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.


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 $M_1=\langle1\rangle$.

  • $\begingroup$ As a follow up, is it possible that there will always exist a filtration without embedded primes? $\endgroup$ – user2520938 Mar 11 at 21:27
  • $\begingroup$ To my follow up question: I think we have a positive answer, by Theorem A.IV.4.11 in Bjork, "Analytic D-modules and applications". Do you agree? $\endgroup$ – user2520938 Mar 11 at 21:43
  • $\begingroup$ Sorry, I don't have access to that book. $\endgroup$ – Friedrich Knop Mar 12 at 14:02
  • $\begingroup$ 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). $\endgroup$ – user2520938 Mar 12 at 14:17

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