Is it true that the Moore spectrum for the group $\mathbb{Z}\_{(p)}$ can be constructed by smashing $\mathbb{S}$ with $q^{-1}\mathbb{S}$ for each $q\neq p$ (here both $q$ and $p$ are primes). It seems we might wish to show this by showing that $[\mathbb{S},\mathbb{S}\_{(p)}\wedge H\mathbb{Z}]\_\ast\cong[\mathbb{S},H\mathbb{Z}\_{(p)}]$ but I cannot see how to show that either.




Thanks for any help on this matter. I apologize if this question is too basic. I have asked it on MSE and not recieved any help.




A Model for the Moore Spectrum of $\mathbb{Z}_{(p)}$
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).


