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> Let $M$ be a holonomic $D\_X$-module. This means that the minimal primes in $\sqrt{\operatorname{Ann}(\operatorname{gr}M)}$ are $n=\dim X$ dimensional, for some (and any) good filtration on $M$. But what about the embedded primes?
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I think there cannot be embedded primes, and the argument would be as follows. Suppose $\mathfrak p\subset \operatorname{gr} D\_X$ is an embedded prime, associated to $m\in \operatorname{gr} M$. Then $(\operatorname{gr} D\_X)/\mathfrak p\subset \operatorname{gr} M$. **Now find an ideal $\tilde{\mathfrak p}\subset D\_X$ such that $\operatorname{gr}(D\_X/\tilde{\mathfrak p})\cong (\operatorname{gr}D\_X)/\mathfrak p$.** Then this means that $\dim \operatorname{Ann}((\operatorname{gr} D\_x)/\mathfrak p)=\dim p\geq n$, contradicting the fact that $\mathfrak p$ was an embedded prime. 


The problem with this argument is that I do not know that this $\tilde{\mathfrak p}$ exists, which inspired [this question](https://math.stackexchange.com/q/3143676/95162). 



For a holonomic $D_X$-module $M$, can $\operatorname{gr}M$ have embedded primes?
Embedded primes may exist and they depend on the good filtration.


Example: Let $X=\mathbf A^1$ be the affine line. Then $D\_X=\mathbb C\langle x,\partial\rangle$. Let $M=D\_X/D\_Xx=\mathbb C[\partial]$. Choose the filtration $M\_n:=\langle 1,\partial,\ldots,\partial^n\rangle$ for $n\ge\mathbf 2$ but put $M\_1:=0$. Then $\mathrm{gr}M=\langle 1,\partial\rangle\oplus\langle\partial^2\rangle\oplus\ldots$. Since $x1=\partial1=0\in\mathrm{gr}M$ we see that the point $(0,0)$ corresponds to an embedded prime. The minimal prime is the ideal generated by $x$.


Clearly that does not happen if one chooses the standard filtration with $M\_1=\langle1\rangle$.


