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The spectrum of a ring $R$ can be defined as $\operatorname{Spec} R := \operatorname{Hom}(R, -)\colon \mathrm{fpRing} \to \mathrm{Set}$ ($\mathrm{fpRing}$ are commutative finitely presentable rings). Is it possible to quickly extract the topos of sheaves $\operatorname{Sh}(\operatorname{Spec} R)$ directly from this definition, bypassing the construction of a locally ringed space?

For example, you can take the slice topos in the topos of sheaves on the Zariski site (hardly this is the answer, but this is an example of what the answer could sound like).

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  • $\begingroup$ Do you mean finitely presented? $\endgroup$
    – Nicolas Hemelsoet
    Commented Sep 20, 2023 at 15:54
  • $\begingroup$ Of course, thank you! Corrected. $\endgroup$
    – Arshak Aivazian
    Commented Sep 20, 2023 at 15:55
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    $\begingroup$ The slice topos produces sheaves on the big Zariski site of the scheme, I think, and the issue is you want sheaves on the small Zariski site (i.e on the underlying topological space) - does that sound right? $\endgroup$
    – Will Sawin
    Commented Sep 20, 2023 at 16:08
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    $\begingroup$ I believe that if you consider the category of Sheaves in a Zariski topology that are over $Spec R$ and locally isomorphic to $Spec R$ (where locally is taken in the sense of the Zariski pretopology) you should get the small Zariski topos (that is, Sh(Spec R) ). That should work with any scheme in fact. $\endgroup$
    – Simon Henry
    Commented Sep 20, 2023 at 17:03
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    $\begingroup$ TeX note: \operatorname should be used in place of \mathrm for things that are semantically operators. Note the difference between $\mathrm{Spec} R$ \mathrm{Spec} R and $\operatorname{Spec} R$ \operatorname{Spec} R. I have edited accordingly. (By the way, there is a special syntax for defining \operatornames to be used repeatedly: \DeclareMathOperator\Spec{Spec}.) $\endgroup$
    – LSpice
    Commented Sep 20, 2023 at 19:14

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