This question has a few parts:


1) Is the Bousfield class of $\langle E\rangle$ the class of $E$-acyclics, i.e. $\langle E\rangle=\left\{ X:E\wedge X=0\right\}$ or is it the class of spectra which are Bousfield equivalent to $E$? It seems that both of these definitions are used. In either case, which is then the correct characterization of the partial ordering on the Bousfield lattice, as the former definition of Bousfield class seems amenable to reverse inclusion, but the latter does not.


2) Secondly, the following statement is made in Hovey and Palmieri's "Structure of the Bousfield Lattice" and I feel that it is probably really obvious, but that I must be missing something:


Given an element of the Bousfield lattice $\langle E\rangle$ such that $\langle E\rangle <\langle H\mathbb{F}\_p\rangle$ we have that $E\wedge H\mathbb{F}\_p=0$ else $\langle E\rangle\geq \langle H\mathbb{F}\_p\rangle$. I cannot see why this is necessarily true in general and assume it must have something to do with the nature of $H\mathbb{F}\_p$ that I am missing. Can anyone help me out? I do not believe any other suppositions are made of $E$ than it is a spectrum. 


So perhaps (2) depends on (1) in some way. H&P say that $\langle E\rangle$ is the class of acyclics, but then how is (2) true in that case? I also believe that, definitions aside, (2) would follow if it was the case that $H\mathbb{F}\_p\wedge H\mathbb{F}\_p=0$ but I don't think that's true....is it?


Please excuse my ignorance!



Bousfield Classes
1) The best convention is probably this. Identify $\langle E\rangle$ with the class of $X$ for which $E\wedge X\neq 0$. The the "Bousfield class of $E$" is the class $\langle E\rangle$, and we say that "$E$ and $F$ have the same Bousfield class" if $\langle E\rangle = \langle F\rangle$. 


With this convention, the partial ordering on Bousfield classes is just class containment: $\langle E\rangle \leq \langle F\rangle$ iff $\langle E\rangle \subseteq \langle F\rangle$.


2) The new ingredient comes from the Kunneth theorem for ordinary homology, which tells you that
$$H\mathbb{F}\_p \wedge X \wedge Y = 0 \quad\text{iff either}\quad H\mathbb{F}\_p\wedge X=0 \quad\text{or}\quad H\mathbb{F}\_p\wedge Y=0.$$
If $\langle E\rangle< \langle H\mathbb{F}\_p \rangle$, then there is an $X$ such that $H\mathbb{F}\_p\wedge X\neq0$ but $E\wedge X=0$. This would satisfy $H\mathbb{F}\_p\wedge E\wedge X=0$, and to be consistent with Kunneth, we must have $H\mathbb{F}\_p\wedge E=0$.


