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I Left/Right Multiplication Maps on Algebras ... Bresar

  1. Dec 3, 2016 #1
    I am reading Matej Bresar's book, "Introduction to Noncommutative Algebra" and am currently focussed on Chapter 1: Finite Dimensional Division Algebras ... ...

    I need help with the proof of Lemma 1.24 ...

    Lemma 1.24 reads as follows:


    ?temp_hash=695688f2306bf2924c46d39e19428801.png




    My questions regarding the proof of Lemma 1.24 are as follows ... ...


    Question 1

    In the above proof by Bresar, we read:

    " ... ... Since ##A## is simple, the ideal generated by ##b_n## is equal to ##A##.

    That is ##\sum_{ j = 1 }^m w_j b_n z_j = 1## for some ##w_j , z_J \in A##. ... ... "


    My question is ... ... how/why does the fact that the ideal generated by ##b_n## being equal to ##A## ...

    imply that ... ##\sum_{ j = 1 }^m w_j b_n z_j = 1## for some ##w_j , z_J \in A## ...?




    Question 2


    In the above proof by Bresar, we read:


    " ... ##0 = \sum_{ j = 1 }^m R_{ z_j } \ ( \sum_{ i = 1 }^n L_{ a_i } R_{ b_i } ) \ R_{ w_j }##


    ##= \sum_{ i = 1 }^n L_{ a_i } \ ( \sum_{ j = 1 }^m R_{ w_j b_i z_j } )##


    ##= \sum_{ i = 1 }^n L_{ a_i } R_{ c_i }##


    ... ... "



    My questions are

    (a) can someone help me to understand how


    ##\sum_{ j = 1 }^m R_{ z_j } \ ( \sum_{ i = 1 }^n L_{ a_i } R_{ b_i } ) \ R_{ w_j }##


    ##= \sum_{ i = 1 }^n L_{ a_i } \ ( \sum_{ j = 1 }^m R_{ w_j b_i z_j } ) ##



    (b) can someone help me to understand how


    ##\sum_{ i = 1 }^n L_{ a_i } \ ( \sum_{ j = 1 }^m R_{ w_j b_i z_j } ) ##


    ##= \sum_{ i = 1 }^n L_{ a_i } R_{ c_i }##




    Help will be appreciated ...

    Peter

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

    *** NOTE ***


    So that readers of the above post will be able to understand the context and notation of the post ... I am providing Bresar's first two pages on Multiplication Algebras ... ... as follows:



    ?temp_hash=695688f2306bf2924c46d39e19428801.png
    ?temp_hash=695688f2306bf2924c46d39e19428801.png
     
    Last edited: Dec 3, 2016
  2. jcsd
  3. Dec 3, 2016 #2

    fresh_42

    Staff: Mentor

    Q1:
    What does it mean to be an ideal ##I## of ##A##?
    For a (two-sided!) ideal, it has to hold, that ##A\cdot I \subseteq I## and ##I\cdot A \subseteq I##. Since ##b_n \in I##, we need to have all left and right multiples to also be in ##I##. So all elements of the form ##w_jb_nz_j## are together with ##b_n## also elements of ##I##.
    Furthermore an ideal is closed under addition, so all summations of elements of ##I## are again in ##I##, esp. any sum ##\sum_{j=1}^{m} w_jb_nz_j##. This is simultaneously the most general form of any element of ##I = <b_n> = A\cdot I \cdot A##.

    Q2:
    We have ##\sum_{i=1}^{n} L_{a_i}R_{b_i} = 0 \; (^*) \;## by assumption.
    Then let us define ##\sum_{j=1}^{m} w_jb_iz_j =: c_i \; (^{**}) \;##, simply as an abbreviation for these sums ##c_1, \ldots , c_n##.
    Because all sums are finite, we won't have to bother any order of summation.
    At last let us assume we have an arbitrary element ##x \in A##.

    Now calculate ##\left( \sum_{j=1}^{m} R_{z_j} \left( \sum_{i=1}^{n} L_{a_i}R_{b_i} \right) R_{w_j} \right)(x)## by using ##(^*)## and ##(^{**})##.
     
  4. Dec 4, 2016 #3
    Thanks again for your assistance, fresh_42 ...

    Most helpful ...

    Peter
     
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