My question is regarding the definition of $\mathcal{D}$-modules of level $m$ given in [this](http://arxiv.org/abs/0811.1168) paper. As an example, let $X=\mathbb{A}^1$ over $S=\text{Spec }\overline{\mathbb{F}\_p}$; I was told that a $\mathcal{D}$-module of level $m$ is a module over $\mathbb{k} \langle x, \partial\_x, \frac{{\partial\_x}^p}{p!}, \cdots, \frac{{\partial\_x}^{p^m}}{(p^m)!} \rangle$; I was wondering how to work this out from first principles, using Definition $2.5$ given in that paper.


Consider the immersion $X \rightarrow X \times\_S X$. Definition $2.1$ from that paper defines what a *`divided power structure of level $m$`* on an immersion is; Definition $2.3$ (and $2.4$) states constructs the *`divided power envelope of level $m$`*, $P\_{X,m}(Y)$. Subsequently, on pg $5$ they define $\mathcal{P}\_{X,m}^n(Y)$; the sheaf of differential operators of level $m$ is defined to be union of the duals of this family of sheaves (as $n$ varies). Most of the details/proofs are done in [this other paper](http://archive.numdam.org/ARCHIVE/ASENS/ASENS_1996_4_29_2/ASENS_1996_4_29_2_185_0/ASENS_1996_4_29_2_185_0.pdf).


I'm having trouble properly understanding these definitions and working out what they are in the case of $X=\mathbb{A}^1, S=\text{Spec } \overline{\mathbb{F}\_p}$. So, my question is what the above objects look like in this particular example, and how to explicitly calculate everything in this example. 




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$\mathcal{D}$-modules of level m
I find it helpful to first work through the definition of multiplication on $\mathcal{D}^{(m)}$ when $m = \infty$, in which case it reduces to the "classical" ring of differential operators in the sense of Grothendieck; read sections 16.7 and 16.8 of [EGA 4, Quatrième partie](http://www.numdam.org/numdam-bin/feuilleter?id=PMIHES_1967__32_).


So let $A$ be a commutative base ring, let $S = Spec(A)$ and let $X = \mathbb{A}^1\_S$ so that $B = A[t] = \Gamma(X, \mathcal{O})$. We want to work out $\mathcal{D}^{(\infty)}(X)$. Let $Y = X \times\_S X$ and $m = \infty$.


Let $n \geq 0$. First we have to work out the global sections of the sheaf $\mathcal{P}^n\_{Xm}(Y)$, considered as a $B$-module. This will turn out to be *dual* (as a $B$-module) to the $B$-module of all Grothendieck differential operators of order at most $n$.


Now $\mathcal{O}(Y) = B \otimes\_A B$ is isomorphic as an $A$-algebra to the polynomial ring $A[t,t']$ where $t \mapsto t \otimes 1$ and $t' \mapsto 1 \otimes t$. The diagonal immersion $X \hookrightarrow Y$ corresponds to the algebra surjection $B \otimes\_A B \to B$ which is just the multiplication map. So the *ideal* of the diagonal, namely the kernel of this map, is generated as an ideal by the element 


$$\tau := t \otimes 1 - 1 \otimes t.$$


Let's view $\mathcal{O}(Y)$ as a $B$-algebra via the map $b \mapsto b \otimes 1$; then $\mathcal{O}(Y) \cong B[\tau]$. By definition ([EGA IV, 16.7.1.1](http://www.numdam.org/numdam-bin/feuilleter?id=PMIHES_1967__32_)), the global sections of $\mathcal{P}^n\_{X\infty}(Y)$ are just


$$P^n := \mathcal{O}(Y) / (\tau^{n+1})$$


--- this is the algebra of functions on the $n$-th infinitesimal neighbourhood of the diagonal $\tau = 0$ inside $\mathcal{O}(Y)$ (hence the $n+1$ in the exponent). So in particular it is a free $B$-module of rank $n+1$ with generators (the images of) $\tau^i$ for $0 \leq i \leq n$. By definition, 


$$\mathcal{D}^{(\infty)}\_n (X) := Hom\_B(P\_n, B) =: D\_n $$


which is again a free $B$-module of rank $n+1$; let $\{ \partial^{[i]}, i=0, \ldots, n\}$ be the dual basis for this module. 


