Lemma 10.133.8. Suppose that we have ring maps $A \to A'$ and $A \to B$ and a $B$-module $M$. Set $B' = B \otimes _ A A'$ and view $M' = M \otimes _ A A'$ as a $B'$-module. The map of Remark 10.133.7 induces an isomorphism $P^ k_{B/A}(M) \otimes _ A A' = P^ k_{B'/A'}(M')$.
Proof. Let $D_{univ} : M \to P^ k_{B/A}(M)$ be the universal differential operator. The induced $A'$-linear map $D_{univ} \otimes 1 : M' = M \otimes _ A A' \to P^ k_{B/A}(M) \otimes _ A A'$ is easily verified to be an order $k$ differential operator for $M'/B'/A'$. We will show that $D_{univ} \otimes 1$ satisfies the universal property. Let $D' : M' \to N'$ be a differential operator of order $k$ into a $B'$-module $N'$. Then the composition $D : M \to M' \to N'$ is a differential operator into $N'$ viewed as a $B$-module. Hence there is a $B$-linear map $\gamma : P^ k_{B/A}(M) \to N'$ such that $D = \gamma \circ D_{univ}$. The reader checks easily that the induced $B'$-linear map $\gamma ' : P^ k_{B/A}(M) \otimes _ A A' \to N'$ satisfies $\gamma ' \circ (D_{univ} \otimes 1) = D'$. We omit the proof that $\gamma '$ is unique. $\square$
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