I am studying $p$-adic Rings and let me explain my understanding and doubt here. As I understood, Let $p$ be a rational prime and $Z$ denotes ring of integers, then form cartesian product
$$P=Z/pZ \times Z/p^2Z......Z/p^nZ \times ...$$ ($n=0,1,2,3.......$) . and let for $r \ge s$ and
$$\phi^r_s: Z/p^{r+1}Z \to Z/p^{s+1}Z$$ be the natural mapping, Then consider subring $Z_p$ of $P$ defined as
$$Z_p=\left\{\alpha =(\alpha_0,\alpha_1,...,\alpha_n,....)\right\}$$ for which $\phi^{n+1}_{n}(\alpha_{n+1})=\alpha_n$ for every $n \ge 0$.
$\textbf{1)- }$Up untill it is clear to me. But the next line in my book says- this means that if $\alpha_i$ is the residue class mod $p^{i+1}$ of the integer $a_i$, then $$a_{n+1} \equiv a_n \pmod{p^{n+1}}\;,\;\; n \ge 0$$
Now $\alpha_i$ is some element of the ring $Z/p^{i+1}Z$ so it is of the form $b_i(p^{i+1}Z)$ for some rational integer $b_i$, so this $a_i$ mentioned above is some integer s.t. $a_i \equiv b_i \hspace{.1cm} mod(p^{i+1}Z)$ ? Does residue class means this? Why calling $\alpha_i$ a residue class, residue class is a set of elements but I guess this is the meaning here, as I interpret it (i.e. calling some representative as the whole class).
$\textbf{2)-}$ Now if let $\alpha=(\alpha_1,\alpha_2,\alpha_3,......)$ such that $\alpha_n=0$, that means $\alpha_n=a_n(p^{n+1}Z)$ where $a_n$ is a multiple of $p^{n+1}$. But why does it imply that $a_n,a_{n+1},...$ are all divisible by $p^{n+1}$ s.t. $\alpha_i$ is the residue class mod $p^i$ of $a_i$. (As used in one of the proofs I was going through).
By our $\phi$, we should have that $\phi^{n}_{n-1}(\alpha_n)=\alpha_{n-1}$ and $\alpha_n$ being zero, should imply that $\alpha_{n-1}=0$ being homomorphic image of $0$ and {$\alpha_0,\alpha_1,....\alpha_n$} these all must be zero.
Please explain my these two doubts, and especially second one.
I think everything becomes clearer if you look at your elements in a $p$-base. Take an $\alpha_n\in \mathbb{Z}/p^n\mathbb{Z}$. Then $\alpha_n$ in $p$ base is some number of the form $a_0 + a_1p+\ldots+ a_{n-1}p^{n-1}$, where the $a_i\in\{0,\ldots,p-1\}$. Note that by extending we can interpret the elements of $\mathbb{Z}_p$ as power series in $p$.
Next, the obvious map to $\mathbb{Z}/p^m\mathbb{Z}$ you talk about (for $m<n$) is $\alpha_n (\mod p^n)\mapsto \alpha_n(\mod p^m)$, or \begin{equation} a_0 + a_1p+\ldots+ a_{n-1}p^{n-1}\stackrel{\phi^{n}_{m}}{\mapsto} a_0 + a_1p+\ldots+ a_{m-1}p^{m-1}. \end{equation} Therefore $a_n = a_m (\mod p^m)$, which answers question 1. I think that with this interpretation you should be able to solve number 2?