Absence of Maps Between p-local and q-local spectra

Suppose $X$ and $Y$ are spectra (or homotopy classes thereof) such that $X$ is p-local and $Y$ is q-local, for primes $p\neq q$. Is it indeed true then, and if so how would one show that $[X,Y]_\ast=0$?

Thanks!

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The rational Eilenberg-Mac Lane spectrum $H\mathbb{Q}$ is $p$-local for every prime $p$, but certainly $[H\mathbb{Q}, H\mathbb{Q}]_*\neq 0$.

If you replace "$p$-local" and "$q$-local" with "$p$-complete" and "$q$-complete", then your conclusion does hold.

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Thanks very much for this. I guess I'm trying to understand this notion of Bousfield's about $E_\ast$-local spectra, the correlation between this notion and the notion of localizing at a prime p. That is, localizing at a prime p by smashing with the sphere spectrum localized at p, and relating this to a localization functor a la Bousfield. What functor does this define and what are its acyclics then? Sorry, perhaps I don't understand the terminology. –  Jon Beardsley Jan 31 '12 at 22:51
Tom Goodwillie's comment got messed up by Latex, so I'm going to try and reproduce it here: The conclusion also holds if you say $p$ local and $q$ complete. Since $p$ complete implies $p$ local this includes the statement that it holds if you say $p$ complete and $q$-complete. In fact, on the homotopy level the $E$ local spectra are what are usually called the $q$ complete spectra while the $E$ acyclic spectra include all the $p$ local spectra for $p\neq q$. And in general a map from an $E$ acyclic spectrum to an $E$ local spectrum is trivial. –  David White Feb 1 '12 at 13:51
It appears all comments will now stretch far off screen. I think this can be fixed if Tom deletes his comment. Incidentally, I think the latex issue was caused by the little dashes after $p$ but before local'' –  David White Feb 1 '12 at 13:53
The issue is that markdown is interpreted before LaTeX, so underscores are interpreted as italics, the solution is to bracket your entry with the little left hand apostrophe thing, `. This exits markdown. Also you can bracket your statement with <p> </p>. –  Jon Beardsley Feb 2 '12 at 2:47
Or at least, that's one possible issue that has caused similar problems for me. –  Jon Beardsley Feb 2 '12 at 2:49

[I wish I could delete my comment to Charles's answer. My comment seems to have exploded and taken your comment down with it.]

Both localization of a spectrum at a prime and completion of spectra at a prime are examples of Bousfield localization. The one is (a little circularly) localization with respect to $p$-localized homotopy theory, while the other is localization with respect to mod $p$ homotopy theory.

Localizing a spectrum at a prime $p$ (or at a set of primes) is the kind of Bousfield localization that is most like localization in algebra. In this case $\pi_n$ of the localization of $X$ is the algebraic localization of $\pi_nX$, i.e. the result of inverting all the primes other than $p$. The local spectra are the spectra $X$ such that for every $n$ the abelian group $\pi_nX$ is one on which every prime other than $p$ acts invertibly, and the acyclic objects are the ones which become trivial upon localizing, i.e. the spectra $X$ such that for every $n$ the group $\pi_nX$ is a torsion group without $p$-torsion. Localization of $X$ is the same as smash product of $X$ with the localization of the sphere spectrum. If you want to describe the acyclic objects as the $E_*$-acyclic objects for some $E$, you can of course let $E$ be the localization of the sphere. Note that we are not starting with an $E$ and using the general machine of Bousfield to make the localization functor, although of course we could.

Completing at a prime is again not too far from being a purely algebraic matter. Here we can choose $E$ to be the mod $p$ Moore spectrum, i.e. the homotopy cofiber of the map $p:S\to S$ from the sphere spectrum to itself. The acyclic objects are the spectra for which each homotopy group has $p$ acting invertibly. The localization $LX$ (which is called the $p$-completion) can be described as the holim, over natural numbers $k$, of the smash product of $X$ with the mod $p^k$ Moore spectrum. If each homotopy group of $X$ is finitely generated, then $\pi_nLX$ can be described as the tensor product of $\pi_nX$ with the group $\mathbb Z_p$ of $p$-adic integers, and also in this case $LX$ is the smash product $X\wedge LS$. But neither of these statements is true for general $X$: for example, the $p$-completion of $H\mathbb Q$ is trivial while $\mathbb Q\otimes \mathbb Z_p$ is the field of $p$-adic numbers and $X\wedge LS=H(\mathbb Q\otimes \mathbb Z_p)$. And (therefore) $LH(\mathbb Q/\mathbb Z)=\Omega H\mathbb Z_p$.

Note that a spectrum which is local at a prime $q$ different from $p$ is trivial mod $p$. In general maps from $E$-acyclic objects to $E$-local objects are trivial, so we see that maps from a $q$-local spectrum to a $p$-complete spectrum are trivial. In particular, as Charles says, maps from a $q$-complete spectrum to a $p$-complete spectrum are trivial.

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Oh wow. This is really fantastic. This is exactly the sort of thing I've been having trouble finding written down as concretely. Thanks! –  Jon Beardsley Feb 1 '12 at 1:59
Additionally, my mistake is in assuming that $p$-local implies $q$-acyclic! Which is of course ridiculous, given Charles' simple example. What I should really be saying is, if a spectrum has all homotopy concentrated in $q$-torsion then it is an acyclic of $S_{(p)}$, the localized sphere spectrum. In that case, assuming $X$ and $Y$ have all homotopy in only $p$ and $q$ torsion, respectively, is it the case that $[X,Y]_\ast=0$? –  Jon Beardsley Feb 1 '12 at 3:16
Oh, right, in fact if $X$ is $p$-torsion and $Y$ is $q$-local (in particular if $X$ is $p$-torsion and $Y$ is $q$-torsion) then you have an $E$ (namely the $q$-local sphere) such that $X$ is $E$-acyclic and $Y$ is $E$-local. So no nontrivial maps from $X$ to $Y$. –  Tom Goodwillie Feb 1 '12 at 4:21
You can delete your comment after Charles's answer. If you scroll all the way to the left there will be a little x in the corner of your comment's box. I think I got all the content from your comment in my comment (with credit given to you, of course) –  David White Feb 1 '12 at 13:54
Done. (The little x is way to the right, not way to the left. The first couple of times that I looked for it I somehow couldn't find it.) –  Tom Goodwillie Feb 1 '12 at 14:27