How do you prove that $\Bbb{Z}_p$ is an integral domain? Let $\Bbb{Z}_p$ be the $p$-adic integers given by formal series $\sum_{i\geq 0} a_i p^i$.  I'm having trouble proving that it's an integral domain.
 A: The p-adic integers are a discrete valuation ring, i.e., with an absolute valuation $\nu$. Hence $0=ab$ means $0=\nu(ab)=\nu(a)\nu(b)$ in the real numbers, which have no zero divisors, hence either $\nu(a)=0$ or $\nu(b)=0$, i.e. $a=0$ or $b=0$.
A: Although the description of $\mathbb Z_p$ as "formal series" resembling power series in $p$ has some appeal, and was part of Hensel's original motivation, this is not a good viewpoint from which to ascertain many of the properties, as it turns out.
A relatively elementary description is that $\mathbb Z_p$ is the metric completion of $\mathbb Z$ with respect to the $p$-adic metric $|a-b|_p$, where the $p$-adic norm $|a|_p$ is $|p^n\cdot c|=p^{-n}$ for integer $c$ relatively prime to $p$. Yes, the exponent has a sign flip, so that highly-divisible-by-$p$ integers are tiny. (One can also, perhaps better-in-the-long-run, describe $\mathbb Z_p$ as the projective limit of $\mathbb Z/p^n$... which also shows that projective limits occur in real life, not just as artifacts... but usually this is too much "load" at the point $p$-adic things are introduced.)
From this viewpoint, it is much more feasible to prove that there are no zero-divisors... (using continuity).
A: As $\Bbb Z_p/p^n\Bbb Z_p\cong\Bbb Z/p^n\Bbb Z$, to show $a,b\ne0\Rightarrow ab\ne0$ it suffices to show that $ab\not\equiv0$ mod $p^n$ for a high enough power $p^n$ given $a,b\ne0$ (see vadim's hint in the comments).
A: The ring of $p$-adic integers $\mathbb{Z}_p$ is the endomorphism ring of $\mathbb{Z}(p^\infty)$ (the Prüfer $p$-group).
Since $\mathbb{Z}(p^\infty)$ is divisible, the image of an endomorphism is divisible, but the only divisible subgroups of $\mathbb{Z}(p^\infty)$ are $\{0\}$ and $\mathbb{Z}(p^\infty)$ itself, because all proper subgroups are finite. Therefore every nonzero endomorphism is surjective. Thus $fg=0$, with $g\ne0$ implies $f=0$.
