# An equivalence between $(\mathcal{D}_X^m)-\text{mod}$ and $(\mathcal{D}_X^{m+1})-\text{mod}$

This question is related to my other question. Consider a scheme $X$ over $S=\text{Spec}(\mathbb{k})$ where $\mathbb{k}=\overline{\mathbb{F}_p})$; let $F: X \rightarrow X$ be the Frobenius $p$-th power map, and let $m$ be a positive integer. Briefly, a $\mathcal{D}$-module of level $m$ over $X=\mathbb{A}^n$ is a module over the algebra $\mathbb{k}[x, \partial_x, \frac{{\partial_x}^p}{p!}, \cdots, \frac{{\partial_x}^{p^m}}{(p^m)!}]$ (for the general definition of $\mathcal{D}$-modules of level $m$ over $S$, see my other question). Denote the category of $\mathcal{D}$-modules of level $m$ by $\mathcal{D}_X^m-\text{mod}$.

Given a $\mathcal{D}$-module of level $m$, $\mathcal{F}$, I was told that $F^* \mathcal{F}$ can be given the structure of a $\mathcal{D}$-module of level $m+1$, and that this construction gives an equivalence $\mathcal{D}_X^m-\text{mod} \simeq \mathcal{D}_X^{m+1}-\text{mod}$. (Maybe a better way of phrasing this statement is that given a crystalline $\mathcal{D}$-module $\mathcal{G}$, i.e. a $\mathcal{D}$-module of level $0$, then $(F^m)^* \mathcal{G}$ can be given the structure of a $\mathcal{D}$-module of level $m$). I was wondering how to prove this statement.

EDIT: So I found the relevant paper by Berthelot mentioned below by David; my understanding is that the above statement actually seems to be true as stated (without imposing the condition that the categories has $0$ ($m$-th order) $p$-curvature); see Theorem $2.3.6$ on pg $47$ of Berthelot's paper for a more precise general statement and proof. This Theorem answers my above question.

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Vinoth - the statement looks wrong to me (even at m=0). Frobenius pullback does indeed give a functor increasing the level of D-modules but it's not essentially surjective, you get only D-modules with a vanishing [m-th order] p-curvature. This is explained in Berthelot's notes on D-modules I believe. –  David Ben-Zvi Sep 21 '12 at 15:29
Thanks David. Oh ok; then perhaps the equivalence is with $\mathcal{D}_X^m$-modules with trivial $p^m$-curvature. Do you mean Berthelot's D-modules arithmetiques I: Operateurs differentiels de niveau fini"? –  Vinoth Sep 22 '12 at 21:21
I think that's right. I think the proof is also fairly easy via faithfully flat descent for the Frobenius morphism - roughly speaking, up to p-curvatures m-th order diffops are given by the groupoid algebra of the equivalence relation defined by the p^mth order neighborhood of the diagonal, and applying Frobenius shifts m to m+1. A similar argument was attempted in my paper with Nevins "Cusps and D-modules" for a kind of descent of differential operators for cusp morphisms. –  David Ben-Zvi Sep 24 '12 at 1:15
I don't know a lot about this story, but you may want to look at two papers by Haastert: "On direct and inverse images of D-modules in prime characteristic" and "Über Differentialoperatoren und D-Moduln in positiver Charakteristik." In the latter paper he considers modules over various such rings $\mathcal D^m_X$ of differential operators and proves that for a smooth variety there is an equivalence of categories between $\mathcal D^m_X-\textrm{mod}$ and $\mathcal O_{X^{(m)}}-\textrm{mod}$. –  Chuck Hague Sep 24 '12 at 17:14
Vinoth -- I'm happy to be wrong (especially since this is far from my characteristic zero comfort zone) but it's not instantly clear to me that the theorem you state says what you want. It concerns not quite the full algebras you want but their specialization acting on a particular module on the base and its F-pullback. So for example if that module is just O you recover the statement about vanishing p-curvature. I still don't know that I believe the you get all D-modules of higher level (without p-curvature restrictions) by acting on modules that are pulled back.. –  David Ben-Zvi Oct 14 '12 at 1:32