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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.) 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

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

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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 $< n$.n$. Let's view$\mathcal{O}(Y)$as a$B$-algebra via the map$b \mapsto b \otimes 1$; then$\mathcal{O}(Y) \cong B[\tau]$. By definition (EGA IV, 16.7.1.1), the global sections of$\mathcal{P}^n_{X\infty}(Y)$are just $$P^n := \mathcal{O}(Y) / (\tau^n)$$ so \tau^{n+1})$$--- this is the algebra of functions on the n-th infinitesimal neighbourhood of the diagonal \tau = 0 inside \mathcal{O}(Y) (hence the n+1 in the exponent). So in particular it is a free B-module of rank n n+1 with generators (the images of) \tau^i for 0 \leq i < \leq n. By definition,$$\mathcal{D}^{(\infty)}_n (X) := Hom_B(\mathcal{O}(Y) / (\tau^n), Hom_B(P_n, B) =: D_n $$which is again a free B-module of rank n; n+1; let \{ \partial^{[i]}, i=0, \ldots, n-1\} n\} be the dual basis for this module. Now the multiplication map D_r \times D_s \to D_{r+s-1} D_{r+s} is the B-module dual of the a map \delta : P^{r+s-1P^{r+s} \to P^r \otimes P^s which is constructed in EGA IV, Lemma 16.8.9.1. Morally \delta sends a \otimes b to a \otimes 1 \otimes 1 \otimes b, as Gros/Le Stum/Quirros mention. This turns out to be a B-algebra homomorphism, and its tts key property is that$$\Gamma(Y, \mathcal{P}^n_{Xm}(Y)) = \oplus_{a=0}^{n-1} oplus_{a=0}^n B \tau^{ \{ a \} } $$6 added 108 characters in body I find it helpful to first work through the definition of multiplication on \mathcal{D}^{(m)} when m = \infty, in which case it reduces to the "classical" ring of differential operators in the sense of Grothendieck; read sections 16.7 and 16.8 of EGA 4, Quatrième partie. So let A be a commutative base ring, let S = Spec(A) and let X = \mathbb{A}^1_S so that B = A[t] = \Gamma(X, \mathcal{O}). We want to work out \mathcal{D}^{(\infty)}(X). Let Y = X \times_S X and m = \infty. Let n \geq 0. First we have to work out the global sections of the sheaf \mathcal{P}^n_{Xm}(Y), considered as a B-module. This will turn out to be dual (as a B-module) to the B-module of all Grothendieck differential operators of order < n. Now \mathcal{O}(Y) = B \otimes_A B is isomorphic as an A-algebra to the polynomial ring A[t,t'] where t \mapsto t \otimes 1 and t' \mapsto 1 \otimes t. The diagonal immersion X \hookrightarrow Y corresponds to the algebra surjection B \otimes_A B \to B which is just the multiplication map. So the ideal of the diagonal, namely the kernel of this map, is generated as an ideal by the element$$\tau := t \otimes 1 - 1 \otimes t.$$Let's view \mathcal{O}(Y) as a B-algebra via the map b \mapsto b \otimes 1; then \mathcal{O}(Y) \cong B[\tau]. By definition, the global sections of \mathcal{P}^n_{X\infty}(Y) are just$$P^n := \mathcal{O}(Y) / (\tau^n)$$so in particular it is a free B-module of rank n with generators (the images of) \tau^i for 0 \leq i < n. By definition,$$\mathcal{D}^{(\infty)}_n (X) := Hom_B(\mathcal{O}(Y) / (\tau^n), B) =: D_n $$which is again a free B-module of rank n; let \{ \partial^{[i]}, i=0, \ldots, n-1\} be the dual basis for this module. Now the multiplication map D_r \times D_s \to D_{r+s} D_{r+s-1} is the B-module dual of the map \delta : P^{r+sP^{r+s-1} \to P^r \otimes P^s. This turns out to be a B-algebra homomorphism and its key property is that$$\delta( \overline{\tau} ) = \overline{ \tau} \otimes 1 + 1 \otimes \overline{\tau}$$(it is a "primitive element" in an appropriate bialgebra)bialgebra --- see EGA IV.4, 16.8.9.4). Let's now work out how to multiply \partial^{[i]} by \partial^{[j]} (drop the bars for clarity):$$(\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-1} 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}.$$The interesting thing is that this structure constant \frac{q_i! q_j!}{q_{i+j}!} \binom{i+j}{i} is always a p-integral rational number (see Lemma 1.1.3(i) in Berthelot's paper), so it makes sense whenever A is an algebra over \mathbb{Z}_{(p)}, say, and in particular if A had characteristic p. Note that if A was a \mathbb{Q}-algebra, then there would be a ring homomorphism from \mathcal{D}^{(m)} to A[t; \partial] which sends$$ \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.

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