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Background:

Definition: A supernatural number $\alpha$ is a formal product

$$\alpha=\prod_{p}p^{n_p},$$

where $p$ runs over the prime numbers and $n_p\in\{0,\infty\}\cup\Bbb N$.

It is claimed in the linked Wikipedia article that

if all $n_p$ are $\infty$, we get zero.

It says a citation is needed and I am skeptical. Hence . . .

The Question:

Is $0=\prod_{p}p^\infty$ accurate in the context of supernatural numbers? Why?

Context:

I was looking up the algebraic closure of a finite field when the book "Field Theory", by Roman introduced Steinitz numbers; they're synonymous with supernatural numbers, I believe.

Why is this relevant to the MSE community?

In lieu of the usual type of context (like an attempt) that I cannot give, I will try to sell this question to the MSE audience.

I think this supposed property is bizarre and kind of fun${}^\dagger$, if true. Also, it looks like it has the potential to be misunderstood if not explained properly. Moreover, there doesn't appear to be anything about this so far on MSE.

$\dagger:$ It made me laugh quite hard when I read the Wikipedia article, and say aloud, sarcastically, "Of course it'd be zero!"

Shaun
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    Since there is no way to add these, there is no way to define what zero would be anyway, so the claim does not seem to make sense. The proposed element does give an absorbing element for the multiplicaiton, though, which might be why they call it zero – Tobias Kildetoft Apr 23 '24 at 15:37
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    It's probably $0$ because if $A=\prod_p p^{\infty}$ then $An=A$ for any supernatural number $n.$ – Thomas Andrews Apr 23 '24 at 15:37
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    That's probably because when you multiply the formal product by a prime, then you get somewhere in the product $p^{\infty+1} = p^\infty$. So for all integer $n$, $n$ times the product is the product, so the product must be zero. – Nolord Apr 23 '24 at 15:38
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    Note that, as defined, $0$ isn't really a supernatural number. But there is only one supernatural number $A$ such that $An=A$ for all $n,$ so we are really just calling this value $0.$ – Thomas Andrews Apr 23 '24 at 15:43
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    Well, putting the cart before the horse, in natural numbers $0$ is the only number that all numbers are divisors of. So here all numbers are divisors of $\prod p^{\infty}$ and $\prod p^{\infty}$ is the only such supernatural number all numbers are divisors of. So If we need that to be zero (rather than say "super-ultimate-divisable-man") we can certainly call it zero. – fleablood Apr 23 '24 at 15:43
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    Essentially, it can only be a definition, because there is no sense in which the natural number $0$ is a supernatural number, so there can be no "proof." – Thomas Andrews Apr 23 '24 at 15:46
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    The more I think about it though, the more sense it makes. $\frac {N}{\omega} = N$ for all supernatural $\omega$ well, that means $N= N\omega$ and if $N\ne 0$ then $1 =\omega$ so $N=0$. (Except I have ambuity about the result of whether $\frac {p^\infty}{p^\infty}$ is equal to $1$ or $p^{\infty}$. If $\frac {p^\infty}{p^\infty}=1$ we get $\frac {\prod p^{\infty}}{7^{\infty}}=\prod_{p\ne 7}p^{\infty} \ne \prod p^\infty}$) – fleablood Apr 23 '24 at 15:49
  • (Except I have ambiguity about the result of whether $\frac {p^\infty}{p^\infty}$ is equal to $1$ or $p^{\infty}$. If $\frac {p^\infty}{p^\infty}=1$ we get $\frac {\prod p^\infty}{7^\infty}=\prod_{p\ne 7}p^{\infty} \ne \prod p^\infty$) – fleablood Apr 23 '24 at 15:55
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    Of course an answer from someone who has actually studied supernatural numbers might be helpful..... – fleablood Apr 23 '24 at 15:57
  • Interesting. $a\cdot 0 = 0$ for all $a$ but $\frac 0a$ need not equal $0$. Likewise $p^k \cdot p^{\infty}=p^{\infty}$ and $\frac {p^{\infty}}{p^k} = p^{\infty}$ but $\frac{p^\infty}{p^\infty} =???$. Neither the wikipedia pager nor this page https://planetmath.org/supernaturalnumber seem to mention it. I can see arguments that we should make it $1$ and arguments we should make it $p^\infty$. Maybe it should be as illegal as dividing by $0$ is. – fleablood Apr 23 '24 at 16:24
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    For further insight you should examine the contexts where they are applied. – Bill Dubuque Apr 23 '24 at 17:45

1 Answers1

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One way to look at this is through $q$-adics. In general the product will diverge, but if only some of the primes have nonzero exponents then we may get a convergent result. For instance the product

$2×3×5×17×17×97×...,$

in which each chosen prime is the smallest one $\equiv1\bmod2^k$ for $k=0,1,2,...$ ($17$ is repeated because it holds for both $k=3$ and $k=4$), will converge $2$-adically:

$2×3×5×17×17×97×...=...1110011110_2.$

One of the cases (the only obvious case) where we get $q$-adic convergence in all bases is when the exponents on all the primes are $\infty$, for then the $q$-adic product is $...00000=0$ in all bases.

Oscar Lanzi
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