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.



> 
> 1. 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?**
> 2. **When does a $D$-module $M$ admit a dual $M^\*$?** (in the sense of monoidal categories).
> 3. **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?**
> 4. Is the $R$-linear dual $Hom\_R(D,R)$ a bialgebra? Is it related to the opposite of $D$?
> 5. **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.
> 
> 
> 



What kind of algebraic object is $\mathcal{D}_X$? (algebra of diifferential operators). What's special about modules over it?
1. 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.
2. 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).
3. 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.
4. 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).
5. I don't really understand what this part is asking.


**Edited** in the light of the comments below by t2suji


