Suppose we are given a thick subcategory of the compact objects in the homotopy category of modules over a ring spectrum $R$. Are there conditions we can place on $R$, or on the category (compact) $R$-modules to ensure that every thick subcategory has a single generator? It seems that, given the existence of finite Bousfield localizations, this would be solved by showing that the associated finite colocalization, which is a functor to the given thick subcategory, preserves compactness. But it doesn't seem that in general this should be true. Certainly localization functors do no always preserve compactness.

My best idea at the moment is to show that finite localizations of compact objects are compact, hence the fiber of the localization morphism $X\to\mathcal{L}^{fin}X$ is a fiber of compact objects, hence compact itself. Then one could apply this to $R$ itself. I am not sure if all of those statements are valid however, and am worried that there are some counterexamples I am not aware of.

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    $\begingroup$ Even in the nicest of cases, finite (co)localizations usually don't preserve compactness. For instance, in (say) $H\mathbb{Z}_{(p)}$-modules, there is a finite localization which inverts $p$; the only compact local object is 0, and the colocalization of a compact object is compact iff the object was already colocal (that is to say, torsion). $\endgroup$ Commented Dec 1, 2012 at 0:43

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When $R$ is the Eilenberg-MacLane spectrum of a Noetherian ring, thick subcategories are in bijection with specialization-closed subsets of $\mathrm{Spec}\ \pi_0(R)$. Such a thick subcategory is generated by a single compact object iff the specialization-closed subset is actually Zariski-closed (and in that case a generator is given by a Koszul complex for generators of an ideal corresponding to the closed set). To say that this holds for all thick subcategories imposes a rather strong condition on the ring $\pi_0(R)$; I believe it's actually equivalent to $\mathrm{Spec}\ \pi_0(R)$ having only finitely many points. (Note that although this condition does hold for $p$-local spectra and the corresponding "Spec" has infinitely many points, one for each height, this situation is badly non-Noetherian.)

The same story holds more generally if the graded ring $\pi_*(R)$ is Noetherian and stratifies the category of $R$-modules, in the sense of Benson-Iyengar-Krause. For instance, this automatically holds if $\pi_*(R)$ is a regular ring concentrated in even degrees.

Here's a sketch of a proof that if $R$ is a Noetherian ring such that any specialization-closed subset of $\mathrm{Spec}\ R$ is closed, then $\mathrm{Spec}\ R$ is finite. A specialization-closed set is just a union of closed sets, so this implies that any set consisting only of closed points is closed. But then quasicompactness implies there can only be finitely many closed points. We can now look at prime ideals of dimension (coheight) 1, and by a similar compactness argument there can be only finitely many of them. Continuing by induction on dimension, we get that for each dimension, $R$ can only have finitely many primes of that dimension. But $\mathrm{Spec}\ R$ has to be finite-dimensional (since, say, there are only finitely many maximal ideal and the localization at any maximal ideal is finite-dimensional), so it can only have finitely many points in total.

Something that might be more reasonable to ask for is that any thick subcategory is a join of irreducible thick subcategories, and that any irreducible thick subcategory is generated by a single object. Here I say a thick subcategory is irreducible if it is not the join of two smaller thick subcategories. In the stratified case, irreducible thick subcategories correspond to irreducible closed subsets of $\mathrm{Spec}\ \pi_*(R)$, and all of these thick subcategories are indeed generated by a single object.

  • $\begingroup$ Eric thanks. Just giving this a quick read. This is exactly the sort of situation I'm trying to think about. Would you mind if I sent you an e-mail about it? $\endgroup$ Commented Dec 1, 2012 at 5:58
  • $\begingroup$ Sure--you should be able to guess my email from my webpage listed on my MO userpage. $\endgroup$ Commented Dec 1, 2012 at 13:04

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