ be the multiplicative group mod $\mathfrak{m}$. I am interested in finite-order characters on the ray class group mod $\mathfrak{m}$, defined as the quotient group
$$
\mathcal{I}_\mathfrak{m} := \frac{J_\mathfrak{m}}{P_\mathfrak{m}^+}.
$$
Specifically, my question is as follows:
Is it true (or, under which circumstances is the following true) that any character $\xi$ on the ray class group $\mathcal{I}_\mathfrak{m}$ may be written as $\xi = \chi \psi$, where $\chi$ is a character on the multiplicative group $(\mathcal{O}_K/\mathfrak{m})^\times$ and $\psi$ is a character on the narrow ideal class group $J/P^+$? Here both characters $\chi$ and $\psi$ are being extended from ideal classes to ideals in the obvious way.
The ray class group $\mathcal{I}_\mathfrak{m}$ is constructed using congruence information ($\alpha \equiv 1 \mod{\mathfrak{m}}$), embedding information ($\alpha$ is totally real), and ideal class information (quotient by principal ideals). The first of these is encoded by the characters $\chi$, and the second and third are encoded by characters $\psi$. As such, it seems that these two sets of characters together should totally describe the characters on $\mathcal{I}_\mathfrak{m}$, but perhaps I'm missing something. Any comments would be most appreciated.
Best Answer
The question makes no sense as it stands, because $\chi$ and $\psi$ are characters on different groups (so it makes no sense to multiply them). Also, I think that the property "$\alpha$ is totally real" should be changed to "$\alpha$ is totally positive", meaning that $\alpha$ is positive in every real embedding $K\hookrightarrow\mathbb{R}$. With all this said, there is a way to interpret the question so that it has a reasonable answer (close to what the OP expects). What I am saying below is standard material, see e.g. Theorem 21.8 in these notes by Andrew Sutherland.
Let $K_\mathfrak{m}^\times$ be the multiplicative group of elements of $K^\times$ coprime to $\mathfrak{m}$, and let $P_\mathfrak{m}$ be the multiplicative group of fractional ideals generated by the elements of $K_\mathfrak{m}^\times$. Let $K_\mathfrak{m,+}^{\times}$ be multiplicative group of totally positive elements of $K^\times$ coprime to $\mathfrak{m}$. Then we have an exact sequence $$1\to\mathcal{O}_K^\times/(\mathcal{O}_K^\times\cap K^\times_{\mathfrak{m},+})\to K^\times_{\mathfrak{m}}/K^\times_{\mathfrak{m},+}\to J_\mathfrak{m}/P_{\mathfrak{m}}^+\to J_\mathfrak{m}/P_{\mathfrak{m}}\to 1.$$ By weak approximation, $J_\mathfrak{m}/P_{\mathfrak{m}}$ is isomorphic to the class group $J/P$, while $K^\times_{\mathfrak{m}}/K^\times_{\mathfrak{m},+}$ is isomorphic to $G=\{\pm 1\}^r\times(\mathcal{O}/\mathfrak{m})^\times$, where $r$ is the number of real places of $K$. So there is a subgroup $H\leq G$ isomorphic to $\mathcal{O}_K^\times/(\mathcal{O}_K^\times\cap K^\times_{\mathfrak{m},+})$ such that there is a short exact sequence $$1\to G/H\to J_\mathfrak{m}/P_{\mathfrak{m}}^+\to J/P\to 1.$$ In other words, the character group $\widehat{J/P}$ embeds into the character group $\widehat{J_\mathfrak{m}/P_{\mathfrak{m}}^+}$, and the corresponding quotient group is isomorphic to $\widehat{G/H}$. Note that $\widehat{G/H}$ is isomorphic to the group of characters of $G$ that are trivial on $H$. In general, the above short exact sequence does not split, so $J_\mathfrak{m}/P_{\mathfrak{m}}^+$ is not isomorphic to $(G/H)\times(J/P)$ even though they have the same order.