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.