The following holds for the ring $ \mathbb{Z}_p, p \in \mathbb{P}$:
The ring $ \mathbb{Z}_p $ is a principal ideal domain, especially an integral domain.
I try to understand the following proof:
Proof:
We will show that $ \mathbb{Z}_p $ is an integral domain. The fact that $ \mathbb{Z}_p $ is a principal ideal domain can be concluded from the proposition "$\mathbb{Z}_p $ contains only the ideals $0$ and $p^n \mathbb{Z}_p $" for $n \in \mathbb{N}_0 $. It holds $ \bigcap_{n \in \mathbb{N}_0} p^n \mathbb{Z}_p=0$ and $ \mathbb{Z}_p / p^n \mathbb{Z}_p \cong \mathbb{Z} \ p^n \mathbb{Z}$. Especially $ p \mathbb{Z}_p$ is the only maximal Ideal. "
Let $ x=(\overline{x_k}), y=(\overline{y_k}) \in \mathbb{Z}_p \setminus \{0\} $ and $ z=(\overline{z_k})=xy=(\overline{x_ky_k})$.
Since $ x,y \neq 0$ there are $ m,n \in \mathbb{N}_0 $ with $x_n \not\equiv 0 \mod{p^{n+1}}$ and $ y_m \not\equiv 0 \mod{p^{m+1}}$. We set $ l:=n+m+1$. From $ x_l \equiv x_n \mod{p^{n+1}}$ and $ y_l \equiv y_m \mod{p^{m+1}}$ we deduce the existence of the partitions $ x_l=u \cdot p^{n'}$ and $ y_l=v \cdot p^{m'}$ with $gcd(u,p)=gcd(v,p)=1$, $ n' \leq n $ and $ m' \leq m$. Then since $ n'+m'<l $ we have that
$ z_l=uvp^{n'+m'} \not\equiv 0 \mod{p^{l+1}}$, and so $ z \neq 0$.
Could you explain me step by step the above proof?
Let $x=(\overline{x_k}), y=(\overline{y_k}) \in \mathbb{Z}_p \setminus \{0\}$. Since $x \neq 0, y \neq 0,$ there exists $n, m \in \mathbb{N}$ such that $x_n \neq 0$ in $\mathbb{Z}/p^{n+1}\mathbb{Z}$ and $y_m \neq 0$ in $\mathbb{Z}/p^{m+1}\mathbb{Z}$ (Choose the smallest integers $m, n$ with the that property). Let $l := m + n + 1.$ Then $x_l = x_n$ in $\mathbb{Z}/p^{n+1}\mathbb{Z}$ and $y_l = y_m$ in $\mathbb{Z}/p^{m+1}\mathbb{Z}.$ By the choice of $m, n$, it follows that $x_l = up^{n}, y_l = vp^{m}$ with gcd$(u, p) = 1$ and gcd$(v, p) = 1.$ Thus $x_ly_l = uvp^{m+n} \neq 0$ in $\mathbb{Z}/p^{l+1}\mathbb{Z}.$