MHB Noetherian Modules - Bland Proposition 4.2.3

[Math Amateur]

I am reading Paul E. Bland's book, "Rings and Their Modules".

I am trying to understand Chapter 4, Section 4.2 on Noetherian and Artinian Modules and need help with fully understanding Proposition 4.2.3, particularly assertion (2).

Bland's statement of Proposition 4.2.3 reads as follows:

View attachment 3733

Consider now Figure 1 below, showing module M with three submodules, $$M_1, M_2 $$and $$M_3$$ respectively:

View attachment 3734

As per Bland's assertion (2) of Proposition 4.2.3 above, the collection of submodules $$\{ M_1, M_2 \}$$ when ordered by inclusion, clearly has a maximal element, namely $$M_1$$.BUT ... ... what is the situation when we consider the collection $$\{ M_2, M_3 \}$$?Are both $$M_2$$ and $$M_3$$ respectively, each maximal elements in the collection $$\{ M_2, M_3 \}$$?

(as you may see from the above I am somewhat unsure as to how to view collections which include disjoint submodules!)

Can someone please help clarify the above issue?
Further to the above analysis, would the module M shown in Figure 1 be noetherian?

It seems to me that M would be noetherian since the only chains of submodules, namely

$$M_2 \subseteq M_1 \subseteq M_1 \subseteq M_1 \subseteq \ ... \ ... $$

and

$$M_3 \subseteq M_3 \subseteq M_3 \subseteq M_3 \subseteq \ ... \ ...
$$

clearly terminate ... ... Can someone please indicate whether this analysis is correct?
Peter

 

[Euge]

Yes, you're right -- $M_2$ and $M_3$ are both maximal elements of $\{M_2,M_3\}$. This is because $M_2$ is not properly contained in any element of $\{M_2,M_3\}$, and similarly for $M_3$. The underlying principle here is that we're dealing with posets, and not particular linearly ordered sets. Therefore, when considering a collection of submodules of $M$, the sets in the collection need not be comparable. So you may find collections of submodules of $M$ that have maximal elements, but no greatest element, such as the case with $\{M_2,M_3\}$. If in your diagram, $M_1$, $M_2$, and $M_3$ are the only proper, nontrivial submodules of $M$, then indeed $M$ is Noetherian.

 

[Math Amateur]

Yes, you're right -- $M_2$ and $M_3$ are both maximal elements of $\{M_2,M_3\}$. This is because $M_2$ is not properly contained in any element of $\{M_2,M_3\}$, and similarly for $M_3$. The underlying principle here is that we're dealing with posets, and not particular linearly ordered sets. Therefore, when considering a collection of submodules of $M$, the sets in the collection need not be comparable. So you may find collections of submodules of $M$ that have maximal elements, but no greatest element, such as the case with $\{M_2,M_3\}$. If in your diagram, $M_1$, $M_2$, and $M_3$ are the only proper, nontrivial submodules of $M$, then indeed $M$ is Noetherian.

Thank you for your help Euge, you are making my quest to understand algebra working by myself without the assistance of a University environment so much easier ...

Peter