D_d(M)/D_{d-1}(M)$ is clean as well.

\[conclusion\] Let $S=K[x_1,\ldots,x_n]$ be a the polynomial ring and $I\subset S$ a monomial ideal. Then the following conditions are equivalent:

1.  $S/I$ is pretty clean;

2.  $S/I$ is sequentially CM, and the non-zero factors in the dimension filtration of $S/I$ are clean;

3.  the non-zero factors in the dimension filtration of $S/I$ are clean.

(a)(b): Since the associated prime ideals of $S/I$ are all generated by subsets of $\{x_1,\ldots,x_n\}$, all hypotheses Theorem \[sequentially\] and Corollary \[interesting\] are satisfied, so that the assertions follow.

(b)(c) is trivial.

(c)(a): The refinement of the dimension filtration by the clean filtrations of the non-zero factors gives us the desired pretty clean filtration of $S/I$.

*Let $S=K[x,z,u,v]$, and consider the ideals $L=(u,v,z)$, $Q_1=(x,z^2)$, $Q_2=(x,v^2,z^3)$ and $I=L\sect Q_1\sect Q_2$. We claim that the module $M=L/I$ is not pretty clean, but that $M$ has a prime filtration $\mathcal F$ with $\Supp({\mathcal F})=\Ass(M)$.*

Note that $(L\sect Q_1)\sect (L\sect Q_2)$ modulo $I$ is an irredundant primary decomposition of $(0)$ in $M$. Hence since $\emptyset \neq \Ass(L/L\sect
Q_i)\subset \Ass(S/Q_i)=\{P_i\}$ with $P_1=(x,z)$ and $P_2=(x,v,z)$ we see that $\Ass(M)=\{P_1,P_2\}$.

It follows from Corollary \[description\] that $D_1(M)=(L\sect Q_1)/I$ and that $D_2(M)=M$. We show that $D_2(M)/D_1(M)=L/L\sect Q_1$ is not clean. Indeed, suppose $L/L\sect Q_1$ is clean. Then, since $\Ass(L/L\sect Q_1)=\{P_1\}$, this module has a filtration with all factors isomorphic to $S/P_1$, and the number of these factors equals the length of the $S_{P_1}$-module $(L/L\sect Q_1)_{P_1}=S_{P_1}/Q_1S_{P_1}$. This length is obviously $2$. On the other hand, since $L/L\sect Q_1$ is generated by 3 elements, it cannot have a filtration with two factors, both of them being cyclic.

Knowing now that $D_2(M)/D_1(M)$ is not clean, we conclude from Corollary \[interesting\] that $M$ is not pretty clean.

Finally we construct a prime f