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Given a global field $F$, we can construct the ring of adeles. Given a general locally compact ring $R$, when is it isomorphic to the ring of adeles of some global field $F$ and how can I find $F$ in $R$?

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Iwasawa gave a characterisation, assuming you are given a subfield F, discrete and such that the quotient is compact. The other conditions are R a semisimple locally compact commutative topological ring with 1 (shared with F). Then R is the adele ring of the global field F.

Edit: I believe it is known that you can't get F from knowledge of R alone. I don't remember details or a reference, but it is something like the fact that the Dedekind zeta function doesn't determine the number field? In other words the ramification degrees e and residue class extension degrees f can be known for each prime, and this will tell you the adele ring R as a restricted product of local fields. But not the field F. Given R, there may be more than one candidate field it contains.

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    $\begingroup$ Link to Iwasawa's paper (On the rings of valuation vectors, Annals 1953) jstor.org/pss/1969863 it seems like knowing the adele ring as a topological ring is much more information than the zeta function. so it would be interesting to know either an example of different F's with the same R! $\endgroup$ – SGP Apr 9 '11 at 21:39
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    $\begingroup$ An example of nonisomorphic number fields with isomorphic adele rings is given by de Smit and Perlis in ams.org/journals/bull/1994-31-02/S0273-0979-1994-00520-8/…. The fields are Q((-33)^(1/8)) and Q((-33*16)^(1/8)). They say that PARI shows these fields have different class number, so in paticular the adele ring does not in general determine the class number. $\endgroup$ – KConrad Apr 10 '11 at 3:06
  • $\begingroup$ @KConrad - thanks. Wonders aloud if there is any extra mileage in Tate's thesis in the situation of one R, two Fs. Someone else's thesis topic? $\endgroup$ – Charles Matthews Apr 10 '11 at 10:27
  • $\begingroup$ @KConrad: nice example! $\endgroup$ – SGP Apr 10 '11 at 12:31
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    $\begingroup$ In Section 2 of de Smit and Perlis' paper there is the following extraordinry sentence, speaking about computations which are in accordance with the main theorem: "Of course this does not qualify as a proof of the theorem. Perhaps rodents in the bowels of the computer center are chewing on wires and altering data". $\endgroup$ – Filippo Alberto Edoardo Jun 3 '18 at 13:51

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