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## 1-dimensional semi-stable Galois representations with coefficients

For any p-adic local field K, all 1-dim semi-stable Galois repn: $G_K \to Q_p^{*}$ are just $Q_p(n)\otimes \mu$, where $Q_p(n)$ is the Tate twist of cyclotomic character, and $\mu$ an unramified charater.

My question is what if we replace the coefficient field to $E \neq Q_p$?

In fact, at the end of the paper by Gerasimos Dousmanis "Rank two filtered $(φ, N)$-modules with Galois descent data and coefficients", the filtered $(\varphi, N)$ modules of all such 1-dim repns are all classified. My question really is, how do we write out the representations explicitly?

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A one-dimensional $p$-adic Galois representation is crystalline if and only if it's semistable, so this is a duplicate of [mathoverflow.net/questions/61998/] – David Loeffler May 27 2012 at 13:16
@David, actually I also found that post, and am now looking at Conrad's paper, but I suspect if the paper has very "explicit" description of what the characters are...Do you know? – 750am May 27 2012 at 13:30

If $E \neq Q_p$ then there may be more $1$-dimensional crystalline representations that the ones you mention. By Lubin-Tate theory, every character of $G_K$ can be written as an unramified character times a character of $O_K^\times$ (after making proper choices and identifications). The algebraic characters of $O_K^\times$ are then crystalline and if $E$ contains $K$, then these provide examples of crystalline characters.

EDIT : oops sorry, I did not read the question carefully enough, I did not see that you were asking for an explicit description of all such representations. As David Loeffler pointed out, the answer is in Brian Conrad's paper (now in appendix B, the paper having grown since last time I mentioned it).

Here is what it basically says, using the same identifications as above: assume that $E$ contains $K^{gal}$ (it's usually harmless to assume that the coefficient field is large enough). If $s$ runs through the set of embeddings $s : K \to E$ and the $a_s$ are integers, then $x \mapsto \prod_s s(x)^{a_s}$ gives rise to a crystalline character of $G_K$ and they're all of this type times an unramified character.

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 Thank you very much！ – 750am May 28 2012 at 4:35

@Berger, just a quick question, so these $a_i$ will be the Hodge-Tate weights? Thanks!

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That should be a comment, not an answer. – Felipe Voloch May 27 2012 at 14:46
Sorry I just realized that, but found out that I cannot delete it... – 750am May 27 2012 at 14:47
yes, the $a_s$ are the HT weights, then. – Laurent Berger May 27 2012 at 15:02