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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

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

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 $< n$.

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 $D_r \times D_s \to D_{r+s-1}$ is the $B$-module dual of the map $\delta : P^{r+s-1} \to P^r \otimes P^s$. This turns out to be a $B$-algebra homomorphism and its key property is that

$$\Gamma(Y, \mathcal{P}^n_{Xm}(Y)) = \oplus_{a=0}^{n-1} B \tau^{ \{ a \} } $$

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 $n$.

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

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

--- 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. 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

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

Now the multiplication map $D_r \times D_s \to D_{r+s}$ is the $B$-module dual of the map $\delta : P^{r+s} \to P^r \otimes P^s$. This turns out to be a $B$-algebra homomorphism and its key property is that

(it is a "primitive element" in an appropriate bialgebra). Let's now work out how to multiply $\partial^{[i]}$ by $\partial^{[j]}$ (drop the bars for clarity):

Now the multiplication map $D_r \times D_s \to D_{r+s-1}$ is the $B$-module dual of the map $\delta : P^{r+s-1} \to P^r \otimes P^s$. This turns out to be a $B$-algebra homomorphism and its key property is that

(it is a "primitive element" in an appropriate 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):