What property identifies the monoid of positive integers?

Question

What abstract algebraic properties uniquely identify the multiplicative monoid of positive integers among all free abelian monoids with infinitely many generators?

i.e. finish the sentence

The positive integers with multiplication are the unique monoid ...

without a circular dependence on the monoid's own properties (such as its prime numbers).

Every multiplicative monoid generated by infinitely many primes is isomorphic. And clearly the positive integers form the unique monoid generated multiplicatively by the prime numbers, but the concept of "the prime numbers" is itself derived from the properties of the monoid.

I'm sure there must be an answer to this question that derives the properties of the additive monoid from the multiplicative one in such a way as to uniquely identify it as the positive integers. If I was to take a stab at it, presumably it's something like:

The positive integers are the unique monoid whose set in union with an absorbing element, is also a monoid by a secondary binary function, over which the first function commutes, when the absorbing element is assumed the identity of the second function.

Answer 1

Note $\mathfrak{C}$ the category

• whose objects are sets $R$ endowed with two binary laws $+$ and $\times$ such that $(R, +)$ is a commutative monoid and such that $(R,\times)$ has a neutral $1$ such that $1 \times r = r \times 1 = r$ for $r\in R$ and $n \times 1_R = \sum_{i=1}^n 1_R$ for all $n\in\mathbf{N}_0$.
• whose arrows (morphisms) are commutative monoid morphisms $\varphi : R' \to R$ such that $\varphi(1)=1$ and $\varphi(r_1 + r_2) = \varphi (r_1) + \varphi (r_2)$ and $\varphi(r_1 \times r_2) = \varphi (r_1) \times \varphi (r_2)$ for $r_1, r_2 \in R'$

Then the $(\mathbf{N_0}, 1, +, \times)$ (with the obvious $1$, $+$ and $\times$) is an initial object of $\mathfrak{C}$, meaning by that that it is a object of $\mathfrak{C}$ such that for each object $R$ in $\mathfrak{C}$, there is a unique morphism (in $\mathfrak{C}$) from $\mathbf{N}_0$ to $R$

Take such a morphism $\varphi$. Then if $p$ is a prime number $\varphi (p) = p \varphi(1) = p 1_R$. Thus $\varphi$ is unique, as any positive integer is product of prime numbers.

Answer 2

OK, let's start with the observation by Max in the comments: The positive integers under multiplication are just the (unique up to isomorphism) free abelian monoid generated by the countably infinite set (“abelian” is just another word for “commutative” which is more commonly used in that context).

As long as you don't add more structure, that's all you can say. But before adding more structure, let's have a closer look at the structure that already is present. And that structure is given by the relation “divides”.

We have by definition that $a$ divides $b$ (written $a\mid b$) if there exists a $c$ such that $ac=b$. Another way to say the same is that $b$ is a multiple of $a$. Note that this is a structure that we get “for free” with the free abelian monoid.

Now looking at the relation $a\mid b$, we see that it is a partial order: It is reflexive ($a\mid a$), transitive (if $a\mid b$ and $b\mid c$, then $a\mid c$) and antisymmetric (if $a\mid b$ and $b\mid a$, then $a=b$).

Moreover, if we look at subsets that are totally ordered by that relation, we see that the largest such subsets are ordered just like the natural numbers (this is known as order type $\omega$).

So given that we have such a partial order, one may ask: Can we extend that to a total order? Maybe even of type $\omega$ again?

Indeed, we want to go even a bit further: The order type $\omega$ is a well order, that means every non-empty set has a smallest element, and as corrolary if an element is not the largest of the set, there is a next-larger element of that set.

And that allows us to demand a sort of uniformity for our order: Given any element $a$ other than $1$, the set of multiples of $a$ shall have the following property: The number of elements strictly between any multiple of $a$ and the next one shall be equal to the number of elements preceding $a$.

So if we denote our free abelian monoid with $P$ and that new order with $\le$, we have two conditions:

• Compatibility with the division order: $a\mid b \implies a\le b$.

• Uniformity: If $a>1$, $M_a = \{x\in P: a|x\}$, $b\in M_a$ and $c = \min\{x\in M_a: x>b\}$, then $\lvert \{x\in P: x < a\}\rvert = \lvert \{x\in P: b < x < c\}\rvert$. Here $\min$ is to be understood according to the order $\le$.

Let's see how this works out:

First, we have $1|a$ for all $a$, therefore $1\le a$ for all $a$, that is, $1$ has to be the smallest element.

So what about the next element, the one that comes after $1$? Let's call it $2$ (because, well, it's what comes after $1$).

Clearly, $2$ has to be a generator, because if it were not a generator, there would be an element $x\ne 1,2$ with $1\mid x$ and $x\mid 2$, and therefore $1<x<2$, in contradiction that $2$ is, by definition, the next element after $1$. Or in other words, we have just proved that $2$ is prime.

Now thanks to uniformity, we know that the next element after $2$, let's call it $3$, is not a multiple of $2$ (because there's exactly one element, $1$, before $2$, there must be one element between $2$ and the next multiple of $2$. Thus $3$ is not a multiple of $2$. Since the only other element below $3$ is $1$, we again have that $3$ has to be a generator. In other words, we've just identified $3$ as the second prime.

Now let's look at the next element after $3$, which we name $4$. This one is, again, a multiple of $2$ (because there's exactly one element, namely $3$, strictly between it and the previous multiple of $2$, namely $2$ itself), but it is not a multiple of $3$. Since larger elements cannot divide it (the compatibility requirement), we know that the complete list of its divisors is $1$, $2$ and $4$ itself. But that is only possible if $4=2^2$.

Well, at that point it should be clear that this imposed order indeed forces the free abelian monoid to be identified with the positive integers. I'm not going to prove it, though.