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Let $X$ be a finite spectrum, and let $N = dim_{\mathbb F_p} H_\ast(X;\mathbb F_p)$. I believe that $p$-completion $X^\wedge_p$ may be built as an $N$-cell complex where the cells are shifts of the $p$-complete sphere $S^\wedge_p$. That is, there is an $N$-step filtration $0 = X_0 \to X_1 \to \cdots \to X_N$ where the associated graded has a shift of $S^\wedge_p$ at each step.

Question 1: Let $X$ be a finite spectrum, and let $N = dim_{K(n)_\ast}K(n)_\ast X$. Then can the $K(n)$-localization $L_{K(n)} X$ be built as an $N$-cell complex where the cells are shifts of the $K(n)$-local sphere $L_{K(n)} S$?

Question 2: Same, but don't assume that $X$ is finite; only assume that $N$ is finite.

The spectra considered in Question 2 are the closure of those considered in Question 1 under retracts.

If it makes a difference to work $T(n)$-locally rather than $K(n)$-locally, that would be interesting.

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    $\begingroup$ A counterexample is given by the "upside-down question mark complex" (the cofiber of the extension of $\eta$ to a map $\Sigma S/(2) \rightarrow S$), whose $K(1)$-localization generates the exotic subgroup of the $K(1)$-local Picard group at $p=2$. I think the right question to ask is: if $dim_{K(n)_\ast} K(n)_\ast X = N$, can $L_{K(n)}X$ be built as an $N$-cell complex, where the cells are elements of the $K(n)$-local Picard group (i.e. have $1$-dimensional $K(n)$-homology)? $\endgroup$
    – William Balderrama
    Commented Dec 15, 2021 at 1:20
  • $\begingroup$ Hovey–Strickland’s Y spectrum is again a problem for Question 2, even allowing for William’s modification, as it was a problem in this older question of yours: mathoverflow.net/questions/363652/… $\endgroup$
    – Eric Peterson
    Commented Dec 15, 2021 at 4:40
  • $\begingroup$ oh wow I duplicated myself. $\endgroup$
    – Tim Campion
    Commented Dec 15, 2021 at 5:15
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    $\begingroup$ I initially deleted this post as a duplicate, but I have referred back to the comments here so many times in the past year that I think I should undelete it and just leave it as CW in case anybody else finds the comments as useful as I have. $\endgroup$
    – Tim Campion
    Commented Dec 22, 2022 at 22:01

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