Noetherian Modules and Short Exact Sequences .... Bland, Corollary 4.2.6 ....

[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 Corollary 4.2.6 ... ...

Corollary 4.2.6 reads as follows:View attachment 8193
Bland gives a statement of Corollary 4.2.6 but does not offer a proof ... ...

Can someone please explain how to prove Corollary 4.2.6 ...Since Corollary 4.2.6 is a corollary to Proposition 4.2.5 I am providing the text of Proposition 4.2.5 and its proof ... as follows ... ... https://www.physicsforums.com/attachments/8194
https://www.physicsforums.com/attachments/8195

 

[steenis]

$$0\rightarrow M_1\overset{f}{\rightarrow}M\overset{g}{\rightarrow}M_2\rightarrow 0$$

is an exact sequence of R-modules, what can you say about the R-maps $f$ and $g$, and the R-modules $M_1$, $M$, and $M_2$ ? (see page 83).

 

[Math Amateur]

$$0\rightarrow M_1\overset{f}{\rightarrow}M\overset{g}{\rightarrow}M_2\rightarrow 0$$

is an exact sequence of R-modules, what can you say about the R-maps $f$ and $g$, and the R-modules $M_1$, $M$, and $M_2$ ? (see page 83).

Thanks steenis ...

I think we can say that ...

\(\displaystyle \text{ I am } f = \text{ Ker } g \text{ at } M\) ...and that \(\displaystyle \text{ I am } 0 = \{ 0 \} = \text{ Ker } f\) at \(\displaystyle M_1\)

so that \(\displaystyle f\) is a monomorphism ... that is \(\displaystyle f\) is injective ...and also \(\displaystyle \text{ I am } g = \text{ Ker } 0 = M_2\) at \(\displaystyle M_2\) ... ...

so that \(\displaystyle g\) is an epimorphism ... that is \(\displaystyle g\) is surjective ...Is that correct?

Peter

 

[steenis]

Yes, that is correct.

What can you say about $M_1$, $M$, and $M_2$. knowing that $g$ is surjective and $f$ is injective, use the first isomorphism theorem.

 

[Math Amateur]

Yes, that is correct.

What can you say about $M_1$, $M$, and $M_2$. knowing that $g$ is surjective and $f$ is injective, use the first isomorphism theorem.

If the module homomorphism \(\displaystyle \phi \ : \ R \to S\) is surjective then \(\displaystyle R/ \text{ Ker } \phi \cong S\) ... so ... in our case ...
\(\displaystyle g\) is a surjective homomorphism so ...

\(\displaystyle M/ \text{ Ker } g \cong M_2 \)

\(\displaystyle \Longrightarrow M/ \text{ I am } f \cong M_2\)
\(\displaystyle f\) is surjective on \(\displaystyle \text{ I am } f\) so ...

\(\displaystyle M_1 / \text{ Ker } f \cong \text{ I am } f\)

\(\displaystyle \Longrightarrow M_1/ \{ 0 \} \cong \text{ I am } f\)

... ... but not sure of the nature of \(\displaystyle M_1/ \{ 0 \}\) and so not sure how to simplify it ...
Hope that is correct ...

Peter

 

[steenis]

$M_1/\{0\}=M_1$

$M/im f \cong \cdots$

MHB is slow, very slow ...

 

[Math Amateur]

$M_1/\{0\}=M_1$

$M/im f = \cdots$

MHB is slow, very slow ...

Thanks steenis ...

Then we have ...

\(\displaystyle f\) is surjective on \(\displaystyle \text{ I am } f\) so ...

\(\displaystyle M_1 / \text{ Ker } f \cong \text{ I am } f\)

\(\displaystyle \Longrightarrow M_1/ \{ 0 \} \cong \text{ I am } f\)

\(\displaystyle \Longrightarrow M_1 \cong \text{ I am } f\) ... ... but now ... can we conclude ...

\(\displaystyle M/ \text{ I am } f = M/M_1\) ... ... but this assumes \(\displaystyle M_1 = \text{ I am } f \) ... and we only have \(\displaystyle M_1 \cong \text{ I am } f\) ...Is it OK to conclude \(\displaystyle M/ \text{ I am } f = M/M_1\) ... ... ?Note that if we can conclude \(\displaystyle M/ \text{ I am } f = M/M_1\) then we can say that ...

\(\displaystyle M/M_1 \cong M_2\) ...
... can you help and clarify the above thinking ...

Peter

 

[steenis]

Actually $M/ \text{ I am } f \cong M/M_1$

Now you have enough to apply proposition 4.2.5. Can you do that?

Suppose $M$ is noetherian, we have $im f \leq M$ then ...

 

[Math Amateur]

Actually $M/ \text{ I am } f \cong M/M_1$

Now you have enough to apply proposition 4.2.5. Can you do that?

Suppose $M$ is noetherian, we have $im f \leq M$ then ...

I think the argument to apply Proposition 4.2.5 so as to prove Corollary 4.2.6 from here would be something like the following ... \(\displaystyle M\) is Noetherian \(\displaystyle \Longrightarrow \text{ I am } f\), being a submodule of \(\displaystyle M\), is Noetherian and \(\displaystyle M/ \text{ I am } f\) is Noetherian (Proposition 4.2.5)


\(\displaystyle \Longrightarrow\) \(\displaystyle M_1\) is Noetherian and \(\displaystyle M/M_1\) is Noetherian since \(\displaystyle M_1 \cong \text{ I am } f\)\(\displaystyle \Longrightarrow M_1\) is Noetherian and \(\displaystyle M_2\) is Noetherian since \(\displaystyle M/M_1 \cong M_2\) ...
Is that correct ... ... ?

Peter

 

[steenis]

That is correct, the converse goes in the same way.