Right Noetherian Rings and Noetherian Modules .... Bland, Proposition 4.2.11 .... ....

[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 4.2.11 ... ...

Proposition 4.2.11 reads as follows:View attachment 8217I need help with the Proof of \(\displaystyle (1) \Longrightarrow (2)\) ...

I am struggling with this proof so forgive me if my questions are possibly formulated badly ...Now in Bland's proof ... given that \(\displaystyle M\) is finitely generated we have by Proposition 2.2.6 that ...

\(\displaystyle \exists \ f\) such that \(\displaystyle f ( R^{ (n) } ) = M\)

for some homomorphism \(\displaystyle f\) ... ... is that correct?... now ...

Bland argues that Corollary 4.2.6 shows that because \(\displaystyle f ( R^{ (n) } ) = M\) then we have that \(\displaystyle M\) is noetherian ... ...

... BUT ...

... how exactly do we use or employ Corollary 4.2.6 to show that \(\displaystyle f ( R^{ (n) } ) = M \Longrightarrow M\) is noetherian ...What would \(\displaystyle M_1\) and \(\displaystyle M_2\) be in this case ... ?
Hope someone can help ...

Peter

=======================================================================***NOTE***

The above post refers to Proposition 2.2.6 and also to Corollary 4.2.6 ... so I am providing the text of each ... as follows:
View attachment 8218
View attachment 8219
View attachment 8220
Hope access to the above text helps ... ...

Peter

 

[steenis]

It is correct that there is an epimorphism $f:R^{(n)} \longrightarrow M$

Now apply the first isomorphism theorem. and then cor.4.2.6. (or prop.4.2.5)

 

[Math Amateur]

It is correct that there is an epimorphism $f:R^{(n)} \longrightarrow M$

Now apply the first isomorphism theorem. and then cor.4.2.6. (or prop.4.2.5)

Thanks steenis ... but not sure if I follow ..

... but will try ... as follows ...We have an epimorphism $f:R^{(n)} \longrightarrow M$

where \(\displaystyle f (( r_i )) = \sum_{ i = 1 }^n x_i r_i\) Hence \(\displaystyle \text{ Ker } f = (0) = (0, 0, \ ... \ ... \ , 0)\) ... \(\displaystyle n\) elements ... ... ... but is this correct?

Now apply First Isomorphism Theorem ...

\(\displaystyle R^{(n)} / (0) = R^{(n)} = \cong M\)

So ... given \(\displaystyle R^{(n)}\) is noetherian ... then so is \(\displaystyle M\) ...... hmmm ... doubt whether my argument is correct ... especially since I did not use Corollary 4.2.6 ...

Also ... not sure how to justify that \(\displaystyle \text{ Ker } f = (0) = (0, 0, \ ... \ ... \ , 0)\) ... n elements ...
Can you help further ...

Peter

 

[steenis]

No, $\text{ker } f = 0$ iff $f$ is a monomorphism.
In this case $f$ is an epimorphism, and $\text{ker } f$ need not be $0$,

Say $N = \text{ker } f$ then $N$ is a submodule of $R^{(n)}$ and $R^{(n)} / N \cong M$.

 

[Math Amateur]

No, $\text{ker } f = 0$ iff $f$ is a monomorphism.
In this case $f$ is an epimorphism, and $\text{ker } f$ need not be $0$,

Say $N = \text{ker } f$ then $N$ is a submodule of $R^{(n)}$ and $R^{(n)} / N \cong M$.

Thanks Steenis ...

You have shown that $R^{(n)} / N \cong M$ where $N = \text{ Ker } f$ ... ... ... ... ... (1)... and we have by Proposition 4.2.5 that:\(\displaystyle R^{(n)}\) is noetherian \(\displaystyle \Longrightarrow R^{(n)} / N\) is noetherian ... ... ... ... ... (2)Now we have that (1) (2) \(\displaystyle \Longrightarrow M\) is noetherian ... Is that correct ... ?

