Let $k$ be an algebraic closed field of zero characteristic and $X$ an affine smooth variety, with $A=\mathcal{O}(X)$ the algebra of regular functions and $\mathcal{V}$ the Lie algebra of vector fields on $X$ ($= \operatorname{Der}_k \, A$). $\mathcal{V}$ is an infinite dimensional simple Lie algebra.
I will begin with my question, and then after formulating it, I will give background on why I think it might be of interest. $U(\mathcal{V})$ is a Hopf algebra, with coproduct $\Delta(v)=v \otimes 1 + 1 \otimes v $ for $v \in \mathcal{V}$. This $\Delta$ can be extended to an algebra homomorphism $U(\mathcal{V}) \rightarrow U(\mathcal{V}) \otimes U(\mathcal{V})$. We will use Sweedler's notation: if $u \in U(\mathcal{V})$, then $\Delta(u)=\sum u_{(1)} \otimes u_{(2)}$. The action of $\mathcal{V}$ on $A$ by derivations allows us to introduce the smash product algebra $\newcommand\hash{\mathbin\#}A \hash U(\mathcal{V})$. It is an associative algebra, which as a vector space is isomorphic to $A \otimes U(\mathcal{V})$. The product is given as follows: $(f\hash u)(g \hash v)= \sum_{(u)} f(u_{(1)}g) u_{(2)} v$. I've only encountered an algebra like that once: in S. Montgomery, 'Hopf algebras and their actions on rings', where further properties were not explored. There are many things I would to ask about this algebra: is it a domain? If, as expected, $U(\mathcal{V})$ is not Noetherian, will the same happen for the smash product algebra? What is its growth function? Is it prime? What is its Jacobson radical?, etc.
Now, the background.
Recently, there has been an intense interest in the study of the representations of the algebras $\mathcal{V}$. In some cases, like the Witt algebra, we have a lot more explicit information, and so things can be done more easily.
However, for general $\mathcal{V}$ they can have a pretty nasty behavior, such as having no semi-simple or nilpotent element. So, there is at the moment no hope of understanding the category of all $\mathcal{V}$-modules.
A subcategory which has been intensivily studied in recent years is the so-called category of $A\mathcal{V}$-modules. They are vector spaces $M$ which are simultaneously $A$-modules and $\mathcal{V}$-modules, with the compatibility relation $v.(f.m)=v(f).m+f.v.m$. If we impose a second relation, $(fv).m=f.v.m$, we just recover the $D(X)$-modules.
Not all $A \mathcal{V}$-modules are $D(X)$-modules: for instance, the adjoint representation of $\mathcal{V}$ with its natural structure of $A$-module is not a $D(X)$-module. There are also $\mathcal{V}$-modules which are not $A\mathcal{V}$-modules.
So we have an inclusion $\text{$D(X)$-modules} \subsetneq A\text{$\mathcal{V}$-modules} \subsetneq \text{$\mathcal{V}$-modules}$.
Note that the largest category is the category of modules for an associative algebra: $U(\mathcal{V})$; for the D-modules this is obvious.
So the question is: is there an associative algebra which controls $A \mathcal{V}$-modules? Yes. These modules are precisely the $A\hash U(\mathcal{V})$-modules.
For this reason, my question. In general, the more we know about the structure of an associative algebra, the more information we gather about its modules.
