# Noetherian Modules - Maximal Condition - Berrick and Keating

1. Nov 9, 2015

### Math Amateur

I am reading the book "An Introduction to Rings and Modules with K-theory in View" by A.J. Berrick and M.E. Keating ... ...

I am currently focused on Chapter 3; Noetherian Rings and Polynomial Rings.

I need someone to help me to fully understand the maximal condition for modules and its implications ...

On page 111, Berrick and Keating state the following:

"The module M is said to satisfy the maximum condition if any nonempty set of submodules of M has a maximal member (with respect to inclusion) "

It seems to me that this definition, when it is satisfied, means that all the submodules of M must be in a chain of inclusions ... so we cannot have a situation like that depicted in Figure 1 below:

Can someone confirm that my basic understanding of the implication of the definition mentioned above is correct ... ... and/or ... ... give a simple explanation of the maximal condition ...

Help will be appreciated ...

Peter

==================================================

*** EDIT ***

Just reflecting a bit on this matter ... ...

Maybe the maximal condition means that for any collection of submodules IF one can arrange a subset into an ascending chain of inclusions then that chain will have a maximal member ... ... ?

Mind you if that is true ... then you can have several maximal members ...

Hope someone can clarify this issue for me ....

Peter

File size:
93.2 KB
Views:
222
2. Nov 9, 2015

### andrewkirk

I'm not familiar with the terminology, but I suspect what you wrote here is correct, because any free module on three or more generators will have pairs of submodules neither of which is contained in the other.

So I suspect the proper definition is that a module satisfies the maximum condition iff any ascending chain of submodules has a maximal member. This then becomes analogous to the least upper bound property of the real numbers.

3. Nov 9, 2015

### andrewkirk

On the other hand, maybe they mean what they say, and they wish to disqualify free modules from having the maximum condition. I suggest checking with one or more alternative sources on the web to see if the definitions are stated differently. With that definition it would imply that for any two submodules, one must be contained in the other.

4. Nov 10, 2015

### Math Amateur

Thanks for for your help Andrew ...

But ... I am still somewhat perplexed and puzzled ...

Please let me know if you come across a definitive answer ...

Thanks again for your help ...

Peter

5. Nov 10, 2015

### Staff: Mentor

The maximum condition is equivalent to the ascending chain condition: Each chain $M_1 ⊆ M_2 ⊆ ...$ of submodules of a module $M$ is getting stationary, i.e. $M_n = M_{n+1} = ...$for some $n ∈ℕ$ Those modules are called Noetherian after Emmy Noether.

You can mirror this. The minimal condition is equivalent to the descending chain condition. Those modules are called Artinian after Emil Artin.

(My source: Introduction to commutative Algebra by M.F.Atiyah and I.G.Macdonald, chapter 6)

Last edited: Nov 10, 2015
6. Nov 10, 2015

### Staff: Mentor

Examples: (Introduction to Commutative Algebra by Atiyah, Macdonald)

1.) A finite abelian group $G = \{a_1, ..., a_n | a_i + a_j = a_j + a_i ∀ 1 ≤ i, j ≤ n\}$ as ℤ-module: $m.a_i = a_i + ... + a_i$ (m-times, m ∈ℤ) is as well Noetherian as Artinian.

2.) ℤ as ℤ-module is Noetherian, not Artinian. $(a) ⊃ (a^2) ⊃ ... ⊃ (a^n) ⊃ ...$ for an $a ≠ 0$

3.) Let $p$ be a fixed prime and $G$ the subgroup of $ℚ/ℤ$ which elements are of order $p^n$ for some $n ∈ ℕ$, i.e. all $x + ℤ ∈ ℚ/ℤ$ with $p^n x ∈ ℤ$.
$G$ has exactly one subgroup $G_n$ of order $p^n$ for each $n ∈ ℕ_0$ and $G_0 ⊂ G_1 ⊂ ... ⊂ G_n ⊂ ...$ shows $G$ is not Noetherian. Since the $G_n$ are the only proper subgroups $G$ is Artinian.

7. Nov 10, 2015

### andrewkirk

I have nothing definitive, but the evidence seems strongly in favour of the first of the two interpretations I suggested - the one in post 2 - which is called the Ascending Chain Condition (ASC). fresh42's posts are in accordance with that definition, with good examples to make it concrete.