Now the multiplication map $D\_r \times D\_s \to D\_{r+s}$ is the $B$-module dual of a map $\delta : P^{r+s} \to P^r \otimes P^s$ which is constructed in [EGA IV, Lemma 16.8.9.1](http://www.numdam.org/numdam-bin/feuilleter?id=PMIHES_1967__32_). Morally $\delta$ sends $a \otimes b$ to $a \otimes 1 \otimes 1 \otimes b$, as Gros/Le Stum/Quirros mention. This turns out to be a $B$-algebra homomorphism, and tts key property is that


$$\delta( \overline{\tau} ) = \overline{ \tau} \otimes 1 + 1 \otimes \overline{\tau}$$


(it is a "primitive element" in an appropriate bialgebra --- see [EGA IV.4, 16.8.9.4](http://www.numdam.org/numdam-bin/feuilleter?id=PMIHES_1967__32_)). Let's now work out how to multiply $\partial^{[i]}$ by $\partial^{[j]}$ (drop the bars for clarity):


$$(\partial^{[i]} \cdot \partial^{[j]})(\tau^k) = (\partial^{[i]} \otimes \partial^{[j]})(\tau \otimes 1 + 1 \otimes \tau)^k = \sum\_{a + b = k} \binom{k}{a} \partial^{[i]}(\tau^a) \partial^{[j]}(\tau^b)$$


which is just $\binom{i+j}{i}\delta\_{k,i+j}$. Since $\binom{i+j}{i} \partial^{[i+j]}$ has the same effect on each $\tau^k$, we deduce that


$$ \partial^{[i]} \cdot \partial^{[j]} = \binom{i+j}{i} \partial^{[i+j]}$$


which is hopefully the familiar rule for multiplying divided powers (morally $\partial^{[i]} = \partial^i/i!$). 


The point of the Berthelot construction is that it is possible to *vary* the divided-power structure on the diagonal, and thereby control just how many divided powers one gets in $\mathcal{D}^{(m)}$. For example, if $m = 0$ then you instead allow *all* divided powers on the ideal of the diagonal (algebraically this means you consider the divided power algebra of the ideal $(\tau)$ in $B[\tau]$ to get $\oplus\_{n=0}^\infty B \tau^{[n]}$), and when you take the $B$-dual, these divided powers in $\tau$ "remove" the divided powers in $\partial$ and you end up with $\mathcal{D}^{(0)}(X) = B[\partial]$, the ring of *crystalline differential operators* (no divided powers).


Now to answer your question, let the level $m \geq 0$ be fixed. Then as Gros/Le Stum/Quirros explain just before Definition 2.5,


$$\Gamma(Y, \mathcal{P}^n\_{Xm}(Y)) = \oplus\_{a=0}^n B \tau^{ \{ a \} } $$


where $\tau^{ \{ a \} }$ is a symbol that "behaves like $\tau^a / q\_a!$" (where $q\_a$ is the integer part of $a / p^m$: thus $a = q\_a p^m + r\_a$ say).


To understand the multiplication of the *dual* vectors to these $\tau^{ \{ a \} }$, namely the $\partial^{ \langle a \rangle }$, we need to understand how to *comultiply* the $\tau^{ \{ a \} }$. So we compute (again dropping bars for convenience)


$$ \delta( \tau^{ \{ a \} }) = \frac{1}{q\_a!} \delta(\tau)^a = \sum\_{i+j = a} \frac{1}{q\_a!} \binom{i+j}{i} \tau^i \otimes \tau^j = \sum\_{i+j = a} \frac{q\_i! q\_j!}{q\_{i+j}!} \binom{i+j}{i} \tau^{ \{ i \} } \otimes \tau^{ \{ j \} }$$ 


and the same computation as above in the case $m=\infty$ shows that 


$$ \partial^{\langle i \rangle} \cdot \partial^{\langle j \rangle} = \frac{q\_i! q\_j!}{q\_{i+j}!} \binom{i+j}{i} \partial^{\langle i + j \rangle}.$$


