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I Simple Modules and quotients of maximal modules, Bland Ex 13

  1. Feb 3, 2017 #1
    I am reading Paul E. Bland's book, "Rings and Their Modules".

    I am focused on Chapter 1, Section 1.4 Modules ... ...

    I need help with the proving a statement Bland makes in Example 13 ... ...

    Example 13 reads as follows:


    ?temp_hash=163dad38af054a70ffbbb915d74f3af9.png



    In the above text from Bland, we read the following:

    " ... If ##N## is a maximal submodule of ##M##, then it follows that ##M/N## is a simple ##R##-module ... ... "


    I do not understand why this is true ... can anyone help with a formal proof of this statement ...



    Hope someone can help ...

    Peter
     

    Attached Files:

  2. jcsd
  3. Feb 3, 2017 #2

    fresh_42

    Staff: Mentor

    What are the submodules of ##M/N\,##? And what is the zero element in this factor module?
     
  4. Feb 3, 2017 #3

    Hi fresh_42 ...

    I cannot answer you with confidence ... which is probably why I do not follow Bland Example 13 ... but ...

    The elements of ##M/N## are the cosets ##\{ x + N \}_{ x \in M }## where ##x + N = \{ x + n \ | \ n \in N \}## ... ...

    ... BUT? ... what are the submodules of ##M/N## ... I am unsure ...

    Zero element would be ##N = \{ 0 + N \}## ...

    Can you help further ... ?

    Peter
     
  5. Feb 3, 2017 #4

    fresh_42

    Staff: Mentor

    Yes, exactly. But zero is in any submodule. So a submodule of ##M/N## as a set ##S := \{x + N \,\vert \, x \in \textrm{ something }\}## has to contain ##N##. Now ##N \subseteq S \subseteq M## is maximal, so ##S## is either equal to ##M## or equal to ##N##. But this means ##S/N = M/N## or ##S/N=N/N=\{0\}## which is the definition of a simple module: ##M/N## has no proper submodules ##S/N##.
     
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