# For a holonomic $D_X$-module $M$, can $gr M$ have embedded primes?

Let $$M$$ be a holonomic $$D_X$$ module. This means that the minimal primes in $$\sqrt{Ann(gr M)}$$ are $$n=\dim X$$ dimensional, for some (and any) good filtration on $$M$$. But what about the embedded primes?

I think there cannot be embedded primes, and the argument would be as follows. Suppose $$p\subset gr D_X$$ is an embedded prime, associated to $$m\in gr M$$. Then $$(gr D_X)/p\subset gr M$$. Now find an ideal $$\tilde{p}\subset D_X$$ such that $$gr(D_X/\tilde{p})\cong (grD_X)/p$$. Then this means that $$\dim Ann((gr D_x)/p)=\dim p\geq n$$, contradicting the fact that $$p$$ was an embedded prime.

The problem with this argument is that I do not know that this $$\tilde{p}$$ exists, which inspired this question.

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 $$M_1=\langle1\rangle$$.
• Ah oke, either way, thanks a lot for your answer. In case you're interested, the theorem states that if $R$ is a filtered ring, such that both $R$ and $gr R$ are Auslander regular, and $M$ is a $k$-pure module (i.e. $Ext_R^i(Ext_R^i(M,R),R)\not=0\Rightarrow i=k$), then there exists a good filtration on $M$ such that $gr M$ is $k$-pure. It is known that over a commutative ring, pure modules and no embedded primes, so the result follows (since holonomic modules are $\dim X$-pure). – user2520938 Mar 12 at 14:17