Is there some example that nicely extends the multiplication of natural numbers?

Question

Motivation: In mathematics, it is natural to decompose a complicated thing into simpler ones. In the system of natural numbers, the process to decompose large number is to factor it. The indecomposable ones, that cannot be further factored, are called primes. But the primes are irregular and the most difficult to understand. So, instead of trying hard to find some pattern of primes, why not consider finding an extension of the integer system so that primes can be further factored into another ones that admit some regularity?

Question: Let $\mathbb N^+$ be the set of positive integers equipped with its canonical addition $+$, multiplication $\times$, and total order $\leq$. Let $\mathbb P$ be the set of all primes. Is there any related work concerning finding (or proving non-existence of) an algebraic system $S$ such that:

(0) there is an binary operation $\cdot : S\times S \to S$ such that $(S,\cdot)$ is a commutative monoid.

(1) there is a monoid embedding $f: (\mathbb N^+, \times) \to (S, \cdot)$.

(2) there is a subset $E := \{e_1, \ldots,e_n, \ldots \} \subset S$ such that

• (2a) any $e \in E$ is not a unit.

• (2b) if $e = a \cdot b$ for any $e \in E$ and $a,b \in S$, then either $a$ or $b$ is a unit.

• (2c) for any $p \in \mathbb P$, the element $f(p)$ has a unique non-trivial factorization $f(p) = \prod_{e \in E} e^{x_e}$ for $x_e \in \mathbb N$.

(3) the structure of $E$ in $S$ is simpler than the structure of $\mathbb P$ in $\mathbb N^+$. (In other words, primes are better understood in the language of $E$. This requires $S$ to have more structure than a monoid)

As many readers said that the condition (3) is vague, I replace it by a more concrete but alternative condition (3'). Keep in mind that this restricts the possibilities:

(3') There is an addition on $S$ making $S$ a semiring and $f$ an embedding of semirings. Moreover,

• (3'a) the set $E$ is countably infinite
• (3'b) there is a polynomial $g \in S[x]$ such that the map $g \circ f: \mathbb N^+ \to E$ is a bijection. (This says that there is a polynomial formula for the new primes $E$)

Edit: refined the description of $S$. Added an alternative condition to (3).

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Note that the extension of the addition operation is not considered yet, so the system we are searching for need not be a ring. It is easy to find some system satisfying (1) and (2). For example, the ring $\mathbb Z[{\sqrt{2}}]$, but factoring numbers to $\{\sqrt{2}, 3, 5 \ldots\}$ doesn't make the situation easier, so the condition (3) is not satisfied. The ring of rationales $\mathbb Q$ doesn't simultaneously satisfy (2) and (3).

I don't expect that any such system can be found easily; since if such one is found, then the study of primes immediately becomes obsolete. Instead, I'd like to discover related researching field and potentially useful connections from established results. So any help will be appreciated. Thanks in advance.

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As an aside, the set of indecomposable elements $E$ cannot be finite. Here is the reason. Assume that there exists a such system with the cardinality of $E$ being $n < \infty$, then by factoring each prime number into products of $e_i \in E$, we associates each prime number to a vector in the $n$-dimensional $\mathbb Z$-module $\mathbb Z^n$, by sending a prime $p$ to $(x_1,\ldots,x_n)$ where $x_i$ is the multiplicity of $e_i$ in the factorization of $p$. Then any $n+1$ many primes are associated to $n+1$ vectors, which are linear dependent. By identifying the addition in that vector space with the multiplication of associated numbers, we obtain that for any $n+1$ primes, some power of some of them are powers of other primes, a contradiction.

This also gives me inspiration. By considering all primes as a set of linear independent vectors in a infinite dimensional $\mathbb Z$-space, the whole natural numbers is considered as its span. Adding the canonical basis into this span is exactly one way to extend the number system.

Answer 1

Edit (following the clarifications of the OP): This is not an answer, it is rather a long comment.

Studying this kind of questions is part of the mission of factorization theory: The language of the theory (at least in its classical incarnation) is the language of commutative (and, to a very large extent, cancellative) monoids, which seems to fit with the OP's idea of a monoid embedding $f$ from the multiplicative monoid of the positive integers into a larger commutative monoid $S$ such that $f(p)$ is a product of atoms${}^{\text{(a)}}$ in an essentially unique way. Note that, in fact, we may assume without loss of generality that $S$ is an atomic monoid (i.e., every non-unit of $S$ factors as a product of atoms).

Now, it is a basic result in the classical theory of factorization that a commutative cancellative monoid $H$ is a unique factorization monoid (i.e., every non-unit of $H$ has an essentially unique factorization into atoms) if and only if $H$ is atomic and every atom is a prime${}^{\text{(b)}}$, if and only the quotient monoid $H/H^\times$ is a free abelian monoid; for a reference, see Theorem 1.2.9 in

A. Geroldinger and F. Halter-Koch, Non-Unique Factorizations. Algebraic, Combinatorial and Analytic Theory, Pure Appl. Math. 278, Chapman & Hall/CRC, 2006).

This means that, if $S$ is a cancellative and commutative monoid with the unique factorization property, then there is nothing new we can hope to learn about the arithmetic (multiplicative) structure of $\mathbb N$ from the (mere) existence of the embedding $f$ — and this is true independently of how condition (3) is interpreted. So, for the question to make sense, we need either to give up the idea that $S$ is a cancellative and unique factorization monoid (which seems to be fine with the OP), or to keep track of the additive structure of $\mathbb N$ and rather look, say, for a semiring embedding of $\mathbb N$ into a larger commutative semiring whose multiplicative monoid satisfies conditions (2) and (3). (Honestly, I think condition (3) is still a bit too vague for any sensible answer to be possible.)

