MHB Noetherian Modules - Cohn Theorem 2.2

[Math Amateur]

I am reading "Introduction to Ring Theory" by P. M. Cohn (Springer Undergraduate Mathematics Series)

In Chapter 2: Linear Algebras and Artinian Rings we find Theorem 2.2 on Noetherian modules. I need help with some aspects of the proof.

Theorem 2 reads as follows:View attachment 3158
View attachment 3157
In the proof of $$(c) \Longrightarrow (d)$$ we read:

If $$N$$ is a submodule of $$M$$, let $$\mathscr{C}$$ be the collection of all finitely generated submodules and choose a maximal term $$N'$$ in $$\mathscr{C}$$.

If $$N' \subset N$$, we can adjoin an element to $$N'$$ to obtain $$N''$$ in $$\mathscr{C}$$ and properly containing $$N'$$, but this contradicts the maximality of $$N'$$.

Hence, $$N' = N$$ and this shows $$N$$ to be finitely generated. … … "
My questions regarding this particular argument are as follows:

1. Why do we need the condition $$N' \subset N$$ in order to be justified in adjoining an element to $$N'$$ to obtain $$N''$$?

2. Why does/how does contradicting the maximality of N' imply that $$N' = N$$? What about the possibility that $$N \subset N'$$?Hoping someone can help.

Peter

 

[Deveno]

To begin with, suppose $N$ is finitely-generated, for any choice of $N$. There's nothing to prove.

So we may as well assume $N$ is NOT finitely-generated, and try to derive a contradiction.

To do this, we need to find some element $a \in N$, but $a \not\in N'$. The assumption we can find such an $a$ is the crux of the argument.

Now if $N' \not\subseteq N$, we could instead consider $L = N+N'$. This still has the element $a$ we need and clearly, $L$ contains $N'$ and $L$ cannot be finitely-generated, or else $N$ as a submodule would surely be.

So without losing any generality (since we could use $L$) we may as well assume $N' \subseteq N$ to start with.

We want to show that this containment cannot be proper (this is where $a$ comes in).

So suppose $N' = \langle S\rangle$ for some finite subset $S$ of $M$ (we can do this, since $N' \in \mathscr{C}$, and is thus finitely-generated).

We set $N'' = \langle S,a\rangle$ (this is generated by the finite set $S \cup\{a\}$, clearly STILL finite).

Now $N' \subseteq N''$, and since $N'$ is maximal (in $\mathscr{C}$), it follows that $N' = N''$.

But this is absurd, by our choice of $a$, we have $a \in N'$, but $a \not\in N$, if the two modules are the same, they have the same elements.

So, we must not be able to find ANY such $a$, which means that $N \subseteq N'$ and thus $N = N'$.

This sort of argument is typical of maximality arguments.

 

[Math Amateur]

To begin with, suppose $N$ is finitely-generated, for any choice of $N$. There's nothing to prove.

So we may as well assume $N$ is NOT finitely-generated, and try to derive a contradiction.

To do this, we need to find some element $a \in N$, but $a \not\in N'$. The assumption we can find such an $a$ is the crux of the argument.

Now if $N' \not\subseteq N$, we could instead consider $L = N+N'$. This still has the element $a$ we need and clearly, $L$ contains $N'$ and $L$ cannot be finitely-generated, or else $N$ as a submodule would surely be.

So without losing any generality (since we could use $L$) we may as well assume $N' \subseteq N$ to start with.

We want to show that this containment cannot be proper (this is where $a$ comes in).

So suppose $N' = \langle S\rangle$ for some finite subset $S$ of $M$ (we can do this, since $N' \in \mathscr{C}$, and is thus finitely-generated).

We set $N'' = \langle S,a\rangle$ (this is generated by the finite set $S \cup\{a\}$, clearly STILL finite).

Now $N' \subset N''$, and since $N'$ is maximal (in $\mathscr{C}$), it follows that $N' = N''$.

But this is absurd, by our choice of $a$, we have $a \in N'$, but $a \not\in N$, if the two modules are the same, they have the same elements.

So, we must not be able to find ANY such $a$, which means that $N \subseteq N'$ and thus $N = N'$.

This sort of argument is typical of maximality arguments.

Thanks so much for the help, Deveno ... Much appreciated!

Just working through the details of the post now ...

Peter

 

[Math Amateur]

To begin with, suppose $N$ is finitely-generated, for any choice of $N$. There's nothing to prove.

So we may as well assume $N$ is NOT finitely-generated, and try to derive a contradiction.

To do this, we need to find some element $a \in N$, but $a \not\in N'$. The assumption we can find such an $a$ is the crux of the argument.

Now if $N' \not\subseteq N$, we could instead consider $L = N+N'$. This still has the element $a$ we need and clearly, $L$ contains $N'$ and $L$ cannot be finitely-generated, or else $N$ as a submodule would surely be.

So without losing any generality (since we could use $L$) we may as well assume $N' \subseteq N$ to start with.

We want to show that this containment cannot be proper (this is where $a$ comes in).

So suppose $N' = \langle S\rangle$ for some finite subset $S$ of $M$ (we can do this, since $N' \in \mathscr{C}$, and is thus finitely-generated).

We set $N'' = \langle S,a\rangle$ (this is generated by the finite set $S \cup\{a\}$, clearly STILL finite).

Now $N' \subseteq N''$, and since $N'$ is maximal (in $\mathscr{C}$), it follows that $N' = N''$.

But this is absurd, by our choice of $a$, we have $a \in N'$, but $a \not\in N$, if the two modules are the same, they have the same elements.

So, we must not be able to find ANY such $a$, which means that $N \subseteq N'$ and thus $N = N'$.

This sort of argument is typical of maximality arguments.

