Skip to main content
3 of 3
added 3 characters in body
Marc Hoyois
  • 9k
  • 1
  • 49
  • 52

This is true, if you define an infinite smash product as a colimit of finite smash products.

If you define a Moore spectrum for the abelian group $A$ to be a spectrum $X$ such that $X\wedge H\mathbb Z=HA$, then obviously $\mathbb S$ is a Moore spectrum for $\mathbb Z$. An arbitrary abelian group can be obtained from $\mathbb Z$ using direct sums (to get free abelian groups), filtered colimits (to get projective abelian groups), and quotient of a group by a subgroup (to get an arbitrary abelian group from projective ones). Since the functor $A\mapsto HA$ preserves sums and filtered colimits and transforms short exact sequences into cofiber sequences, you have a recipe to build any Moore spectrum from $\mathbb S$.

For example, $\mathbb Z_{(p)}$ is the colimit of the filtered diagram consisting of all the multiplication maps $n: \mathbb Z\to\mathbb Z$ for $n$ not divisible by $p$; replacing $\mathbb Z$ by $\mathbb S$ in this diagram and taking the (homotopy) colimit gives you the Moore spectrum for $\mathbb Z_{(p)}$.

To get the description you're interested in, note that $A\mapsto HA$ also transforms tensor products into smash products. EDIT: it transforms derived tensor products (of chain complexes) into derived smash products of spectra (the underived statement can be made true with strict models of EML spectra, but it's not very relevant).

Marc Hoyois
  • 9k
  • 1
  • 49
  • 52