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By the fundamental theorem of arithmetic, we can identify a natural number $x$ with the sequence $(a_n)$ of exponents in its prime factorization $x=\prod_np_n^{a_n}$, where $p_n$ is the $n$th prime. A sequence obtained in this way will of course have only finitely many nonzero elements.

However, multiplication/$\gcd$/$\operatorname{lcm}$ operations on natural numbers correspond respectively to pointwise addition/$\min$/$\max$ of these sequences, and these operations make sense on arbitrary sequences. So I wonder: How much arithmetic can we do with these sequences, thinking of them as formal infinite products of prime powers?

Specifically:

  • Is there a semiring structure (without $0$) on $R=(\Bbb Z_{\ge 0})^\Bbb N$ where $\times$ is pointwise addition and $+$ coincides with normal addition on finite natural numbers (represented as described above)?

If so, I'm also interested in whether we can get other familiar properties.

  • Does $\gcd(a,a+b)=\gcd(a,b)$ hold (where $\gcd$ = pointwise $\min$)?
  • Is $<$, defined by $a<b\iff\exists c.a+c=b$, a total order?
  • If we include $0$ and negatives (e.g. by considering equivalence classes of formal differences), does Bézout's lemma hold?
  • Can we in fact get a (non-standard) model of arithmetic? Are these the hyperintegers or something?

Edit: Commenters have pointed out that the answer to the last bullet is no, since this structure lacks large primes and includes elements with no largest prime divisor (and both of these properties are inconsistent with the axioms of arithmetic). The latter issue prevents this structure from even being a sub-semiring of a model of arithmetic.

Karl
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    Note that this structure has no nonstandard primes, so it can't be a nonstandard model of arithmetic. – Noah Schweber Mar 24 '21 at 23:15
  • I had to look it up (don't know the terminology in English), this seems to be related to division lattices. – user3733558 Mar 24 '21 at 23:16
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    For your final bullet, you can prove in arithmetic that any number has a largest prime divisor. That won't be true in your proposed semiring, so you won't get a a model of arithmetic. (I'd be surprised if the semiring structure you are looking for exists, but I don't know how to prove that off the top of my head) – Rob Arthan Mar 24 '21 at 23:16

1 Answers1

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The comments supplied some intuition about why this construction doesn't come close to modeling arithmetic or even supporting addition in a "reasonable" way.

For one thing, we always expect $x$ and $x+1$ to have no common prime divisor, but this forces $\prod p_n+1=1$. This is undesirable since $x+a$ should be infinite whenever $x$ is infinite, and $a+b=a$ should never be true if we're trying to represent positive values.

This essentially answers my question, even though I think a subtraction axiom is needed to turn the argument into an actual contradiction.

I briefly considered avoiding this problem by restricting to sequences that have infinitely many $0$s, but this doesn't work since we still get e.g. $\prod p_{2n}+\prod p_{2n-1}=1$. We could try harder by requiring the density of nonzero entries to be small, but I doubt this will work. In fact, "thinning out" the infinite numbers seems like the opposite of what we need, since the infinite products are already all "highly composite" and therefore too sparse for addition to work. As the comments suggest, if the goal is to model arithmetic, we should add nonstandard primes and consider only finite products (edit: bounded or "internal" products) of them.

Karl
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    "consider only finite products of them." It's actually substantially more complicated than that! Bounded infinite products (or at least, some of them) are fine, and indeed necessary: think about things like $n!$ for $n$ nonstandard. On the other hand, no nonstandard number can have as its prime factors exactly the standard primes, by overspill. Nonstandard models of arithmetic are extremely bizarre things (see e.g. here). – Noah Schweber Mar 25 '21 at 03:59
  • Wow! I actually wondered whether I should say "bounded" instead, but since even a bounded set of primes might not have a largest element, I went with "finite" to make sure Rob's condition was met. Makes sense that this turns out to be too restrictive. – Karl Mar 25 '21 at 04:15
  • Is the overspill argument that if $z$ is the product of exactly the standard primes, then the formula "every prime divisor of $x$ is a divisor of $z$" would define the cut corresponding to the standard natural numbers? – Karl Mar 25 '21 at 04:19
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    The right notion is actually "internal" (roughly, "coded by some element in the structure"). Every finite set is internal and every internal set is bounded and has a top element, while not every bounded set has a top element. The standard primes, for instance, are a bounded non-internal set without a top element. As to your second comment, that wouldn't work - consider $2^n$ for a nonstandard $n$. Instead, use the ordering: $x$ is standard if $x$ is less than some prime dividing $z$. (Indeed the multiplicative structure alone will not be enough here.) – Noah Schweber Mar 25 '21 at 05:18