Let $A_0$ be an Artinian ring, $M$ a free $A_0$-module. Then, is the length of the composition series of $M$ identical to the number of its bases?
It seems to me that it is not. If $\mathfrak a$ is an ideal of $A_0$ and $e$ is a basis of $M$ then ${\mathfrak a}e$ is a submodule of $A_0 e$. Therefore the chain has to have the length more than the number of bases. But I know I'm wrong, because there is an example shown in Introduction to Commutative Algebra by Atiyah and McDonald:
Let $A=A_0[x_1,\cdots,x_s]$, where $A_0$ is an Artinian ring and the $x_i$ are independent indeterminates. Then $A_n$ is a free $A_0$-module generated by the monomials $x_1^{m_1}\cdots x_s^{m_s}$ where $\sum m_i =n$; there are $\left( \begin{smallmatrix} s+n-1 \\ s-1 \end{smallmatrix} \right)$ of these, hence $P(A,t)=(1-t)^{-s}$.
The Poincaré series $P(M,t)$ is defined for a finitely-generated graded $A$-module $M$ ($A$ is a Noetherian graded ring) and an additive function $\lambda$ as follows:
$$ P(M,t) = \sum_{n=0}^\infty \lambda(M_n) t^n \;\; \in {\mathbb Z}[[t]]. $$
In the example, $\lambda(M)$ is the length $l(M)$ of a finitely-generated $A_0$-module $M$.
What is wrong with my reasoning ?