Structure of $(a,b)$-adic natural numbers

Question

Introduction

One way to define the $n$-adic integers is to start from the ring of modular integers $\mathbb{Z}/n\mathbb{Z}$, the quotient of $\mathbb{Z}$ by the ideal $n\mathbb{Z}$. There is an obvious sequence of ring homomorphisms

$$\cdots \to \mathbb{Z}/n^r\mathbb{Z} \cdots \to \mathbb{Z}/n^3\mathbb{Z} \to \mathbb{Z}/n^2\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$$

given by expressing the integer modulo $n^r$ in base $n$ notation and taking the last $r-1$ digits (i.e. reducing modulo $n^{r-1}$).

The $n$-adic integers $\mathbb{Z}_n$ can be defined as the inverse limit of this sequence. The standard intuitive picture is that of "integers" with infinitely many digits to the left in base $n$.

Similarly, we can define what I will call the $(a,b)$-adic natural numbers starting from a semiring of the form $\mathbb{N}/(a\mathbb{N}+b)$, the quotient of the nonnegative integers $\mathbb{N}$ by the congruence $a\mathbb{N}+b$ (where we identify two numbers whenever they are both greater or equal to $b$ and differ by a multiple of $a$). There is again a sequence of semiring homomorphisms

$$\cdots \to \mathbb{N}/(a^3\,\mathbb{N}+b(a^2+a+1)) \to \mathbb{N}/(a^2\,\mathbb{N}+b(a+1)) \to \mathbb{N}/(a\mathbb{N}+b)$$

given by expressing the numbers in a certain modified base $a$ notation with $b$ extra digits (for $b=1$ it is called bijective numeration, but I'm not aware of any standard name for it when $b>1$) and truncating. For example, one can express the elements of $\mathbb{N}/(16\mathbb{N}+10)$ with two-digit representatives in base $4$ with $2$ extra digits:

$$0,1,2,3,4,5,12,13,14,15,22,23,24,25,32,33,34,35,42,43,44,45,52,53,54,55$$

and the homomorphism to $\mathbb{N}/(4\mathbb{N}+2)$ is given by taking the units digit.

The semiring of $(a,b)$-adic naturals $\mathbb{N}_{a,b}$ can be then defined analogously as the inverse limit of this sequence. As before, one can think of them as "naturals" with infinitely many digits expressed in this generalized base $a$ notation. Note that the special case $\mathbb{N}_{a,0}$ is isomorphic to $\mathbb{Z}_a$.

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Motivation

My motivation to do this is that the exponentiation function $(m,n)\mapsto m^n : \mathbb{N}^2 \to \mathbb{N}$ (where we take $0^0=1$) descends to the quotient $\mathbb{N}/(a\mathbb{N}+b)$ whenever $\lambda(a)\mid a$ and $\nu(a)\le b$, where $\lambda$ is the Carmichael function and $\nu(a)$ denotes the highest exponent appearing in the prime factorization of $a$. This follows from this property and the fact that $k^b>b$ for any $b$ and $k>1$.

Since these two conditions imply that $\lambda(a^r)\mid a^r$ and $\nu(a^r)\le b(a^{r-1}+\cdots+a+1)$ for any $r$, exponentiation is well-defined for the entire sequence (in a compatible way, though I won't prove that here), and hence is well-defined when we take the limit. This way we get semirings $\mathbb{N}_{a,b}$ admitting an everywhere defined exponentiation function $(m,n)\mapsto m^n : \mathbb{N}_{a,b}^2 \to \mathbb{N}_{a,b}$, which together with the nonnegative reals $\mathbb{R}^+$ are the only exponentiation-preserving "completions" of $\mathbb{N}$ that I know of.

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Question

I would like to know the basic structure of $\mathbb{N}_{a,b}$. It is known that the $n$-adic integers $\mathbb{Z}_n$ can be decomposed as a direct product of irreducible $\mathbb{Z}_p$ factors, one for each prime $p$ appearing in the prime factorization of $n$.

Is there an analogous factorization of $\mathbb{N}_{a,b}$ with $b>0$ in terms of some special irreducible $\mathbb{N}_{x,y}$? If not, can these semirings be expressed in terms of other more familiar objects?

I am especially interested in factorizations that preserve the exponential structure whenever it's there, but I would be also satisfied with a general statement for any $a, b$.

Answer 1

As Keith has mentioned in the comments, by chance we have recently been thinking about the version of this where rather than fixing $a$, you let it be arbitrary as well. The description in this case is entirely analogous, so let me present it.

Let $N(m,k)$ be the quotient of $\mathbb N$ by the congruence generated by $m\sim m+k$, that is, $\mathbb N/(k\mathbb N+m)$. It consists of the initial segment $0,\dots,m-1$ followed by a length $k$ loop $m,\dots,m+k-1$. Let me call elements of the initial segment "small" elements of $N(m,k)$, and the remaining elements "large".

For $b>0$, $\mathbb N_{a,b}$ is the inverse limit of $N(m,k)$ for $k$ ranging over powers of $a$ and $m$ arbitrary (it's not hard to see that your restriction to $k=b(a^l+\dots+1)$ doesn't change anything, in particular it is independent of $b$). Elements of the inverse limits are given by tuples of $x_{m,k}\in N(m,k)$ which are compatible under maps corresponding to coarsening the quotients. There are two possibilities:

1. Suppose some $x_{m,k}\in N(m,k)$ is small, given by some element $n\in\mathbb N$. Then for any other $N(m',l)$ which maps onto $N(m,k)$, we must have $x_{m',l}=n$ too - this is because in the quotient of $\mathbb N$ defining $N(m,k)$, no other element is identified with $n$, so there can't be any other such element in a finer quotient either. So these elements correspond precisely to elements of $\mathbb N$.

2. All $x_{m,k}\in N(m,k)$ are large. If we consider the elements $x_{0,k}\in N(0,k)=\mathbb Z/k$, those define an element of $\mathbb Z_a$. And further, any such element uniquely extends to the whole inverse system, since there is a unique large element in $N(m,k)$ which maps to $x_{0,k}$ in $N(0,k)$. So those elements correspond to elements of $\mathbb Z_a$.

Thus we see that as a set, $\mathbb N_{a,b}$ is in a natural bijection to $\mathbb N\sqcup\mathbb Z_a$. We can describe its monoid structure quite explicitly - both $\mathbb N$ and $\mathbb Z_a$ are subsemigroups (but the latter is not a submonoid! the identity is different), and the sum of $n\in\mathbb N$ and $x\in\mathbb Z_a$ corresponds to the usual sum $n+x$ viewed in $\mathbb Z_a$.

Being an inverse limit of finite sets, this structure also has a natural topology, which we can describe as follows: $\mathbb Z_a$ is a closed subset, and it is enough to characterize convergent sequences of elements $\mathbb N$. And this is fairly easy - if a sequence in $\mathbb N$ is not eventually constant, then it is convergent iff it is convergent in $\mathbb Z_a$ and its limit is what we expect.