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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!

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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.

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  • $\begingroup$ thank you for the answer. This construction won't work for $F=Q$; would it be trivial then? $\endgroup$
    – Tian An
    Commented Jul 22, 2015 at 15:43
  • $\begingroup$ @TAWong No, there's the obvious counterexample--see revised answer. $\endgroup$
    – Kimball
    Commented Jul 23, 2015 at 2:57
  • $\begingroup$ I see now. Thanks! I was just being muddled with unitary vs unramified. $\endgroup$
    – Tian An
    Commented Jul 23, 2015 at 20:45

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