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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?

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2 Answers 2

up vote 12 down vote accepted

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.

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Thanks so much! –  Jon Beardsley Feb 23 '12 at 21:22
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Neil's answer is great. I just wanted to add that in fact the Bousfield classes of the Morava $K$-theories are minimal non-zero classes in the Bousfield lattice. In particular, $\langle K(n) \rangle$ and $\langle K(n-1) \rangle$ are not comparable.

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Thanks John. So we might say something like $\langle K(n)\rangle$ are the atoms of the Bousfield lattice? –  Jon Beardsley Feb 23 '12 at 21:59
    
Sorry that probably seems like a completely random question. I'm trying to look at either BA or DL from the point of view of order theory and that kind of stuff and figure out things like what are the prime ideals and so forth. In a Boolean algebra an element generates a prime ideal if and only if its complement is an atom. So, it appears that there are prime ideals of the form ↓a⟨K(n)⟩. –  Jon Beardsley Feb 23 '12 at 22:04
    
Look at my paper with Hovey, "The structure of the Bousfield lattice." We don't address prime ideals, but we discuss a conjectured description of the atoms for BA. (The atoms for the whole Bousfield lattice could be more complicated. The Brown-Comenetz dual of the sphere might be one.) –  John Palmieri Feb 24 '12 at 15:49
    
Thankyou! I'm looking at it now. –  Jon Beardsley Feb 24 '12 at 21:18
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