Proposition 3.1.2 in B&K: Help Needed on Exact Sequences/Noetherian Rings

[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 yet another aspect of the proof of Proposition 3.1.2.

The statement and proof of Proposition 3.1.2 reads as follows (pages 109-110):https://www.physicsforums.com/attachments/4873
https://www.physicsforums.com/attachments/4874I need help with an aspect of the above proof ... indeed, I have two questions ...Question 1

In the above text from B&K write the following:

" ... ... Conversely, consider a submodule \(\displaystyle N\) of \(\displaystyle M\). Let \(\displaystyle N' = N \cap \alpha M'\) and let \(\displaystyle N''\) be the image of \(\displaystyle N\) in \(\displaystyle M''\), so that there is an exact sequence ... ... "

My question is as follows:

How do we know such an exact sequence exists ... that is, how do we demonstrate, formally and rigorously, that such a sequence exists ... ... ?Question 2

In the above text we read:

" ... ... Since both \(\displaystyle N'\) and \(\displaystyle N''\) are finitely generated, so also is \(\displaystyle N\). ... ... "

How can we demonstrate, formally and rigorously that \(\displaystyle N'\) and \(\displaystyle N''\) being finitely generated, imply that \(\displaystyle N\) is finitely generated ... ...Hope someone can help with these two questions ... ...

Peter

 

[Deveno]

$N\cap \alpha M'$ is a submodule of $N$ so we have an injective homomorphism:

$i: N' \to N$ given by $i(n') = n'$.

So let $\beta' = \beta|_N$.

Then $\beta'(i(n')) = \beta(n') = \beta(\alpha(m')) = 0_{M''}$ for some $m' \in M'$ since $N' = N \cap \alpha M' \subseteq \text{im }\alpha $, by the exactness of our original sequence.

This shows that $\text{im }i \subseteq \text{ker }\beta'$.

On the other hand, if, for $n \in N$, we have $\beta'(n) = 0_{M''}$, then since $\beta'$ is just a restriction of $\beta$, it follows that $\beta(n) = 0_{M''}$, and by the exactness of our original sequence $n \in \text{im }\alpha$.

Hence $n \in N \cap \alpha M' = \text{im }i$, so $\text{im } i = \text{ker }\beta'$.

Now $N'' = \beta'(N) = \beta(N)$ so it is immediate that $\beta':N \to N''$ is surjective and our sequence is short exact.

*******************

I'll work on question 2, later.

 

[Math Amateur]

$N\cap \alpha M'$ is a submodule of $N$ so we have an injective homomorphism:

$i: N' \to N$ given by $i(n') = n'$.

So let $\beta' = \beta|_N$.

Then $\beta'(i(n')) = \beta(n') = \beta(\alpha(m')) = 0_{M''}$ for some $m' \in M'$ since $N' = N \cap \alpha M' \subseteq \text{im }\alpha $, by the exactness of our original sequence.

This shows that $\text{im }i \subseteq \text{ker }\beta'$.

On the other hand, if, for $n \in N$, we have $\beta'(n) = 0_{M''}$, then since $\beta'$ is just a restriction of $\beta$, it follows that $\beta(n) = 0_{M''}$, and by the exactness of our original sequence $n \in \text{im }\alpha$.

Hence $n \in N \cap \alpha M' = \text{im }i$, so $\text{im } i = \text{ker }\beta'$.

Now $N'' = \beta'(N) = \beta(N)$ so it is immediate that $\beta':N \to N''$ is surjective and our sequence is short exact.

*******************

I'll work on question 2, later.

Thanks Deveno ... appreciate your help ...

Will be working through your post shortly ...

Thanks again,

Peter