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I Finitely Generated Modules ... another question

  1. Oct 2, 2016 #1
    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:



    ?temp_hash=686be1ce874012e5acd1f87513653b2c.png




    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
     

    Attached Files:

  2. jcsd
  3. Oct 2, 2016 #2

    andrewkirk

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    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##.
     
  4. Oct 2, 2016 #3

    fresh_42

    Staff: Mentor

    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##.
     
    Last edited: Oct 2, 2016
  5. Oct 2, 2016 #4
    Thanks ... very clear and very helpful ...

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

    Peter
     
  6. Oct 3, 2016 #5

    fresh_42

    Staff: Mentor

    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.
     
  7. Oct 3, 2016 #6
    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
     
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