Let R be a commutative ring, and look at its unbounded derived category from the viewpoint of stable homotopy theory. In particular, we can define its Bousfield lattice: first, two objects X and Y are Bousfield equivalent if
. This defines an equivalence relation on the objects of
; the equivalence class of an object X is called its Bousfield class and is written
. We can define a partial ordering on these Bousfield classes, by setting
if and only if
. (Therefore,
is the largest equivalence class and
is the smallest.) The Bousfield lattice
is the resulting partially ordered “set”.
Question 1. Is the Bousfield lattice a set?
It is known to be a set if the ring R is countable or noetherian, but not in general.
Question 2. Is every object in Bousfield equivalent to a module?
This is known in the noetherian case. Indeed, the noetherian case is pretty well understood, by work of Amnon Neeman: if R is noetherian, then is isomorphic to the lattice of subsets of Spec(R). For any prime ideal
in R, define an R-module
to be
; then any object X in the derived category is Bousfield equivalent to
, which in turn is Bousfield equivalent to
, where supp(X) is the set of primes
for which
.
Question 3. What does the Bousfield lattice look like for non-noetherian rings?
This is likely to be quite complicated, but perhaps some examples could be understood. Neeman has shown that the Bousfield lattice for the ring has cardinality at least
. Note that this ring has exactly one prime ideal, so the lattice of subsets of Spec has two elements, and so this is very different from the noetherian situation…