[Math] finite length and finitely generated

modules

it seems that a module over a noetherian ring R is finitely generated if and only if it has finite length (sorry, it turns out to be false! i must have had a misunderstanding!)

but why in the following two cases, we have some extra assumptions?

Example (i) A module M over a (commutative) noetherian ring R has finite length if and only if it
is finitely generated and AssM consists of only maximal ideals, where AssM is the set of associated primes for M.

(definition: We say that a prime ideal p is an associated prime for M if there
exists m in M such that p = ann m.)

Example (ii) in this thesis:
http://digitalcommons.mcmaster.ca/opendissertations/3531/
Page 12 Lemma 2.15: A finitely generated $A$-module is $\Sigma$-torsion if and only if it has finite length.

As i understand: $A$ is "weyl algebra" which is noetherian.
so why do we need the extra assumption "$\Sigma$-torsion"?

Best Answer

A (long) comment: the definition of an associated prime for $M$ you are mentioning is for the case of commutative rings. There are several problems with this definition, if $R$ is noncommutative. This is discussed in the following thesis: http://math.fullerton.edu/sannin/Research/thesis2.pdf.

The author uses the following definition instead (Def. $14$): Let $R$ be a ring, not necessarily commutative. Let $M$ be an $R$-module. An Ideal $P$ is called an associated prime of $M$ if there exists a prime submodule $N\subseteq M$ such that $P = ann(N)$. Here a nonzero $R$-module $N$ is called prime if $ann(N) = ann(N′)$ for every nonzero submodule $N′ ⊆ N$.

Edit: As to the implications of finitely generated and finite length for $M$, for $R$ being noncommutative: if $R$ is left-Artinian (hence also left-Noetherian) then these are still equivalent.