# 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?

#### Solutions Collecting From Web of "The $p$-adic integers as a profinite group"

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.

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