I am given this definition of the field of fractions of the p-adic integers:

$$Q_p=\left\{ \frac{r}{s} \mid r, s \in Z_p, s \neq 0\right\}$$

How can I show that:

$Q_p$ consists of the sums of the form $\sum_{i=-k}^{\infty} a_ip^i$ ?


A $p$-adic integer can be expressed uniquely in the form $$ \sum_{i=0}^{\infty}a_{i}p^{i} $$ where $a_i$ is an integer with $0\le a_i<p$. Moreover, any $p$-adic integer can be written in a unique way as $$ up^k $$ with $u$ invertible and $k\ge0$. In particular, $$ \frac{r}{s}=\frac{r}{up^k}=\frac{u^{-1}r}{p^k} $$ Write $u^{-1}r=\sum_{i=0}^{\infty}a_{i}p^{i}$ and you're done.

  • $\begingroup$ So, is it like that? $\displaystyle{ \frac{r}{s}=\frac{u^{-1}r}{p^k}=\frac{\sum_{i=0}^{\infty} a_ip^i}{p^k}}=p^{-k} \sum_{i=0}^{\infty} a_ip^i=\sum_{i=0}^{\infty} a_{i}p^{i-k}=\sum_{i=-k}^{\infty} a_{i+k} p^i$ $\endgroup$ – evinda Oct 17 '14 at 22:12
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    $\begingroup$ @evinda Yes, good. $\endgroup$ – egreg Oct 17 '14 at 22:14
  • $\begingroup$ So, do we have to set $b_i=a_{i+k}$ ? $\endgroup$ – evinda Oct 17 '14 at 22:17
  • $\begingroup$ @evinda Yes, but it's just bookkeeping. $\endgroup$ – egreg Oct 17 '14 at 22:20
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    $\begingroup$ @evinda No, $p$ is a counterexample. $\endgroup$ – egreg Oct 18 '14 at 5:23

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