I am having trouble seeing why zero divisors exist in the $n$-adic numbers. Consider $\mathbb{Z}_{10}$ where here I want to think of this ring as the inverse limit of the rings $\mathbb{Z}/10^{n}\mathbb{Z}$, so a subset of their direct product. If $(a_{0},a_{1},a_{2},...)(b_{0},b_{1},b_{2},...)=(0,0,0,...)$ then we must have $a_{0}=2$ and $b_{0}=5$ (or vice versa), but then the only possible pairs for $(a_{1},b_{1})$ are $(50,2)$, $(25,4)$, $(20,5)$, $(5,20)$, $(4,25)$, and $(2,50)$, but none of these would give elements of $\mathbb{Z}_{10}$ ($a_{1}$ has to be congruent to $2$ mod $10$, so it must be $2$, but then $b_{1}$ is $50$ which is not congruent to $5$ mod $10$).
EDIT: As pointed out in the comments $(2,5)$ is not the only possibility, but it is one possibility: assuming $a_{n-1}$ and $b_{n-1}$ are relatively prime and $a_{n-1}b_{n-1}\equiv 0$ (mod $10^{n}$), we want to show that there exists $a_{n}$ and $b_{n}$ such that $a_{n}=10^{n}s_{n}+a_{n-1}$ and $b_{n}=10^{n}t_{n}+b_{n-1}$ and $a_{n}b_{n}\equiv 0$ (mod $10^{n+1}$). But the remainder part of $a_{n}b_{n}$ (mod $10^{n+1}$) is $10^{n}(s_{n}b_{n-1}+t_{n}a_{n-1}+k)$ and $s_{n}$ and $t_{n}$ can be chosen so that $(s_{n}b_{n-1}+t_{n}a_{n-1}+k)$ is any number. How to show that $a_{n}$ and $b_{n}$ are also relatively prime?
