1
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $M_1=\langle1\rangle$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.