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Following Jean Pierre Serre's way of introducing $p$-adic numbers in his Cours d'arithmétique (1970), referring to the footnotes of the post for some possibly useful details :

$\forall n\geq 1$, let $$A_n:=\mathbb Z/p_n\#\mathbb Z$$ This is the ring of integer classes ($\mod p_n\#$). An element of $A_n$ clearly defines an element of $A_{n-1}$; this results in a homomorphism $$\varphi_n:A_n\to A_{n-1}$$ which is surjective, and Kernel $p_{n-1}\#A_n$ The sequence $$...\to A_n\to A_{n-1}\to ...\to A_2\to A_1$$ forms a projective system, indexed by integers $\geq 1$

Notation. - I will note $A$, he projective limit of the system $(A_n,\varphi_n)$ defined above.

By definition, one element of $$\boxed{A=\varprojlim(A_n,\varphi_n)}$$ is a sequence $x=(...,x_n,...,x_1)$, with $$x_n\in A_n\land \varphi_n(x_n)=x_{n-1}\text{ if }n\geq 2$$

  1. I made sure in a previous post that $A$ is well constructed. But perhaps you will raise objections.
  2. With your help, I would like to construct examples of elements of $A$ and maybe see how to add and multiply them.
  3. And, I hope I'll be excused for this somewhat vague and naïve question, for me who is new to $p$-adic integers, what is $A$?

Notes :

  • $p_1=2, p_2=3, p_3=5, p_4=7,p_5=11,... $ are the primes;
  • $p_1\#:=p_1=2, p_2\#:=p_1\times p_2=2\times 3=6,p_3\#:=p_1\times p_2\times p_3=2\times 3\times 5=30,...$ are the primorials;
  • with notations above $\varphi_2:\mathbb Z/6\mathbb Z\to \mathbb Z/2\mathbb Z$ $$0\mapsto 0$$ $$1\mapsto 1$$ $$2\mapsto 0$$ $$3\mapsto 1$$ $$4\mapsto 0$$ $$5\mapsto 1$$
  • Using Chinese Remainder Theorem (CRT), ie isomorphism $$\mathbb Z/2\times...\times \mathbb Z/p_n\approx A_n$$ $\varphi_n((\xi_1,...,\xi_n)):=(\xi_1,...,\xi_{n-1})$, where $x=(\xi_1,...,\xi_n)\in A_n$ [$\varphi_n$ is simply a projection.]
Stéphane Jaouen
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    $A$ is isomorphic to the ring product of $\mathbb Z/p\mathbb Z$ over all primes $p.$ So we get an element $(a_p)_{p\in P}$ where $a_p\in \mathbb Z/p\mathbb Z.$ – Thomas Andrews Apr 03 '24 at 17:56
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    Given such an $(a_p)_{p\in P},$ we can find the corresponding element of $A_n$ by solving the Chinese remainder theorem problem for $x\equiv a_i\pmod {p_i}$ for $i=1,2,\dots,n.$ – Thomas Andrews Apr 03 '24 at 17:59
  • What does the "not quite" refer to, @stéphane? $5\notin A_2.$ – Thomas Andrews Apr 03 '24 at 18:03
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    No, it is surjective. I have shown you how to get the inverse. – Thomas Andrews Apr 03 '24 at 18:18
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    It wouldn't be true if it was $\prod Z/p_n#Z,$ but there is no sharp in my product. – Thomas Andrews Apr 03 '24 at 18:20
  • Looks right. Are you still arguing if the correspondence is a bijection? @Stéphane – Thomas Andrews Apr 03 '24 at 19:51
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    I am curious how this relates to the $p$-adics, which refers to a specific prime $p$, and a set of numbers that differ for each prime $p$. The $2$-adics are a different number system than the $3$-adics, which are both different from the $5$-adics, etc. In this construction, there is no given $p$ used in defining $A$ which is singled out and would produce a different $A$ than some other prime. This is an interesting construction, but I see no relation to the $p$-adics. – Paul Sinclair Apr 04 '24 at 18:14
  • @PaulSinclair: There is no relationship other than using the Serre construct to define the "3-adics", to define $A$ – Stéphane Jaouen Apr 04 '24 at 18:29
  • $(1+2Z,0+3Z,1+5Z,0+7Z,10+11Z,8+13Z,4+17Z,2+19Z,21+23Z,21+29Z,21+31Z,...)\longleftrightarrow (...,21+2310Z,21+210Z,21+30Z,3+6Z,1+2Z)$. Using Primorial number system, it can be designed by $(...:0:0:0:0:3:1:1)$(see https://math.stackexchange.com/questions/4893261/help-to-mathematically-theorize-the-eventual-value-of-primorial-number-system) – Stéphane Jaouen Apr 06 '24 at 05:48
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    It’s similar, but doesn’t have most of the nice properties of $p-adics$. It’s not a field. You can’t embed the rational numbers in it. It is a metric space (if you do the same thing as the p-adics) which is nice. It vaguely reminds me of adeles. – Eric Apr 13 '24 at 03:29

1 Answers1

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3.- $\forall x\in A$

$x=(...,x_n,...,x_1)$, with $\forall n\ge 1,$ $$x_n\in A_n\land \varphi_n(x_n)=x_{n-1}\text{ if }n\geq 2$$

