4-adic numbers and zero divisors

Question

The $p$-adic numbers form an integral domain provided that $p$ is prime.

Let's look at the $n$-adic numbers when $n$ is not prime.

Case $n = 10$

There are zero divisors. See this previous question.

Case $n = pq$ where $p$ and $q$ are coprime (not necessarily prime but not $1$).

There are also zero divisors. A similar construction works.

Case $n = p^k$ where $p$ is prime and $k > 1$

I have not figured this one out yet, not even the simplest case of $n = 4$. The construction in the previous question above does not work and I have not found an alternative yet. Looking at approximations in $\mathbb{Z}_4$, $\mathbb{Z}_{16}$, $\mathbb{Z}_{64}$, etc just leads me to zeros divisors ending in zeroes suggesting, but not proving, that there are none.

Note that I am using $\mathbb{Z}_n$ for the integers modulo $n$ and not the $n$-adic numbers. I think that I have seen it used for both. What is usual if you want discuss both at the same time?

Another previous question asks why $4$-adic numbers are not possible. The answer seems to be that they are possible but a norm cannot be defined. So, it leaves the existence of zero divisors open.

Are there zero divisors in the $4$-adic numbers? Are there idempotents in the $4$-adic numbers?

I have not looked at $9$-adic or other prime powers yet.

Please don't answer directly but some hints would be appreciated.

Answer 1

If one defines the $4$-adic numbers as the inverse limit $$\Bbb Z_4\cong\lim_{\longleftarrow}(\Bbb Z/4^n\Bbb Z)$$ then $\Bbb Z_4\cong\Bbb Z_2$, the $2$-adic numbers.

In general $\Bbb Z_{p^k}\cong\Bbb Z_p$.

Answer 2

In "p-adic Analysis Compared with Real" from Svetlana Katok it is proven that for distinct primes $p_1,\ldots,p_k$ we have $\mathbb{Q}_g = \mathbb{Q}_{p_1} \oplus \ldots \oplus \mathbb{Q}_{p_k}$. She doesn't mention the case of non-distinct primes. Moreover she mentions that Hensel proved the above fact.