MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $q^{-1}\mathbb Z$? – Tom Goodwillie Feb 2 '12 at 2:13
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. – Jon Beardsley Feb 2 '12 at 3:24
up vote 7 down vote accepted

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).

share|cite|improve this answer
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? – Jon Beardsley Feb 2 '12 at 3:25
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. – Marc Hoyois Feb 2 '12 at 5:52
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? – Jon Beardsley Feb 2 '12 at 21:28
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 $colim MA_i=M(colim A_i)$. – Marc Hoyois Feb 2 '12 at 23:39
Of course! Thanks so much! – Jon Beardsley Feb 3 '12 at 1:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.