*All rings are assumed to have unity.*

Let $k$ be a field. Recall the definition of Grothendieck's ring of ($k$-linear) differential operators $D(R;k)$ of a commutative $k$-algebra $R$:

Definition.Set $D_0(R;k)$ to be $R$ viewed as a subring of $\operatorname{End}_k(R)$ via multiplication, then for all $i>0$, define $$ D_{i+1}(R;k) = \left\{P\in\operatorname{End}_k(R)\,\middle|\, [P,r]\in D_i(R;k)\text{ for all }r\in D_0(R;k)\right\}.$$ Then we set $$D(R;k) = \bigcup_{i=0}^\infty D_i(R;k).$$

### Question Context

Remark 17.7 of *Twenty-Four Hours of Local Cohomology* by Iyenger, et al. states the following without proof or citation:

Let $I$ be an ideal of a commutative $k$-algebra $R$, and set $S=R/I$. One can identify $D(S;k)$ with the subring of $D(R;k)$ consisting of operators that stabilize $I$, modulo the ideal generated by $I$.

I've interpreted this as follows:

Let $\operatorname{Stab}(I) = \{P\in D(R;k) \mid P(I) \subseteq I\}$, and let $\Phi\colon \operatorname{Stab}(I) \to \operatorname{End}_k(S)$ be the natural map. Then (i) $\operatorname{im}\Phi = D(S;k)$, and (ii) $\ker\Phi$ is the (two-sided) ideal generated by $I$.

### Question

Can anyone point me to and/or outline a proof of this fact? Specifically, how does one prove that $\operatorname{im}\Phi \supseteq D(S;k)$? (The other inclusion is immediate).

The proof of (ii) is not difficult: one direction is immediate. For the other direction, equip $\operatorname{End}_k(R)$ with the left $R\otimes R$-mod structure $(x\otimes y)\varphi = x\varphi + \varphi y$. Let $N=I\otimes R + R\otimes I$ be the kernel of the natural map $R\otimes R\to S\otimes S$. Let $P\in \operatorname{ker}\Phi$, and look at the $R\otimes R$-module $M$ it generates. Then $$ 0 = (S\otimes S)\Phi(P) = M/NM,$$ so $M=NM=NP=IP + PI$, which is contained in the ideal of $\operatorname{Stab}(I)$ generated by $I$. In particular, $P$ is contained in this ideal.

rightideal generated by $I$, not two-sided. $\endgroup$ – Ketil Tveiten Jan 19 '15 at 10:09