# Subset of differential operators is a finitely generated module?

I was reading about differential operators, and there is a small claim I don't understand.

First, let $A$ be a commutative algebra over $k$, a field. We have the recursive definition for the algebra $D(A)$ of differential operators given by $$D_0(A)=\operatorname{End}_A(A)=A$$ and $$D_i(A)=\{d\in\operatorname{End}_k(A):[d,D_0]\subset D_{i-1}\}$$ and finally $\displaystyle D(A)=\bigcup_{i=0}^\infty D_i(A)$.

So if $A$ is finitely generated as an algebra over $k$, then for each $i>0$, $D_i$ is a finitely generated $A$-module. Why does this follow? Is there a nice proof of this?

I'm aware that $D_jD_k\subset D_{j+k}$, and so $D_0D_k\subset D_k$, so that in particular each $D_i$ is a left $A$-module. Perhaps I'm missing something obvious to see that it's also finitely generated. Thank you for any explanations.

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Suppose $a_1,\dots,a_n$ generate $A$.

Show that an element $L\in D_k$ is determined by its values $L(a_{i_1}\cdots a_{i_l})$ on all products of at most $k$ of the $a_i$s, so that the function $$D_k\to A^{N}$$ which takes $L$ to the tuple of all those values (here $N$ is the number of all those products) is injective.

The map is in fact a map of left $A$-modules. Since $A$ is noetherian, $D_k$ is finitely generated.

Later. For each $a\in A$ let $m_a:b\in A\mapsto ab\in A$.

A map $L:A\to A$ is in $D_1$ iff $[L,m_a]\in D_0$ for all $a\in A$, and this happens iff $[[L,m_a],m_b]=0$ for all $a, b\in A$. Writing this out, this means that $$L(abc)-aL(bc)-bL(ac)+baL(c)=0$$ or, equivalently, $$L(abc)=aL(bc)+bL(ac)-baL(c).$$ If $a$ and $b$ are two elements of the generating set and $c$ is a monomial in the generating set of length $k$, the left hand side is $L$ evaluated at a monomial of length $k+2$, whie at the right we only have $L$ evaluated at monomials of length $<k+2$. Using this, we can see inductively that $L$ is completely determined by its values at all monomials of length $\leq 2$ in the generating set.

If we start with $L\in D_2$, something similar can be done, and the general case can be done by induction.

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Thanks Mariano. So if $\{a_1,\dots,a_n\}$ is a generating set of $A$, the words of length at most $k$ from that set generate $D_k$, and hence $L$ is uniquely determined by its value on those products. Also, is the map supposed to be $D_k\to A^N$ instead of $D_k\to A^I$? I hope I am not way off the mark. – Kally Apr 11 '12 at 1:23
You write «the words of length at most $k$ from that set generate $D_k$» but that is not true (it does not even make sense :) ) What I meant is: an element of $D_k$ is completely determined by its values in those words. – Mariano Suárez-Alvarez Apr 11 '12 at 1:26
Oops sorry, I'm talking nonsense. I think I meant that these words generate the domain of such $L\in D_k$, and that's why these values uniquely determine $L$? – Kally Apr 11 '12 at 1:29
The domain of all elements of $D_k$ is $A$. – Mariano Suárez-Alvarez Apr 11 '12 at 1:32
So is that all there is to it, since the endomorphisms are determined by their image on a generating set? – Kally Apr 11 '12 at 1:39