MHB 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 ...

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

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

so that $$g$$ is an epimorphism ... that is $$g$$ 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 $$\phi \ : \ R \to S$$ is surjective then $$R/ \text{ Ker } \phi \cong S$$ ... so ... in our case ...
$$g$$ is a surjective homomorphism so ...

$$M/ \text{ Ker } g \cong M_2 $$

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

$$M_1 / \text{ Ker } f \cong \text{ I am } f$$

$$\Longrightarrow M_1/ \{ 0 \} \cong \text{ I am } f$$

... ... but not sure of the nature of $$M_1/ \{ 0 \}$$ 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 ...

$$f$$ is surjective on $$\text{ I am } f$$ so ...

$$M_1 / \text{ Ker } f \cong \text{ I am } f$$

$$\Longrightarrow M_1/ \{ 0 \} \cong \text{ I am } f$$

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

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

$$M/M_1 \cong M_2$$ ...
... 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 ... $$M$$ is Noetherian $$\Longrightarrow \text{ I am } f$$, being a submodule of $$M$$, is Noetherian and $$M/ \text{ I am } f$$ is Noetherian (Proposition 4.2.5)


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

Peter

 

[steenis]

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