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All valuations of $\mathbb{Q}$ are associated to either a p-adic field or the field of real numbers. Now, the classification of fields gives that every local field in characteristic $0$ is a finite extension of these.

Is there another notion for a topology on $\mathbb{Q}$ than coming from a valuation, which allows more general completions? Preferably, their additive groups should have Haar measures, but are not necessarily locally compact?

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If you want to look at completions, you should look at metrics on $\mathbb Q$. For example, you can take the discrete topology, and the completion is $\mathbb Q$. –  Kimball Jun 24 '11 at 11:19
    
Presumably, you want the topology to respect the algebra, that is, you want the field operations to be continuous, right? –  Gerry Myerson Jun 24 '11 at 12:51
    
In most generality, I guess you can ask if there are any metrics on $\mathbb{Q}$ which are compatible with the algebraic operations (as Gerry mentioned) and have a completion which is neither $\mathbb{Q},\mathbb{R}$ or $\mathbb{Q}_p$ for $p$ prime. –  Mark Jun 24 '11 at 13:06
    
I was hoping for some definitions which allow more structure at infinity, since I heard somebody saying that Arakelov geometry provides things like this. –  plusepsilon.de Jun 24 '11 at 16:25

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up vote 5 down vote accepted

You can take the "finite adele" topology, which comes from pulling back the dense diagonal copy of $\mathbb{Q}$ in the ring $\prod_p \mathbb{Z}_p + \mathbb{Q}$ of finite adeles. You can also call it the "Furstenburg topology" if you want to annoy BCnrd.

There is an analogous construction using any subset of the primes, and indeed, you may leave out any nonempty subset of finite or infinite places from the adeles to get a completion of a non-discrete topology on $\mathbb{Q}$. All such additive groups are locally compact, and admit Haar measures.

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don't tell B, but...+1. –  Pete L. Clark Jun 24 '11 at 19:35
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Good example +1; But It might be worth mentioning though that under this topology, $\mathbb{Q},$ is a topological ring but not a topological field. –  jspecter Jun 25 '11 at 2:49

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