Submodules and Factor Modules of a Noetherian Module ....

[Math Amateur]

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

I am focused on Section 4.2: Noetherian and Artinian Modules and need some help to fully understand the proof of part of Proposition 4.2.5 ... ...

Proposition 4.2.5 reads as follows:

My questions are as follows:Question 1

In the above text from Bland we read the following:

" ... ... Conversely, suppose that $N$ and $M/N$ are noetherian. Let$M_1 \subseteq M_2 \subseteq M_3 \subseteq \ ... \ ...$be an ascending chain of submodules of $M$. Then $M_1 \cap N \subseteq M_2 \cap N \subseteq M_3 \cap N \subseteq \ ... \ ...$ ... ... "My question is ... what about the case where all the $M_i$ fail to intersect with N ... is this possible? ... if so how does the proof read then ...?
Question 2

In the above text from Bland we read the following:

" ... ... If $i \ge n$ and $x \in M_i$ then $x + N \in (M_i + N)/N = (M_n + N)/N$ ... ... "My question is ... why does $x \in M_i \Longrightarrow x + N \in (M_i + N)/N$ ... ... is it because ...

$x \in M_i \Longrightarrow x + 0_N + N \in (M_i + N)/N$ ...

... and $x + 0_N + N = x + N$ ... ... ?
Hope someone can help ...

Peter

 

[member 587159]

Question 1: Don't all modules on the same set share the null element? Not that it matters for the proof.

Question 2: That's correct. Without reading the proof (so I might be incorrect!), I guess they do that because $N$ is not necessarily contained in $M_i$, so writing the quotient $M_i/N$ makes no sense. Therefore, since clearly $N \subseteq M_i + N$, we can write the quotient module $(M_i + N)/N$

 

[Math Amateur]

Thanks Math_QED ...

Appreciate your help ...

Peter

 

[mathwonk]

[this post has been edited to remove errors pointed out below]:
a basic piece of advice is to somewhat ignore the actual proof give and try to give your own. in this (noetherian) case: if N is a submodule of M and both N and M/N are noetherian, show M is noetherian also. that is trivial (I hope): i.e. let K be a submodule of M, we must show K is finitely generated. consider the image of K in M/N, which we know is finitely generated, say by [a1],...,[ar], where a1,...ar, are elements of M. Next consider also KintersectN which is also finitely generated, say by b1,...,bs. Then we hope that K is generated by a1,...,ar, b1,...,bs. What else do we have to work with? I leave this as an exercise to the OP. hint if x is any element of K, then map x into M/N and express the image as a linear combination of [a1],...,[ar]. then we know the difference of x and the same linear combination of the aj must map to zero in M/N (why?). hence the diffeence of x and that linear combination is an element of N (why?), hence can be written as a linear combination of the bi. done. write it up.

 

[mathwonk]

just for fun let's see if the version given in bland is equally easy. take an increasing sequence of submodules of M. the sequence of their intersections with N must terminate. so what ? also the sequence of their images in M/N must terminate, so what? I guess we should show that a submodule of M is determined by its image in M/N and its intersection with N. can we do this? probably but it seems harder than the easy proof we just gave. so what, we try anyway. ok suppose K is strictly contained in L and L is contained in M. then suppose also that L has the same image in M/N as does K and also that L has the same intersection with N as does K. can we get a contradiction? i think so. let x be an element of L that is outside K. now map x into M/N with image [x], and assume that [x] = [y] with y in K. Then by definition of M/N we have x-y = z belongs to N. hence x = y+z and hence x belongs to K, since y does and z belongs to the submodule N. contradiction. but this is not as intuitively easy a proof in my opinion. what do you think?

 

[steenis]

a basic piece of advice is to somewhat ignore the actual proof give and try to give your own. in this (noetherian) case: if N is a submodule of M and both N and M/N are noetherian, show M is noetherian also. that is trivial (I hope): i.e. let K be a submodule of M, we must show K is finitely generated. consider the image of K in M/N, which we know is finitely generated, say by [a1],...,[ar], where a1,...ar, are elements of M. Next consider also KintersectN which is also finitely generated, say by b1,...,bs. Then we hope that M is generated by a1,...,ar, b1,...,bs. What else do we have to work with? I leave this as an exercise to the OP. hint if x is any element of M, then map x into M/N and express the image as a linear combination of [a1],...,[ar]. then we know the difference of x and the same linear combination of the aj must map to zero in M/N (why?). hence the diffeence of x and that linear combination is an element of N (why?), hence can be written as a linear combination of the bi. done. write it up.

