How to show if $x\in\mathbb{Q}$ ,then there exists $N \ge0$ such that $p^Nx\in\mathbb{Z}_p$? Or are there any references?

  • 1
    $\begingroup$ You just have to clear $p$ from the denominator. Every integer prime to $p$ is invertible in $\mathbb Z_p$. $\endgroup$ – lulu Apr 20 '17 at 1:10
  • $\begingroup$ @lulu how to show every integer prime to $p$ is invertible in $Z_p$? $\endgroup$ – Katherine Apr 20 '17 at 1:14
  • 2
    $\begingroup$ Every integer,$n$ prime to $p$ is invertible $\pmod {p^n}$. Convince yourself that we can string the inverses together in a sequence $\{a_1,a_2,\cdots \}$ such that $na_i\equiv 1 \pmod {p^i}$ and $a_i\equiv a_{i-1}\pmod {p^{i-1}}$. Then this sequence defines a $p$-adic inverse to $n$. $\endgroup$ – lulu Apr 20 '17 at 1:17

Let $q$ be a prime different from $p$. Here’s an explicit way of finding $1/q$ as a $p$-adically convergent series of ordinary integers:

Since $q\not\equiv0\pmod p$, you get $q^{p-1}\equiv1\pmod p$, in other words $q^{p-1}=1+mp$ for some ordinary integer $m$. Thus we have \begin{align} \frac1{q^{p-1}}&=1-mp+m^2p^2-\cdots=\sum_{i\ge0}(-mp)^i\\ \frac1q&=q^{p-2}\sum_{i\ge0}(-mp)^i\,, \end{align} where the infinite sums are convergent, the common ratio being $p$-adically smaller than $1$.

You see that primality of $q$ was not used here: the argument is valid for any integer not divisible by $p$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.