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I Artinian Modules - Bland - Proposition 4.24

  1. Oct 29, 2016 #1
    I am reading Paul E. Bland's book, "Rings and Their Modules".

    I am focused on Section 4.2: Noetherian and Artinian Modules and need some help to fully understand the proof of part of Proposition 4.2.4 ... ...

    Proposition 4.2.4 reads as follows:


    ?temp_hash=05dde20a7081ba4eb52a559921ca1799.png
    ?temp_hash=05dde20a7081ba4eb52a559921ca1799.png




    I need help to fully understand Part of the proof proving that ##(2) \Longrightarrow (3)## ...


    In that part of the proof Bland seems to be assuming that

    ## \bigcap_F M_\alpha = N ##

    if and only if

    ## \bigcap_F (M_\alpha / N ) = 0 ##



    In other words, if ##F = \{ 1, 2, 3 \}## then

    ##M_1 \cap M_2 \cap M_3##

    if and only if

    ##M_1 / N \cap M_2 / N \cap M_3 / N##


    But why exactly is this the case ... ...

    ... ... how do we formally and rigorously demonstrate that this is true ...


    Hope someone can help ...

    Peter



    ====================================================

    Proposition 4.2.4 refers to the (possibly not well known) concept of cogeneration so I am providing Section 4.1 Generating as Cogenerating Classes ... ... as follows ...


    ?temp_hash=05dde20a7081ba4eb52a559921ca1799.png
    ?temp_hash=05dde20a7081ba4eb52a559921ca1799.png
    ?temp_hash=05dde20a7081ba4eb52a559921ca1799.png
     

    Attached Files:

  2. jcsd
  3. Oct 29, 2016 #2

    andrewkirk

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    Start by observing that
    $$(\bigcap_{\alpha\in F} M_\alpha)/N=N/N$$
    which is the zero element of the module ##M/N##.

    So we have to show that
    $$\bigcap_{\alpha\in F} (M_\alpha/N)=(\bigcap_{\alpha\in F} M_\alpha)/N\ \ \ \ \ \ \ (1)$$

    First observe that if ##m\in \bigcap_{\alpha\in F} M_\alpha## then ##\forall\alpha\in F:\ m\in M_\alpha##
    so that ##\forall\alpha\in F:\ m+N \in M_\alpha/N## whence
    ##m+N \in \bigcap_{\alpha\in F}M_\alpha/N##.

    Hence we have
    $$(\bigcap_{\alpha\in F} M_\alpha)/N\subseteq \bigcap_{\alpha\in F} (M_\alpha/N)$$

    For the other direction of inclusion, consider an arbitrary element ##m+N## of the LHS of (1). That is, ##m+N\in \bigcap_{\alpha\in F} (M_\alpha/N)##. Then for any ##\alpha\in F## we have ##m+N\in M_\alpha/N##, so there exists ##m'\in M_\alpha## and ##n\in N## such that ##m=m'+n##, which is in ##M_\alpha## since ##n\in N\subseteq M_\alpha##. Hence ##m\in \bigcap_{\alpha\in F} M_\alpha## and therefore ##m+N## is an element of the RHS of (1).

    This is not as elegant as I would like because it is late at night and I am tired. But hopefully it makes sense.
     
  4. Oct 29, 2016 #3
    Thanks for the help Andrew ... much appreciated...

    Will be working through your post very shortly...

    Thanks again ...

    Peter
     
  5. Oct 29, 2016 #4
    Thanks Andrew ... found that very clear and convincing ...

    Peter
     
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