# Ring of differential operators of a quotient ring

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.

• The correct statement would be that (in your notation) $\ker\Phi$ is the right ideal generated by $I$, not two-sided. Jan 19, 2015 at 10:09
• If you try to do it with e.g. $R=k[x]$, you'll see why: the Weyl algebra is a simple ring, so there are no nontrivial two-sided ideals. Jan 19, 2015 at 10:11
• @KetilTveiten if it were just a right ideal, you wouldn't get a quotient ring. As to your other comment, it's not an ideal of the full ring of differential operators but of the subring stabilizing $I$. Jan 19, 2015 at 18:23
• I removed the last comment, which was somewhat nonsensical; I've been having my mind stuck in the case where the formula in the question actually works, for $R$ a polynomial ring. $IW$ is a left ideal in the Weyl algebra, but a two-sided ideal in $Stab(I)$, is the thing. Sorry for confusing things. Jan 19, 2015 at 19:32

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 $\Delta:R\to R$.

• Interestingly enough, if you do the proposed construction, you get the ring $k\langle x,x^a\partial_x^b|a\neq 0\rangle$, which is the ring of differential operators on the punctured affine line, i.e. the smooth locus of the $\{y=0\}$ component of $\{xy=0\}$. It would be interesting to see if that holds in general. Jan 19, 2015 at 17:02
• Maybe the question is true when $R = k[x_1, ..., x_n]$ a polynomial ring. Anh it is enough to understand the rings of diffirential operators of finite $k$-algebra. Jan 20, 2015 at 3:49
• I don't think the hypothesis of characteristic zero is necessary in the addendum. Jan 20, 2015 at 12:40
• @KetilTveiten you might be right, I haven't checked. Jan 21, 2015 at 8:17
• @MichaelBächtold the formula for $D(S)$ certainly works in characteric p, at least. I can't see anything in your argument that relies on the characteristic (though I may be wrong there?). Jan 21, 2015 at 8:21

EDIT: Michael Bächtold's answer shows that the claim is false in general (sloppy of me not to remember that; I worked out the exact same counterexample last year when writing a paper on a related topic). It is true for $R=k[x_1,\ldots,x_n]$ however, as you can check using the arguments below. This is probably the context the original source had in mind, as this is the most common way to produce explicit representations for rings of differential operators.

This is probably more of a sketch/outline than a proper proof, but should give you what you're after.

The statement you want to prove is that $D(S;k) = \{\delta\in D(R;k)|\delta(I)\subset I\}/ID(R;k)$. You don't need to go by way of Grothendieck's definition (which is just clunky for most practical purposes), just think about how the $D$'s act on $R$ and $S$ (they are left modules).

It's clear that $D(S)$ has to be a subquotient of $D(R)$, so you just need to think of what elements of $D(R)$ descend to $D(S)$ (those that preserve $I$), and how to handle the "modulo $I$" part, i.e. which operators on $R$ descend to the zero operator on $S$ (the right ideal $ID(R)$).

You could equivalently phrase this in terms of lifting an element $\delta'\in D(S)$ to an element $\delta\in D(R)$ such that $\delta'\pi = \pi\delta$ (where $\pi:R\to S/I$ is what you think). If you really really want to do it in terms of the Grothendieck definition you could try to see how $[\delta,r+I]$ behaves for $\delta\in D_i(R)$; here the salient point is that $[\delta,I]$ is in $ID(R)$ if and only if $\delta(I)\subset I$.

I found a reference, with proof: it's Theorem 15.5.13 in Noncommutative Noetherian Rings by J.C. McConnell and J.C. Robson; here it is made explicit that $R$ is a polynomial ring.

They mention that the result was first proved in Differential operators on an affine curve by S.P. Smith and J.T. Stafford, but that "it was known earlier".