TITLE: What kind of algebraic object is $\mathcal{D}_X$? (algebra of diifferential operators). What's special about modules over it?
QUESTION [11 upvotes]: Let $R$ be a regular ring over a field of char 0. Let $X=Spec R$ and $D=\mathcal{D}_X$
the algebra of differential operators over it. 
The overall vague question is what kind of algebraic object is $D$ and what kind of category is the category of its modules? Here are some points whose answers could together be considered an answer to this.


When (meaning for what kind of $D$-modules $M,N$ left or right) is there a $D$-module structure on $Hom_R(M,N)$? On $M\otimes_RN$? Does this make
  $D$-mod into an abelian monoidal category with fiber functor to
  $R$-mod? If so is it closed monoidal?
When does a $D$-module $M$ admit a dual $M^*$? (in the sense of monoidal categories).
Is the abelian category of $D$-modules isomorphic to $D^{op}$-mod? (before deriving). If not how are they related? Maybe the correct
  thing to consider is the opposite co-opposite $D^{op}_{cop}$? Specifically Why does the dualizing sheaf $\omega_X$ pop up in this context?
Is the $R$-linear dual $Hom_R(D,R)$ a bialgebra? Is it related to the opposite of $D$?
How do I derive correctly the category of $D$-modules? Suppose $M$ and $N$ are $D$-modules and suppose $Hom_R(M,N)$ is the "correct" internal Hom in the abelian monoidal category of $D$-modules. If we hope to have a fiber functor between the derived categories $Hom_R(Q,N)$ (with $Q \to M$ a resolution as a $D$-module) should go to (something quasi isomorphic) to $Hom_R(P,N)$ (with $P \to M$ a resolution as an $R$-module). It doesn't seem to follow easily from the rest of the structure.

REPLY [10 votes]: Proposition 1.2.9 of http://math.columbia.edu/~scautis/dmodules/hottaetal.pdf explains that if $M$ and $N$ are both left $D$-modules and $M'$ and $N'$ are both right $D$-modules then
(a) $M\otimes_{R} N$ is naturally a left $D$-module;
(b) $M'\otimes_{R} N$ is naturally a right $D$-module;
(c) $\mathrm{Hom}_{R}(M,N)$ is naturally a left $D$-module;
(d) $\mathrm{Hom}_{R}(M',N')$ is naturally a left $D$-module;
(e) $\mathrm{Hom}_{R}(M,N')$ is naturally a right $D$-module.
This Proposition also gives explicit formulae that explain why each of these is the case. Remark 1.2.10 explains why $M'\otimes_{R}N'$ is not naturally a left or right $D$-module in general. 
Section 2 of https://arxiv.org/abs/dg-ga/9702008 fits these results into a wider framework. The thing about $D$ that makes these things true is that it is the universal enveloping algebra of a $(k,R)$ Lie-Rinehart algebra. 
I think that it is clear from what I've written that $D$-mod (=cat of left $D$-modules) is a (symmetric) monoidal category with $\otimes=\otimes_{R}$ and  the forgetful functor $D$-mod to $R$-mod preserves the monoidal product. I'm not sure if a fiber functor requires more than this. 
As noted by t3suji in the comments below the internal Hom $\mathrm{Hom}_R(-,-)$ on $R$-mod which induces an internal Hom on $D$-mod via part (c) above does make $D$-mod into a closed monoidal category. Once again this observation works for the enveloping algebra of any Lie-Rinehart algebra. 
I think it is probably the case that $M$ has a dual precisely if it is a projective $R$-module of finite rank (that is also a left $D$-module). In this case $M^\ast$ should just be $\mathrm{Hom}_R(M,R)$ which is a left $D$-module by 1(c). 
Yes. The reason that $\omega_X$ crops up is that it is a right $D$-module that is a rank $1$ projective $R$-module. It follows that tensoring with it defines an auto-equivalence of categories on $R$-mod (the inverse is given by $\mathrm{Hom}_R(\omega_X,-)\cong \mathrm{Hom}_R(\omega_X,R)\otimes_R-$. It follows from 1(b) that this auto-equivalence sends left $D$-modules to right $D$-modules and from 1(d) that its inverse sends right $D$-modules to left $D$-modules. 
Since $D$ is not a finitely generated $R$-module, $\mathrm{Hom}_R(D,R)$ is pretty badly behaved. In particular it is unlikely that the algebra structure on $D$ will induce a coalgebra structure on its 'dual'. Incidentally I'm not sure what you mean by saying that $D$ is an $R\otimes R$ bialgebra (in particular I don't know what the coalgebra structure is nor the counit). 
I don't really understand what this part is asking. 

Edited in the light of the comments below by t2suji