I want to prove $ \Bbb{Z}_p$ is isomorphic to $\lim_{←}\Bbb{Z}_p/p^n \Bbb{Z}_p$ as a ring. Here, $\Bbb{Z}_p$ is completion (as metric space) of $\Bbb{Z}$ with p adic metric.
My try:
Let define natural ring hom $ \Bbb{Z}_p→\lim_{←}\Bbb{Z}_p$ by $a→(\cdots,a \pmod {p^2},a \pmod p)$. This is clearly injective since Ker is $0$ (note that $ \bigcap {p^n\Bbb{Z}_p}＝0$).
But how can I prove surjectivity ? I guess I use completeness of $\Bbb{Z}_p$(every caushy sequence converges), but I don't have confident. Thank you for your help.
Of course the answer depends on the definitions, and in your case, the completeness is indeed the key.
Given any $(a_1+p\mathbb Z, a_2+p^2\mathbb Z, \cdots)\in \lim \mathbb Z/p^n\mathbb Z$, we show that $\{a_n\}\in\mathbb Z\subset\mathbb Z_p$ is a Cauchy sequence. Due to compatibility, we know that $a_n\equiv a_m \mod p^{m}$ for all $n\ge m$, hence $|a_n-a_m|_p\le p^{-m}$ which is small for sufficiently large $m$.
Let $a=\lim a_n$, we have the image of $a$ is $(a_i+p^i\mathbb Z)$. Indeed, let $n\rightarrow\infty$ in $|a_n-a_m|_p\le p^{-m}$ we have $|a-a_m|_p\le p^{-m}$, i.e. $a\equiv a_m \mod p^m$.