I'm trying to understand why the definition of a regular holonomic D-module is a good generalisation of the usual definition of a regular singular point for a differential equation. More precisely, here is the definition of regular holonomic D-module I want to use

A D-module $M$ on a complex manifold $X$ is holonomic if $\text{char}(M)$ has the same dimension as $X$. Moreover, it is regular if there is a good filtration $\{M_j\}_{j\in \mathbb{Z}}$ such that $$f\cdot\text{gr}(M)=0$$ for all $f\in \text{gr}(D_X)$ vanishing on $\text{char}(M).$

Now I'd like to test this definition on an example. I consider a Bessel equation $$P(f) :=(z^2 \partial_z^2 + z\partial_z + (z^2-4))(f)=0.$$ The point $z=0$ is regular singular thanks to the $z^2$. Of course if the leading coefficient is replaced by $z^3$ it becomes irregular singular. Now I consider the D-module $$M := D_{\mathbb{C}}/D_{\mathbb{C}} P$$ naturally attached to this equation. We should normaly be able to prove that this D-module on $\mathbb{C}$ is regular holonomic. (And not regular if we put $z^3$ as a leading coefficient) First the characteristic variety is given by the null-set of $(z,w)\mapsto z^2w^2$ and so $$\text{char}(M) = \mathbb{C} \cup T_0^{*}\mathbb{C}$$ which has dimension $1$, so the module is holonomic. Now as a candidate for the good filtration, I set $M_0 = D_{\mathbb{C}}/D_{\mathbb{C}} (z^2-4), M_1 = D_{\mathbb{C}}/D_{\mathbb{C}} (z\partial_z+(z^2-4)), M_2 = M$ and $M_j=0$ for other $j$. If I'm not mistaken this verifies the axiom of a good filtration.

Now I'm getting lost, how can I prove that $f\cdot\text{gr}(M)=0$ for all $f\in \text{gr}(D_X)$ vanishing on $\text{char}(M) \,?$ It doesn't make much sense to me.

Thank you for any help.

I'm trying to understand why the definition of a regular holonomic D-module is a good generalisation of the usual definition of a regular singular point for a differential equation. More precisely, here is the definition of regular holonomic D-module I want to use

A D-module $M$ on a complex manifold $X$ is holonomic if $\text{char}(M)$ has the same dimension as $X$. Moreover, it is regular if there is a good filtration $\{M_j\}_{j\in \mathbb{Z}}$ such that $$f\cdot\text{gr}(M)=0$$ for all $f\in \text{gr}(D_X)$ vanishing on $\text{char}(M).$

Now I'd like to test this definition on an example. I consider a Bessel equation $$P(f) :=(z^2 \partial_z^2 + z\partial_z + (z^2-4))(f)=0.$$ The point $z=0$ is regular singular thanks to the $z^2$. Of course if the leading coefficient is replaced by $z^3$ it becomes irregular singular. Now I consider the D-module $$M := D_{\mathbb{C}}/D_{\mathbb{C}} P$$ naturally attached to this equation. We should normaly be able to prove that this D-module on $\mathbb{C}$ is regular holonomic. (And not regular if we put $z^3$ as a leading coefficient) First the characteristic variety is given by the null-set of $(z,w)\mapsto z^2w^2$ and so $$\text{char}(M) = \mathbb{C} \cup T_0^{*}\mathbb{C}$$ which has dimension $1$, so the module is holonomic. Now as a candidate for the good filtration, Simon Wadsley proposes to take $$M_n=D_n.(1+D_\mathbb{C}P).$$

Now I'm getting lost, how can I prove that $f\cdot\text{gr}(M)=0$ for all $f\in \text{gr}(D_X)$ vanishing on $\text{char}(M) \,?$ It doesn't make much sense to me.

Thank you for any help.