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Let $p$ be a prime number and $d$ a positive integer. If $n\in \mathbb{Z}_p$ (the ring of $p$-adic integers) how would one define $d^n$?

Under what conditions would this be an element of $\mathbb{Z}_p$?

Any help would be very much appreciated.

Josh F.
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    Usually you would want $d \in 1 + p \mathbb{Z}$ to define this, is this your case? – Cocopuffs Jul 05 '13 at 19:24
  • Yes, probably, although the article on which my question is based doesn't impose any conditions on $d$. Under your assumption how would you first define $d^n$ and then show that it lies in $\mathbb{Z}_p$? Thanks – Josh F. Jul 05 '13 at 19:43

3 Answers3

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This is defined if (and only if) $d$ is congruent to 1 mod $p$. You can see this by thinking about the limit (in the $p$-adic topology) of $d^{n_i}$, where $n_i$ is some sequence of integers converging $p$-adically to $n$. If $d$ is is $1 \pmod p$, then the limit exists in $\mathbb{Z}_p$, and this is the natural definition of $d^n$.

David Loeffler
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  • I see, so if $d=1+kp$ would $v_p(d^{n})$ be defined as $\lim_{i\rightarrow \infty}v_p(d^{n_i})$? This would then equal $\lim_{i \rightarrow \infty}n_iv_p(d)\ge 0$ since $v_p(d)=v_p(1+kp)= 0$? – Josh F. Jul 05 '13 at 20:06
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    Yes, if $d \in 1 + p\mathbb{Z}_p$ then $d^n \in 1 + p\mathbb{Z}_p$, so $v_p(d^n) = 0$. – David Loeffler Jul 06 '13 at 08:30
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Assume that $d\equiv1\pmod p$ as others have pointed out. Let $$ n=\sum_{i=0}^\infty a_ip^i, $$ with $a_i\in\{0,1,\ldots,p-1\}$. The upshot is that (binomial theorem!) $$ d^{p^i}\equiv 1\pmod{p^{i+1}} $$ for all $i$. Therefore also $x_i:=d^{a_ip^i}\equiv1\pmod{p^{i+1}}$. This implies in turn that the infinite product $$ \prod_{i=0}^\infty x_i $$ converges (it is clear that the sequence of partial products stabilizies modulo $p^{i+1}$ after $i$ factors). There's your definition: $$ d^n=\lim_{k\to\infty}\prod_{i=0}^kx_i. $$ With a little bit extra work we see that this turns $1+p\mathbb{Z}_p$ into a $\mathbb{Z}_p$-module, which comes in handy sometimes.

Jyrki Lahtonen
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  • So would the sequence of partial products form a Cauchy sequence wrt the $p$-adic valuation and therefore converge in $\mathbb{Z}_p$ since the latter is a complete ring? – Josh F. Jul 05 '13 at 20:32
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    @user68418: That's correct. – Jyrki Lahtonen Jul 05 '13 at 20:49
  • I have one more question- isn't $1+p\mathbb{Z}_p$ a multiplicative abelian subgroup of $\mathbb{Q}_p^{\times}$ and not an additive one? I can't quite see how you can make the former into a $\mathbb{Z}_p$-module. Thank you. – Josh F. Jul 05 '13 at 21:09
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    That's exactly the beauty of it! Multiplication is the "addition" of this module, and raising to $n$th power (as we just saw) is the "scalar multiplication". Some structure theory of the resulting module is known. See e.g. Weil's "Basic Number theory" for details. – Jyrki Lahtonen Jul 05 '13 at 21:18
  • Thanks- I'll definitely look it up. Your explanations and comments have been very helpful. – Josh F. Jul 05 '13 at 21:36
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In addition to the other good answers, it is sometimes nice to realize that the usual power series for $\log(1+px)$ and $e^{px}$ converge $p$-adically for $x\in\mathbb Z_p$ and $p\not=2$.

paul garrett
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