Skip to main content
1 of 3
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 $H\mathbb Z_{(p)}$.

To get the description you're interested in, note that $A\mapsto HA$ also transforms tensor products into smash products.

Marc Hoyois
  • 9k
  • 1
  • 49
  • 52