Let $E $ be a (possibly nonconnective) spectrum. Suppose $E \wedge K = 0$ (where $K$ is complex $K$theory). Does it follow that $E = 0$?
Take the 2minute tour
×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.
Sure. Smashing a based space with a spectrum is equivalent to smashing its suspension spectrum with that spectrum. So it suffices to give a nontrivial space whose reduced $K$homology is trivial. A classical example due to Luke Hodgkin is $Coker J$. See Hodgkin, Luke; Snaith, Victor. The Ktheory of some more wellknown spaces. Illinois J. Math. 22 (1978), no. 2, 270–278. 


Another type of example: Let $E$ be the mod $p$ homology spectrum. The integral homology groups of the periodic $K$theory spectrum are rational vector spaces, i.e. the integral homology groups of the mod $p$ spectrum are zero. 

