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Let $R$ be a (possibly noncommutative, but unital) ring. There are several finiteness conditions on a left $R$-module that we can consider. The ones I have come across include

  • finitely generated
  • finitely presented
  • Noetherian
  • Artinian
  • finite length

There are several relations between these that hold in general. For instance

  • finitely presented $\Rightarrow$ finitely generated
  • Noetherian $\Rightarrow$ finitely generated (actually Noetherian $\Leftrightarrow$ every submodule is finitely generated)
  • Noetherian and Artinian $\Leftrightarrow$ finite length

Yet the finiteness conditions above do not coincide in general. For example, there are rings which have Artinian but non-Noetherian left modules, etc.

However, if the ring $R$ itself satisfies some similar conditions for its (left) ideals, then some of the finiteness conditions coincide for left $R$-modules. I am interested in the following problem:

Given a set of finiteness conditions, classify the rings where these conditions coincide for left $R$-modules.

I guess the most elementary example of this is the following:

1) $R$ is left Noetherian $\Leftrightarrow$ "finitely generated $\equiv$ Noetherian" for left $R$-modules.
The nontrivial part of $\Rightarrow$ -finitely generated left modules over a left Noetherian ring are Noetherian- is proved in probably every introductory textbook in module theory and the other direction comes for free by considering $R$ as a left module over itself.

Perhaps less well known, we also have:

2) $R$ is left Noetherian $\Leftrightarrow$ "finitely generated $\equiv$ finitely presented" for left $R$-modules.
This can be shown by using the following lemma (this is stated in Matsumura's book Commutative Ring Theory):
Lemma: Let $R$ be any ring. Given a short exact sequence of left $R$-modules $$0\longrightarrow M_1 {\longrightarrow}M_2{\longrightarrow}M_3\longrightarrow 0 $$ if $M_3$ is finitely presented and $M_2$ is finitely generated, then $M_1$ is finitely generated.
Proof. There is an exact sequence of the form $$R^m {\longrightarrow}R^n{\longrightarrow}M_3\longrightarrow 0 $$ and as it involves projective modules, there are maps $f:R^m \rightarrow M_1$ and $g: R^n \rightarrow M_2$ such that together with the identity map $M_3 \rightarrow M_3$, they connect the above two exact sequences in a commutative diagram. Now by the snake lemma, we get that $\text{coker}\,f \cong \text{coker}\,g$ is finitely generated. $\text{im} \, f$ is also finitely generated, hence $M_1$ is finitely generated.

Proof of (2): $\Rightarrow$ is easy, we can surject on a finitely generated module by a finitely generated free module and the kernel will be finitely generated also by the Noetherian assumption. For $\Leftarrow$, let $I$ be a left ideal of $R$, then the above lemma applies for the short exact sequence $$0\longrightarrow I {\longrightarrow}R{\longrightarrow}R/I\longrightarrow 0 $$ and we get that $I$ is finitely generated.

$R$ is said to be semiprimary if its Jacobson radical $\text{rad} \, R$ is nilpotent and $R / \text{rad} \, R$ is semisimple. The Hopkins-Levitzki theorem stated as in T.Y. Lam's book$\,$ A First Course in Noncommutative Rings says that

3) $R$ is semiprimary $\Rightarrow$ "Noetherian $\equiv$ Artinian $\equiv$ finite length" for left $R$-modules.

I don't know whether (3) can be made into an if and only if. So I'd like to know of a counterexample or a proof to the reverse implication. Note that since being semiprimary is a left-right symmetric notion, (3) is also true for right $R$-modules. This makes me believe that the converse of (3) is not true.

Now if $R$ is left Artinian, it is semiprimary and hence left Noetherian by (3). Combining (1), (2) and (3) we see that all the five finiteness conditions I listed above coincide for left Artinian rings. Also this happens only for left Artinian rings by considering $_R R$. That is, we have

4) $R$ is left Artinian $\Leftrightarrow$ "finitely generated $\equiv$ finitely presented $\equiv$ Noetherian $\equiv$ Artinian $\equiv$ finite length" for left $R$-modules.

I'm most interested in knowing for exactly which rings the Noetherian and Artinian conditions coincide on left modules. But answers considering other interesting combinations of finiteness conditions are also welcome.

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1 Answer

The coincidence of Artinian and Noetherian are discussed within the 2009 article Rings Over which the Krull Dimension and the Noetherian Dimension of All Modules Coincide by Hashemia, Karamzadeh and Shiralia.

From the intro to the article:

Our aim in this article is to characterize rings R over which k-dim M = n-dim M for any R-module M with Krull dimension. Since every Artinian (Noetherian) R-module has Krull dimension, we infer that Artinian modules coincide with Noetherian modules over the rings that we are looking for. These rings are called amen (i.e., Artinian modules equals Noetherian) in Tanabe (1994) and are also studied in Facchini (1980). Our main result in Section 3, shows that in fact amen rings are the only rings over which the above equality between the two dimensions occurs.

The full names of those other references are:

Facchini , C. ( 1980 ). Loewy and Artinian modules over commutative rings . . CXXVIII: 359 – 374 .

Tanabe , K. ( 1994 ). On rings whose Artinian modules are precisely Noetherian modules . Comm. Algebra 22 : 4023 – 4032 .

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Your question is what led me to the article, and I haven't read it enough to know whether or not 'amen' rings are semiprimary, but I have a feeling that your conjecture that they are not the same is true. –  rschwieb Feb 26 '13 at 22:12
    
Thanks for the reference! –  Cihan Feb 28 '13 at 0:29
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