*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.



Ring of differential operators of a quotient ring
I believe the statement is wrong. Here is a counterexample: Let $R=k[x,y]/(xy)$ (the algebra of polynomial functions on a cross) and let $I$ the ideal generated by $y$ in $R$. Then the quotient $S$ is $k[x]$ (affine line) on which we have the standard vector field $\partial\_{x}$. I claim that there is no extension of this operator to a differential operator on $R$.


I have a proof along the following lines. Check that differential operators on $k[x,y]$ which preserve the ideal $(xy)$ are of the form
$$f\_{0}(x,y)+x\sum\_{i>0}a\_{i}(x,y)\partial\_{x}^{i}+y\sum\_{j>0}b\_{j}(x,y)\partial\_{y}^{j}+xy\sum\_{i,j>0}c\_{i,j}(x,y)\partial\_{x}^{i}\partial\_{y}^{j}$$ with coefficients $f,a,b,c$ being arbitrary polynomials. Also check that differential operators mapping any polynomial $\phi\in k[x,y]$ to the ideal $J:=(xy)$ are those with coefficients in $(xy)$. It follows, assuming the claim is right, that any differential operator on $R$ is the restriction of a unique differential operator on $k[x,y]$ of the form
$$C\_0+x\sum\_{i\geq 0}a\_{i}(x)\partial\_{x}^{i}+y\sum\_{j\geq 0}b\_{j}(y)\partial\_{y}^{j}$$
Now an extension $\Delta$ of $\partial\_x$ to $R$ should satisfy
$$\Delta(x)=1 \mod (y)$$
but using the representation given above this implies $C\_0 x+x^2 a\_0(x)+a\_1(x) x= 1 \mod (y)$ which is not possible.


**Addendum**: as mentioned in the comments and in Ketil Tveiten' answer, the claim is true when $R=k[x\_1,\ldots,x\_n]$ and $k$ is a field of characteristic zero. This should yield a proof: let $\pi:R\to R/I$ denote the projection and let $\nabla:R/I\to R/I$ be a differential operator of order $\leq l$. We need to find a lift of $\nabla$ to a differential operator $\Delta$ on $R$ such that $\nabla \circ \pi =\pi \circ \Delta$. Now $\nabla \circ \pi$ is a differential operator from $R$ to $R/I$ (understood as an $R$-module) of order $\leq l$ and any such factors uniquely through the universal differential operator out of $R$, sometimes denoted with $j\_l:R\to\mathcal{J}^l(R)$ where $\mathcal{J}^l(R)$ is the $R$-module of jets of order $l$ of elements of $R$. So $\nabla \circ \pi=\phi \circ j\_l$ where $\phi:\mathcal{J}^l(R)\to R/I$ is an $R$-module homomorphism. In case $R$ is a polynomial algebra the module $\mathcal{J}^l(R)$ is free and hence projective, so there is lift of $\phi$ to an $R$-module morphism $\tilde{\phi}:\mathcal{J}^l(R)\to R$ which composed with $j\_l$ gives the desired differential operator $\Delta:R\to R$.