Using Chinese Remainder Theorem (CRT), ie isomorphism $$\mathbb Z/2\times...\times \mathbb Z/p_n\approx A_n$$ $$(\xi_{1},......,\xi_{n})=x_n\in A_n$$ $$(\xi_{1},...,\xi_{n-1})=x_{n-1}\text{ if }n\ge2$$

So, let's consider $$f:A\to \prod_{n=1}^{+\infty}\mathbb{Z}/p_n\mathbb{Z}$$ $$x\mapsto(\xi_1,\xi_2, ..., \xi_n,...)$$

According to CRT, $\forall m\ge 1, \exists ! x_m\in \mathbb{Z}/p_1...p_m\mathbb{Z},$ $$\forall x\in \mathbb Z, \begin {cases} x\in \xi_1 \\ ...\\ x\in \xi_m \end{cases}\iff x\in x_m$$

So, $f$ is an isomorphism of rings and we have $$\boxed{A\approx \prod_{n=1}^{+\infty}\mathbb{Z}/p_n\mathbb{Z}}$$

2.- Following comments in other posts on similar topics and here, I've constructed a few examples:

2.1- $(1+2Z,0+3Z,1+5Z,0+7Z,10+11Z,8+13Z,4+17Z,2+19Z,21+23Z,21+29Z,21+31Z,...)\longleftrightarrow (...,21+2310Z,21+210Z,21+30Z,3+6Z,1+2Z)$

Using Primorial number system, it can be designed by $$(...:0:0:0:0:3:1:1)_{Primorial}$$ Note that $(3:1:1)_{Primorial}:=3\times 6+1\times 2+1\times 1$

2.2- $$\color{red}{carry........1..1...}$$ $$(...:0:0:0:3:1:1)$$ $$+$$ $$(...:0:0:0:0:1:1)$$ $$=(...:0:0:4:0:0)$$

2.3- $P=\alpha\delta$, with $$\alpha=(...:c:b:a:0:1:9:6:2:2:1)_{Primorial}$$$$\equiv (1,2,2,1,8,3,11,...)\in\mathbb{Z}/2\times\mathbb{3}\times \mathbb{Z}/5\times \mathbb{Z}/7\times \mathbb{Z}/11\times \mathbb{Z}/13\times \mathbb{Z}/17\times... $$ and $$\delta=(...:\gamma:\beta:\alpha:0:3:4:4:0:1:1)_{Primorial}$$$$\equiv (1,0,1,4,4,6,4,...)\in \mathbb{Z}/2\times\mathbb{3}\times \mathbb{Z}/5\times \mathbb{Z}/7\times \mathbb{Z}/11\times \mathbb{Z}/13\times \mathbb{Z}/17\times... $$.

$P=\color{blue}1+2k_0. 1+2k_0=0 \text{ in }Z/3\implies k_0=\color{blue}1+3k_1. P=1+2(1+3k_1)=3+6k_1. P=3+1k_1 \text{ in }Z/5\implies k_1=3+5k_2. P=\color{blue}1\times 1+\color{blue}1\times 2+\color{blue}3\times 6 +30k_2. P=0+2k_2 \text{ in }Z/7\implies k_2=2+7k_3. P=\color{blue}1\times 1+\color{blue}1\times 2+\color{blue}3\times 6 +\color{blue}2 \times 30+210k_3...$

So, $$P=(...:\color{blue}{2:3:1:1})_{Primorial}$$ (example constructed with $4\,397\times 7\,893=34\,705\,521$)

Note that it is not only the pleasure of intellectual construction work that has driven me to build these examples, but these decompositions intervene in an algorithm for searching for the decomposition of a semi-prime number $N$ into the product $$N=pq$$

2.4- Here is an illustration of the embedding of $\mathbb Z$ into $\prod_{n=1}^{+\infty}\mathbb{Z}/p_n\mathbb{Z}$ via $$n\to (n,n,n,n,n,n,...)$$ $$$$\begin{bmatrix}\mathbb{Z}/2 & \mathbb{Z}/3 & \mathbb{Z}/5 & \mathbb{Z}/7 & ...&... \\ ..&..&..&..&..&...\\-1&-1&-1&-1&-1&...\\0 & 0 & 0 & 0 & 0&... \\ 1& 1 & 1 & 1 & 1 &...\\0 & 2 & 2 & 2 & 2 &...\\1 & 0 & 3 & 3 & 3 &...\\0 & 1 & 4 & 4 & 4 &...\\1 & 2 & 0 & 5 & 5 &...\\0 & 0 & 1 & 6 & 6 &...\\1 & 1 & 2 & 0 & 7 &...\\0 & 2 & 3 & 1 &8&...\\1&0&4&2&9&...\\0&1&0&3&10&...\\1&2&1&4&0&...\\..&..&..&..&..&...\end{bmatrix}$$$$

This question had benefited from a bounty but it was exhausted without any answer being provided. I hope to renew interest in the search for examples in $(A,+,\times)$. It’s very simple, especially for addition in primorial numeration, or through writing $$\boxed{A\approx \prod_{n=1}^{+\infty}\mathbb{Z}/p_n\mathbb{Z}}$$ where, as Serre wrote about $\mathbb Z_p$, addition and multiplication are defined ‘coordinate by coordinate’.

Stéphane Jaouen
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