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.