2
$\begingroup$

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 ?

$\endgroup$
  • $\begingroup$ What are the $A_n$? $\endgroup$ – k.stm Apr 11 '14 at 7:15
  • $\begingroup$ $A$ is a Noetherian graded ring $A=\oplus_{n=0}^\infty A_n$. $M$ is a finitely-generated graded $A$-module $M=\oplus_{n=0}^\infty M_n$. $\endgroup$ – Aki Apr 11 '14 at 7:20
  • $\begingroup$ In the example, $A_n$ is the set of the homogeneous polynomials of degree $n$. $\endgroup$ – Aki Apr 11 '14 at 7:25
  • 1
    $\begingroup$ Okay, it seems to me that you are right. For $λ(M) = l(M)/l(A)$ it’ll work, though, wouldn’t it? $\endgroup$ – k.stm Apr 11 '14 at 7:41
  • 1
    $\begingroup$ It is the page 118 of 1969 version. $\endgroup$ – Aki Apr 11 '14 at 10:10

Your Answer

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

Browse other questions tagged or ask your own question.