The Bousfield Class of the Infinite Wedge of Telescopes of Finite Spectra - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T02:01:13Z http://mathoverflow.net/feeds/question/93037 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93037/the-bousfield-class-of-the-infinite-wedge-of-telescopes-of-finite-spectra The Bousfield Class of the Infinite Wedge of Telescopes of Finite Spectra Jon Beardsley 2012-04-03T20:39:34Z 2012-04-04T00:00:21Z <p>The spectrum $T(n)$ which is the telescope of a finite spectrum of type n along its self-map, has a unique Bousfield class $\langle T(n)\rangle$ which only depends on $n$. It is also known, from Ravenel's paper "Localization with respect to certain periodic homology theories" that $\langle S\rangle=\langle T(0)\rangle\vee\langle T(1)\rangle\vee\ldots\vee\langle T(n-1)\rangle\vee\langle F(n)\rangle$ where $F(n)$ is a finite spectrum of type n. Is it known whether or not $\langle S\rangle=\bigvee_n\langle T(n)\rangle$? </p> <p>In the same vein, is it known whether or not $\bigvee_n\langle K(n)\rangle=\langle S\rangle$?</p> http://mathoverflow.net/questions/93037/the-bousfield-class-of-the-infinite-wedge-of-telescopes-of-finite-spectra/93053#93053 Answer by John Palmieri for The Bousfield Class of the Infinite Wedge of Telescopes of Finite Spectra John Palmieri 2012-04-04T00:00:21Z 2012-04-04T00:00:21Z <p>The spectrum $H\mathbf{F}_p$ is an acyclic for both $\bigvee \langle T(n) \rangle$ and $\bigvee \langle K(n) \rangle$. Therefore neither of these wedges equals $\langle S \rangle$.</p>