Suppose we have $\langle K(n)\rangle$ and $\langle K(n-1) \rangle$ for some fixed prime $p$. do we know whether or not $\langle K(n) \rangle \geq \langle K(n-1) \rangle$ or $\langle K(n-1) \rangle \geq \langle K(n) \rangle$?


It seems to me that the first relation is accurate if we somehow restricted ourselves to finite spectra (from Ravenel's "Localization at Certain Periodic Homology Theories"). However, we obviously aren't doing that (right?). 


It also seems to me that such a relationship would be incredibly problematic, mainly because we'd have the following. Assume the first relationship held and we restrict ourselves to operating within the distributive sub-lattice of the Bousfield lattice. Then $\langle K(n)\rangle \wedge\langle K(n-1)\rangle=\langle K(n-1)\rangle = \langle 0\rangle$ and this would seem to me to be incredibly problematic. Is this an accurate assessment of the situation, or have I missed something?



Relationship of Bousfield Classes of Morava K-theories
It is standard that $K(n)\wedge K(m)=0$ for $n\neq m$. One way to think about this is as follows: if $E$ and $F$ are complex oriented ring spectra then the corresponding formal group laws become isomorphic over $\pi\_\*(E\wedge F)$, but it is easy to see that formal group laws of different heights can only become isomorphic over the zero ring.


On the other hand, as $K(n)$ is a ring spectrum we have maps $K(n)\xrightarrow{\eta}K(n)\wedge K(n) \xrightarrow{\mu} K(n)$ whose composite is the identity, so $K(n)\wedge K(n)$ is nonzero.


This means that we cannot have $\langle K(n)\rangle\leq\langle K(m)\rangle$ unless $m=n$. Indeed, if $m\neq n$ then we saw that $K(n)$ is $K(m)$-acyclic. If we had $\langle K(n)\rangle\leq\langle K(m)\rangle$ we could conclude that $K(n)$ was also $K(n)$-acyclic, or in other words $K(n)\wedge K(n)=0$, which is false.


It is true that when $n\leq m$ we have 
$$\{K(n)-\text{acyclic finite spectra}\}\supseteq\{K(m)-\text{acyclic finite spectra}\}$$
which might suggest that $\langle K(n)\rangle\leq\langle K(m)\rangle$, but that is only a suggestion and it does not work out to be true.


