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.

  • $\begingroup$ I'm not sure which parts of the story you know and which parts you don't know. Do you already understand that $q^{-1}\mathbb S$ is a Moore spectrum for the group $q^{-1}\mathbb Z$? $\endgroup$ Feb 2, 2012 at 2:13
  • $\begingroup$ Like the thing above, I do "know" this I guess, but I can't see off the top of my head how to prove it. $\endgroup$ Feb 2, 2012 at 3:24

1 Answer 1


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).

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    $\begingroup$ Wow thanks so much. Ultimately my question really concerned the behavior of the functor H, and what you told me makes that really clear! Is there a good reference that proves that fact? $\endgroup$ Feb 2, 2012 at 3:25
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    $\begingroup$ The fact about tensor products (which I just edited, I was being a bit hasty) is in EKMM, Theorem 2.1. I can't think of a reference for the others, but the fact about sums and filtered colimits is seen just by looking at homotopy groups, and the fact about exact sequences follows from the fact that $HA$ represents ordinary (co)homology which transforms s.e.s. of coefficients into l.e.s. $\endgroup$ Feb 2, 2012 at 5:52
  • $\begingroup$ Been thinking about this a little more. You seem to imply that what you say following "For example..." follows from the fact that $H$ preserves sums and filtered colimits. It seems that what we really need is the functor $M:A\to M(A,0)$ which takes $A$ to its Moore spectrum, to commute with filtered colimits, at least in this case, since we're replacing $\mathbb{Z}$ with $\mathbb{S}$, not $H\mathbb{Z}$. Is this correct? $\endgroup$ Feb 2, 2012 at 21:28
  • $\begingroup$ Yes, but it suffices to know that $H$ has these properties: $(colim MA_i)\wedge HZ=colim (MA_i\wedge HZ)=colim HA_i=H(colim A_i)$, hence $colim MA_i=M(colim A_i)$. $\endgroup$ Feb 2, 2012 at 23:39
  • $\begingroup$ I think since projective abelian groups are just free, you mention about filtered colimits is not about projectives, but to view infinite direct sums as a filtered colimit of finite direct sums. $\endgroup$
    – Lao-tzu
    Jul 19, 2021 at 9:19

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