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I Module Over a Division Ring - Blyth Theorem 1.1, Part 4

  1. Aug 9, 2016 #1
    I am reading T. S. Blyth's book "Module Theory: An Approach to Linear Algebra" ... ... and am currently focussed on Chapter 1: Modules, Vector Spaces and Algebras ... ...

    I need help with an aspect of Theorem 1.1 part 4 ...

    Theorem 1.1 in Blyth reads as follows:


    ?temp_hash=130091d5b5e518dfcff28c7ba4b6c3d1.png




    In the above text, in part 4 of the Theorem we read:


    " ... ... when ##R## is a division ring

    (4) ##\lambda x = 0_M## implies ##\lambda = 0_R## or ##x = 0_M## ... ... "


    Blyth proves that if ##R## is a division ring and ##\lambda x = 0_M## with ##\lambda \neq 0_R## then we have that ##x = 0_M## ... ...


    But ... ... Blyth does not show that if ##R## is a division ring and ##\lambda x = 0_M## with ##x \neq 0_M## then we have that ##\lambda = 0_R## ... ...


    Can someone please help me to prove this ...

    Peter
     

    Attached Files:

  2. jcsd
  3. Aug 9, 2016 #2

    micromass

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    Contrapositive.
     
  4. Aug 9, 2016 #3
    Sorry Micromass ... I do not follow you ...

    Can you explain what you mean more explicitly ... ?

    My apologies for not following you ...

    Peter
     
  5. Aug 9, 2016 #4

    micromass

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  6. Aug 9, 2016 #5
    Hi Micromass ... I understand the meaning of "contrapositive" ... but fail to see exactly how it applies to the problem that I posed ...

    Peter
     
  7. Aug 9, 2016 #6
     
  8. Aug 9, 2016 #7

    micromass

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    Come on, think a bit about it. You answered merely 6 minutes after me. I'm not here to spoonfeed you the answer.
     
  9. Aug 9, 2016 #8

    fresh_42

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    Proof of (4).
    Case 1: ##\lambda = 0_R## ##\checkmark## by (2)
    Case 2: ##\lambda \neq 0_R## - see Blyth's proof
     
  10. Aug 10, 2016 #9

    Well ... thanks fresh_42 ... that is an easy resolution to the issue ...

    BUT ... still reflecting and wondering about Case 1 ...

    .... indeed ... does ##( 0_R x = 0_M )## ...

    ... imply (or mean) that ...

    ##( \lambda x = 0_M \text{ and } x \neq 0_M \Longrightarrow \lambda 0_R )##

    is true ...


    But why exactly ... ???


    (Not quite sure why this logic bothers me ...)

    Peter
     
    Last edited: Aug 10, 2016
  11. Aug 10, 2016 #10
    Thanks for th

    Thanks for the hint, micromass ...

    in (4) Blyth has shown the following:

    ##( \lambda x = 0_M \text{ and } \lambda \neq 0_R ) \longrightarrow (x = 0_M)## ... ... ... (1)

    So ... contrapositive of (1) is as follows:

    ##\neg ( x = 0_M) \longrightarrow \neg ( \lambda x = 0_M \text{ and } \lambda \neq 0_R )##

    But ... how do I progress from here ...

    Can someone help further ...?

    Peter
     
  12. Aug 10, 2016 #11

    fresh_42

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    We have ##\lambda x = 0_M## as a fact.
    (2) is needed to show that ##\lambda = 0_R## is a solution at all.
    Since there are only the two possibilities for ##\lambda## - either being zero or not - and the first solves the condition according to (2), it is sufficient to consider the other case, ##\lambda \neq 0_R##.

    You may write the whole thing as:

    ##[\lambda x = 0_M] ##
    ##= \; (\lambda x = 0_M) \, ∧ \, \text{ (true) }##
    ##= \; (\lambda x = 0_M) \, ∧ \, [(\lambda = 0_R)\, ∨ \, (\lambda \neq 0_R)]##
    ##= \; [(\lambda x = 0_M) \, ∧ \, (\lambda = 0_R)] \, ∨ \, [(\lambda x = 0_M) \, ∧ \, (\lambda \neq 0_R)]##

    → (now I use (2), i.e. if ##\lambda = 0_R## is true then ##\lambda x = 0_M## is also true)

    ##= \; (\lambda = 0_R)\, ∨ \, [(\lambda x = 0_M) \, ∧ \, (\lambda \neq 0_R)]##

    Therefore if ##(\lambda = 0_R) = \text{ (true) }## we are done and we may ask when ##[(\lambda x = 0_M) \, ∧ \, (\lambda \neq 0_R)]## will be true without ##(\lambda = 0_R)## being true, i.e. ##(\lambda \neq 0_R)##. Now Blyth has shown

    $$[(\lambda x = 0_M) = \text{ (true) }] \,⇔\, [(\lambda = 0_R)=\text{ (true) }]\, ∨ \, [((\lambda x = 0_M) \, ∧ \, (\lambda \neq 0_R)) \,=\, \text{ (true) }]\, \overset{Blyth}⇒\, [(\lambda = 0_R)=\text{ (true) }]\, ∨ \, \; [(x=0_M) \,=\, \text{ (true) }]$$
     
    Last edited: Aug 10, 2016
  13. Aug 20, 2016 #12
    Thanks for that fresh_42 ... most helpful ...

    Much appreciate your assistance ...

    Peter
     
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