 In particular, all graded pretty clean filtrations of $S/I$ have the same length, namely $\sum_{i=0}^ne(\Ext_S^i(S/I,S))$, and this number equals the arithmetic degree of $S/I$.

The number in question equals $h_{ki}$, the $k$-th coefficient of the $h$-vector of $D_i(S/I)/D_{i-1}(S/I)$. Hence this number only depends on $S/I$. Moreover, it follows that the length of a graded pretty clean filtration of $S/I$ equals $\sum_{i=0}^nQ_i(1)$. As a consequence of Proposition \[hilbert\] and [@BH Proposition 4.1.9] we have that $Q_i(1)=e(\Ext_S^i(S/I,S))$. In [@NR Theorem 3.11] Nagel and Römer have shown that the arithmetic degree of a sequentially Cohen-Macaulay $R$-module $M$ equals the number $\sum_{i=0}^ne(\Ext_R^i(M,\omega_R))$. Since by Theorem \[sequentially\], $S/I$ is sequentially CM, all assertions follow.

We would like to remark that the fact that the length of all pretty clean filtrations of $S/I$ have length equal to the arithmetic degree of $S/I$ also follows from Lemma \[allthesame\].

Suppose that $I\subset S$ is a monomial ideal, and that $\mathcal F$ is a multigraded prime filtration of $S/I$ with factors $(S/P_i)(-a_i)$, $i=1,\ldots,r$, where $a_i\in \NN^n$. Then this filtration decomposes $S/I$ as a multigraded $K$-vectorspace, that is, we have $$S/I\iso \Dirsum_{i=1}^rS/P_i(-a_i).$$ Each module $M_i$ in the filtration $\mathcal F$ is of the form $I_i/I$ where $I_i$ is a monomial ideal. The monomials not belonging to $I_i$ form a $K$-basis of $(S/I)/M_i=S/I_i$, and so $S/I=(S/I)/M_i\dirsum M_i$ decomposes naturally as a $K$-vectorspace. Identifying $S/P_i(-a_i)=M_i/M_{i-1}\subset S/I_{i-1}$ with its image in $S$, we get $S/P_i(-a_i)=u_iK[Z_i]$ where $u_i=\prod_{j=1}^nx_j^{a_i(j)}$ and $Z_i=\{x_j\: j\not\in P_j\}$. Thus $$S/I= \Dirsum_{i=1}^ru_iK[Z_i].$$ Any decomposition of $S/I$ as a direct sum of $K$-vectorspaces of the form $uK[Z]$ where $Z$ is a subset of $X=\{x_1,\ldots,x_n\}$ and $u$ is a monomial of $K[X]$ is called a [*Stanley decomposition*]{}. Stanley decompositions have been studied in va