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If $p$ is prime and not a divisor of $q$, are there any non-trivial rational integers in the $p$-adic closure of the set or powers of $q$?

Edit: $q$ is also a (rational) integer, not a $p$-adic.

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Could you please add the definition of a rational integer? – Asaf Karagila May 5 '11 at 17:05
It's just a synonym for "integer." There are things called $p$-adic integers, so in a context like this, we use "rational integer" to be clear. – Thomas Andrews May 5 '11 at 17:08
Ah. Very well. Thank you! – Asaf Karagila May 5 '11 at 17:12
up vote 3 down vote accepted

If $p>2$ then there is an integer $q$ not divisible by $p$, with the properties: $q^k\not\equiv 1$ mod $p$ for $1\leq k\leq p-2$ and $q^{p-1}\not\equiv 1$ mod $p^2$. Under these conditions the $p$-adic closure of $\{1,q,q^2,\dots\}$ is the whole $\mathbb{Z}_p^\times$ - in particular it contains all the rational integers not divisible by $p$.

For $p=2$ and $q\equiv 5$ mod $8$ then the closure is $1+4\mathbb{Z}_2$ - i.e. it contains all the integers which are $1$ mod $4$.

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I know we can find a generator, $q$, for the multiplicate group, $\pmod p$, but how do we know that that we can find one with $q^{p-1}\not\equiv 1 \pmod {p^2}$? – Thomas Andrews May 5 '11 at 19:29
@Thomas Andrews: $(q+ap)^{p-1}=q^{p-1}+(p-1) q^{p-2} a p +\dots$, so with a suitable $a$ we certainly have $(q+ap)^{p-1}\not\equiv 1$ mod $p^2$. – user8268 May 5 '11 at 19:34
Is it true, in general, if $q^{p-1}\not\equiv 1 \pmod {p^2}$ then rational integer $n$ is in the closure if and only if $n\equiv q^k \pmod p$ for some $k$? – Thomas Andrews May 5 '11 at 19:40
Yes. (The fact that the closure $\{1,c,c^2,\dots\}$, where $c=q^{p-1}$, is $1+p\mathbb{Z}_p$, can be shown e.g. using $\log$.) – user8268 May 5 '11 at 19:48

You never said what $q$ is (or what a nontrivial rational integer is :)). I presume you meant that $q$ is some $p$-adic integer and you want an integer outside the sequence of nonnegative powers of $q$ which is a $p$-adic limit of a subsequence of the nonnegative integral powers of $q$.

In that case, there are lots of examples: if $p > 2$ and $q = -1 \bmod p$ then the subsequence $q^{p^k}$ tends to -1 as $k$ tends to infinity. More generally, for any $p$-adic integer $q$, the subsequence $q^{p^k}$ tends to the unique solution of $t^p = t$ in the $p$-adic integers satisfying $t = q \bmod p$. So when $p$ is odd and $q = -1 \bmod p$, $t = -1$.

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Sorry, $q$ was meant to be a rational integer, and I was including $q^0=1$ in the set of powers of $q$. – Thomas Andrews May 5 '11 at 17:21
This also means that if any power of $q$ is $-1\pmod p$, then -1 is in the closure, of course. – Thomas Andrews May 5 '11 at 17:30

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