# Why is the kernel of an algebraic Hecke character open in the ideles?

I've been reading about algebraic Hecke characters, and how one obtains one dimensional $p$-adic representations from them. I have a question about why the kernel of a map defined on the ideles is open. Let me state the conventions I'm using and then ask the precise question.

Let $K$ be a number field, $\mathfrak{c}\subset\mathcal{O}_K$ an ideal of $K$, and $\phi = \sum_\sigma n_\sigma\sigma\in \mathbb{Z}[Hom_{Ring}(K,\overline{\mathbb{Q}})]$. Furthermore, let $I(\mathfrak{c})$ be the group generated by ideals of $K$ prime to $\mathfrak{c}$ and $P_1(\mathfrak{c})$ the group generated by principal ideals prime to $\mathfrak{c}$ which have a generator congruent to $1\bmod\mathfrak{c}$. Then one may define an algebraic Hecke character of $K$ with infinity type $\phi$ and modulus $\mathfrak{c}$ as a homomorphism $$\chi:I(\mathfrak{c})\longrightarrow\overline{\mathbb{Q}}^\times$$ such that for all $(\alpha)\in P_1(\mathfrak{c})$, $\alpha\equiv 1\bmod\mathfrak{c}$, $$\chi((\alpha)) = \alpha^\phi = \prod_\sigma \sigma(\alpha)^{n_\sigma}$$

Using $\chi$, one may then define a group homomorphism on the ideles, which I will denote $\chi_\mathbb{A}$, as follows. Let $\mathbb{A}_K^\times$ be the ideles of $K$ and let $$U_1(\mathfrak{c}) = \{ x = (x_v)\in\mathbb{A}_K^\times : x_v\equiv 1\bmod \mathfrak{c}, \forall \text{ finite } v, x_v>0, \forall \text{ real } v\}$$ Let $x\in\mathbb{A}_K^\times$. To define $\chi_\mathbb{A}(x)$, let $\alpha\in K$ such that $\alpha x\in U_1(\mathfrak{c})$ (such an $\alpha$ exists by the Chinese remainder theorem), and define $$\chi_\mathbb{A}(x) = \chi(i(\alpha x))/\alpha^\phi$$ where $i(\alpha x)$ is the ideal of $K$ obtained from the idele $\alpha x$. $\chi_\mathbb{A}$ is well-defined by the definition of $\chi$.

My question is, how does one prove that the kernel of $\chi_\mathbb{A}$ is open in $\mathbb{A}_K^\times$?

I understand that the set $$U(\mathfrak{c}) = \{x = (x_v)\in U_1(\mathfrak{c}): x_v\in\mathcal{O}_{K_v}^\times,\forall \text{ finite } v\}$$ is contained in the kernel of $\chi_\mathbb{A}$ since for $x\in U(\mathfrak{c})$, $$\chi_\mathbb{A}(x) = \chi(i(x)) = \chi((1)) = 1$$ and that $U(\mathfrak{c})\subset\mathbb{A}_K^\times$ is open. But I don't see how to prove that either the whole kernel of $\chi_\mathbb{A}$ is open. It seems to me that the kernel may be more than just $U(\mathfrak{c})$.

Thanks for any help!

• Once the kernel contains an open subgroup $U$, it is necessarily open, since it is the union of translates of $U$. – John Binder May 4 '16 at 19:20
• I'm voting to close this question because it has been answered in a comment. – Stefan Kohl May 5 '16 at 13:09