Lemma 10.133.9. Let $R \to S$ be a ring map. Let $M$ be an $S$-module. Let $J = \mathop{\mathrm{Ker}}(S \otimes _ R S \to S)$ be the kernel of the multiplication map. There is a canonical isomorphism of $S$-modules
\[ P^ k_{S/R}(M) \longrightarrow (S \otimes _ R M)/J^{k + 1}(S \otimes _ R M) \]
where $s \in S$ acts on the target via multiplication by $s \otimes 1$ and such that the universal differential operators of order $k$ to the map given by $m \mapsto \text{class of }1 \otimes m$.
Proof.
Consider the map $T : M \to S \otimes M$, $m \mapsto 1 \otimes m$ Since $T$ is $R$-linear and since
\begin{align*} g_0g_1\ldots g_ k T(m) - \sum g_0 \ldots \hat g_ i \ldots g_ k T(g_ im) + \ldots +(-1)^{k + 1}T(g_0\ldots g_ km) \\ = \prod _{i = 0,\ldots ,k} (g_ i \otimes 1 - 1 \otimes g_ i) 1 \otimes m \end{align*}
for $m \in M$ and $g_0, \ldots , g_ k \in S$, we conclude that the rule $m \mapsto \text{class of }1 \otimes m$ in the statement of the lemma is a differential operator of order $k$ (see proof of Lemma 10.133.3). By the universal property of $P^ k_{S/R}(M)$ we obtain the arrow in the statement of the lemma. On the other hand, if $D : M \to N$ is a differential operator of order $k$, then we can consider the $S$-linear map $L : S \otimes _ R M \to N$, $g \otimes m \mapsto gD(m)$. The reader checks, by a computation similar to the one above and using that $J$ is generated by elements of the form $g \otimes 1 - 1 \otimes g$, that $L$ annihilates $J^{k + 1}(S \otimes _ R M)$. In particular, if we apply this to the universal differential operator $D_{univ} : M \to P^ k_{S/R}(M)$, then we obtain an $S$-linear map $(S \otimes _ R M)/J^{k + 1}(S \otimes _ R M) \to P^ k_{S/R}(M)$. We leave it to the reader to see that this map is the inverse to the map in the statement of the lemma.
$\square$
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