... BUT as an aside ... Bland suggests we could use Corollary 4.2.6 to show \(\displaystyle M\) is noetherian ...But how do we do this ... ... ?We would need to find a short exact sequence $$0\rightarrow M_1\overset{ \psi }{\rightarrow}M\overset{ \phi }{\rightarrow}M_2\rightarrow 0$$where \(\displaystyle \text{ I am } \psi = \text{ Ker } \phi\) But ... if we use \(\displaystyle M_1 = R^{(n)}\) ...... we then have \(\displaystyle \text{ Ker } \psi = N\) (see above) and \(\displaystyle \text{ I am } 0 = 0\) ... so sequence is not exact at \(\displaystyle M_1\) ... ?
How do we proceed ... ?Can you help in this matter ...

Peter

 

[steenis]

We have an $R$-epimorphism $f:R^{(n)} \rightarrow M$.

$R^{(n)}$ is noetherian.

Then $\text{ker } f \leq R^{(n)}$ and $R^{(n)} / \text{ker } f$ are noetherian by prop.4.2.5.

Thus $M \cong R^{(n)} / \text{ker } f$ is noetherian.

To prove that $M$ is noetherian using cor.4.2.6. you do not need $\text{ker } f$ (my mistake, sorry).

You can prove easily that this series is exact:

$0\rightarrow \text{ker } f \overset{i}{ \rightarrow}R^{(n)} \overset{f}{ \rightarrow}M \rightarrow 0$

$R^{(n)}$ is noetherian, mow you can apply cor.4.2.6.

 

[Math Amateur]

We have an $R$-epimorphism $f:R^{(n)} \rightarrow M$.

$R^{(n)}$ is noetherian.

Then $\text{ker } f \leq R^{(n)}$ and $R^{(n)} / \text{ker } f$ are noetherian by prop.4.2.5.

Thus $M \cong R^{(n)} / \text{ker } f$ is noetherian.

To prove that $M$ is noetherian using cor.4.2.6. you do not need $\text{ker } f$ (my mistake, sorry).

You can prove easily that this series is exact:

$0\rightarrow \text{ker } f \overset{i}{ \rightarrow}R^{(n)} \overset{f}{ \rightarrow}M \rightarrow 0$

$R^{(n)}$ is noetherian, mow you can apply cor.4.2.6.

Thanks steenis ... most helpful ...

Can see that the short exact sequence

$0\rightarrow \text{ker } f \overset{i}{ \rightarrow}R^{(n)} \overset{f}{ \rightarrow}M \rightarrow 0$

will work ... why didn't I see that ... ... :( ... ...

We have \(\displaystyle \text{ I am } 0 = 0 = \text{ Ker } i\) ...

... and \(\displaystyle \text{ I am } i = \text{ Ker } f \) ...

... and \(\displaystyle \text{ I am } f = M = \text{ Ker } 0' \) ... where \(\displaystyle 0' : M \to 0\) and \(\displaystyle 0' (x) = 0\) for all \(\displaystyle x \in M\) ... ... Then since \(\displaystyle R^{(n)}\) is noetherian then by Corollary 4.2.6 \(\displaystyle \text{ Ker } f\) and \(\displaystyle M\) are noetherian ...Thanks steenis for all the help!

Peter

 

[steenis]

sorry, wrong answer

 

[Math Amateur]

sorry, wrong answer

Sorry Steenis ... I don't understand you ...

Can you give me a hint as to what is wrong ...?

Peter

 

[steenis]

O, I am sorry. not you gave a wrong answer, I did gave a wrong answer and I do not know how to delete a submitted post in which I wrote rubbish.

Your answer in post #7 is correct.

 

[Math Amateur]

O, I am sorry. not you gave a wrong answer, I did gave a wrong answer and I do not know how to delete a submitted post in which I wrote rubbish.

Your answer in post #7 is correct.

Thanks steenis ...

No worries at all ...

Thanks for all your help ...

Peter