# How can i show $\mathbb{Z}_p/p^n\mathbb{Z}_p \approx \mathbb{Z}/p^n\mathbb{Z}$

I would like to show $$\mathbb{Z}_p/p^n\mathbb{Z}_p \approx \mathbb{Z}/p^n\mathbb{Z}$$.

I have this proof but i dont know how can i finish it:

I know $$\forall x\in\mathbb{Z}_p$$ $$\exists !(\alpha_n)$$ Cauchy sequence converging to $$x$$ such as:

$$\cdot$$ $$\alpha_n\in\mathbb{Z}:0\leq\alpha_n\leq p^{n}-1$$.

$$\cdot$$ $$\alpha_n\equiv \alpha_{n-1}(mod\hspace{0.1cm}p^n)$$ para todo $$n=1,2,...$$

I define $$\psi_n:\mathbb{Z}_p\longrightarrow \frac{\mathbb{Z}}{p^n\mathbb{Z}}:\psi(x)= \alpha_n\hspace{0.2cm} mod \hspace{0.1cm}p^n$$

It is easy to check $$ker(\psi_n)=p^n\mathbb{Z}_p$$ and sobrejective. My problem is to show $$\psi_n$$ is a homomorphism of ring.

For that: Let be $$x,y\in\mathbb{Z}_p$$ and $$(\alpha_n)$$, $$(\beta_n)$$ associated with $$x$$ and $$y$$. I showed $$\alpha_n +\beta_n \longrightarrow x+y$$ and $$\alpha_n\beta_n \longrightarrow xy$$ and $$\alpha_n+\beta_n\equiv \alpha_{n-1}+\beta_{n-1}(mod\hspace{0.1cm}p^n)$$, $$\alpha_n\beta_n\equiv \alpha_{n-1}\beta_{n-1}(mod\hspace{0.1cm}p^n)$$ $$\forall n=1,2,...$$

But i couldnt show $$0\leq\alpha_n\beta_n\leq p^{n}-1$$, and $$0\leq\alpha_n+ \beta_n\leq p^{n}-1$$.

So i cant say $$\alpha_n+\beta_n$$ is the correspondence Cauchy sequence for x+y or $$\alpha_n\beta_n$$ is the correspondence Cauchy sequence for $$xy$$.

• Well, how are addition and multiplication defined on $\mathbb Z_p$ for you to begin with? Because you have the problem that your set of representing Cauchy sequences is not closed under those anyway, before even trying to show some ring homomorphism. (As an example, take $\alpha_n=p^n-1$ and add it to itself.) Commented May 13, 2020 at 14:50