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Let $X$ be a scheme and conisder the bounded derived category $D^b(X)=D^b(\textbf{Coh}(X))$.

Assume that $D^b(X)\cong D^b(Y)$ then $\dim X=\dim Y$ by Serre duality when $X$ and $Y$ are smooth projective varieties.

I wonder if we expect this is to be true for more general schemes?

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    $\begingroup$ This works for Gorenstein varieties over a field for example, as you have Serre duality there and the same argument than for smooth projective varieties should follow. $\endgroup$
    – AT0
    Commented Mar 16, 2023 at 16:28

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For a topologically Noetherian scheme $X$, you can recover $X$ from the tensor triangulated category of perfect complexes $D^{\rm perf}(X)$ as the Balmer spectrum. But in general you can't reconstruct it from $D^b(X)$ unless $X$ is smooth, in which case these two categories are equivalent. This is explained in detail with references to proofs and counterexamples in the introduction to Balmer, "Presheaves of triangulated categories and reconstruction of schemes" and expanded on in Balmer, "The spectrum of prime ideals in tensor triangulated categories".

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    $\begingroup$ There are non-isomorphic smooth schemes whose derived categories are equivalent as triangulated categories. $\endgroup$
    – Jason Starr
    Commented Mar 16, 2023 at 23:25
  • $\begingroup$ Are they topologically Noetherian? How does this fit with Theorem 6.3 of the second paper of Balmer quoted above? $\endgroup$
    – Dave Benson
    Commented Mar 16, 2023 at 23:28
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    $\begingroup$ Doesn't the Balmer spectrum require not just the triangulated structure, but also the monoidal structure on it? $\endgroup$
    – R. van Dobben de Bruyn
    Commented Mar 16, 2023 at 23:33
  • $\begingroup$ Yes it does. So I guess I should have said that you can recover $X$ from $D^{\mathrm{perf}}$ as a tensor triangulated category. For smooth algebraic varieties with ample canonical or anticanonical sheaf, Bondal and Orlov prove this without using the tensor structure. So do you know smooth examples where the tensor structures are different? $\endgroup$
    – Dave Benson
    Commented Mar 16, 2023 at 23:37
  • $\begingroup$ I've edited my answer to reflect this conversation. $\endgroup$
    – Dave Benson
    Commented Mar 16, 2023 at 23:39

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