# Intuition - difference between Moore spectrum and Eilenberg-Mac Lane spectrum

I know very little about algebraic topology, and more about $$k$$-linear stable $$\infty$$-categories (i.e. homological algebra).

Given an abelian group $$A$$, there is the Eilenberg-Mac Lane spectrum $$HA$$, for which: $$\pi_{< 0}(\cdot) = 0$$, $$\pi_0 (\cdot) = A$$, $$\pi_{>0}(\cdot) = 0$$. There is also the Moore spectrum $$SA$$, for which $$\pi_{<0}(\cdot) = 0$$, $$\pi_{=0} (\cdot)= A$$, $$H_{> 0} (\cdot , \mathbb{Z}) = 0$$.

I more-or-less have some feeling for $$HA$$ (under Dold-Kan, it is simply $$A$$, so one simply gets connected deloopings $$BA$$, $$BBA$$, etc. making the infinite loop space structure explicit).

I have no feeling what-so-ever about $$SA$$.

But reading about Bousfield localization, it seems that most the standard examples are with $$SA$$. How to understand this for a novice? Say, for $$p$$-localization and $$p$$-completion, how do I understand that I "want" to use the Moore spectrum and not the Eilenberg-Mac Lane spectrum?

Or, at least, what is the rationale for introducing Moore spectra?

Thank you

• Denis's answer is 100% the reason why the specific object S/p appears in the theory of localization. You could still ask whether you could use HZ/p, and here the answer is somewhat unintuitive: for connective spectra, localization at S/p and at HZ/p agree, and so you cannot tell the difference before passing to nonconnective spectra, which are somewhat exotic objects. The spectrum $KU := \Sigma^\infty_+ \mathbb CP^\infty[\beta^{-1}]$ is a concrete example of a spectrum with $H_*(KU; Z)$ rational, hence $H_*(KU; Z/p)$ and $L_{HZ/p} KU$ null, but $L_{S/p} KU$ nontrivial. – Eric Peterson Feb 2 at 13:26
• @EricPeterson: Thank you! I will have to think about it. – Sasha Feb 2 at 19:40

I'm not sure I can answer to the general question, but I can explain why the Moore spectrum $$\mathbb{S}/p$$ shows up in the discussion of Bousfield localizations.
This is just the cofiber of multiplication by $$p$$ from $$\mathbb{S}$$ to itself. Hence saying that for a spectrum $$X$$ we have $$X\wedge \mathbb{S}/p=0$$ means simply that multiplication by $$p$$ acts on $$X$$ as an equivalence. So $$X$$ is $$\mathbb{S}[1/p]$$-local iff it is $$\mathbb{S}/p$$-acyclic. From standard localization arguments (plus the fact that $$\mathbb{S}/p$$ is a finite spectrum) you get the fracture square relating $$\mathbb{S}[1/p]$$-localization and $$p$$-completion (i.e. the "complementary localization" of $$\mathbb{S}[1/p]$$-localization).
So, essentially, the reason it shows up is because the Moore spectrum is the layer of the diagram $$\mathbb{S} \xrightarrow{p} \mathbb{S}\xrightarrow{p}\mathbb{S}\xrightarrow{p}\mathbb{S}\xrightarrow{p}\cdots$$ that computes $$\mathbb{S}[1/p]$$-localization. From this it is clear that $$\mathbb{S}/p^n$$ is also going to be relevant to the study of $$p$$-completion (since these are just the "higher" layers of the tower), and from there to $$\mathbb{S}/p^\infty = \mathbb{S}\mathbb{Q}_p/\mathbb{Z}_p = \mathrm{colim}\,\mathbb{S}/p^n$$ and $$\mathbb{S}\mathbb{Q}/\mathbb{Z}=\bigvee_p \mathbb{S}/p^\infty$$ is just a small step.
• Thank you! This now makes perfect sense. But for p=2, it seems that the cofiber of multiplication by 2 on the sphere spectrum has $\pi_0$ with four elements, so this should differ from the Moore spectrum as defined in my question, no? – Sasha Feb 2 at 19:40
• @Sasha No, $π_0\mathbb{S}/2=H_0\mathbb{S}/2=\mathbb{Z}/2$. Maybe you're confounding it with $[\mathbb{S}/2,\mathbb{S}/2]=\mathbb{Z}/4$? (Yes, it is a bit counterintuitive but this nontrivial extension is secretly coming from $\pi_1\mathbb{S}$) – Denis Nardin Feb 2 at 19:41
• I don't quite have a say, since this is so common in homological algebra (representaiton theory etc.) that I can't go against it. But perhaps it makes sense to put $\pi_1$ in negative degree, since it represents "smaller" stuff? (like the cardinality of $BG$ is $|G|^{-1}$ etc.) Anyway, probably one can find justification for each. – Sasha Feb 2 at 19:52