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There seems to be an importance to the ring of adeles for the rational numbers (discussed here), with valuations for every $\mathbb{Q}_p$, but also one "infinite" valuation "$\mathbb{Q}_∞$", seemingly equal to $\mathbb{R}$.

Why would something like $\mathbb{Q}_∞$ be used in the first place, and how is that equal to the reals? Is there something like a $∞$-adic metric that works like the usual one?

Moreover, it seems to suggest that $∞$ here is a sort of an infinite prime number, i.e. the real prime, having some occult-sounding books written about it. So, does it exist as some sort of a describable object here, or is it just notation?

IS4
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  • "Why would it be used in the first place?" Because it makes sense to summarize all valuations. "How is this equal to the reals?" Just by definition: $\Bbb Q_{\infty}:=\Bbb R$. Of course, $p=\infty$ is not a rational prime. – Dietrich Burde Nov 12 '20 at 11:29
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    According to some mathematicians like John Conway, sometimes it makes sense to identify the "real prime" with the number $-1$, see for example this answer. – pregunton Nov 12 '20 at 11:46
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    To me the idea makes no sense, if we look at the first $p-1$ integers $|1|_p = |2|_p = \cdots = |p-1|_p = 1$ and so taking the limit as $p \to \infty$ would imply the "infinite" case is actually the trivial absolute value. – Merosity Nov 13 '20 at 04:20

2 Answers2

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It is a formal notation. We treat $|\cdot|$ as an absolute value $|\cdot|_{\infty}$ coming from an “infinite prime”, so that we obtain, among other things, a product formula $$ \prod_{p\le \infty} |\alpha|_p=1 $$ for every $\alpha\in \Bbb Q^{\times}$. Of course, $p=\infty$ is not really a prime. So $\Bbb Q_{\infty}:=\Bbb R$ is just a notation. We can summarize all completions of $\Bbb Q$ by

$$ \Bbb Q_2,\Bbb Q_3,\Bbb Q_5,\cdots ,\Bbb Q_{\infty} $$

Dietrich Burde
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    So why $∞$? Why not have $\mathbb{Q}0$, $\mathbb{Q}{-1}$, or $\mathbb{Q}_$ for another new or unique subscript? – IS4 Nov 12 '20 at 12:02
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    $p=\infty$ is better for the "intuition" than $p=-1$ or something else. One reason comes from the Hasse-Minkowski theorem, where we have a rational solution, if we have a solution for all primes $p$ and a real solution - conventiently denoted by $p=\infty$. On the other hand, the characteristic of a field, which is not prime, is not always denoted by $\infty$, but rather by $0$ now. So it is perhaps mostly convention. – Dietrich Burde Nov 12 '20 at 12:19
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    @IllidanS4wantsMonicaback The notation $\infty$ comes from the analogy with the function field case, where the "extra prime" can actually be identified with the $\infty$ coming from the projective line. But not everybody accepts that convention: as I mentioned Conway uses $-1$, and the book cited in the question uses $\eta$. Taking the analogy with $\infty$ too far can sometimes lead to confusion, see e.g. here. – pregunton Nov 12 '20 at 12:22
  • The product formula is only valid for $\alpha\neq0$. It could be rewritten for all $\alpha\in\mathbb Q$ as $$\prod_p|\alpha|_p=|\alpha|_0$$ (the trivial absolute value; but here also the subscript seems arbitrary). – mr_e_man Feb 27 '23 at 17:54
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The real prime, typically marked as $\eta$, is an object on its own. It is not a sort of abstraction or something arriving from limiting process. A completion of $\mathbb{Q}$ with respect to $\eta$ is $\mathbb{R}$. So it is the notation $\mathbb{Q}_{\infty}$ that is creating a confusion over what would be the purpose of such conundrum.

Basically $\mathbb{Q}_{\infty} = \mathbb{Q}_{\eta}$, if we want to be precise, should mean precisely "a closure over $\mathbb{Q}$ with respect to $\eta$". This object $\eta$ is somewhat mysterious, that is true, and we find its properties by defining it as something that closes $\mathbb{Q}$ so that we get $\mathbb{R}$. This fact does not make it in any way less concrete than, say, imaginary unit $i$.

How relevant is this real prime? Quite relevant. For example we can write for so called completed Riemann Zeta function:

$$\zeta_{\mathbb{A}}(s)=\prod_{p \geq \eta} \zeta_p(s), \quad \Re(s)>1$$ $$\zeta_p(s) = \left(1-p^{-s}\right)^{-1}$$ $$\zeta_\eta(s)=\pi^{-s / 2} \Gamma(s / 2) \big( = -\pi \Gamma\left(\frac{s}{2}+1\right) \big )$$

This is a definition, even though each factor is derived from the Mellin transform in the corresponding field, but it gives a clear picture over what Riemann Zeta function really is. Without this background it is difficult to understand the origin of the factor $\pi^{-s / 2} \Gamma(s / 2)$ in the classical case. Basically in classical derivation of the the extension for Riemann Zeta function we implied the reduction of the completed product over the prime $\eta$ even though we never specified it. That is to say the factor $p=\eta$ is removed from the product implicitly, but it appears later on.

So our classical Riemann zeta is only

$$\zeta(s) = \prod_{p \neq \eta} \zeta_p(s)$$

However, the functional equation is valid only for the full product $\zeta_{\mathbb{A}}(s)$ not just for this last product that does not include $\eta$. This is why we are talking about completed Zeta function.

I hope this convince you that real prime is not just a passing index. It is an object on its own, and it poses many questions that are not easy to answer. For example what $\mathbb{Z}_{\eta}$ is. What would be the completion of $\mathbb{Z}$ over this real prime? And so on.

(Interestingly, the real prime is the smallest of all primes. That much about assumed infinity.)

And just to complete the answer. This is the connection between the real prime and its Zeta factor. $\mathcal{M}_x(s)$ is Mellin transform.

$$\mathcal{M}_x\left[e^{-\pi x^2}\right](s)=\frac{1}{2} \pi^{-s / 2} \Gamma\left(\frac{s}{2}\right)$$

and $e^{-\pi x^2}$ is basically the version of standard normal distribution, which on its own is saying tons over what we are actually doing by deriving various forms of Riemann Zeta using this transform.