There are a number of notions of "cover" for a scheme: etale, faithfully flat, fpqc, fppf, Zariski, Nisnevich, etc. Most of these have a nice property, which is that a cover of that type satisfies effective descent for modules or quasicoherent sheaves. For me it will be easiest to state this just for rings (hence for affine objects): A morphism $f:R\to S$ of commutative rings satisfies effective descent for modules if the category of $R$-modules can be recovered from the category of $S$-modules with descent data. There are many interpretations of descent data, but it should basically be thought of as gluing data for $Spec(S)\times_{Spec(R)} Spec(S)$.

Grothendieck showed that faithfully flat morphisms always satisfy this condition, and Joyal and Tierney showed that a more general class of morphisms, called pure morphisms, completely classifies the morphisms which satisfy this condition (i.e. a morphism satisfies effective descent for modules if and only if it is "pure"). Does anyone know what the "pure" site looks like? Is it subcanonical? How come nobody ever uses it for anything?

More generally, suppose I have some other stack (i.e. not the stack of modules, but something else). I can associate to it a topology where the covers are exactly the morphisms along which this stack descends. What does this topology look like? Has anyone studied it?

  • 1
    $\begingroup$ Is the question "is the pure site subcanonical?" equivalent to the question "does any morphism which satisfies effective descent for modules also satisfy effective descent for morphisms to any target ring $T$ (meaning $\mathop{\mathrm{Spec}}T$-points)?"? I'm trying to understand the question and I'm lost in a maze of twisty little conditions, all alike. ☹ $\endgroup$ – Gro-Tsen Apr 7 '16 at 22:00
  • $\begingroup$ Yeah basically. You'd need to at least prove that given a pure morphism $Spec(R)\to Spec(S)$ and a map $Spec(R)\to Spec(T)$ such that the pair of pulled back maps $Spec(R\otimes_S R)\to Spec(T)$ agree, you get a unique map $Spec(S)\to Spec(T)$. $\endgroup$ – Jonathan Beardsley Apr 8 '16 at 23:04
  • $\begingroup$ It may be that the induced topology is just not subcanonical, since it would be quite strong (stronger than fpqc I think). $\endgroup$ – Jonathan Beardsley Apr 8 '16 at 23:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.