Finitely Generated Modules .... another question

[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:

In the above text (in the proof) by Berrick and Keating we read the following:"... ... Any submodule properly containing ##X'## must contain ##x_0##, also, so ##X'## is maximal among all the proper submodules of ##M##. ... ... My first question is ... why must any submodule containing ##X'## also contain ##x_0## ... ?

My second question is ... why does ##X##' containing ##x_0## allow us to conclude that
##X'## is maximal among all the proper submodules of ##M##. ... ... ?Hope someone can help ...

Peter

 

[andrewkirk]

Let ##N## be a submodule of ##M## that properly contains ##X'##.
Then we have ##x_1R+...+x_sR+L\subseteq X'\subsetneq N##
So if ##x_0\notin N##, we have ##N\in S##, which contradicts our assumption that ##X'## is maximal in ##S##.

Hence we must have ##x_0\in N##.

But that means that ##N\supseteq x_0R+(x_1R+...+x_sR+L)=M##. So ##M## is maximal because any submodule ##N## of ##M## that properly contains it must be equal to ##M##.

 

[fresh_42]

"... ... Any submodule properly containing ##X'## must contain ##x_0##, also, so ##X'## is maximal among all the proper submodules of ##M##. ... ...My first question is ... why must any submodule containing ##X′## also contain ##x_0## ... ?

My second question is ... why does ##X'## containing ##x_0## allow us to conclude that
##X'## is maximal among all the proper submodules of ##M##. ... ... ?

Let us proceed by the pattern I told you.

Firstly ##X'## is chosen as a maximal element of the set
$$S := \{ N \,\vert \, N\subsetneq M \text{ submodule and } x_1R+\dots +x_sR+L \subseteq N \text{ and } x_0 \notin N \}$$
Secondly, we know that ##M## is finitely generated by ##\{x_0,x_1, \dots , x_s\}##.

Now the pattern goes:

Assume we have a submodule ##P##, which lies between ##X'## and ##M##, that is ##X' \subseteq P \subseteq M##.
We have to show that either ##P=X'## or ##P=M##.

1.) If ##X' \subsetneq P## is a proper inclusion, then ##x_0## has to be in ##P## because all other generators are already in ##X'## and ##x_0## is the only one which is left and can be added to create a proper inclusion. But then ##P=M##. *)

2.) If ##X' \subseteq P## and ##P \subsetneq M## is a proper submodule of ##M##, then ##P## belongs to ##S##. But ##X'## is a maximal element of ##S##. So the only way out is ##P=X'##.

Thus we have shown that there cannot be a submodule ##P## between ##X'## and ##M##, which makes ##X'## a maximal submodule.

*) Edit: In contrast to the example with the integers, you can't escape by taking elements like ##x_0^2## or similar instead of ##x_0##, because ##x_0## belongs to ##M## and there is no multiplication within ##M## and ##x_0## is not in ##R##.

 

[Math Amateur]

Thanks ... very clear and very helpful ...

Pity the text couldn't be as clear ...

Peter

 

[fresh_42]

Thanks ... very clear and very helpful ...

Pity the text couldn't be as clear ...

Peter

I'm afraid I made a little mistake in my cases. Given ##X' \subseteq P \subseteq M## then one has to distinguish
1.) ##x_0 \in P## and
2.) ##x_0 \notin P##.

##X' \subsetneq P## being a proper submodule does not imply ##x_0 \in P##. E.g. ##2x_0## could be a generator of ##P## which is not ##x_0## ; and ##2## might not be a unit of ##R##, so it cannot be divided.

However, the cases above work. And one of them has to be true: ##x_0## is either in ##P## or it is not.
So if 1.) ##x_0 \in P## then ##M = X' + x_0R \subseteq P## and ##P=M##,
and if 2.) ##x_0 \notin P## then ##P \in S## and ##X' \subseteq P## implies ##P=X' ## by the maximality of ##X'## in ##S##.

Thus we have ##X' \subseteq P \subseteq M\, \Longrightarrow \, P=X' \text{ or } P=M##, i.e. ##X' \subseteq M## is maximal.

 

[Math Amateur]

Well ... have to confess ... I did not notice that subtlety ...

... ... hmmm ... still reflecting on it ...

... well ... learning quite a lot from you ...

Thanks so much for your help and insights ...

Peter