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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?

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    $\begingroup$ 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
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    $\begingroup$ 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
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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 $p$.

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