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I Multiplication Maps on Algebras ... Further questions ...

  1. Dec 4, 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 some further help with the statement and proof of Lemma 1.24 ...

    Lemma 1.24 reads as follows:


    ?temp_hash=092a942624b9c0487a3dfca35ccaad07.png



    My further questions regarding Bresar's statement and proof of Lemma 1.24 are as follows:


    Question 1

    In the statement of Lemma 1.24 we read the following:

    " ... ... Let ##A## be a central simple algebra. ... ... "


    I am assuming that since ##A## is central, it is unital ... that is there exists ##1_A \in A## such that ##x.1 = 1.x = 1## for all ##x \in A## ... ... is that correct ... ?



    Question 2

    In the proof of Lemma 1.24 we read the following:

    " ... ... Suppose ##b_n \ne 0##. ... ... "


    I am assuming that that the assumption ##b_n \ne 0## implies that we are also assuming

    that ##b_1 = b_2 = \ ... \ ... \ = b_{n-1} = 0## ... ...

    Is that correct?




    Question 3

    In the proof of Lemma 1.24 we read the following:

    " ... ...where ##c_i = \sum_{ j = 1 }^m w_j b_i z_j## ; thus ##c_n = 1## for some ##w_j, z_j \in A## ... ...

    This clearly implies that ##n \gt 1##. ... ... "


    My question is ... why/how exactly must ##n \gt 1## ... ?

    Further ... and even more puzzling ... what is the relevance to the proof of the statements that ##c_n = 1## and ##n \gt 1## ... ?

    Why do we need these findings to establish that all the ##b_i = 0## ... ?




    Hope someone can help ...

    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=092a942624b9c0487a3dfca35ccaad07.png
    ?temp_hash=092a942624b9c0487a3dfca35ccaad07.png
     
    Last edited: Dec 4, 2016
  2. jcsd
  3. Dec 4, 2016 #2

    fresh_42

    Staff: Mentor

    Central means the center ## C(A) ## of ##A## is equal to the corresponding scalar domain, i.e. a field or a division ring at least. Since the center is part of the algebra, the field ##\mathbb{F} =C(A) ## is contained in the algebra. Then ##1_\mathbb{F} \in A##. So ##1_\mathbb{F}=1_A##, because it does what a one has to do: ##1_\mathbb{F}\cdot a = a## and there cannot be two different ones: ##1=1\,\cdot\,1'= 1'##.
    No. We only assume at least one ##b_i \neq 0## to derive a contradiction. So without loss of generality, we assume it to be ##b_n##, for otherwise we would simply change the numbering. We don't bother ## b_1, \ldots ,b_{n-1}##. They may be equal to zero or not. Only the case in which all are zero is ruled out, for then we would have nothing to show: the ##a_i## would be linear independent.
    If ##n=1## then ##\left( \sum_{i=1}^n L_{a_i}R_{c_i} \right)(x) = (L_{a_1}R_{c_1})(x)=a_1 \cdot x \cdot c_1=a_1 \cdot x=0## for all ##x \in A##, because ##c_n=c_1=1##. Especially ##a_1\cdot 1_\mathbb{F} = a_1 = 0##, but the ##a_i## are linear independent, so they cannot equal ##0##.
    It is important for the sum in ##0=\sum_{i=1}^{n-1} L_{a_i}R_{xc_i-c_i x}## being a real sum and not only trivially fulfilled.
    They are used to show all ##c_i \in \mathbb{F}##. Then ##c_1a_1+\ldots c_na_n## is not only an equation in ##A## but a linear one with coefficients in ##\mathbb{F}##. Now if ##0=L_{c_1a_1+\ldots c_na_n}## then ##0=L_{c_1a_1+\ldots c_na_n}(1_\mathbb{F})=(c_1a_1+\ldots c_na_n)\cdot 1_\mathbb{F}=c_1a_1+\ldots c_na_n## and the ##a_i## are linear dependent over ##\mathbb{F}##, because at least ##c_n \neq 0## and we have our contradiction.
     
  4. Dec 4, 2016 #3
    Thanks fresh_42 ...

    That post was INCREDIBLY helpful!!

    Peter
     
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