TITLE: Dedekind-Like construction of p-adic numbers
QUESTION [21 upvotes]: Recently I've been studying p-adic numbers. I understand the idea of a cauchy completion of the rationals with respect to the metric defined by the norm $\vert\vert \cdot \vert \vert_p $. When I was studying the real numbers I found that understanding both the cauchy completion and the explicit construction using Dedekind cuts to be very enlightening.
My question is: Is there a explicit set theoretic construction of the p-adic numbers analagous to the construction of the real numbers using Dedekind cuts. 

Edit: To clarify my question.

I know that the p-adic numbers are not an ordered field.
I am not asking if there is a construction of the p-adic numbers that uses the ordering of the rationals (which is how Dedekind's approach to $\mathbb{R}$ works). 
What I want is a construction of the p-adic numbers that is more than just "$\mathbb{Q}_p$ is the cauchy completion of $\mathbb{Q}$ with respect to $d(x,y)=\vert \vert x-y \vert \vert_p $". The sentence is true but in my opinion not very enlightening since the same sentence would work for the completion of any metric space.

REPLY [5 votes]: Here’s yet another construction. I’m going to exhibit $\mathbb Z_p$ as the endomorphism ring of a certain abelian group, $T$ (for “torsion”), which I’ll describe in two ways. First definition of $T$ is that it’s $\mathbb Z[1/p]\,/\,\mathbb Z$, and by $\mathbb Z[1/p]$ I mean the rationals with only powers of $p$ in the denominator. You can see that this is a divisible group (for every $z\in T$ and every $m>0$, there are $z'\in T$ with $mz'=z$).
The other definition of $T$ is just the group of all $p$-power roots of unity in $\mathbb C$. Now $T$ is written multiplicatively, of course, and you can get an isomorphism between the two constructed groups by taking the rational number $r/p^n$ to $e^{2ri\pi/p^n}$.
Maybe it’s easier to think of $T$ as the rationals with only $p$-powers in the denominator, modulo $\mathbb Z$. Each subgroup $C_n=(p^{-n}\mathbb Z)/\mathbb Z$ is characteristic and must be mapped into itself by any endomorphism of $T$, so we get something that gives consistent endomorphisms of these cyclic groups of order $p^n$. But the endomorphism ring of a cyclic group $C_n$ of order $p^n$ is just $\mathbb Z/(p^n)$. You look at things, and see that an endomorphism of $T$ is just an element of $\text{projlim}_n(\mathbb Z/(p^n))$, with respect to the natural maps $\mathbb Z/(p^{n+1})\to\mathbb Z/(p^n)$
There’s no question that this is harder to see than the Cauchy-sequence construction, but you stare at things for a while, and see that they’re the same.