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 $$N$$ of $$M$$. Let $$N' = N \cap \alpha M'$$ and let $$N''$$ be the image of $$N$$ in $$M''$$, 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 $$N'$$ and $$N''$$ are finitely generated, so also is $$N$$. ... ... "

How can we demonstrate, formally and rigorously that $$N'$$ and $$N''$$ being finitely generated, imply that $$N$$ 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