The interesting thing is that this structure constant $\frac{q\_i! q\_j!}{q\_{i+j}!} \binom{i+j}{i}$ is always a *$p$-integral* rational number (see Lemma 1.1.3(i) in [Berthelot's paper](http://archive.numdam.org/ARCHIVE/ASENS/ASENS_1996_4_29_2/ASENS_1996_4_29_2_185_0/ASENS_1996_4_29_2_185_0.pdf)), so it makes sense whenever $A$ is an algebra over $\mathbb{Z}\_{(p)}$, say, and in particular if $A$ had characteristic $p$. Note that if $A$ was a $\mathbb{Q}$-algebra, then there would be a ring homomorphism from $\mathcal{D}^{(m)}$ to $A[t; \partial]$ which sends 


$$ \partial^{\langle i \rangle} \mapsto \frac{q\_i!\partial^i}{i!} $$


since


$$ \left(\frac{q\_i! \partial^i}{i!}\right) \cdot \left(\frac{q\_j! \partial^j}{j!}\right) = \frac{q\_i! q\_j!}{q\_{i+j}!} \binom{i+j}{i} \left( \frac{q\_{i+j}! \partial^{i+j}}{(i+j)!}\right).$$


Thus morally $\partial^{\langle i \rangle}$ should be thought of as "modified divided powers" $q\_i! \partial^i / i!$. 


Finally, you don't need *all* of the $\partial^{\langle i \rangle}$ to generate $\mathcal{D}^{(m)}$. As is well-known, the full ring of Grothendieck differential operators in characteristic $p$ can be generated by the divided powers $\partial^{[p^a]}$ (for all $a \geq 0$). Since $q\_i = 0$ for $i < p^m$ and $q\_{p^m} = 1$, the modified divided powers $\partial^{\langle p^i \rangle}$ are equal to the "true" divided powers $\partial^{[p^i]}$ for $0 \leq i \leq m$. If $a > m$ then since $q\_{p^a} = p^{a-m}$,


$$\partial^{\langle p^a \rangle} = \frac{ p^{a-m}! }{ p^a! } \partial^{p^a} = \left(\frac{ p^{a-m}! (p^m!)^{p^{a-m}} }{p^a!} \right) (\partial^{\langle p^m \rangle})^{p^{a-m}}$$


shows that $\partial^{\langle p^a \rangle}$ is a $p$-adic unit times a power of $\partial^{\langle p^m \rangle}$ for $a \geq m$ since the $p$-adic valuation of that big fraction is 


$$\frac{p^{a-m}-1}{p-1} + p^{a-m} \frac{p^m-1}{p-1} - \frac{p^a-1}{p-1} = 0.$$


So we see that $\mathcal{D}^{(m)}(\mathbb{A}^1\_S)$ in this case is the $A$-algebra generated by $t$ and the divided powers $\partial^{[p^0]}, \partial^{[p^1]}, \ldots, \partial^{[p^m]}$, subject to appropriate natural relations.


**Edit:** To see what the map $X \to P\_{Xm}(Y)$ looks like in the case $X = \mathbb{A}^1\_S$, it is enough to describe the corresponding map $C := \mathcal{O}(P\_{Xm}(Y)) \to B = \mathcal{O}(X)$ on functions, because everything in sight is affine. $C$ is a $B$-algebra, generated by symbols $\tau^{ \{a \} }$ for all $a \geq 1$ subject to the relations 


$$\tau^{ \{a \} } \cdot \tau^{ \{b \} } = \frac{q\_{a+b}!}{q\_a!q\_b!} \tau^{ \{ a + b \} }$$


(note that the structure constant $\frac{q\_{a+b}!}{q\_a!q\_b!}$ is actually an *integer*, this again follows from Lemma 1.1.3(i) in [Berthelot's paper](http://archive.numdam.org/ARCHIVE/ASENS/ASENS_1996_4_29_2/ASENS_1996_4_29_2_185_0/ASENS_1996_4_29_2_185_0.pdf).) The map $C \to B$ sends all of the generators $\tau^{ \{a \} }$ to zero, and the map $C \to P^n$ which corresponds to the closed subscheme $P^n\_{Xm}(Y)$ of $P\_{Xm}(Y)$ sends all of the $\tau^{ \{ a \} }$ to zero for $a \geq n+1$.


This description makes it easy to see that the algebra of functions $C$ on $P\_{Xm}(Y)$ is isomorphic to the polynomial ring $B[\tau]$ when $m = \infty$ (since we can take $q\_a$ to be always zero in this case), and to the "free" divided-power algebra $B[\tau^{[n]} : n\geq 1]$ that Gros/Le Stum/Quirros call $\Gamma\_\bullet(B \tau)$ when $m = 0$. This is because $q\_a = a$ in this case, so the defining relations between the $\tau^{ \{a \}}$ reduce to


$$\tau^{ \{a \} } \cdot \tau^{ \{b \} } = \binom{a+b}{a} \tau^{ \{ a + b \} }.$$