• Notes.

(a) An atom of a monoid $H$ is a non-unit $a \in H$ such that $a \ne xy$ for all non-units $x, y \in H$.

(b) A prime of a monoid $H$ is a non-unit $p \in H$ such that, if $p \mid_H xy$ for some $x, y \in H$, then $p \mid_H x$ or $p \mid_H y$ (here $\mid_H$ is the divisibility preorder on $H$, so $a \mid_H b$, for some $a, b \in H$, if and only if $b \in HaH$).

Answer 2

It is not completely clear to me what you are looking for, but here is an argument that it probably does not exist. What makes primes interesting and difficult to study is that they can be added as well as multiplied. If there were no such thing as addition, and the only thing you could do with primes is multiply them, then the primes are already as simple as you can get—they form a free abelian group. The only thing simpler would be a free abelian group of finite rank, but you've already excluded that.

If addition is part of the picture, then suppose $p$ and $q$ are distinct primes. Because $p$ and $q$ are (relatively) prime, there exist integers $a$ and $b$ such that $ap+bq=1$. Now suppose that there is some other thing $x$ such that $x\mid p$ and $x\mid q$. Then I would think that we could infer that $x \mid ap+bq$. But that means that $x\mid 1$, meaning that there exists $y$ such that $xy = 1$. So $x$ is a unit. When we talk about unique factorization, it can be unique only up to multiplication by units, so one typically regards units as trivial factors. This means that there can't be any nontrivial $x$ that contributes to the factorization of different primes, and that fact already places severe limits on the possibilities. There still exist interesting examples; for example, if you take the ring of integers in an algebraic number field, then primes can indeed split into products of other things, but you've already rejected $\mathbb{Z}[\sqrt{2}]$ as "not simpler."

I can imagine ways in which one can get around the arguments in the preceding two paragraphs, but I think you need to be clearer about what you are really looking for and why it's not ruled out by what I've just said.

For example, here is one thing that number theorists do in order to come to grips with the complexity of the primes: they turn some of the primes into units. A general method for doing this is called localization. You can increase the set of units to a set of $S$-units. Or, if you pick a single prime $p$ and turn all the other primes into units, and in addition to localizing you also pass to a completion, then you get $p$-adic numbers. Factorization in the ring $\mathbb{Z}_p$ of $p$-adic integers is easy; everything has the form $p^nu$ where $n\in\mathbb{N}$ and $u$ is a unit. Can't get much simpler than that! This might seem too simple but $p$-adic numbers are one of the most powerful tools in all of number theory.

Answer 3

[I'm adding a new (non-)answer since what follows has almost nothing to do with my previous one.]

Fix a real number $\alpha \ge 1$ and set $S_\alpha := \mathbb N \cup \mathbb R_{\ge \alpha}$. Of course, $S_\alpha$ is a commutative semidomain under the operations of addition and multiplication inherited from the real field, and the inclusion map $\mathbb N \to S_\alpha$ is a (unit-preserving) semiring embedding. Further, the only unit of $S_\alpha$ is the (multiplicative) identity $1$; and it is easily checked that, for $\alpha > 1$, (the multiplicative monoid of the non-zero elements of) $S_\alpha$ is BF, by which we mean here that (i) every non-zero element of $S_\alpha$ is a product of atoms and (ii) the atomic factorizations of each element are bounded in length. In particular, it is seen that, if $\mathbb P \subseteq \mathbb N$ is the set of (natural) prime numbers, then the set $\mathscr A(S_\alpha)$ of atoms of $S_\alpha$ is contained in the union of the (left-closed, right-open) interval $[\alpha, \alpha^2[$ with $[2, \alpha] \cap \mathbb P$: Most notably, we have that $\mathscr A(H) = \emptyset$ (and hence $S_\alpha$ is not even atomic) when $\alpha = 1$, and $\mathscr A(S_\alpha) = [\alpha, \alpha^2[$ when $1 < \alpha \le 2$.

Now suppose that $1 < \alpha \le \sqrt{2}$. The set of atoms of $S_\alpha$ has then a "very smooth structure" (so I would consider it to be "simple" in the vague sense of condition (3) of the OP), and each $p \in \mathbb P$ has a non-trivial factorization into atoms of $S_\alpha$ (since $\alpha \le \sqrt{2}$ and the only unit is the identity, $p$ is not an atom of $S_\alpha$); however, it is not clear to me if one can tune $\alpha$ so as to guarantee that such a factorization is essentially unique (possibly after replacing $\mathbb R_{\ge \alpha}$ with something smaller). In fact,

Note that, in general, $S_\alpha$ is not FF, which means here that each non-zero element has only finitely many atomic factorizations that are pairwise non-equivalent in the obvious sense: If, for instance, $\alpha = \sqrt{2}$, then $a$ and $3a^{-1}$ are atoms of $S_\alpha$ (and hence the $S_\alpha$-word $a \ast 3a^{-1}$ is a length-$2$ atomic factorization of $3$ in $S_\alpha$) for all $a \in \bigl]\frac{3}{2}, 2\bigr[$ (see my answer here if something in the terminology or in the notation doesn't sound familiar).