Noetherian Modules - Maximal Condition - Berrick and Keating Ch. 3, page 111

[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 \(\displaystyle M\) is said to satisfy the maximum condition if any nonempty set of submodules of \(\displaystyle M\) has a maximal member (with respect to inclusion)"It seems to me that this definition, it 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:https://www.physicsforums.com/attachments/4883Can someone confirm that my basic understanding of the implication of the definition mentioned above is correct?

Peter

 

[steenis]

The attachment with figure 1 has disappeared, I try to give an answer anyway.
If you want to apply Zorn’s Lemma on a collection $\hat S$ of sets, which is ordered by inclusion, then you have to prove that every nonempty chain in $\hat S$ has an upper bound. So, if $\hat C$ is a chain in the set $\hat S$, then there must be an $S \in \hat S$ such that for every $C \in \hat C$ we have $C \subset S$. Then $S$ is an upper bound of $\hat C$ and $S$ does NOT need to be an element of $\hat C$.
A maximal element is something else.
If $\hat A$ is a collection of sets (for instance, submodules of a module $P$), then $M \in \hat A$ is maximal in $\hat A$ if the condition [$X \in \hat A$ AND $M \subset X$] implies $X=M$. So notice that the maximal element $M$ of $\hat A$ is an element of $\hat A$ and that $\hat A$ need not be a chain.

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). So if $\hat S$ is the collection of all submodules of $M$, ordered by inclusion, and every nonempty subset $\hat A \subset \hat S$ has a maximal element, then the module $M$ is said to satisfy the maximal condition.

I am sorry for my clumsy notation, I was looking for $\mathscr{C}$ and $\displaystyle \mathscr{S}$, but could not find them.

 

[Math Amateur]

The attachment with figure 1 has disappeared, I try to give an answer anyway.
If you want to apply Zorn’s Lemma on a collection $\hat S$ of sets, which is ordered by inclusion, then you have to prove that every nonempty chain in $\hat S$ has an upper bound. So, if $\hat C$ is a chain in the set $\hat S$, then there must be an $S \in \hat S$ such that for every $C \in \hat C$ we have $C \subset S$. Then $S$ is an upper bound of $\hat C$ and $S$ does NOT need to be an element of $\hat C$.
A maximal element is something else.
If $\hat A$ is a collection of sets (for instance, submodules of a module $P$), then $M \in \hat A$ is maximal in $\hat A$ if the condition [$X \in \hat A$ AND $M \subset X$] implies $X=M$. So notice that the maximal element $M$ of $\hat A$ is an element of $\hat A$ and that $\hat A$ need not be a chain.

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). So if $\hat S$ is the collection of all submodules of $M$, ordered by inclusion, and every nonempty subset $\hat A \subset \hat S$ has a maximal element, then the module $M$ is said to satisfy the maximal condition.

I am sorry for my clumsy notation, I was looking for $\mathscr{C}$ and $\displaystyle \mathscr{S}$, but could not find them.

Thanks for the help, Steenis ...

Just reflecting on what you have written and revising the issue now ...

Peter