So, I am reading Masaki Kashiwara's book "$D$-modules and microlocal calculus" and I have a question about a definition that he makes in the begining of a proof. I will post a screen capture of the page in question below:
To lay a bit of context, here $\mathcal{G}$ and $\mathcal{F}$ are arbitray $\mathcal{O}_{X}$-modules (where $X$ is a complex manifold) and $D_{X}$ is the sheaf of rings of differential operators. $Diff(\mathcal{F}, \mathcal{G})$ is a sheaf generated by the presheaf consituting of morphisms of sheaves $f$ between $\mathcal{F}$ and $\mathcal{G}$ which satisfy the fact that, for each section $s$ of $\mathcal{F}$, there are finite sections $v_i$ of $\mathcal{G}$ and $P_i$ of $D_{X}$ such that $f(as) = \sum_{i} P_i(a) v_i$.
The main issue I have in this proof is actually in the definition of $F_{m}(K)$. As I checked, I'm pretty sure it is well defined if and only if $v_{\alpha} \otimes \partial^{\alpha}_{x} = 0 \implies v_{\alpha} = 0$, which I cannot understand how it can be true for any $\mathcal{O}_{X}$-module $\mathcal{G}$.
Thank you so much for all the help! :)
I worked it out. I guess it is just a property that comes from $D_X$. I'll specify:
So, we have a canonical $D_X$-module morphism $D_{X} \to \mathcal{O}_{X}$ given by $P \mapsto P(1)$. Now, if $\mathcal{G}$ is an $\mathcal{O}_{X}$-module, we can define $\mathcal{G} \otimes_{\mathcal{O}_{X}} D_{X}$, which we equip with the structure of right $\mathcal{O}_{X}$-module, by defining $(g \otimes P ) a := g \otimes (P \circ a)$ which is different than the natural left $\mathcal{O}_{X}$-module structure.
We can now use the functoriality of $\mathcal{G} \otimes_{\mathcal{O}_{X}} \bullet$ to define $P_{\mathcal{G}}: \mathcal{G} \otimes_{\mathcal{O}_{X}} D_{X} \to \mathcal{G}$ as $g \otimes P \mapsto g P(1)$.
Now, using this map, we can now prove easily that $g \otimes \partial^{\alpha}_{x} = 0 \implies g=0$.
If $g \otimes \partial^{\alpha}_{x} = 0$, for any $a \in \mathcal{O}_{X}$, $(g \otimes \partial^{\alpha}_{x})a = 0$, and so $P_{\mathcal{G}}((g \otimes \partial^{\alpha}_{x})a) = P_{\mathcal{G}}(g \otimes (\partial^{\alpha}_{x} \circ a)) = g (\partial^{\alpha}_{x}a) = 0$. But choosing $a = \frac{1}{\alpha!} x^{\alpha}$, we will have $g = 0$.
(Here, $\alpha = (\alpha_{1},...,\alpha_{n})$, $\alpha! = \alpha_{1}!...\alpha_n!$ and $x^{\alpha} = x_{1}^{\alpha_{1}}... x_{n}^{\alpha_{n}}$).
So, at least locally, $\mathcal{G} \otimes_{\mathcal{O}_{X}} D_{X}$ is a pretty well behaved right $\mathcal{O}_{X}$-module.