Do $D$-modules necessarily have the "composition" condition?

Question

This might be a stupid question. Every bone in my body tells be that if I have a ring $R$ and a right $R$-module $M$ then we should have $(m\cdot r)\cdot s= m\cdot(rs).$ However, I have just done this exercise for $D$-modules: If $X$ is a smooth variety, $\mathcal D_X$ the sheaf of differential operators on $X,$ $\mathcal M$ and $\mathcal N$ left and right $\mathcal D_X$-modules respectively, then $\mathcal N\otimes_{\mathcal O_X}\mathcal M$ naturally has a right $\mathcal D_X$-module structure. I did this by defining $(n\otimes m)\cdot P=nP\otimes m-n\otimes Pm$ and verifying the three conditions (locally, of course)

1. $(n\otimes m)(P\cdot f) = ((n\otimes m)\cdot P)\cdot f\quad$ for every $f\in \mathcal O_X$ and $P\in \mathcal D_X$
2. $((n\otimes m)f)\cdot P = ((n\otimes m)\cdot P)\cdot f+(n\otimes m)\cdot (f\cdot P)\quad$ for every $f\in \mathcal O_X$ and $P\in \mathcal D_X$ (here I am using the right $\mathcal D_X$-module structure on $\mathcal O_X$ given by changing signs).
3. $(n\otimes m)\cdot [P,Q] = ((n\otimes m)\cdot P)\cdot Q - ((n\otimes m)\cdot Q)\cdot P\quad$ for every $P,Q\in \mathcal D_X$

Maybe I am wrong here but all the notes I can find tell me that this is enough to guarantee a right $\mathcal D_X$-module structure. I noticed when verifying the third point that $$ ((n\otimes m)\cdot P)\cdot Q = nPQ\otimes m - nP\otimes Qm - nQ\otimes Pm + n\otimes QPm$$ which is not equal to $$(n\otimes m)\cdot (PQ) = nPQ\otimes m-n\otimes PQm.$$ Is there some mistake in all this? Maybe the three points to verify are not enough? Or maybe I have the wrong candidate $\mathcal D_X$-module structure. The noncommutativity of everything is really messing with my head...

Answer 1

I talked to a friend and he pointed out that the three conditions I listed above are conditions on a map $\mathcal T_X\otimes_{\mathcal O_X} (\mathcal N\otimes_{\mathcal O_X}\mathcal M)\to\mathcal N\otimes_{\mathcal O_X}\mathcal M,$ where $\mathcal T_X$ is the tangent sheaf of $X.$ If these three conditions are satisfied then there is a unique right $\mathcal D_X$-module structure on $\mathcal N\otimes_{\mathcal O_X}\mathcal M$ extending this map (since $\mathcal D_X$ is generated as an $\mathcal O_X$-algebra by $\mathcal T_X$ and $\mathcal O_X$). So the point is that the formula I used to define my module structure only holds for $P,Q\in \mathcal T_X,$ and for higher order operators you just do the obvious thing (what I called the "composition law"). Anyway, that answers my question.