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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:

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

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

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

##x_0 R + \ ... \ ... \ , x_s R + L = M##?

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

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

Lemma 1.2.21 and its proof reads as follows:

__In the above text by Berrick and Keating, we read the following:"... ... Since ##M## is finitely generated, there is a minimal subset ##\{ x_0, \ ... \ ... \ , x_s \}## of ##M## such that__**Question 1**##x_0 R + \ ... \ ... \ , x_s R + L = M. \ ... \ ... \ ...## "My problem is as follows:

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

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

##x_0 R + \ ... \ ... \ , x_s R + L = M##?

__In the above text by Berrick and Keating, we read the following:"... ... Let ##S## be the set of submodules ##X## of M that contain ##x_1 R + \ ... \ ... \ , x_s R + L## but do not contain ##x_0##. It is obvious that ##S## is inductive ... ...Can someone please explain exactly why ##S## is inductive ... ... ?Hope someone can help ...__**Question 2**Peter========================================================================B&K's definition of "inductive" is contained in section 1.2.18 ... ... . which reads as follows: