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It is true that considering the archimedean places as well is more general, but that still doesn't explain why it is more natural. If we consider both the definitions of an absolute value and that of a discrete valuation, they both seem arbitrary. However, discrete valuations have the extra appeal of being related to geometric objects (they arise as local rings at non-singular codimension 1 subschemes of integral schemes, and in various other situations: local rings at non-singular codimension 1 subschemes of a scheme birational to the original scheme, or more abstract situations like Zariski-Riemann spaces). It seems (to me at least) that the archimedean places don't enjoy a similar geometric analogy.

This leads me to ask: why are the archimedean places natural? Why do they always feel like they're "missing points" of rings of integers?

I should note in the interest of full disclosure that I'm unfamiliar with the ways of Arakelov theory.

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Tag: Soft question? –  Martin Brandenburg Nov 27 '10 at 18:30
James, why do the definitions of absolute value and discrete valuation seem arbitrary to you? They both come out of core examples: real and p-adic numbers for absolute values and order-of-vanishing functions on Riemann surfaces for discrete valuations. Any definition that has no context will look arbitrary, but absolute values and discrete valuations have very basic contexts which explain how to think about them. –  KConrad Nov 27 '10 at 19:50
I find this question somewhat perplexing: I feel like I know enough about absolute values and discrete valuations to answer it, but I don't know enough about what it means for one mathematical object to be "more natural" than another. (Certainly both parties in question feel quite natural to me -- is it really a competition?) I wonder if anyone else feels the same way. –  Pete L. Clark Nov 28 '10 at 1:33
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Perhaps the main reason why people care about archimedean places is that many results in algebraic number theory become much neater if you treat them on equal footing with finite primes, even before you get to Arakelov theory. E.g. zeta- and L-functions have contributions from infinite primes, Dirichlet's S-unit theorem is nicer to state, theorems in class field theory become nicer, and the whole adelic business becomes possible. As a concrete example, Kevin Buzzard's answer and especially Matt Emerton's comment in

Positive polynomial having roots modulo any integer

relies on the fact that you can treat complex conjugation like any other conjugacy class of Frobenius elements.

On the other hand, when there is no need to care about infinite places, many (most?) people will definitely state what they want in terms of discrete valuations rather than absolute values, because, as you say, they are nicer and more intuitive. I don't think many will say "because $4\in{\mathbb Z}[i]$ has $(1+i)$-adic absolute value $1/16$...", they will say "because $4\in{\mathbb Z}[i]$ has valuation $4$ at $(1+i)$".

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I agree that they appear in many places (no pun intended), but what is the reason that we get contributions from the infinite primes? Why does adding those contributions lead to a more natural theory? –  James D. Taylor Nov 27 '10 at 18:45
James, look at the product formula. I think that is a succinct reason why you must take all absolute values into account. In fact Artin and Whaples derived all of classical algebraic number theory from a small list of axioms centering around the product formula. –  KConrad Nov 27 '10 at 19:09
Dear James: Here's another take on the product formula. The "diagonal" map $K \rightarrow \mathbf{A}_K$ into the locally compact adele ring has discrete image (analogous to discreteness of $O_K$ in the Euclidean space $\mathbf{R}\otimes_{\mathbf{Q}} K = \prod_{v|\infty} K_v$). But if you remove a single factor $K_v$ from the adele ring then the diagonal imageof $K$ is dense ("weak approx."). The adelic appartus (for Tate's thesis, automorphic forms, etc.) treats all places uniformly, needs all of them to make sense, and allows one to argue in ways that are uniform across all $K$. –  BCnrd Nov 27 '10 at 19:33
The product formula also highlights something that, depending on your background, may look idiosyncratic: the correct "absolute value" to use for complex places is the square of the usual absolute value. Strictly speaking the squared absolute value on C is not an absolute value (although you get a triangle inequality if you throw a factor of 4 on the right side: |z+w|^2 <= 4(|z|^2 + |w|^2)). On the ideles a similar thing happens: in the idelic norm you need the contribution at complex places to be the squared absolute value (and that can be conceptually explained there using Haar measure). –  KConrad Nov 27 '10 at 19:45
@BCnrd: You have high standards indeed: what you call "weak approximation", I call "strong approximation"! –  Pete L. Clark Nov 28 '10 at 9:38
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