Lemma 10.133.10. Let $A \to B$ be a ring map. Let $M, N$ be $B$-modules. Let $S \subset B$ be a multiplicative subset. Any differential operator $D : M \to N$ of order $k$ extends uniquely to a differential operator $E : S^{-1}M \to S^{-1}N$ of order $k$.

**Proof.**
By induction on $k$. If $k = 0$, then $D$ is $B$-linear and hence we get the extension by the functoriality of localization. Given $b \in B$ the operator $L_ b : m \mapsto D(bm) - bD(m)$ has order $k - 1$. Hence it has a unique extension to a differential operator $E_ b : S^{-1}M \to S^{-1}N$ of order $k - 1$ by induction. Moreover, a computation shows that $L_{b'b} = L_{b'} \circ b + b' \circ L_ b$ hence by uniqueness we obtain $E_{b'b} = E_{b'} \circ b + b' \circ E_ b$. Similarly, we obtain $E_{b'} \circ b - b \circ E_{b'} = E_ b \circ b' - b' \circ E_ b$. Now for $m \in M$ and $g \in S$ we set

To show that this is well defined it suffices to show that for $g' \in S$ if we use the representative $g'm/g'g$ we get the same result. We compute

which is the same as before. It is clear that $E$ is $R$-linear as $D$ and $E_ g$ are $R$-linear. Taking $g = 1$ and using that $E_1 = 0$ we see that $E$ extends $D$. By Lemma 10.133.9 it now suffices to show that $E \circ b - b \circ E$ for $b \in B$ and $E \circ 1/g' - 1/g' \circ E$ for $g' \in S$ are differential operators of order $k - 1$ in order to show that $E$ is a differential operator of order $k$. For the first, choose an element $m/g$ in $S^{-1}M$ and observe that

which is a differential operator of order $k - 1$. Finally, we have

which also is a differential operator of order $k - 1$ as the composition of linear maps (multiplication by $1/g'$ and signs) and $E_{g'}$. We omit the proof of uniqueness. $\square$

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