Numbers in $\mathbb{Q}_p$ can be written uniquely as $\sum_{i=k}^\infty \alpha_i p_i$

Question

I'm studying the completion $\mathbb{Q}_p$ of $\mathbb{Q}$, through the classic approach of building equivalence classes of Cauchy sequences on $\mathbb{Q}$, with the $p$-adic metric.

At this point some articles I've read says:

It can be shown that every element in $\mathbb{Q}_p$ can be written in a unique way as $\sum_{i=k}^\infty \alpha_i p_i$, where $k$ is some integer such that $\alpha_k\neq 0$ and each $\alpha_i \in \{0,1,\ldots,p-1 \}$.

but I can't figure out how from this completion can be inferred the statement above.

Answer 1

Here’s what you need for the slightly simpler case where your Cauchy sequence consists of integers only. Consider a $p$-adic number, represented by a Cauchy sequence $\{a_n\}_n$. Show for yourself that, given this sequence, for every $m$ there is an $N_m$ with the property that the sequence is constant modulo $p^m$. That is, $\forall n\ge N_m$, you get $a_n\equiv a_{N_m}\pmod{p^m}$.
Then make up a new Cauchy sequence out of the given one, $\{a_{N_1},a_{N_2},a_{N_3},\cdots\}$, which you show is equivalent to the given one.
Now, for each $m\ge1$, set $b_m$ to be the integer with $0\le b_m<p^m$ and $b_m\equiv a_{N_m}\pmod{p^m}$. Each of these is writable as $b_m={_0b}_m+{_1b}_mp+\cdots +{_{m-1}}b_mp^{m-1}$ with each $_ib_m$ in the range $0,\cdots,p-1$, uniquely. And you check that the $_0b_m$’s are all the same, the $_1b_m$’s are all the same, etc. This gives you your expansion $_0b+{_1b}p+{_2b}p^2+\cdots$

Answer 2

This is a classical result about $p$-adic numbers, which are the completion of $\mathbb{Q}$ with respect to the $p$-adic metric. A proof is given, for example, in Theorem $3.4$ of the lecture notes by Evertse. One needs to show first that such a series converges. The uniqueness comes from Lemma $3.2$, namely that for every $\alpha\in \mathbb{Z}_p$ and every positive integer $m$ there is a unique $a_m\in \mathbb{Z}$ such that $\alpha\equiv am \bmod p^m$ with $ 0\le a_m <p^m$.