Thanks Deveno … I have to say that your argument is so much clearer than Cohn's text … …

However, I still need a further clarification (having followed most of your argument) … …You write:

" … … Now $N' \subseteq N''$, and since $N'$ is maximal (in $\mathscr{C}$), it follows that $N' = N''$. … … "

But surely we actually need $$N' = N$$ not $$N' = N''$$ in order to conclude:

" … … But this is absurd, by our choice of $a$, we have $a \in N'$, but $a \not\in N$, if the two modules are the same, they have the same elements. … … "

Can you please clarify?

Peter

***EDIT*** I have been reflecting further on your post and I think (but not sure …) I see your argument … ...

We have assumed that $$a \in N$$ but $$a \notin N'$$ and we have assumed that $$N' \subseteq N$$ (actually I think the assumption possibly should be $$N' \subset N$$?) … …… …

then we show - through constructing $$N' = \langle S, a \rangle$$ that $$N' = N'' $$which means that $$a \in N'$$ which contradicts our assumption that $$a \notin N$$' … …

So … … it seems that we cannot find any a such that $$a \in N$$ and $$a \notin N'$$ so $$N' \subseteq N$$ is in fact $$N' = N$$ … …

… … and hence N is finitely generated … …

Can you confirm that my reasoning is correct?

… … …

I note that you mention that this sort of argument is typical of maximality arguments … so then your post has been helpful beyond the details of this particular theorem …I found your argument so much clearer than Cohn's … … so as I have said many times before … if you ever decide to write a text … or a book of examples … then please let me know …Peter

 

[Deveno]

Thanks Deveno … I have to say that your argument is so much clearer than Cohn's text … …

However, I still need a further clarification (having followed most of your argument) … …You write:

" … … Now $N' \subseteq N''$, and since $N'$ is maximal (in $\mathscr{C}$), it follows that $N' = N''$. … … "

But surely we actually need $$N' = N$$ not $$N' = N''$$ in order to conclude:

" … … But this is absurd, by our choice of $a$, we have $a \in N'$, but $a \not\in N$, if the two modules are the same, they have the same elements. … … "

Can you please clarify?

Peter

***EDIT*** I have been reflecting further on your post and I think (but not sure …) I see your argument … ...

We have assumed that $$a \in N$$ but $$a \notin N'$$ and we have assumed that $$N' \subseteq N$$ (actually I think the assumption possibly should be $$N' \subset N$$?) … …… …

then we show - through constructing $$N' = \langle S, a \rangle$$ that $$N' = N'' $$which means that $$a \in N'$$ which contradicts our assumption that $$a \notin N$$' … …

So … … it seems that we cannot find any a such that $$a \in N$$ and $$a \notin N'$$ so $$N' \subseteq N$$ is in fact $$N' = N$$ … …

… … and hence N is finitely generated … …

Can you confirm that my reasoning is correct?

… … …

I note that you mention that this sort of argument is typical of maximality arguments … so then your post has been helpful beyond the details of this particular theorem …I found your argument so much clearer than Cohn's … … so as I have said many times before … if you ever decide to write a text … or a book of examples … then please let me know …Peter

You have it right.

Note $N'$ is some specific submodule (which could be hard to exhibit explicitly, depending on the construction of $M$), and $N$ is an arbitrary submodule.

There are a number of possibililities for the relationship of $N'$ and $N$:

1. $N$ lies in $N'$. In this case, $N$ is clearly finitely-generated, since $N'$ is. Suppose $S$ generates $N'$. Can you find (a) set that generates $N$ (generating sets are not typically unique: for example $[1]_4$ and $[3]_4$ both generate $\Bbb Z_4$).

2. $N'$ lies in $N$. This is the case we want to show does not happen.

3. $N' = N$. See (1) above.

4. None of the above. In this case, we use $L = N + N'$ to reduce this to case 2.

Interesting side-note: for some modules, proof of the EXISTENCE of $N'$ from $\mathscr{C}$ relies on Zorn's lemma-that any partially ordered set in which any linear chain (ordered by inclusion, in this case) that is bounded above has a maximal element. $M$ itself might be infinite, and conceivably have infinitely many submodules. For a nice discussion of this see here:

commutative algebra - Is every Noetherian module finitely generated? - Mathematics Stack Exchange

in particular, Arturo Magidin's answer, which indicates which of the steps requires Zorn's lemma/axiom of choice, and which strength. Recognition of the importance of the ascending chain condition for rings/modules was first grasped by the great mathematician Emma Noether, in whose honor such structures are named. One of her students was B.L. van der Waerden, whose seminal text on Algebra is justifiably considered a classic.

A word about how/why this ACC comes to be considered:

Suppose we have an integer, $k$. We can factor this integer into primes. For every factor we "divide out", we get a larger ideal that contains $(k)$. For example:

$(532) \subseteq (266) \subseteq (133) \subseteq (19)$

Note how we are substituting "divisibility" with "ideal inclusion". So when we talk about "maximal ideals", what we are really trying to find is "minimal elements" of some sort. We're looking for "basic building blocks" that we can consider, instead of having to look at the ENTIRE ring/module. It's much easier to deal with a set of generators, than an entire ring/module. If the set of generators is FINITE, it's "the best possible situation". Ideally, it would be nicer if the finite set was even simpler: a single element. So quite a bit of work goes into finding minimal generating sets, and a lot of work goes into finding (when such generating sets are not singletons) decompositions into "aggregates" (in the best-case scenario, a direct sum/product) of ideals/submodules that are generated by a single element. This let's us "filter out" a lot of complexity, and study our structures "one piece at a time".

While the integers are a fairly simple ring, with many special properties, many of the "more advanced rings" are classified by HOW MANY of these "nice properties" we keep. We're pretty good at "counting things", so it's nice to have "something reasonable to count".