Quotient of the algebra of differential operators by an ideal generated by functions

Question

Say $X$ is a smooth algebraic variety, and write $\mathcal D_X$ for the ring of differential operators on $X$ with the order filtration $F_{\bullet} \mathcal D_X$.

My problem is the following : I have a (coherent) ideal sheaf $\mathcal I \subseteq \mathcal O_X$, defining a subvariety $Y \subseteq X$. Is it true that $\text{Gr}^F(\mathcal D_X \cdot \mathcal I)$ can be identified with $\text{Gr}^F \mathcal D_X \cdot \mathcal I$ ? My intuition tells me it has to be the case, since as a $\mathcal D_X$-module it corresponds to a system whose solutions are intuitively functions that vanish everywhere outside $Y$.

For more context : I am trying to adapt the proof of proposition 3 from this article to the Fourier transformed module. The case I'm really interested in is $X = \mathbb C^n$ and my ideal $I$ is the left $D_{\mathbb C^n}$ ideal generated by some homogeneous polynomials $(f_\alpha)_\alpha$ and operators of the form $E_i = \sum_j a_{i,j} x_j \partial_{x_j}$, and I want to prove that $\text{Gr}^F I$ is generated by the highest symbols of these generators. This should be doable by a Koszul complex argument applied to the $E_i$, but for this I need to consider the ring $\text{Gr}^F \big(D_{\mathbb C^n} / (\sum_\alpha D_{\mathbb C^n} f_\alpha)\big)$.

I was able to prove it if $\mathcal I$ is locally generated by one element because for $f$ regular function, $P$ differential operator of order $k$, $Pf = fP + $ some element of $F_{k-1} \mathcal D_X$, meaning that when we identify $\text{Gr}^F \mathcal D_X \cdot \mathcal I$ with an ideal of $\text{Gr}^F \mathcal D_X$, we obtain $(f)$.

This reasoning fails if there are multiple generators, since the highest order terms could cancel each other, and then I have no clue what the second highest order terms look like. If it is not true, are there any nice things that can be said about $\text{Gr}^F (\mathcal D_X \cdot \mathcal I)$, especially in the case where $X$ is the affine $n$-space ?

Answer 1

It is true that $\mathrm{Gr}(\mathcal{D}_X \cdot \mathcal I) = \mathrm{Gr}(\mathcal{D}_X) \cdot \mathcal I$. The point is that since $X$ is smooth, $\mathcal{D}_X$ is a locally free right $\mathcal O_X$-module. If $\{\partial_1,\ldots,\partial_n\}$ is a local basis for $T_X$, then a local basis for $\mathcal D_X$ as a right $\mathcal O_X$-module is the set of monomials $\partial_1^{\alpha_1}\cdots \partial_n^{\alpha_n}$. An operator $P = \sum_\alpha \partial^\alpha f_\alpha$ lies in $F_i\mathcal{D}_X$ if and only if all nonzero terms $\partial^\alpha f_\alpha$ have $|\alpha| \leq i$, while $P$ lies in $\mathcal D_X \cdot \mathcal I$ if and only if $f_\alpha \in \mathcal I$ for all $\alpha$. Thus $(F_i \mathcal D_X) \cdot \mathcal I = (F_i\mathcal D_X) \cap \mathcal D_X \cdot \mathcal I$, and the claim about associated graded follows.