Fix some field $k$ of characteristic $0$, a smooth $k$-scheme $X$, a commutative and unital ring $R$ with nilradical $I$, and two infinitesimally close $R$-points: $$x,y: \mathrm{Spec}R \rightrightarrows X$$ We define the de Rham prestack to be the following functor: $$X^{dR}: \mathsf{CRing} \to \mathsf{Sets}: R \mapsto X^{dR}(R) := X(R/I)$$ Then, is there is the following equivalence of categories: $$\mathsf{CrysQCoh}(X) \cong \mathsf{QCoh}(X^{dR})$$ between the categories of crystals of quasi-coherent sheaves over $X$ and of quasi-coherent sheaves over the de Rham prestack $X^{dR}$ ? Pointwise, the de Rham prestack is the coequaliser of the diagram: $$X(x), X(y): X \rightrightarrows X(R)$$ by the definition of infinitesimally close points. This suggests to me that if given some quasi-coherent sheaf $\mathcal{F}$ on $X$, it can then be pulled back along $x, y$, and along the canonical map $i: \mathrm{Spec} R/I \to \mathrm{Spec} R$ to yield an isomorphism of quasi-coherent $R/I$-modules: $$i^*x^*\mathcal{F} \cong i^*y^*\mathcal{F}$$ Thus, I suspect the equivalence to be true, but I'm having trouble articulating my thoughts.

  • $\begingroup$ According to de Rham space in nLab the big site of $X_{dR}$ is the crystaline site of $X$. $\endgroup$ – Niels May 13 at 20:27
  • $\begingroup$ @Niels Could you elaborate ? $\endgroup$ – Dat Minh Ha May 13 at 21:37
  • 1
    $\begingroup$ Your coequalizer diagram looks rather ill-defined. How are $x$ and $y$ maps from the scheme $X$ to the set $X(R)$? I thought they were infinitesimally nearby elements of $X(R)$, meaning composition with the map from $\operatorname{Spec} R/I$ yields identical elements of $X(R/I)$. If $X$ is smooth, then you can use the coequalizer of the two projections from the completion of the diagonal in $X \times X$. $\endgroup$ – S. Carnahan May 15 at 13:02
  • $\begingroup$ @S.Carnahan I applied the functor $h_X$ to the first diagram. $\endgroup$ – Dat Minh Ha May 15 at 13:04
  • 1
    $\begingroup$ Presumably it goes something like the following: u have a universal pair of inf close points given by the two projections from the formal neighborhood of the diagonal, and this formal nhd is precisely the fiber product of X with itself, taken relative to the dR stack. A suitable version of descent should identify the two categories now. $\endgroup$ – EBz May 15 at 13:55

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.