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

2012-09-18T19:43:01Z

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.

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

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