Is the endomorphism of $\mathbb{Z}_{p}$ induced by multiplication by $p^{n}$ surjective? Let $p$ be a prime number. Is it true that $p^{n}\mathbb{Z}_{p}\cong\mathbb{Z}_{p}$ as additive groups for any natural number $n$ and if so, why? Here, $\mathbb{Z}_{p}$ denotes the ring of $p$-adic integers.
Any help would be very much appreciated.   
 A: The multiplication by $p^n$ is injective - after all, $\mathbb{Z}_p$ is an integral domain - and so it gives an isomorphism $$\mathbb{Z}_p \cong p^n \mathbb{Z}_p$$ to its image. That doesn't make it surjective as an endomorphism, though, and it isn't, because $p^n$ isn't invertible in $\mathbb{Z}_p$.
A: To elaborate a little, note elements $\mathbb{Z}_p$ are sequences $(a_0,a_1,...,a_n,...)$ with $a_i\equiv a_j\mod (p^i)$ whenever $i<j$. Denote by $p(n)$ the greatest number such that $p^{p(n)}|n$ ($+\infty$ if $n=0$). Now from the relation we see $p(a_j)=p(a_i)$ if $a_i\ne 0$. So if $a,b\ne 0\in \mathbb{Z}_p$, $p(a_i)$ and $p(b_i)$ will be finite and stationary for sufficiently large $i$, so will it be for $ab$, which proves that $\mathbb{Z}_p$ is an integral domain.
A: No (for the question in title) and yes (for the question in the body).
There is a ring morphsism $\newcommand\Z{\Bbb Z}\Z_p\to \Z/p\Z$, because of the inverse limit construction that defines the $p$-adic integers (there are similarly morphisms to any $\Z/p^k\Z$). Under this morphism multiplication by $p^n$ transforms into multiplication by $0^n$ and since that is not surjective in $\Z/p\Z$ (unless $n=0$ of course) it follows that multiplication by $p^n$ cannot be surjective in $\Z_p$.
However multiplication by any nonzero element, such as $p^n$, in the integral domain $\Z_p$ is injective and thus induces an isomorphism of Abelian groups from $\Z_p$ to the image of the multiplication (a principal ideal of $\Z_p$), in particular here $\Z_p\cong p^n\Z_p$ as Abelian groups (or as $\Z_p$-modules if you like, but not as rings since $p^n\Z_p$ is not a (unitary) ring at all).
