Let $\Gamma$ be Galois group of cyclotomic $\mathbb{Z}_p$ extension over $\mathbb{Q}$. Consider the function $G$ which sends each finite order character $\chi$ of $\Gamma$ to the Gauss sum $G(\chi)$, view $G(\chi)$ as an element of $\overline{\mathbb{Q}}_p$. My question is if $G$ is a $p$-adic measure on $\Gamma$?
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1$\begingroup$ Dear user57657, can you please write down the formula defining $G(\chi)$ to ensure that everyone understands the same thing with this notation? $\endgroup$– JoëlCommented Sep 12, 2014 at 14:49
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1$\begingroup$ Please explain what you mean by a $p$-adic measure on $\Gamma$. $\endgroup$– GH from MOCommented Sep 12, 2014 at 23:31
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$\begingroup$ @Joël We can view $\chi$ as a Dirichlet character, then the definition is Well-known, for instance, see page 4 of Bump "automorphic forms and representations". We note that the conductor of $\chi$ in my question is always a power of $p$. $\endgroup$– user57657Commented Sep 13, 2014 at 6:04
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1$\begingroup$ Your citation of Bump's book is not good. There he defines the complex Gauss sums, for which there is a canonical additive character. You are asking about $p$-adic valued Gauss sums, and there your notation in fact is incomplete. Such a sum relies on a choice of additive character as well, since there is not a canonical choice. Therefore I agree with Joël: you should have written out a formula for your Gauss sum instead of just saying it is well-known. $\endgroup$– KConradCommented Sep 13, 2014 at 11:19
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1$\begingroup$ What I wrote does not imply that different choices must change the $p$-adic valuation, but only that it might change it. You need to compute the sums to find out. In fact Odoni's paper, which I mentioned in a comment to David's answer, shows that the story of the $p$-adic valuation for Gauss sums mod $p^n$ is a lot simpler than for Gauss sums on finite fields, where the $p$-adic valuation is very sensitive to the choice of character, as described by Stickelberger's congruence (or the Gross-Koblitz formula). $\endgroup$– KConradCommented Sep 13, 2014 at 12:28
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No. The Gauss sum is not a $p$-adic measure. One cheap way to see this is as follows: if $\chi$ has conductor $p^n$, the $p$-adic valuation of $G(\chi)$ is $n/2$. But if $\mu$ is a measure, the asymptotics of $\int \chi \mathrm{d}\mu$ for $\chi$ of increasing $p$-power conductor, are governed by the $\lambda$ and $\mu$ invariants of $\mu$, and in particular the valuations of these numbers must tend to a limit -- they cannot tend to $\infty$.
answered Sep 13, 2014 at 5:04
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$\begingroup$ Can you give a reference on determine the $p$-adic valuation of $G(\chi)$? The well-know fact is the complex norm of $G(\chi)$ is $\sqrt{N}$, if $N$ is the conductor of $G(\chi)$. Another result is Stickelberger's congruence. For me, it is not trivial to know the $p$-adic valuation of $G(\chi)$. $\endgroup$ Commented Sep 13, 2014 at 6:14
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1$\begingroup$ Here is a reference. See R. Odoni, "On Gauss sums (mod $p^n$), $n \geq 2$," Bulletin London Math. Soc. 5 (1973), 325-327. $\endgroup$– KConradCommented Sep 13, 2014 at 11:11
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2$\begingroup$ From the identity $G(\chi)G(\chi^{-1}) = N$ one can deduce that at least one of the factors has large valuation and that suffices for the above argument. $\endgroup$ Commented Sep 13, 2014 at 16:30