1,665
edits
DavidBrooks (talk  contribs) m (Fix references) 
(added more on Pontryagin duality) 

where <math>\varprojlim \mathbb{Z}/n\mathbb{Z}</math> indicates the [[profinite completion]] of <math>\mathbb{Z}</math>, the index ''p'' runs over all [[prime number]]s, and <math>\mathbb{Z}_p</math> is the ring of [[padic integer''p''adic integers]].
Concretely the profinite integers will be the set of
Example: Let <math>\overline{\mathbf{F}}_q</math> be the [[algebraic closure]] of a [[finite field]] <math>\mathbf{F}_q</math> of order ''q''. Then <math>\operatorname{Gal}(\overline{\mathbf{F}}_q/\mathbf{F}_q) = \widehat{\mathbb{Z}}</math>.<ref>{{harvnbMilne2013loc=Ch. I Example A. 5.}}</ref>
The tensor product <math>\widehat{\mathbb{Z}} \otimes_{\mathbb{Z}} \mathbb{Q}</math> is the [[ring of finite adeles]] <math>\mathbf{A}_{\mathbb{Q}, f} = \prod_p{}^{'} \mathbb{Q}_p</math> of <math>\mathbb{Q}</math> where the prime ' means [[restricted product]].<ref>[https://math.stackexchange.com/q/233136 Questions on some maps involving rings of finite adeles and their unit groups].</ref>
The set of profinite integers has a topology in which it is a [[compact]] [[Hausdorff space]], coming from the fact that it can be seen as a closed subset of the product <math>\prod_n \mathbb{Z}/n\mathbb{Z}</math>, which is compact with its product topology by [[Tychonoff's theorem]]. Addition of profinite integers is continuous, so <math>\widehat{\mathbb{Z}}</math> becomes a compact Hausdorff abelian group, and thus its [[Pontryagin dual]] must be a discrete abelian group.
In fact the Pontryagin dual of <math>\widehat{\mathbb{Z}}</math> is the discrete abelian group <math>\mathbb{Q}/\mathbb{Z}</math>. This fact is exhibited by the pairing
:<math>\mathbb{Q}/\mathbb{Z} \times \widehat{\mathbb{Z}} \to U(1), \, (q, a) \mapsto \chi(qa)</math><ref>{{harvnbConnesConsani2015loc=§ 2.4.}}</ref>
where <math>\chi</math> is the character of <math>\mathbf{A}_{\mathbb{Q}, f}</math> induced by <math>\mathbb{Q}/\mathbb{Z} \to U(1), \, \alpha \mapsto e^{2\pi i\alpha}</math>.<ref>K. Conrad, [http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/characterQ.pdf The character group of '''Q''']</ref>
== See also ==
