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.

  • 2
    $\begingroup$ I would expect that we have something like $\mathbb{Z}_{p^k} = \mathbb{Z}_p$ and $\mathbb{Z}_{ab} = \mathbb{Z}_a \times \mathbb{Z}_b$, but I didn't check it. $\endgroup$
    – user301452
    Sep 22, 2018 at 7:48
  • $\begingroup$ @JyrkiLahtonen That suggests another question. What are the possibilities? To avoid additional questions and answers hidden in comments, I will post another question unless you can point me to an existing answer. $\endgroup$
    – badjohn
    Sep 22, 2018 at 7:52
  • $\begingroup$ as a hint, if you imagine converting from base 4 to base 2, not much changes. $\endgroup$
    – user207119
    Sep 23, 2018 at 10:16
  • 1
    $\begingroup$ @Steven Thanks. I figured that out now. I briefly considered the 100-adic numbers and it was obviously that they were essentially the same as the 10-adic. $\endgroup$
    – badjohn
    Sep 23, 2018 at 10:56
  • $\begingroup$ In answer to your question I understand $\Bbb Z/n\Bbb Z$ is better for cyclic groups, especially when used in the setting of p-adic numbers. $\endgroup$ Dec 26, 2019 at 20:03

2 Answers 2


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

  • $\begingroup$ Thanks - that's an answer that I had not suspected yet. It explains why I could not find any zero divisors. Also, reviewing my attempts at finding them, I see that maybe I should have suspected this. $\endgroup$
    – badjohn
    Sep 22, 2018 at 7:54
  • 2
    $\begingroup$ It’s fairly simply seen: the families $\{p^n\Bbb Z\}$ and $\{p^{2n}\Bbb Z\}$ are coterminal: each member of the one family contains a member of the other; and vice versa. $\endgroup$
    – Lubin
    Sep 22, 2018 at 18:38

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.

  • $\begingroup$ Thanks. We also have, from Lord Shark, $\Bbb Z_{p^k}\cong\Bbb Z_p$ so between the two results, we have the whole story. $\endgroup$
    – badjohn
    Oct 2, 2018 at 13:02
  • 2
    $\begingroup$ I would pay a bit attention there. The $g$-adic numbers in Katok are defined via a pseudo-norm which is defined analogously to the $p$-adic norm. I am not sure if one can obtain the $g$-adic integers like this if you define them via the inverse limit. $\endgroup$
    – user301452
    Oct 2, 2018 at 14:21
  • $\begingroup$ Thanks. I have some more study to do. $\endgroup$
    – badjohn
    Oct 2, 2018 at 14:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .