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

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In summary: So we can say that if $M$ is noetherian, then $M_1$ and $M_2$ are also noetherian. In summary, Corollary 4.2.6 states that if $M$ is a Noetherian module and $0 \rightarrow M_1 \overset{f}{\rightarrow} M \overset{g}{\rightarrow} M_2 \rightarrow 0$ is an exact sequence of $R$-modules, then $M_1$ and $M_2$ are also Noetherian. This can be proven using Proposition 4.2.5, which states that if $M$ is Noetherian then any submodule and quotient
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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

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

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).

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

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

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.

steenis said:
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

$M_1/\{0\}=M_1$

$M/im f \cong \cdots$

MHB is slow, very slow ...

Last edited:
steenis said:
$M_1/\{0\}=M_1$

$M/im f = \cdots$

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

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 ...

steenis said:
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

Last edited:
That is correct, the converse goes in the same way.

## 1. What is a Noetherian module?

A Noetherian module is a module over a ring that satisfies the ascending chain condition, which means that any increasing chain of submodules eventually stabilizes.

## 2. How is a Noetherian module related to a short exact sequence?

A Noetherian module can be used to construct a short exact sequence, where the first and third terms are Noetherian modules and the second term is a submodule of the first term.

## 3. What is Corollary 4.2.6 of Bland's theorem on Noetherian modules?

Corollary 4.2.6 states that if a module is Noetherian, then any submodule and quotient module are also Noetherian.

## 4. How does Bland's theorem apply to short exact sequences?

Bland's theorem states that a short exact sequence involving Noetherian modules remains exact when any of the modules is replaced by a submodule or quotient module. This allows for the construction of new short exact sequences using Noetherian modules.

## 5. What are some applications of Noetherian modules and short exact sequences in mathematics?

Noetherian modules and short exact sequences have applications in various areas of mathematics, such as commutative algebra, algebraic geometry, and representation theory. They are also used in the study of algebraic number theory and algebraic topology.

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