The wikipedia article on Noetherian modules gives the integers $\mathbb{Z}$, as a module over the ring $\mathbb{Z}$ (sometimes written as ${}_\mathbb{Z}\mathbb{Z}$, as an example of a Noetherian Module. That module does not obey the definition in the OP and post 3, because $\{3\mathbb{Z},2\mathbb{Z}\}$ is a collection of submodules that does not have a maximal element. But it does obey the ASC.

I suggest provisionally using the ASC as the definition, subject to reconsidering later on if you run into something that credibly suggests otherwise.

8. Nov 10, 2015

### Staff: Mentor

The maximal condition in the OP is defined sloppyly(?).

The set of submodules has to be required partially ordered (by inclusion ⊆). The same holds for the minimum condition. Therefore the ℤ-modules {3ℤ, 2ℤ} cannot serve as counter example. The only partially ordered sets here are {2ℤ}, {3ℤ} which trivially contain a maximal submodule.
ℤ as ℤ-module is Noetherian as mentioned in example 2, i.e. it fulfills the maximal condition! For example {2ℤ, 3ℤ, 6ℤ} has {2ℤ, 6ℤ} and {3ℤ, 6ℤ} as partially ordered subsets, each containing a maximal submodule.

Rem.: A set Σ is partially ordered (⊆) if for M, N, P ∈ Σ it's reflexive (M ⊆ M), transitive (M ⊆ N ∧ N ⊆ P ⇒ M ⊆ P)
and symmetric ( M ⊆ N ∧ N ⊆ M ⇒ M = N).

Last edited: Nov 10, 2015
9. Nov 10, 2015

### andrewkirk

It's a counter-example to the claim that the maximum condition is as defined in the OP, not to the claim that ${}_\mathbb{Z}\mathbb{Z}$ is a Noetherian Module.

10. Nov 10, 2015

### andrewkirk

Unfortunately, Wikipedia gives the same definition here. It even uses essentially the same words, which makes me wonder whether it has been copied from the same original, poorly written, source.

To be fair, the wikipedia entry qualifies that statement by saying that that definition is only equivalent to the ASC if we also assume Axiom of Choice. But I can't see how the Axiom of Choice could make the counterexample of the last few posts work. Do you think the wikipedia statement is wrong? I'm inclined to edit it but I'd like to be more sure of myself first.

11. Nov 10, 2015

### Staff: Mentor

Yep. That's why the maximum condition is defined incompletely. I'm 100% certain he meant Noetherian and Artian modules since he pretty much follows exactly Atiyah, Macdonald which is a standard textbook here.´
Noetherian modules are kind of "common", i.e. give me any module and there's a good chance it'll be Noetherian. Artinian is harder to find. Therefore example 3 is a little bit more complicated than example 2.

Last edited: Nov 10, 2015
12. Nov 10, 2015

### Staff: Mentor

Your example {3ℤ,2ℤ} nevertheless is interesting, because both submodules are maximal in their own partial ordered subset. The maximal submodule itself does not have to be unique!

The Axiom of choice is similar to the maximal condition. It's equivalent to Zorn's Lemma: Every nonvoid partially ordered set in which each chain has an upper bound has a maximal element. (The module itself can serve as upper bound.)
Caution: the maximal condition for modules does not imply the axiom of choice, which is more general.
(Real and Abstract Analysis by E. Hewitt and K. Stromberg, Chp. 1, Theorem 3.12)

The equivalence proof from Atiyah, Macdonald is indirect and uses an induction to construct a non terminating increasing sequence. Without getting any deeper here that heavily smells like the axiom of choice, which you can get rid of if the Modules are finitely generated.

Last edited: Nov 10, 2015
13. Nov 10, 2015

### mathwonk

indeed, in this context, and usually, "maximal" submodule, means one not contained in another strictly larger submodule (of the collection). this word essentially always means this. Thus in the OP's picture, N1, N2, and N3, are all maximal.

the topic is discussed on p. 14 of these free algebra notes:

http://alpha.math.uga.edu/~roy/844-1.pdf