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I Simple Matrix Rings ...

  1. Nov 27, 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 some aspects of Bresar's Example 1.10 on a simple matrix ring over a division ring ...

    Example 1.10, including some preamble, reads as follows:


    ?temp_hash=0d1733a7c5582f06a2878772fa2eb20b.png


    In the above text from Bresar we read the following:

    " ... and hence also ##(d a_{jk}^{-1} ) E_{ii} \cdot a_{jk} E_{il} = d E_{il} \in I ## for every ##d \in D##. Consequently, ##I = M_n(D)##. ... ... "


    My questions are as follows:


    Question 1


    I am assuming that ##(d a_{jk}^{-1} ) E_{ii} \cdot a_{jk} E_{il} = d E_{il}## because you can take the "scalars" out of the multiplication and multiply them as in

    ##c_1 (a_{ij} ) \cdot c_2 (b_{ij} ) = c_1 c_2 (a_{ij} ) \cdot (b_{ij} )##


    Is that correct?


    (Note: why we are messing with multiplications by scalars in a problem on rings, I don't know ... we seem to be treating the ring ##M_n (D)## as an algebra over ##D## ... )




    Question 2


    Bresar seems to be assuming that ##d E_{il} \in I ## for all ##1 \le i, l \le n## and for every ##d \in D## implies that ##I = M_n (D)## ...

    But ... ... why exactly is this true ...


    My thoughts ... maybe it is true because the ##E_{il}## generate the ring ##M_n (D)## ... or to put it another way ... any element in ##I## or ##M_n (D)## can be written uniquely in the form ##\sum_{i, j = 1}^n d_{ij} E_{ij}## ...and further, that all the ##E_{ij}## belong to I ...


    Help with these questions will be appreciated ...

    Peter


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

    So that readers of the above post can appreciate the relevant context of the post, I am providing the introduction to Section 1.3 Simple Rings ... as follows:



    ?temp_hash=0d1733a7c5582f06a2878772fa2eb20b.png
    ?temp_hash=0d1733a7c5582f06a2878772fa2eb20b.png
     

    Attached Files:

  2. jcsd
  3. Nov 27, 2016 #2

    fresh_42

    Staff: Mentor

    Yes.
    Yes. If you consider ##(n \times n)-##matrix multiplication then it is defined with ##n## scalar multiplications in ##n^2## positions - a total of ##n^3## scalar multiplications (*). Therefore it is essential how scalars multiply with each other. Imagine our scalar multiplication was defined by ##a\cdot a=0## for all scalar ##a##. We'd get a pretty different matrix ring. (I didn't write ##a \in D## for it would have suggested ##D## to be a division ring, which can't be with such a definition of multiplication. Therefore I simply wrote "scalar ##a##".)
    With ##dE_{il} \in I## we have all scalar multiples in ##I##, esp. ##d^{-1}(dE_{il})=E_{il}## and all of them, i.e. the entire basis or ##M_n(D)=\sum_{ij}a_{ij}E_{ij} \subseteq \sum D\cdot I \subseteq I \subseteq M_n(D)##.
    Yes.

    (*) I know it can be done with less than ##n^3## and that the current record holder is est. ##\omega = 2.37_3##.
     
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