Rings of p adic integers contains copy of the integers I need to prove that p adic integers contains copy of integers. I proved this way:
For integer $n=n_0$, we can define $n_i$ as $n_i = n_{i+1} p^{i+1}+ a_i$ where $a_i$ is between $0 $ and $p-1$. Hence all integers are in p adic integer ring. 
Is this correct proof? Actually, i can't understand what 'contains copy of integers' means. What is the correct definition?
 A: Usually, the definition of $p$-adic integers is not given in a way, that $\mathbb{Z}$ is a subset of $\mathbb{Z}_p$. Indeed, if you define the $p$-adic integers by the inverse limit $\varprojlim \mathbb{Z}/(p^k)$, then this is not the case. There is, however, an injective homomorphism
$$
\pi : \mathbb{Z} \hookrightarrow \varprojlim \mathbb{Z}/(p^k),
$$
given by the universal property of inverse limits. Hence $\mathbb{Z}_p$ contains a copy of $\mathbb{Z}$.
Remark: One can show that $\mathbb{Z}_p$ also contains a copy of $\mathbb{Z}[x]$,  with $x$ represented by some element of $\mathbb{Z}_p$ that is transcendental over  $\mathbb{Z}$.
A: Here is a concrete realization of this injective homomorphism $\mathbb{Z} \hookrightarrow \mathbb{Z}_p$. Let's assume that we have defined the $p$-adic integers as an inverse limit, specifically
$$
\mathbb{Z}_p = \varprojlim_n \mathbb{Z}/p^n \mathbb{Z},
$$
then elements of $\mathbb{Z}_p$ correspond to "coherent sequences" in the infinite product $\prod_{n=1}^{\infty} \mathbb{Z}/p^n \mathbb{Z}$; that is, an infinite tuple $(a_1,a_2,\ldots) \in \prod_{n=1}^{\infty} \mathbb{Z}/p^n \mathbb{Z}$ is in $\mathbb{Z}_p$ iff $a_{n+1} \equiv a_n \bmod p^n$ for all $n$. Intuitively, if you take the term $a_{n+1}$ in this tuple, then it must reduce to $a_n$ mod $p^n$, and to $a_{n-1}$ mod $p^{n-1}$, and so on.
In this case, an explicit injective homomorphism $\mathbb{Z} \hookrightarrow \mathbb{Z}_p$ is given by sending an integer $m$ to the "constant" tuple $(m,m,\ldots)$ consisting of the reductions of $m$ mod $p^n$ for each $n$.
A: A sketch: take an integer in $\Bbb Z$. Write it in base $p$ as $b_0 + b_1 p + b_2 p^2 + \dots + b_n p^n$, with $0 \le b_i < p$. Consider the sequence
$$b_0 \in \Bbb Z _p, \space b_0 + b_1 p \in \Bbb Z _{p^2}, \space b_0 + b_1 p + b_2 p^2 \in \Bbb Z _{p^3}, \dots, \space b_0 + b_1 p + \dots + b_n p^n \in \Bbb Z _{p^n} ,$$
and from here upwards this stays constant, i.e. $b_0 + b_1 p + \dots + b_n p^n \in \Bbb Z _{p^k}, k \ge n$.
You send each integer into its representation in base $p$ and this gives you an embedding of $\Bbb Z$ in $\Bbb Z _p$.
A: What "contains a copy of the integers" means is that there is a subring $R$ of $\mathbb Z_p$ which is isomorphic to $\mathbb Z$ as a ring. For $p$-adics, people usually think of $R$ and $\mathbb Z$ as being equal, instead of just isomorphic.
The proof you're getting at uses the fact that every integer can be written in base $p$. The tricky thing, though, is that in the $p$-adics, negative integers have infinite expansions, like
$$
-1 = (p-1) + (p-1)p + (p-1)p^2 + \cdots.
$$
Your argument would have to take these expansions into account.
(I suppose that if you use the inverse limit definition of $\mathbb Z_p$ then the problematic expansion of $-1$ is immaterial because we can also express $-1$ as $(-1,-1,-1,\dots)$)
A more highbrow way to show that $\mathbb Z\subset\mathbb Z_p$ is just to observe that $\mathbb Z_p$ has characteristic zero, which means that
$$
\underbrace{1 + 1 + \cdots + 1}_{n\text{ times}} = 0
$$
if and only if $n=0$. Any ring with this property contains a unique subring isomorphic to $\mathbb Z$, namely the subring generated by $1$. If this is not obvious to you, try proving it.
