Can we have $n$-adic integers?

Question

I am starting to read about $p$-adic integers and one thing struck me. I have some related questions below. Can anybody help me out?

Can we have $n$-adic integers if we try to construct it in the same way? Like if we take $n=4$, then the number $10=2+4.2$, in $4$-adic integer becomes $22$?

If $n=pq$, does numbers represented in this $n$-adic form split into a number in $\mathbb Z_p\times\mathbb Z_q$, where $\mathbb Z_p,\mathbb Z_q$ are fields of $p$ and $q$-adic integers? The reason I am asking it is, clearly we can represent numbers in this form which I call $n$-adic integers, in which the number $a_0a_1...a_k$, $a_i\in \mathbb{Z}_n$, can be splitted into a tuple in $\mathbb Z_p\times\mathbb Z_q$, by mapping each $a_i$ to $a_i\;mod\;(p)$ and $a_i\;mod\;(q)$. Am I correct?

related to https://math.stackexchange.com/questions/2019647/the-n-adic-integers-are-isomorphic-to-the-product-of-the-p-i-adic-integers-w — , Dec 13 '20 at 07:56
$\Bbb{Z}_p = \varprojlim \Bbb{Z/p^j Z}$ is the ring of limits of sequences of integers that converge $\bmod p^j$ for all $j$. It is the completion (limits of Cauchy sequences) of $\Bbb{Z}$ for the $|p^j a|_p = p^{-j}$ absolute value ($p\nmid a$).
It is not a field, $p$ has not inverse, the field is $\Bbb{Q}_p = \Bbb{Z}_p[p^{-1}]$.

Then $\Bbb{Z}n= \varprojlim \Bbb{Z/n^j Z}$ is the ring of limits of sequences of integers that converge $\bmod n^j$ for all $j$. It is the completion for the norm $ |a|_n=\sum{p|n}|a|p$. Yes $\Bbb{Z}_n\cong \prod{p|n} \Bbb{Z}_p$. — reuns, Dec 13 '20 at 08:17
Thanks.. So it has applications then? And we can define all properties of $n$-adic similar to $p$-adic? And any computation in $Z_n$ can be splitted in the decomposed rings? — roydiptajit, Dec 13 '20 at 08:40
Also in the same way can we define $\mathbb{Q}_n=Z_n[n^{-1}]$? And $\mathbb{Q}_n≅∏p|n\mathbb{Q}_p$? I am not sure about these... @reuns — roydiptajit, Dec 13 '20 at 08:45
Note: if $n$ is a prime power, $n=p^k$, then the ring of $n$-adic integers is isomorphic to the ring of $p$-adic integers. If $n \ge 6$ is not a prime power, then the ring of $n$-adic integers has zero divisors. Math students should prove these once in their life. — GEdgar, Dec 13 '20 at 11:45
So, $\mathbb{Z}_{p^k}[p^{-k}]$ is a field? and it's exactly $\mathbb{Q}_p$?@GEdgar — roydiptajit, Dec 13 '20 at 13:41
Your $10=22$ thing doesn't make sense. Is it clear to you that my first comment defines the $p$-adic integers in two ways ? One of those ways works for any $n$ and it is obvious that it gives the same for $p$ and $p^k$. — reuns, Dec 13 '20 at 14:03
Yes, that was a mistake. It males sense from your explanation — roydiptajit, Dec 13 '20 at 18:01
Does this answer your question? How do p-adic fields degenerate for non-prime $p$? — Torsten Schoeneberg, Nov 30 '22 at 02:03

Can we have $n$-adic integers?

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