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Fix a prime p. The Teichmuller representative associated to a p-adic integer a is the unique root of x^p - x in Zp congruent to a mod p. One can identify this representative with the limit, as n tends to infinity, of a^{p^n}.

Now let a1, a2, ... ak be the roots of an irreducible monic polynomial in Zp[x]. One can show that the limit, as n tends to infinity, of a1^{p^n} + a_2^{p^n} + ... + a_k^{p^n} also exists as a p-adic integer. Is there a characterization of this p-adic integer analogous to the above characterization?

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I think I can give a characterization of your limit as a sum of Teichmüller representatives.

Let q = p^f be some power of p. Let Z_q = W(F_q) be the valuation ring of the unramified extension of Q_p of degree f. Then for any a in Z_q, there is a unique root of x^q - x in Zq congruent to a mod p. One can identify this with the limit, as n tends to infinity, of a^{q^n}.

I've never seen this before, but I guess you can do the same thing even if your extension is ramified. Let R be some finite extension of Z_p. Let F_q denote its residue field. Then for any a in R, there is a unique root of x^q - x in R congruent to a mod p^(1/e), where e is the ramification index. Again, it can be idenified with the limit of a^{q^n}.

Assuming the limit you mentioned exists, it is the same as the limit of a_1^{q^n} + ... + a_k^{q^n}. And then this limit is the sum of the Teichmüller representatives that I just described.

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Right, but it has a stronger property than that, namely being in Z_p and not just Z_q. What characterizes this element of Z_p? (I guess it's the trace of left multiplication by the Teichmuller representative of a on Z_q. Is that all one can say?) –  Qiaochu Yuan Oct 14 '09 at 22:32
You're right about the trace description. I don't see anything better than that. Since Teichmüller reps aren't well-behaved with respect to addition, I wouldn't be surprised if that's the best you can do. If you changed addition to multiplication, then you'd probably get that the limit was the Teichmüller rep of the norm. Out of curiosity, how do you know that the limit exists? For my limits, I was using that if a = b mod p (resp. mod p^{1/e}), then a^{p^n} = b^{p^n} mod p^{1+n} (resp. mod ...). But I don't see how to use that for this sum. –  CJD Oct 15 '09 at 5:13
One can use the fact that for an integer matrix A we have tr(A^{p^k}) = tr(A^{p^{k-1})) mod p^k, which has a fairly straightforward proof using group actions. –  Qiaochu Yuan Oct 24 '09 at 1:47
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