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$\DeclareMathOperator\Spec{Spec}\DeclareMathOperator\SH{SH}\DeclareMathOperator\Mod{Mod}\newcommand{\fin}{\mathrm{fin}}\newcommand{\cc}{\mathrm{c}}$I'm mainly interested in the dynamics of the Balmer spectrum construction at the moment. I know that famously the Balmer spec of a derived category of a large number of schemes recovers the original space, but it is known that the construction only results in a quasi-compact locally ringed space, which might not be a scheme for an arbitrary tensor triangulated category. Balmer gives the following result in

Balmer, Paul, Spectra, spectra, spectra – tensor triangular spectra versus Zariski spectra of endomorphism rings, Algebr. Geom. Topol. 10, No. 3, 1521-1563 (2010). ZBL1204.18005.

Proposition 9.7. The locally ringed space $\Spec(\SH^{\fin})$ is not a scheme. Nor is any of the local ones $\Spec(\SH^{\fin}_{(p)})$, for any prime p.

Where $\SH^{\fin}$ is the stable homotopy category of finite spectra.

My question is: Let $D=D(\Mod(\Spec(\SH^{\fin}))$ be the derived category of sheaves of modules over $\Spec(\SH^{\fin})$, this category is a tensor triangulated category. Is it true that $\Spec(D^{\cc})\cong \Spec(\SH^{\fin})$?

Here $D^{\cc}$ is the subcategory of compact objects which I assume is also a tensor triangulated category.

I suspect that the answer is either no or that this is too hard to calculate in particular. The spirit of my question, however, is more in the lines of the following:

Let $T$ be a compactly generated tensor triangulated category and $K=T^{\cc}$, when is $\Spec(D(\Spec(K)^{\cc})\cong \Spec(K)$? Obviously if $D(\Spec(K))^{\cc}\cong K$ as tensor triangulated categories this works, but are there other situations?

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    $\begingroup$ What a collision in naming! $\endgroup$
    – Buzz
    Aug 24, 2022 at 19:35
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    $\begingroup$ It looks like the Ballmer spectrum of the stable category is the Zariski spectrum of the stack of formal groups. So the question probably reduces to the similar question for stacks. The simplest relevant stack is the moduli of filtered vector spaces: $\mathbb A^1/\mathbb G_m$. What are the modules on its Ballmer spectrum? $\endgroup$ Aug 24, 2022 at 21:41
  • $\begingroup$ @BenWieland Ah thanks for the observation. My ( very amateur ) understanding of this is that the derived category of M_fg is not the same as SH^fin, no? So I think the same must be true for the Zariski space? I need to think of this and about that moduli space you propose. $\endgroup$
    – AT0
    Aug 24, 2022 at 22:53

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