# Unramified Hecke character

I'm looking for a reality check here:

Let $\chi$ be a character of $(F^\times\backslash \mathbb A_F^\times)^1$ where $F$ is a number field. Call $\chi$ unramified at a place $v$ if $\chi(a_v)=|a_v|^s$ for some $s$, i.e., it is trivial on units.

Does it follow that if $\chi$ is unramified everywhere then $\chi$ must be trivial? If not what should it be, and does it make a difference if $\chi$ is unitary?

Thanks!

No, you can just take $\chi(a) = |a|_{\mathbb A}^s = \prod_v |a|_v^s$ for some $s$, and choose $s$ to be purely imaginary to make it unitary.
Here's another construction. Suppose $F$ has two places $v_1$, $v_2$ such that $F_{v_1} \simeq F_{v_2}$. Fix $s \in \mathbb C^\times$. Take $\chi_v$ to be trivial if $v \not \in \{ v_1, v_2 \}$, $\chi_{v_1} = | \cdot |^s$ and $\chi_{v_2} = | \cdot |^{-s}$. Then $\chi = \otimes_v \chi_v$ is a character of $\mathbb A_F^\times$ which is trivial on $F^\times$, but nontrivial on global norm 1 elements. We can make it unitary by choosing $s$ to be purely imaginary.
• thank you for the answer. This construction won't work for $F=Q$; would it be trivial then? Commented Jul 22, 2015 at 15:43