This entire question takes place in the $HF_p$-local category of $p$-local spectra, i.e. the essential image of $HF_p$-localization on the stable homotopy category. $HF_p$ itself is in there, and of course so is the $HF_p$-local sphere $L_{HF_p}S^0$. Let $loc(X)$ denote the smallest localizing subcategory containing an object $X$.

Question: Is there a way to show that $L_{HF_p}S^0$ is NOT in $loc(HF_p)$?

I'm imagining there might be some property $HF_p$ has that's preserved by triangles and coproducts, but that $L_{HF_p}S^0$ does not have. (But remember that coproducts here are not the same as in the full category of spectra; likewise the smash product).

Here's why I'm interested. If the answer is yes, then this local category has at least three localizing subcategories: $loc(0)$, $loc(HF_p)$, and $loc(L_{HF_p}S^0)$, the last one being the entire category. (I'm pretty sure $loc(HF_p)$ is minimal, but that's irrelevant.) This would be exciting, because I recently found out that the Bousfield lattice of this category has exactly two elements. If the answer to my question above is yes, then there's a localizing subcategory, namely $loc(HF_p)$, that isn't a Bousfield class. (The only other time that this is known to happen is in the derived category of a certain bizarre type of non-Noetherian ring; see Stevenson's http://www.arxiv.org/abs/1210.0399).

If one took a $K(n)$ instead of $HF_p$, the answer to my question is no. Hovey and Strickland ("Morava $K$-theories and localization") classified localizing subcategories in the $K(n)$-local category, and there are only two (ditto with Bousfield classes). But their proof uses that $L_{K(n)}F(n)$ is a small graded weak generator in the $K(n)$-local category. Later in the same paper they show that the $HF_p$-local category has no nonzero small objects, so the same trick won't work.

Or, on the other hand, if anyone knows how to build $L_{HF_p}S^0$ from $HF_p$, that's great too. But that seems harder.

  • $\begingroup$ Fascinating question Luke. I wish I knew the answer! $\endgroup$ Dec 15, 2013 at 19:30

1 Answer 1


First note that it is equivalent to ask whether the mod p Moore spectrum $M(p)$ is in the localizing subcategory (in the local sense) generated by $HF_p$. Indeed, the fiber $C_0 S$ of the map $S \xrightarrow{} H\mathbb{Q}$ is in the localizing subcategory generated by $M(p)$ (in the usual sense), and $L_{HF_p}C_0 S=L_{HF_p}S$. So if $M(p)=L_{HF_p}M(p)$ is in the localizing subcategory generated by $HF_p$ in the local sense, then so is $L_{HF_p}S$.

Now consider the class of all $HF_p$-local spectra $X$ such that $[X,M(p)]_*=0$. This contains $HF_p$ because $HF_p$ is dissonant and $M(p)$ is harmonic. It is obviously a thick subcategory. I claim it is also closed under $HF_p$-local coproducts. Indeed, suppose $[X_i, M(p)]_*=0$ for all $i$. Then $[\coprod X_i, M(p)]_*=0$. But $M(p)$ is $HF_p$-local, so

$ [L_{HF_p} (\coprod X_i), M(p)]_+ = [\coprod X_i, M(p)]_* =0. $

So we have a localizing subcategory in the local sense that contains $HF_p$ but not $M(p)$, like you wanted.

So there is a localizing subcategory that is not a Bousfield class! Very nice!

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    $\begingroup$ Man, that is really cool! $\endgroup$ Dec 16, 2013 at 2:35
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    $\begingroup$ Hmm, and apparently also a cohomological Bousfield class that isn't a (homological) Bousfield class. So the next question is do you think the CBC of $M(p)$ is the same as $loc(HF_p)$, or might this be a loc subcat that isn't a CBC either? Is it bad to ask new questions in the comments? $\endgroup$ Dec 16, 2013 at 6:15
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    $\begingroup$ I guess what I think is that there are tons of cohomological Bousfield classes among $HF_p$-local spectra. My guess is that $loc(HF_p)$ is a cohomological Bousfield class, but more likely of $BP_p$ (which is smaller than the CBC of $M(p)$ since there are no nonzero maps from $BP_p$ to $M(p)$). $HF_p$-localization makes the Bousfield lattice trivial, but you probably have to go to $I$-localization to make the localizing subcategory lattice trivial, where $I$ is the Brown-Comenetz dual of $S$. $\endgroup$
    – Mark Hovey
    Dec 16, 2013 at 12:27
  • $\begingroup$ That was the next one I was looking at. I was thinking maybe the CBC of $M(p)$ in the $I$-local category is a nonzero proper localizing subcategory, so we get a similar situation. But it might just be zero; I don't know that many $I$-local spectra. $\endgroup$ Dec 16, 2013 at 15:46

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