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I've been watching Alex Kontorovich's lectures on Analytic number theory, and he often references a "prime at infinity" or a similar sounding concept. For example, in lecture 15 at 45:21 he says:

So there are local obstructions that are not just the prime at infinity.

For an example that doesn't require the context of multiple lectures to understand, he explains $\xi(s)$ (Symmetric zeta function) in terms of the prime at infinity. I can't find the specific quote where he says this (as it would involve re-watching multiple hours of lectures,) however here's roughly what he said. Suppose we wrote $\xi(s)$ as follows:

$$\xi(s)=\pi^{-\tfrac{s}{2}}\Gamma(\frac{s}{2})\prod_{p\text{ prime}}(1-p^{-s})^{-1}$$

He refers to the factors in the infinite product coming from each prime (which makes sense) and the $\pi^{-\tfrac{s}{2}}\Gamma(\frac{s}{2})$ factor coming from the prime at infinity. Can someone explain what this object is, or perhaps point me to a place where I can learn about it?

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A prime in this context is a non-trivial absolute value on $\mathbb Q$. By Ostrowski's theorem the valuations are given by either the Euclidean absolute value $|\cdot|_\infty$ or, for a rational prime $p$ (i.e., $2,3,\dots$), by $$|x|_p:=p^{-v}$$ where $v$ is an integer such that $x=p^{v}\cdot a/b$ where $a$ and $b$ are integers, neither of which are divisible by $p$.

By a slight abuse, the absolute value $|\cdot|_p$ is often identified with the rational prime $p$ itself. In this way, the Euclidean absolute value $|\cdot|_\infty$ can be seen as some "infinite" prime.

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    $\begingroup$ This misses the point: why is the archimedean abs. value written as $|\cdot|_\infty$? Among nontrivial abs. values on $\mathbf Q$, the archimedean one is unlike the rest in a few ways, like being unbounded on $\mathbf Z$. Historically, $\mathbf Z$ in $\mathbf Q$ was like $\mathbf C[z]$ in $\mathbf C(z)$, and among abs. values on $\mathbf C(z)$ trivial on $\mathbf C$, only the "negative degree" abs. value, where $|f| = (1/2)^{-\deg f} = 2^{\deg f}$, is unbounded on $\mathbf C[z]$. That abs. value is naturally associated to $\infty$ on the Riemann sphere: its uniformizer $1/z$ is 0 at $\infty$. $\endgroup$
    – KCd
    Sep 8, 2022 at 4:27

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