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Jul 9, 2013 at 19:00 comment added Pig Thanks for looking into it! I appreciate that. Your answer is the same as the highest-rated answer in this thread, or the reference mentioned in that answer. (Chapter 1 of Hida's book). For the sake of clarity let me restate my question: I would like to know if $\chi^{-1}$ should be employed in the lifting rather than $\chi$. Thanks again!
Jul 9, 2013 at 15:17 comment added Marc Palm You are certainly right. My earlier comment was wrong. So I deleted. I will look for the mistake later this week and withdraw or edit my answer. Thank you.
Jul 9, 2013 at 15:12 comment added Pig The last term should be $1/l$ (typo), but then the lifted character acting on this seems to be $\chi(1/l)$.
Jul 9, 2013 at 14:57 comment added Pig As for the lifted character, $(1,...,l,,.1) = l \cdot (1, 1/l,\cdots, 1 ,\cdots, 1/l) \cdot l$, where $l \in \mathbb{Q}^{\times}$, the second term is in $\prod \mathbb{Z}_p^{\times}$, where the first entry is in $\mathbb{R}$, the 1 is in $l$-th entry. In particular the $p$-th entry is $1/l$. The last term is in $\mathbb{R}^{\times}$.
Jul 9, 2013 at 14:53 comment added Pig but why is that? For classical Dirichlet character, the unramified L factor should be $(1 - \chi(l)l^{-s})^{-1}$, or am I wrong?
Jul 9, 2013 at 1:07 comment added Pig Sorry for bumping this old question, but I am confused. Consider a character on $\chi:\left(\mathbb{Z}/p^k\mathbb{Z}\right)^{\times}$. If we follow this construction, then when we restrict the lifted character $c$ to $c_l: \mathbb{Q}_l^{\times} \to \mathbb{C}$ for some prime $l \neq p$, we get $c_l(l) = \chi(l)^{-1}$. However I thought this must be $\chi(l)$ instead, for otherwise the $L$-function associated to the $\chi$ and $c$ seems to be different. Where did I make a mistake?
Apr 7, 2012 at 16:57 history edited KConrad CC BY-SA 3.0
added 13 characters in body; added 15 characters in body; deleted 11 characters in body
Apr 7, 2012 at 15:11 history edited Marc Palm CC BY-SA 3.0
added 3 characters in body
Apr 7, 2012 at 14:52 history answered Marc Palm CC BY-SA 3.0