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Is there a classification of the compact MU or BP modules in any category of spectra? Can the periodicity theorem be finagled to give a MU-module structure on finite spectra?

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What does the second sentence mean? – Dylan Wilson Dec 14 '12 at 21:53
My suspicion is that there are almost no finite spectra which admit MU-module structures. – Tyler Lawson Dec 14 '12 at 23:47
up vote 10 down vote accepted

No nonzero finite spectrum admits an $MU$-module structure. Indeed, suppose $F$ is a finite spectrum with an $MU$-module structure. Then for all $n$, $F$ has a map $v_n:\Sigma^{2p^n-2}F\to F$, which induces an isomorphism on $K(n)_*F$ (there's a subtlety here in that it's not obvious that the $v_n$ map on $F$ and the $v_n$ map on $K(n)$ give rise to the same map on $K(n)\wedge F$; see eg the end of the proof of Lemma 7 of Thus for each $n$, $F/v_n$ is $K(n)$-acyclic. But $F/v_n$ is finite, so this implies it is also $K(m)$-acyclic for all $m<n$, and so $v_n$ is also an isomorphism on $K(m)_*F$. But by finiteness of $F$, for any $m$ sufficiently large we can find $n>m$ for which $v_n$ must be $0$ on $K(m)_*F$ just for reasons of degree (by the AHSS for $K(m)_*F$). Thus $K(m)_*F=0$ for all sufficiently large $m$, which implies $F=0$.

I would also add that even if you did have a finite spectrum with an $MU$-module structure, it could not possibly be compact as an $MU$-module. Indeed, if it were, after smashing with $H\mathbb{Z}$ it would be a compact $H\mathbb{Z}\wedge MU$-module. But $\pi_*(H\mathbb{Z}\wedge MU)$ is a polynomial ring on infinitely many generators, and so all but finitely many of those generators have to act non-nilpotently on any compact module (basically, any "finite presentation" of a compact module can only involve finitely many of the polynomial generators). Since $\pi_*(H\mathbb{Z}\wedge F)=H_*(F)$ vanishes in all but finitely many degrees for $F$ finite, this is impossible.

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Eric, having trouble fully understanding this proof, will have to look at it again later. Do you think there are no compact objects at all in the homotopy category of MU-modules though? Or just no modules which are finite spectra? – Jon Beardsley Dec 17 '12 at 0:32
For any $A_\infty$ ring spectrum $R$, $R$ itself is compact as an $R$-module. More generally, for instance, for any finite spectrum $F$, $R\wedge F$ is a compact $R$-module. You should think of "compact" as meaning something like "finitely presented". Just as there's no reason to expect a finitely generated module over a ring to be finitely generated as an abelian group, there's no reason to expect a compact $R$-module to be a finite spectrum (=compact $S$-module). – Eric Wofsey Dec 17 '12 at 1:20
Ok yeah, thanks. Interesting. – Jon Beardsley Dec 17 '12 at 5:34

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