# The $p$-adic integers as a profinite group

How to prove that if $\mathbb{Z}_p$ is the set of $p$-adic integers then $\displaystyle{\mathbb{Z}_p=\varprojlim\mathbb{Z}/p^n\mathbb{Z}}$ where the limit denotes the inverse limit?

$\mathbb{Z}_p$ is the inverse limit of the inverse system $(\mathbb{Z}/p^n\mathbb{Z}, f_{mn})_{\mathbb{N}}$, but I don't know what the $f_{mn}$ are.

Can someone help me?

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The $f_{mn}$ are the "obvious" maps $\mathbb{Z}/p^m\mathbb{Z}\to\mathbb{Z}/p^n\mathbb{Z}$, where $m\gt n$. Note that $p^m\mathbb{Z}\subseteq p^n\mathbb{Z}$, so that $\mathbb{Z}/p^n\mathbb{Z}$ is a quotient of $\mathbb{Z}/p^n\mathbb{Z}$. The connective maps are the quotient maps. –  Arturo Magidin Jun 11 '12 at 1:15
How are you defining the $p$-adics, if not as the inverse limit? I'm asking because for many, the $p$-adics are the inverse limit (by definition), so asking how to prove that they are the same would be asking how to "prove" a definition. So obviously, you must have a different definition of $\mathbb{Z}_p$, and you probably should specify it. –  Arturo Magidin Jun 11 '12 at 1:16
Now that I think about it, after you state your definition of $\Bbb Z_p$ (I guess it is with power series expansions), you can also mention if you want to show isomorphisms as additive groups, as rings, and if you want topology in the mix as well. I was hasty adding the (topological-groups) tag; I imagine you just want an isomorphism as bare groups. –  anon Jun 11 '12 at 1:28
You can view a $p$-adic integer $a_0 + a_1 p + ...$ as a sequence $(..., a_2 p^2 + a_1p + a_0, a_1 p + a_0, a_0)$, which is an element of the inverse limit. This is how the isomorphism goes I believe. –  fretty Jun 11 '12 at 11:54

## 2 Answers

Let me guess the definition of $p$-adic integers you got in mind is some $p$-power series. Now, forget it and let us consider some topology:

We DEFINE the $p$-adic integers to be the completion of certain metric on $\mathbb{Z}$, then the $p$-power series is a way to make completion, and the inverse limit is another way. However, the completion of a metric space is unique up to unique isometry. Now you can check the isometry is a ring isomorphism by hands.

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Yes, but proving that the inverse limit is the completion of $\mathbb{Z}$ with respect to p-adic metric is not so trivial and you are suppose to do it. –  Makoto Kato Jun 11 '12 at 20:43
@Makoto Kato, by writting explicitly, a small element in the inverse limit is something of the form $(0,0,...,0,a_1,a_2,...)$ and elements in $\mathbb{Z}$ are of the form $(a,a,a...)$ (maybe mod something in the first few terms). Thus $\mathbb{Z}$ is dense in the inverse limit and every Cauchy sequence in $\mathbb{Z}$ converges in the inverse limit, and therefore the inverse limit is a completion. Anyway, you have already gave a detailed proof in your answer, right? –  Ch Zh Jun 11 '12 at 23:33
and every Cauchy sequence in $\mathbb Z$ converges in the inverse limit, How do you prove this? –  Makoto Kato Jun 14 '12 at 23:48
@Makoto Kato, let $A_1,...A_N,...$ be a Cauchy sequence in $\mathbb{Z}$, then for every positive integer $n$, the $n$-th terms in the sequence becoming stable for big enough $N$ in this sequence, denote it by $b_n$. Then $(b_1,b_2,...)$ is an element in the inverse limit and the sequence $A_i$ converges to it. –  Ch Zh Jun 15 '12 at 6:05
What is $N$? What is the relation between $n$ and $N$? –  Makoto Kato Jun 15 '12 at 12:21

Let $p$ be a prime number. Let $x$ be a non-zero element in $\mathbb{Z}$. There exist integers $n$ and $a$ such that $x = p^na$, $(p, a)$ = 1. $n$ is uniquely determined by $x$. We denote $|x|_p = p^{-n}$. We define $|0|_p = 0$. For $x, y\in \mathbb{Z}$, we denote $d(x, y) = |x - y|_p$. $d$ is a metric on $\mathbb{Z}$. With this metric $d$, $\mathbb{Z}$ becomes a topological ring. We define $\mathbb{Z_p}$ as the completion of $\mathbb{Z}$ with respect to $d$. $\mathbb{Z_p}$ is a topological ring which contains $\mathbb{Z}$ as a dense subring.

$(p^n\mathbb{Z_p}), n = 1, 2, ...$ is a fundamental system of neighbourhoods of 0 in $\mathbb{Z_p}$. Each $\mathbb{Z_p}/p^n\mathbb{Z_p}$ is isomorphic to $\mathbb{Z}/p^n\mathbb{Z}$. Hence it suffices to prove that $\mathbb{Z_p} \cong \varprojlim\mathbb{Z_p}/p^n\mathbb{Z_p}$. Let $f_n: \mathbb{Z_p} \rightarrow \mathbb{Z_p}/p^n\mathbb{Z_p}$ be the canonical map for each n. When $n ≧ m$, $p^n\mathbb{Z_p} ⊂ p^m\mathbb{Z_p}$. Hence we can define a ring homomorphism $f_{mn}: \mathbb{Z_p}/p^n\mathbb{Z_p} \rightarrow \mathbb{Z_p}/p^m\mathbb{Z_p}$ by $f_{mn}(f_n(x)) = f_m(x)$. Let $A = \varprojlim\mathbb{Z_p}/p^n\mathbb{Z_p}$. Since $f_{mn}f_n = f_m$, we can define a map $f: \mathbb{Z_p} \rightarrow A$ by $f(x) = (f_n(x))$ for each $x \in \mathbb{Z_p}$. It's easy to see that $f$ is a continuous ring homomorphism. Since $∩p^n\mathbb{Z_p} = 0$, $f$ is injective.

Let $x = (x_n) \in A$. For each n, choose $a_n \in \mathbb{Z_p}$ such that $f_n(a_n) = x_n$. Since $a_n ≡ a_{n+1}$ (mod $p^n\mathbb{Z_p}$), $(a_n)$ is a Cauchy sequence in $\mathbb{Z_p}$. Hence there exists $a$ = lim $a_n$ in $\mathbb{Z_p}$. It's easy to see that $f(a) = x$. Hence $f$ is surjectve.

It remains to prove that $f$ is an open map. Let $π_n:A \rightarrow \mathbb{Z_p}/p^n\mathbb{Z_p}$ be the projection map. Since $f_n = π_nf$, $p^n\mathbb{Z_p} = (f_n)^{-1}(0) = f^{-1}((π_n)^{-1}(0))$. Hence $f(p^n\mathbb{Z_p}) = (π_n)^{-1}(0)$. Hence $f(p^n\mathbb{Z_p})$ is open. QED

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