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I Triangular Matrix RIngs ... Another Question

  1. Sep 17, 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 yet another aspect of Example 1.14 ... ...

    Example 1.14 reads as follows:



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    ?temp_hash=be8a623e9154ee8ca2b39fa65a5078a6.png


    Near the end of the above text from T. Y. Lam we read the following:

    " ... ... Moreover ##R \oplus M## and ##M \oplus S## are both ideals of ##A##, with ##A / (R \oplus M ) \cong S## and ##A / ( M \oplus S ) \cong R## ... ... "


    Can someone please help me to show, formally and rigorously that ##A / (R \oplus M ) \cong S## and ##A / ( M \oplus S ) \cong R## ... ...


    My only thought so far is that the First Isomorphism Theorem for Rings may be useful ...

    Hope someone can help ... ...

    Peter
     

    Attached Files:

    Last edited: Sep 18, 2016
  2. jcsd
  3. Sep 18, 2016 #2
    First isomorphism theorem. Try it. What would an obvious surjection ##A\rightarrow S## be?
     
  4. Sep 19, 2016 #3
    Thanks for the help, micromass ... ...

    Use of the First Isomorphism Theorem for Rings in order to show that R/ ( R \oplus M ) \cong S would proceed as follows:

    Define a surjection ##\phi \ : \ A \longrightarrow S## ... ...

    ... where ##\phi## is defined as the map ##\begin{pmatrix} r & m \\ 0 & s \end{pmatrix} \ \mapsto \ \begin{pmatrix} 0 & 0 \\ 0 & s \end{pmatrix}##


    ##\phi## is clearly an epimorphism with kernel ... :

    ##\text{ ker } \phi = \begin{pmatrix} r & m \\ 0 & 0 \end{pmatrix} \ \cong \ R \oplus M##


    So by the First Isomorphism Theorem for Rings we have the following

    ##A / \text{ ker } \phi \ \cong \ S## ...

    ... that is ...

    ##A / ( R \oplus M ) \ \cong \ S## ...


    Is the above analysis correct?

    If it is correct ... then ##A / ( M \oplus S ) \ \cong \ R## ... follows similarly ...

    [Note: I think we could have proved the above by only invoking the First Isomorphism Theorem for Groups ... ]

    Any comments critiquing the above analysis are welcome ...

    Peter
     
  5. Sep 19, 2016 #4

    fresh_42

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    Staff: Mentor

    Looks correct, although I think you should check (as an exercise) that your homomorhisms are actually ring homomorphims.
    And this is the reason, why the group isomorphism theorem would not be enough. You need ring homomorphisms and ideals ##M \oplus S## and ##R \oplus M##. It should be easy to check whether they are really ideals in ##A##.

    Ideals and kernels of ring homomorphisms correspond 1:1.
     
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