Finitely Generated Modules and Their Submodules .... Berrick and Keating ....

[Math Amateur]

I am reading An Introduction to Rings and Modules With K-Theory in View by A.J. Berrick and M.E. Keating (B&K).

I need help with the proof of Lemma 1.2.21 ...

Lemma 1.2.21 and its proof reads as follows:
View attachment 6037Question 1In the above text by Berrick and Keating, we read the following:"... ... Since \(\displaystyle M\) is finitely generated, there is a minimal subset \(\displaystyle \{ x_0, \ ... \ ... \ , x_s \}\) of \(\displaystyle M\) such that

\(\displaystyle x_0 R + \ ... \ ... \ , x_s R + L = M. \ ... \ ... \ ... \)" My problem is as follows:

I cannot see exactly why there exists a minimal subset \(\displaystyle \{ x_0, \ ... \ ... \ , x_s \}\) of \(\displaystyle M\) such that

\(\displaystyle x_0 R + \ ... \ ... \ , x_s R + L = M\). ... ... ... Can someone please demonstrate, rigorously and formally, that there exists a minimal subset \(\displaystyle \{ x_0, \ ... \ ... \ , x_s \}\) of \(\displaystyle M\) such that

\(\displaystyle x_0 R + \ ... \ ... \ , x_s R + L = M\)?

Question 2In the above text by Berrick and Keating, we read the following:"... ... Let \(\displaystyle S\) be the set of submodules \(\displaystyle X\) of \(\displaystyle M\) that contain \(\displaystyle x_1 R + \ ... \ ... \ , x_s R + L\) but do not contain \(\displaystyle x_0\). It is obvious that \(\displaystyle S\) is inductive ... ..." Can someone please explain exactly why \(\displaystyle S\) is inductive ... ... ?Hope someone can help ...

Peter========================================================================B&K's definition of "inductive" is contained in section 1.2.18 ... ... . which reads as follows:https://www.physicsforums.com/attachments/6038
View attachment 6039

 

[caffeinemachine]

I am reading An Introduction to Rings and Modules With K-Theory in View by A.J. Berrick and M.E. Keating (B&K).

I need help with the proof of Lemma 1.2.21 ...

Lemma 1.2.21 and its proof reads as follows:
Question 1In the above text by Berrick and Keating, we read the following:"... ... Since \(\displaystyle M\) is finitely generated, there is a minimal subset \(\displaystyle \{ x_0, \ ... \ ... \ , x_s \}\) of \(\displaystyle M\) such that

\(\displaystyle x_0 R + \ ... \ ... \ , x_s R + L = M. \ ... \ ... \ ... \)" My problem is as follows:

I cannot see exactly why there exists a minimal subset \(\displaystyle \{ x_0, \ ... \ ... \ , x_s \}\) of \(\displaystyle M\) such that

\(\displaystyle x_0 R + \ ... \ ... \ , x_s R + L = M\). ... ... ... Can someone please demonstrate, rigorously and formally, that there exists a minimal subset \(\displaystyle \{ x_0, \ ... \ ... \ , x_s \}\) of \(\displaystyle M\) such that

\(\displaystyle x_0 R + \ ... \ ... \ , x_s R + L = M\)?

Question 2In the above text by Berrick and Keating, we read the following:"... ... Let \(\displaystyle S\) be the set of submodules \(\displaystyle X\) of \(\displaystyle M\) that contain \(\displaystyle x_1 R + \ ... \ ... \ , x_s R + L\) but do not contain \(\displaystyle x_0\). It is obvious that \(\displaystyle S\) is inductive ... ..." Can someone please explain exactly why \(\displaystyle S\) is inductive ... ... ?Hope someone can help ...

Peter========================================================================B&K's definition of "inductive" is contained in section 1.2.18 ... ... . which reads as follows:

The existence of a minimal $S$ is rather obvious. Since $M$ is finitely generated, one can find a finite set $S$ such that $L+\langle S\rangle =M$. Take a smallest subset of $S$ which generates the same submodule as $S$. This is a required minimial subset.

I will post about the inductiveness of $S$ when I have some more time. But at first look it semmed like this too follows immediately from the definitions.

 

[Math Amateur]

The existence of a minimal $S$ is rather obvious. Since $M$ is finitely generated, one can find a finite set $S$ such that $L+\langle S\rangle =M$. Take a smallest subset of $S$ which generates the same submodule as $S$. This is a required minimial subset.

I will post about the inductiveness of $S$ when I have some more time. But at first look it semmed like this too follows immediately from the definitions.

Thanks for the help, caffeinemachine BUT ... I do not follow ...

You write:"... ... Since $M$ is finitely generated, one can find a finite set $S$ such that $L+\langle S\rangle =M$ ... ... BUT ... why exactly is this true ... can you be more explicit ...

Peter