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One way to define the p-adic integers is as the $p$-adic completion of $\mathbb{Z}$. With some additional work, it can be shown that this is isomorphic to $\mathbb{Z}[[x]]/(x-p)$.

Now, I know that another approach is to define $\mathbb{Z}_p$ as the ring of power series with powers of $p$ and coefficients from ${0, 1,..., p-1}$.

My question is: How can we see that the first definition of $\mathbb{Z}_p$ coincides with the second definition? and how can we find an explicit isomorphism from $\mathbb{Z}[[x]]/(x-p)$ to the ring of power series described above?

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The second definition is incomplete. You haven't specified either the addition or the multiplication. – Qiaochu Yuan Jul 24 '11 at 22:55
Is it obvious how to define the ring structure for your second construction (without referencing the usual tacks of completing the valued field or taking an inverse limit)? I've heard that this is possible, but have never seen it done. – Dylan Moreland Jul 24 '11 at 23:00
@Qiaochu: The arithmetic operations are simply defined as in the case of formal power series, when p replaces x. – Alexander Jul 24 '11 at 23:03
@Alexander: that is incorrect. You need to take into account carrying or else you get either something that is not well-defined or $\mathbb{F}_p[[x]]$, which has characteristic $p$ rather than $0$. – Qiaochu Yuan Jul 24 '11 at 23:19
@Alexander : Well it's not exactly like formal power series in $x$ because of carry over (the rule looks more like the usual rule for addition and multiplication in base $p$). For a more precise description, you might have a look at . – Joel Cohen Jul 24 '11 at 23:24

You want a map from ${\bf Z}[[x]]$ to that ring of coefficient-restricted power series, with kernel generated by $p-x$. Here's a start on constructing such.

Any positive integer can be written in base $p$ as $a=a_0+a_1p+\cdots+a_rp^r$ with $0\le a_i\le p-1$ for each $i$. This gives you a map $\phi$ from positive integers to (coefficient-restricted) polynomials by $\phi(a)=a_0+a_1x+\cdots+a_rx^r$. Now all you have to do is extend the domain from the positive integers to ${\bf Z}[[x]]$.

Start with the negative integers. In fact, start with $-1$; $$-1=(p-1)+(p-1)p+(p-1)p^2+\cdots$$ so $$\phi(-1)=(p-1)+(p-1)x+(p-1)x^2+\cdots$$ Now you can get $\phi(n)$ for any negative integer $n$ as a coefficient-restricted power series - details left to the reader.

The more complicated problem is what to do after you've applied $\phi$ to each coefficient of an element of ${\bf Z}[[x]]$ and because of the interaction between coefficients you still have some (perhaps infinitely many) coefficients outside the desired range. Well, apply $\phi$ again, and again, and again. After $k$ applications, at least the first $k$ coefficients will be OK, and they will stay OK forever after, so in the limit, you have your isomorphism.

There's probably a way of stating this in finite terms, but I'm not seeing it right now.

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The problem of showing that $\mathbb Z[[x]]/(x-p)$ coincides with the $p$-adic completion of $\mathbb Z$ has been addressed a couple of times already on this site: see here and here.

Recall that the $p$-adic completion of $\mathbb Z$ is the projective limit $\mathbb Z_p := \varprojlim_n \mathbb Z/p^n$.
The isomorphism of $\mathbb Z[[x]]/(p-x)$ with $\mathbb Z_p$ is obtained as the projective limit of the maps $\mathbb Z[[x]]/(x-p) \to \mathbb Z[[x]]/(x-p,x^n) = \mathbb Z/p^n$, the first map being the natural surjection, and the second map being obtained by setting $x = p$.

The isomorphism of $\mathbb Z_p$ with the "power series" that you describe is pretty straightforward: an element of $\mathbb Z/p^n$ has a unique representative in the form $a_0 + a_1 p + \cdots a_{n-1} p^{n-1}$ with $0 \leq a_i < p$. An element of $\mathbb Z_p$ thus can be described in a unique way as a "power series" $\sum_{i \geq 0} a_i p^i$ with $0 \leq a_i < p$.

Composing the two indicated isomorphisms gives the isomorphism you asked about. It is pretty explicit: given an element of $\mathbb Z[[x]]/(x-p)$, represented by some power series $f(x)$, to get the first $n$ terms of the associated "power series", I take the degree $\leq n-1$ part of $f(x)$, replaced $x$ by $p$, reduce modulo $p^n$, and then choose the coset representative of the form $a_0 + \cdots a_{n-1} p^{n-1}$ as above.

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