---
tag: TAPCWE72R0
kind: definition
name: Ring of differential operators
msc: [16S32, 13N10]
defines: GBWQF3HZG0
uses: [04C5ZBG1WT, 0310REJJY8, AWSBZWWMG9, 3J89REV9WQ, 98QE7KY55W, 09EGC6F8PT]
sources:
  - url: https://en.wikipedia.org/wiki/Differential_operator
---

## Definition

Let $k$ be a [[field|04C5ZBG1WT]] and let $A$ be a [[commutative|0310REJJY8]] $k$-algebra.
For $a \in A$, write $a$ also for the map $A \to A$, $f \mapsto af$.
For a $k$-linear map $\theta \colon A \to A$, write $[\theta, a] = \theta \circ a - a \circ \theta$.
Define subspaces $D^m(A)$ of the [[endomorphism ring|AWSBZWWMG9]] $\End_k(A)$ by induction on $m \ge 0$:

1. $D^0(A)$ is the set of maps $f \mapsto af$ for $a \in A$.
2. For $m \ge 1$, $D^m(A)$ is the set of $k$-linear maps $\theta \colon A \to A$ with $[\theta, a] \in D^{m-1}(A)$ for every $a \in A$.

The *ring of differential operators* of $A$ is $D(A) = \bigcup_{m \ge 0} D^m(A)$, and the elements of $D^m(A)$ are the differential operators of order at most $m$.

## Remarks

$D(A)$ is a subring of $\End_k(A)$, because $D^m(A) \circ D^l(A) \subseteq D^{m+l}(A)$.
Every $k$-linear [[derivation|3J89REV9WQ]] $\partial$ of $A$ lies in $D^1(A)$, since $[\partial, a]$ is multiplication by $\partial(a)$.
Example: let $k$ have characteristic $0$ and $A = k[x_1, \dots, x_n]$.
Then $D(A)$ is generated by the multiplications by $x_i$ and the [[partial derivatives|09EGC6F8PT]] $\partial/\partial x_i$, and it is isomorphic to the [[Weyl algebra|98QE7KY55W]] $A_n(k)$.
Non-example: let $k = \Q$, $A = \Q[x]$ and let $\tau$ be the shift $f(x) \mapsto f(x + 1)$.
Then $[\tau, x] = \tau$, since $\tau(xf) = (x+1)\tau(f)$.
If $\tau$ were in $D^m(A)$, then $\tau = [\tau, x]$ would be in $D^{m-1}(A)$, and by induction $\tau$ would be in $D^0(A)$, a multiplication map; this is false, so $\tau \notin D(A)$.
In characteristic $p > 0$ the ring $D(k[x])$ is not generated by $x$ and $d/dx$, since $(d/dx)^p = 0$ while the operator $x^p \mapsto 1$, $x^j \mapsto \binom{j}{p} x^{j-p}$ lies in $D^p(k[x])$.