Bousfield, in his paper "The Boolean algebra of spectra" (Comm Math Helv 54, 368–377 (1979), https://doi.org/10.1007/BF02566281), defined $\mathbf{DL}$, a sublattice of the Bousfield lattice, to consist of all Bousfield classes $\langle X \rangle$ such that $\langle X \rangle \wedge \langle X \rangle = \langle X \rangle$. He pointed that if $X$ is a wedge of ring spectra, then $\langle X \rangle \in \mathbf{DL}$. Are there classes in $\mathbf{DL}$ which are known not to be Bousfield equivalent to a wedge of ring spectra? If the telescope conjecture fails, then that would yield examples. Are there others?

## 1 Answer

My paper *A combinatorial model for the known Bousfield classes* defines an complete ordered semiring $\mathcal{A}$ and a homomorphism from $\mathcal{A}$ to the Bousfield lattice mod the telescope conjecture, whose image contains most of the Bousfield classes that have been named and studied. In $\mathcal{A}$, all elements satisfying $x\wedge x=x$ correspond to wedges of unital ring spectra. (In fact, they all correspond to unital ring spectra, except for the case of $\bigvee_{i\in U}K(i)$, which is not obviously a unital ring if $U$ is infinite.) Although the main results are stated in a quotient of the Bousfield lattice where the telescope conjecture is forced to be true, many intermediate results are stated in the Bousfield lattice itself. I am therefore fairly confident that the literature does not contain any unconditional counterexamples for your question.