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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 http://math.harvard.edu/~lurie/252xnotes/Lecture33.pdf). 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 (since by the AHSS for $K(n)_*F$ degenerates). 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. 2 added 686 characters in body 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 http://math.harvard.edu/~lurie/252xnotes/Lecture33.pdf). 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 (since the AHSS for$K(n)_*F$degenerates). 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. 1 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 http://math.harvard.edu/~lurie/252xnotes/Lecture33.pdf). 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 (since the AHSS for$K(n)_*F$degenerates). Thus$K(m)_*F=0$for all sufficiently large$m$, which implies$F=0\$.