How Does Corollary 4.2.6 Prove M is Noetherian?

[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 $$(1) \Longrightarrow (2)$$ ...

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

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

for some homomorphism $$f$$ ... ... is that correct?... now ...

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

... BUT ...

... how exactly do we use or employ Corollary 4.2.6 to show that $$f ( R^{ (n) } ) = M \Longrightarrow M$$ is noetherian ...What would $$M_1$$ and $$M_2$$ 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 $$f (( r_i )) = \sum_{ i = 1 }^n x_i r_i$$ Hence $$\text{ Ker } f = (0) = (0, 0, \ ... \ ... \ , 0)$$ ... $$n$$ elements ... ... ... but is this correct?

Now apply First Isomorphism Theorem ...

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

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

Also ... not sure how to justify that $$\text{ Ker } f = (0) = (0, 0, \ ... \ ... \ , 0)$$ ... 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:$$R^{(n)}$$ is noetherian $$\Longrightarrow R^{(n)} / N$$ is noetherian ... ... ... ... ... (2)Now we have that (1) (2) $$\Longrightarrow M$$ is noetherian ... Is that correct ... ?

... BUT as an aside ... Bland suggests we could use Corollary 4.2.6 to show $$M$$ 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 $$\text{ I am } \psi = \text{ Ker } \phi$$ But ... if we use $$M_1 = R^{(n)}$$ ...... we then have $$\text{ Ker } \psi = N$$ (see above) and $$\text{ I am } 0 = 0$$ ... so sequence is not exact at $$M_1$$ ... ?
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 $$\text{ I am } 0 = 0 = \text{ Ker } i$$ ...

... and $$\text{ I am } i = \text{ Ker } f $$ ...

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