2
$\begingroup$

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"?

Improve this question
$\endgroup$
8
  • $\begingroup$ It would have been best not to accept Dietrich's interesting comment as an answer. As your own comment to it makes evident, it does not answer your question! $\endgroup$
    – Mariano Suárez-Álvarez
    Commented May 30, 2013 at 16:45
  • $\begingroup$ but at least he has shown me that this definition of an associated prime is for the case of commutative rings, something i was not aware of. i don't know if R needs to be commutative, but at least the author of the statement as it is, meant it in the commutative environment. $\endgroup$
    – Paslig Keir
    Commented May 30, 2013 at 16:54
  • $\begingroup$ Mariano, you are right. I should have used the "Add comment" field. $\endgroup$
    – Dietrich Burde
    Commented May 30, 2013 at 18:28
  • 1
    $\begingroup$ Dear Paslig. If your question is whether for modules $M$ over a left-Artinian ring $R$, finite length and finite generation are equivalent properties, then the answer is yes. This is a basic fact. Any module of finite length is finitely generated. As you remark, being left-artinian, $R$ is also left-noetherian, hence has finite length itself. And so are $R^n$, for any integer~$n$, and its quotients. So a finitely generated $R$-module has finite length. This is explained in (almost) every book on algebra, eg, Lang, Jacobson, or Bourbaki (Algèbre, chapter 8). $\endgroup$
    – ACL
    Commented May 30, 2013 at 19:30
  • 2
    $\begingroup$ Dear Paslig, Unless I am very confused, your first sentence is quite wrong. E.g. $\mathbb Z$ is Noetherian, and is f.g. (indeed cyclic) as a module over itself, but is not finite length as a module over itself. The same remark will apply with $\mathbb Z$ replaced by any Noetherian ring that is not Artinian. That is why all the other hypotheses are necessary. Regards, $\endgroup$
    – Emerton
    Commented May 30, 2013 at 22:41

1 Answer 1

Reset to default
2
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thank you for your reply again. :) i appreciate it very much. what is the relationship between finite length and finitely generated? when does one imply the other and vice versa? $\endgroup$
    – Paslig Keir
    Commented May 30, 2013 at 16:30
  • $\begingroup$ (@Dietrich, regarding your comment aboveÑ I think your use of an answer to write your comment is quite fine, as it is often a pain to type long comments!) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented May 30, 2013 at 23:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .