# 4-adic numbers and zero divisors

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.

• 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. – Paul K Sep 22 '18 at 7:48
• @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. – badjohn Sep 22 '18 at 7:52
• as a hint, if you imagine converting from base 4 to base 2, not much changes. – user207119 Sep 23 '18 at 10:16
• @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. – badjohn Sep 23 '18 at 10:56
• 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. – samerivertwice Dec 26 '19 at 20:03

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 $$\Bbb Z_{p^k}\cong\Bbb Z_p$$.
• 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. – Lubin Sep 22 '18 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.
• 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. – badjohn Oct 2 '18 at 13:02
• 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. – Paul K Oct 2 '18 at 14:21