The nilpotence theorem of Devinatz, Hopkins, and Smith says that if $f: F \to X$ is a map out of a finite spectrum, and if $1_{MU} \wedge f = 0: MU \wedge F \to MU \wedge X$, then $f^{\wedge n} = 0$ for some $n \in \mathbb N$.
Here's a stronger claim:
Dubious claim: If $1_{MU} \wedge f = 0$, then $f = 0$.
Note that the condition that $1_{MU} \wedge f = 0$ is stronger than saying that $f$ induces the zero map on $MU$-homology, so the dubious claim is not quite as dubious as it might appear at first glance (although it's still quite dubious).
"Proof" of dubious claim:
The map of cosimplicial spectra $f \wedge 1_{MU^{\wedge \bullet+1}}: F \wedge MU^{\wedge \bullet+1} \to X \wedge MU^{\wedge \bullet+1}$ is the zero map, and its totalization if the completion of $f: F \to X$ at $MU$, which is the same as the localization of $f$ at $MU$, which is just $f$ itself. The totalization of a zero map is zero, so $f=0$.
Question: Where does this "proof" go wrong?
The nlab seems to indicate that the totalization of these complexes really should recover $F$ and $X$. The theorem of Bousfield stated as Proposition 3.3 there seems to validate the claim that the completion and localization agree, and I'm sure I've been told that the Bousfield class of $MU$ is maximal. Is the mistake in asserting that a totalization of zero maps is zero? Or are the conditions on $F$, $X$ for these totalizations to work out more restrictive than I'm realizing?
After some discussion with a friend, I think I know where it goes wrong. The mistake is in asserting that the natural transformation $f\wedge M^{\wedge \bullet+1}$, as a map of cosimplicial spectra, is zero. It is true that this map is zero at the level of objects, but in order to be a zero map of cosimplicial spectra, all of the higher commutativities of this natural transformation must also be null. But there's no reason to believe this is the case. In terms of spectral sequences, nontriviality of this map may be detected in a higher filtration.