# In what senses are archimedean places infinite?

According to Bjorn Poonen's notes here (§2.6), we should add the archimedean places of a number field $K$ to $\operatorname{Spec} \mathscr{O}_K$ in order to get a good analogy with smooth projective algebraic curves. This suggests that the archimedean places are infinite in the sense of being ‘points at infinity’ and not just because rational integers become arbitrarily ‘large’.

1. Is there any reason to expect this, i.e. was there some intuition that the archimedean places were the points needed to make $\operatorname{Spec} \mathscr{O}_K$ complete in some sense? I am aware of the construction of abstract smooth projective curves from function fields via DVRs (e.g. as described in Hartshorne [Ch. I, §6]), but the set $$\{ x \in K : \left| x \right|_v \le 1 \}$$ fails to be a subring, let alone a DVR, for any archimedean place $v$. (At any rate, archimedean absolute values do not correspond to valuations.)

2. Was the term ‘infinite prime’ used for an archimedean place before these analogies were known? If so, why?

• I always understood that the places of a function field are precisely the points of the corresponding projective curve. Is there more to say than that? – Qiaochu Yuan Dec 15 '11 at 8:02
• @Qiaochu: Yes, but there it suffices to consider the places coming from discrete valuations; so, in particular, archimedean absolute values don't come into the picture. I am asking if there was some heuristic reasoning which suggested that we needed to expand our notion of ‘place’ to include archimedean ones. – Zhen Lin Dec 15 '11 at 8:17
• One reason is the product formula: $$\prod_v |x|_v =1$$ that holds for all rational functions $x$, when the product ranges over all the valuations $v$ of the function field. For a similar formula to hold in the case of rationals (resp. its extension fields), we need to include the usual absolute value (resp. the infinite places). This does not explain, why they should be named "the infinite place(s)". – Jyrki Lahtonen Dec 15 '11 at 8:26

## 2 Answers

Over the function field $K(x)$ there are places coming from discrete valuations attached to every irreducible polynomial in $K[x]$, and then there is also a place coming from the valuation which assigns to a rational function the degree of its pole at $\infty$. These places, suitably normalized, satisfy the product formula $$\prod_v |x|_v = 1.$$

Over the number field $\mathbb{Q}$ there are places coming from discrete valuations attached to every prime in $\mathbb{Z}$, and then there is also a real archimedean place. These places, suitably normalized, again satisfy the product formula $$\prod_v |x|_v = 1.$$

The two product formulas suggest that one should think of the degree of a polynomial as analogous to the logarithm of the absolute value of an integer, and since the former can be thought of in terms of the point at infinity, why not think of the latter in terms of a point at infinity as well?

There are good reasons to think that this analogy is sound coming from, for example, the fact that using this convention makes the "prime number theorem over function fields" (take $K$ to be a finite field) take the correct form. That is, consider the absolute value $e^{\deg f}$ on polynomials $f$ over a finite field, and let $\pi(x)$ denote the number of monic irreducible polynomials of absolute value less than or equal to $x$. Then using the standard closed formula for the number of irreducible polynomials of degree $n$ it is straightforward to show that asymptotically $$\pi(x) \sim \frac{x}{\ln x}.$$

• +1: But the number of irreducible polynomials of degree exactly $N$ is asymptotically $q^N/N$, where $q=|K|$. IMHO a better analogy here is that a positive rational integer $n$ is prime with probability $1/\ln |R|$ (asymptotically) whereas a monic polynomial of degree $N$ is irreducible with probability approximately $1/\log_q(|R|)$. Here $R$ is the number of elements of the residue class ring, so $|R|=n$ for a positive integer $n$, and $|R|=q^N$ for a polynomial of degree $N$. – Jyrki Lahtonen Dec 15 '11 at 8:38

This is a comment more than an answer, but is a bit long for the comment box:

My memory is that in Hilbert'z Zahlbericht, he denotes the archimedean primes as $1, 1', 1'',$ etc. I forget how he describes them in words, but you might look in there to get some idea of how he uses them. One thing you will see is that the condition of being positive (or not) at a real prime is one that has always been important in algebraic number theory. (I think that this is the main way that they --- that is, the archimedean primes --- intervene in Hilbert.)

You might also consider the role of the archimedean primes in the proof of Dirichlet's unit theorem (and compare with the role of the primes at infinity in determining the units in the affine ring of an affine curve over a finite field).

In any case, I think that the analogy you ask about has been known in some form for more than a hundred years, going back at least to Hilbert, and based on Dedekind and Weber's original analogy between algebraic number theory and the theory of curves.

Finally, I should mention that the analogy finds its most precise modern form in Arakelov theory.