Ring of differential operators

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 $\partial$ 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 $\partial/\partial x_i$, and it is isomorphic to the Weyl algebra $A_n(k)$. 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 $D^p(k[x])$.