Theorem 3.1.4 - Berrick and Keating - Noetherian Rings and Modules

[Math Amateur]

I am reading the book "An Introduction to Rings and Modules with K-theory in View" by A.J. Berrick and M.E. Keating ... ...

I am currently focused on Chapter 3; Noetherian Rings and Polynomial Rings.

I need help with the proof of Theorem 3.1.4.

A brief explanation of the proof precedes the statement of Theorem 3.1.4 In Berrick and Keating and reads as follows (page 110):https://www.physicsforums.com/attachments/4878
I do not follow the authors brief explanation of the proof ... which is given in a few lines above the statement of the Theorem ... ...

Can someone please give a more explicit, clear and easy to follow explanation?

Hope someone can help as at present the explanation of B&K means little to me ...

Peter
***EDIT 1***

The explanation of the proof refers to Lemma 2.5.7 whch reads as follows:View attachment 4879***EDIT 2***

The explanation of the proof also refers to "the last two results" ... ... that is Proposition 3.1.2 and Corollary 3.1.3 ... these two results read as follows ... ... View attachment 4880
View attachment 4881View attachment 4882

 

[steenis]

Proof of Theorem 3.1.4 in “Berrick - An Introduction to Rings and Modules; With K-theory in View (2000)”. In the proof, we use Lemma 2.5.7, Prop. 3.1.2, and Corol. 3.1.3:
Lemma 2.5.7, p.92
$M$ is a fingen (finitely generated) R-module with n generators if and only if there is a surjective homomorphism from $R^{(n)}$ to $M$.

(((Note that $R^{(n)}$ is the direct sum, where $R^n$ is the direct product of $n$ copies of $R$.)))

Prop. 3.1.2, p.109/110
Let

$0 \longrightarrow M’ \longrightarrow^ \alpha M \longrightarrow^ \beta M” \longrightarrow 0$

be a SES of right $R$-modules $M’$, $M$, and $M”$.
Then $M$ is noetherian if and only if both $M’$ and $M”$ are noetherian.

Corol. 3.1.3, p.110
A free right module of finite rank over a right noetherian ring must be noetherian.
So if $R$ is a right noetherian ring (this means that $R$ is noetherian as a right $R$-module) then for each integer $n \geq 1$, the free right $R$-module $R^{(n)}$ must be noetherian.

Theorem 3.1.4, p.110:
Suppose that $R$ is a right noetherian ring and that $M$ is a fingen right R-module. Then $M$ is noetherian.
Proof:
Given: $R$ is a right Noetherian ring and $M$ is a fingen right R-module.
By lemma 2.5.7 there is an integer $n \geq 1$ such that $M$ is the homomorphic image of $R^{(n)}$. Let $p: R^{(n)} \longrightarrow M$ be this homomorphism and $p$ is surjective. This means that

$0 \longrightarrow ker(p) \longrightarrow^i R^{(n)} \longrightarrow^p M \longrightarrow 0$

is a SES
By Corol. 3.1.3 $R^{(n)}$ is noetherian, and $M$ is noetherian by Prop. 3.1.2.

 

[Math Amateur]

Proof of Theorem 3.1.4 in “Berrick - An Introduction to Rings and Modules; With K-theory in View (2000)”. In the proof, we use Lemma 2.5.7, Prop. 3.1.2, and Corol. 3.1.3:
Lemma 2.5.7, p.92
$M$ is a fingen (finitely generated) R-module with n generators if and only if there is a surjective homomorphism from $R^{(n)}$ to $M$.

(((Note that $R^{(n)}$ is the direct sum, where $R^n$ is the direct product of $n$ copies of $R$.)))

Prop. 3.1.2, p.109/110
Let

$0 \longrightarrow M’ \longrightarrow^ \alpha M \longrightarrow^ \beta M” \longrightarrow 0$

be a SES of right $R$-modules $M’$, $M$, and $M”$.
Then $M$ is noetherian if and only if both $M’$ and $M”$ are noetherian.

Corol. 3.1.3, p.110
A free right module of finite rank over a right noetherian ring must be noetherian.
So if $R$ is a right noetherian ring (this means that $R$ is noetherian as a right $R$-module) then for each integer $n \geq 1$, the free right $R$-module $R^{(n)}$ must be noetherian.

Theorem 3.1.4, p.110:
Suppose that $R$ is a right noetherian ring and that $M$ is a fingen right R-module. Then $M$ is noetherian.
Proof:
Given: $R$ is a right Noetherian ring and $M$ is a fingen right R-module.
By lemma 2.5.7 there is an integer $n \geq 1$ such that $M$ is the homomorphic image of $R^{(n)}$. Let $p: R^{(n)} \longrightarrow M$ be this homomorphism and $p$ is surjective. This means that

$0 \longrightarrow ker(p) \longrightarrow^i R^{(n)} \longrightarrow^p M \longrightarrow 0$

is a SES
By Corol. 3.1.3 $R^{(n)}$ is noetherian, and $M$ is noetherian by Prop. 3.1.2.

Many thanks Steenis ...

Working through your proof now ...

Peter