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(Apparently) simple question rearding module homomorphisms

  1. May 9, 2014 #1
    I am reading Dummit and Foote Chapter 10: Introduction to Module Theory.

    I am having difficulty seeing exactly why a conclusion to Proposition 27 that D&F claim is "immediate":

    I hope someone can help.

    Proposition 27 and its proof read as follows:

    attachment.php?attachmentid=69619&stc=1&d=1399695567.jpg

    In the first line of the proof (see above) D&F state the following:

    "The fact that [itex] \psi [/itex] is a homomorphism is immediate."

    Can someone please explain exactly why [itex] \psi [/itex] is a homomorphism?

    Would appreciate some help.

    Peter
     

    Attached Files:

  2. jcsd
  3. May 10, 2014 #2

    micromass

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    Any thoughts? What is it exactly that you need to prove?

    Also, could you please shrink the images next time. It's really annoying.
     
  4. May 10, 2014 #3
    Tried to resize image - new image is displayed below.

    attachment.php?attachmentid=69655&stc=1&d=1399778625.png

    Peter
     

    Attached Files:

  5. May 10, 2014 #4
    Another size option would be as follows:

    attachment.php?attachmentid=69656&stc=1&d=1399778901.png

    Peter
     

    Attached Files:

  6. May 11, 2014 #5
    I'd recommend you try to figure it out for yourself, using the definition of phi' they give at the top.
     
  7. May 11, 2014 #6
    Thanks.

    Problem is now solved.

    Peter
     
  8. May 11, 2014 #7

    adjacent

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    If you don't mind,can you show us your answer?
     
  9. May 11, 2014 #8
    I received help on the MHB forum.

    The solution was as follows:

    "We have to verify that for:

    [itex] [/itex]

    [itex] \psi'(f+g) = \psi'(f) + \psi'(g)[/itex] in [itex]\text{Hom}_R(D,M)[/itex].

    To do this, let's take an arbitrary element [itex]d \in D[/itex].

    Then:

    [itex] (\psi'(f+g))(d) = (\psi \circ (f+g))(d) = \psi((f+g)(d)) = \psi(f(d)+g(d)) = \psi(f(d)) + \psi(g(d))[/itex] (since [itex]\psi[/itex] is a module homomorphism)

    [itex]= (\psi \circ f)(d) + (\psi \circ g)(d) = (\psi'(f))(d) + (\psi'(g))(d) = (\psi'(f) + \psi'(g))(d)[/itex].

    Since these two functions are equal for every [itex]d \in D[/itex] they are the same element of [itex] \text{Hom}_R(D,M)[/itex]."

    The solution is due to Deveno on the Math Help Boards, Linear and Abstract Algebra forum.

    Peter
     
    Last edited: May 11, 2014
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