Measure of "adeles minus ideles"

2010-12-18T15:41:06Z

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.

Answer by S. Carnahan for Measure of "adeles minus ideles"

2010-12-20T08:54:41Z

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.

Measure of "adeles minus ideles" - MathOverflow

Measure of "adeles minus ideles"

Answer by S. Carnahan for Measure of "adeles minus ideles"