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Hi, I am interested in the set $\mathbb A-\mathbb A^\times$ i.e. the complement of ideles in the adele ring of a number field.

Is it measurable, and what is its volume, with respect to the standard measure of adeles?
("standard" means the same as in Tate's thesis)

Thank you.

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I forget the definition of the measure on adeles, but it seems that the Borel sigma algebra (generated by opens) is the same as the one generated by direct products of measurable sets in each component, so that there's no confusion. It is probably measurable. Reason: it is a countable union indexed by S (finite set of ...), so suffices to consider the direct product of p-adic units inside the direct product of p-adic integers, which is measurable since each component is. The measure seems to be zero: product of 1-1/p=1/\zeta(s\to 1+)=0. Correct me if I'm wrong. –  shenghao Dec 18 '10 at 16:42
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Yes, when regarded as a subset of the adeles, the ideles have measure zero. –  Pete L. Clark Dec 18 '10 at 23:02
    
@ Shenghao: how to deal with that finite many $\mathbb Q_p$ in a direct product? @ Pete: Could you explain in some detail? –  user4245 Dec 19 '10 at 9:03
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You deal with those copies of $\mathbb{Q}_p$ by viewing them as countable unions of translates of $\mathbb{Z}_p$. You get a countable union of measure zero sets. –  S. Carnahan Dec 19 '10 at 9:33
    
@Scott: I see, thanks –  user4245 Dec 19 '10 at 10:18
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1 Answer 1

up vote 8 down vote accepted

This is a bookkeeping post, since the answer seems to have been resolved in the comments. Somebody please vote this up once so this question leaves the "unanswered" queue.

Shenghao's answer is essentially that you can view the ideles as a countable union of translates of $\widehat{\mathbb{Z}}^\times$, which has measure zero, so the ideles have measure zero. The measure zero property of $\widehat{\mathbb{Z}}^\times = \prod_p \mathbb{Z}_p^\times$ arises from the fact that for each prime $p$, the set $\mathbb{Z}_p^\times$ has volume $\frac{p-1}{p}$ in $\mathbb{Z}_p$, and the product of these numbers over primes $p$ absolutely converges to zero. Very loosely speaking, we have $\prod_p \frac{p-1}{p} = \frac{1}{\zeta(1)}$ and the latter term is morally zero, since it is the reciprocal of the harmonic series.

In conclusion, the adeles minus the ideles is a measurable set with infinite volume, because the volume of the adeles is infinite and the measure of the ideles is zero.

Apparently, this fact was a primary flaw in a serious attempt at a proof of the Riemann Hypothesis from a couple years ago. The mathematician in question tried to manipulate the integral (over the adeles) of a function supported on the ideles as if it were nonzero.

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