You want to prove that $K$ is fingen, that I can follow: if every submodule of $M$ is fingen, then $M$ is noetherian. Then you say something about $K \cap N$ and the image of $K$ in $M/N$. And then you switch to prove that $M$ is fingen. You lost me there. Where is the proof that $K$ is fingen, or am I missing something ?

EDIT: you meant K instead of M, then I can follow your proof. It can be done because Bland proves first that "M is noetherian iff every submodule of M is fingen" and then he proves prop.4.2.5. Other authors change this order. The conditions are equivalent.
What can I say? There are many roads that lead to Rome ...

 

[Math Amateur]

a basic piece of advice is to somewhat ignore the actual proof give and try to give your own. in this (noetherian) case: if N is a submodule of M and both N and M/N are noetherian, show M is noetherian also. that is trivial (I hope): i.e. let K be a submodule of M, we must show K is finitely generated. consider the image of K in M/N, which we know is finitely generated, say by [a1],...,[ar], where a1,...ar, are elements of M. Next consider also KintersectN which is also finitely generated, say by b1,...,bs. Then we hope that M is generated by a1,...,ar, b1,...,bs. What else do we have to work with? I leave this as an exercise to the OP. hint if x is any element of M, then map x into M/N and express the image as a linear combination of [a1],...,[ar]. then we know the difference of x and the same linear combination of the aj must map to zero in M/N (why?). hence the diffeence of x and that linear combination is an element of N (why?), hence can be written as a linear combination of the bi. done. write it up.

Hi mathwonk ... Like steenis I am struggling somewhat when you write:

" ... ... Then we hope that M is generated by a1,...,ar, b1,...,bs. ... ... "

Indeed ... did you mean " ... ... Then we hope that K is generated by a1,...,ar, b1,...,bs. ... ... "

Peter

 

[mathwonk]

yes! that is what i meant. hooray! , now i think you are understanding the proof. this is to me an example of how you learn more from doing it yourself, in this case correcting my errors, than just reading a proof. i have gone back now and tried to correct the errors of mine that you guys pointed out. both of you are practicing excellent participatory reading, i.e. asking yourself what must have been meant as opposed to just reading passively. this is great. you will learn a lot by always reading this way.

 

[steenis]

I leave posting the proof to Math Amateur, it is a good exercise for him.

 

[steenis]

[this post has been edited to remove errors pointed out below]:
a basic piece of advice is to somewhat ignore the actual proof give and try to give your own. in this (noetherian) case: if N is a submodule of M and both N and M/N are noetherian, show M is noetherian also. that is trivial (I hope): i.e. let K be a submodule of M, we must show K is finitely generated. consider the image of K in M/N, which we know is finitely generated, say by [a1],...,[ar], where a1,...ar, are elements of M. Next consider also KintersectN which is also finitely generated, say by b1,...,bs. Then we hope that K is generated by a1,...,ar, b1,...,bs. What else do we have to work with? I leave this as an exercise to the OP. hint if x is any element of K, then map x into M/N and express the image as a linear combination of [a1],...,[ar]. then we know the difference of x and the same linear combination of the aj must map to zero in M/N (why?). hence the diffeence of x and that linear combination is an element of N (why?), hence can be written as a linear combination of the bi. done. write it up.

What is the relevance of $K \cap N$ ? Or do you mean $N$ instead ?

 

[Math Amateur]

yes! that is what i meant. hooray! , now i think you are understanding the proof. this is to me an example of how you learn more from doing it yourself, in this case correcting my errors, than just reading a proof. i have gone back now and tried to correct the errors of mine that you guys pointed out. both of you are practicing excellent participatory reading, i.e. asking yourself what must have been meant as opposed to just reading passively. this is great. you will learn a lot by always reading this way.

Been a couple of sticking points ... especially showing that $x - \sum a_i r_i \in K$...

Anyway ... my proof ... as developed so far ... is as follows: (Thanks to Steenis for discussion, hints and help ...)Problem/Exercise

$M$ is an R-module.
$N$ is a submodule of $M$.
$N$ and $M/N$ are Noetherian

Show that $M$ is Noetherian ...

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

We note that if a R-module $L$ is Noetherian then every submodule of $L$ (including, of course, $L$ itself) is finitely generated ( 'fingen' ) ...

The proof focuses on $K \cap N$ ... hint by mathwonk ...Let $K$ be a submodule of $M$ ... must show $K$ is fingen ...Consider $x \in K$ ... then $\overline{x} \in M/N$ ...

But ... $M/N$ is fingen ... so $\overline{x} = \sum \overline{ a_i } r_i$ where $\overline{ a_i } \in M/N$ so that $a_i \in M$ ...

