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I Example on Triangular Rings - Lam, Example 1.14

  1. Sep 15, 2016 #1
    I am reading T. Y. Lam's book, "A First Course in Noncommutative Rings" (Second Edition) and am currently focussed on Section 1:Basic Terminology and Examples ...

    I need help with an aspect of Example 1.14 ... ...

    Example 1.14 reads as follows:


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    ?temp_hash=0f58364433f8d45cf4a489c5179e1889.png



    I cannot follow why the results in Table 1.16 follow ...

    For example, according to Table 1.16 ...

    ##mr = 0## for all ##m \in M## and ##r \in R## ... but why?


    Similarly I don't follow the other entries in the Table ...

    Can someone please help ...

    help will be much appreciated ...

    Peter
     

    Attached Files:

  2. jcsd
  3. Sep 16, 2016 #2

    andrewkirk

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    The ##R,M,S## in that table refer not to the original algebraic entities with those names but to isomorphs of them in ##A##.

    Let's put primes on the ones in the table to avoid confusion. Then :
    $$R'\triangleq \left\{\begin{pmatrix}r&0\\0&0\end{pmatrix}\ :\ r\in R\right\}$$
    $$M'\triangleq \left\{\begin{pmatrix}0&m\\0&0\end{pmatrix}\ :\ m\in M\right\}$$
    $$S'\triangleq \left\{\begin{pmatrix}0&0\\0&s\end{pmatrix}\ :\ s\in S\right\}$$
    If you replace the ##R,M,S## in the table and its column and row labels by their primed versions, and do the matrix multiplications, you will see that the table works.

    The trick is the words 'it is convenient to identify'. I understand why authors do identification. It can save on notation. But for the confusion it causes, I really don't think it's worth it. Anyway, keep an eye out for that deadly word 'identify' and its derivatives. At least this author admitted that he did it. Sometimes they don't.
     
  4. Sep 16, 2016 #3
    Thanks Andrew ... that certainly clarifies things a lot ...

    I think you caution regarding "identification" is absolutely necessary ...

    Will now continue to work on Lam's example ...

    Thanks again,

    Peter
     
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