Thus $\overline{x} - \sum \overline{ a_i } r_i = \overline{ 0 }$

Hence $x - \sum a_i r_i \in N$ ... ... ... ... ... (1)But it also follows that $x \in K$ ...

... implies $\overline{x} \in K/N$ ... and so $\sum \overline{a_i} r_i \in K/N$

$\Longrightarrow \sum a_i r_i \in K$ ... ...

$\Longrightarrow x - \sum a_i r_i \in K$ since $K$ is submodule ...

$\Longrightarrow x - \sum a_i r_i \in K \cap N$ ... ... see (1) above ...Now ... $K \cap N$ is fingen because it is a submodule of $N$ which is Noetherian ...

So ... $x - \sum a_i r_i = \sum b_i r'_i$ ...

so ... $x = \sum a_i r_i + \sum b_i r'_i$ ...

that is $K$ is finitely generated by $a_1, a_2, \ ... \ ... \ , a_r, b_1, b_2, \ ... \ ... \ , b_s$ ... ...

so ... $M$ is Noetherian ...But ... unsure of my demonstration that $x - \sum a_i r_i \in K$ ...'

Peter

 

[Math Amateur]

There is an obvious error in my post above in the demonstration that $x - \sum a_i r_i \in K$ ... the argument involves $K/N$ but one cannot form $K/N$ unless $N \subset K$ (indeed unless $N$ is a submodule of $K$!) ... hmmm .. too desperate for a result! ...

Steenis pointed this error out ... but it should have been obvious to me ... apologies ...

Peter

 

[steenis]

I made this mistake in the first place, I should have known better ...

 

[Math Amateur]

Been a couple of sticking points ... especially showing that $x - \sum a_i r_i \in K$...

Anyway ... my proof ... as developed so far ... is as follows: (Thanks to Steenis for discussion, hints and help ...)Problem/Exercise

$M$ is an R-module.
$N$ is a submodule of $M$.
$N$ and $M/N$ are Noetherian

Show that $M$ is Noetherian ...

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

We note that if a R-module $L$ is Noetherian then every submodule of $L$ (including, of course, $L$ itself) is finitely generated ( 'fingen' ) ...

The proof focuses on $K \cap N$ ... hint by mathwonk ...Let $K$ be a submodule of $M$ ... must show $K$ is fingen ...Consider $x \in K$ ... then $\overline{x} \in M/N$ ...

But ... $M/N$ is fingen ... so $\overline{x} = \sum \overline{ a_i } r_i$ where $\overline{ a_i } \in M/N$ so that $a_i \in M$ ...

Thus $\overline{x} - \sum \overline{ a_i } r_i = \overline{ 0 }$

Hence $x - \sum a_i r_i \in N$ ... ... ... ... ... (1)But it also follows that $x \in K$ ...

... implies $\overline{x} \in K/N$ ... and so $\sum \overline{a_i} r_i \in K/N$

$\Longrightarrow \sum a_i r_i \in K$ ... ...

$\Longrightarrow x - \sum a_i r_i \in K$ since $K$ is submodule ...

$\Longrightarrow x - \sum a_i r_i \in K \cap N$ ... ... see (1) above ...Now ... $K \cap N$ is fingen because it is a submodule of $N$ which is Noetherian ...

So ... $x - \sum a_i r_i = \sum b_i r'_i$ ...

so ... $x = \sum a_i r_i + \sum b_i r'_i$ ...

that is $K$ is finitely generated by $a_1, a_2, \ ... \ ... \ , a_r, b_1, b_2, \ ... \ ... \ , b_s$ ... ...

so ... $M$ is Noetherian ...But ... unsure of my demonstration that $x - \sum a_i r_i \in K$ ...'

Peter

To finish Problem/Exercise ...

Steenis has pointed out that a focus on N is the way to go ... ...

Indeed once one has established that $x - \sum a_i r_i \in N$ ... the solution is immediate ...

We note, first that N is Noetherian ... and hence fingen ...

... so ... $N = \langle b_1, \cdots, b_s \rangle$, say ...Then ... $x - \sum a_i r_i \in N$ ... established in Post #11 above ... see (1) ...

$\Longrightarrow x - \Sigma a_i r_i = \Sigma b_j s_j$

$\Longrightarrow x = \Sigma a_i r_i + \Sigma b_j s_j$

$\Longrightarrow K = \langle a_1, \dots, a_r, b_1, \cdots, b_s \rangle$

$\Longrightarrow K$ is fingen

$\Longrightarrow M$ is Noetherian ...Peter