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Math Help - Herstein's matrix proof

  1. #1
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    Cool Herstein's matrix proof

    Herstein's abstract algebra: Chapter 4 section 3 number 19

    If R = {( a b ) | a, b, c real} and I = {( 0 b ) | b real }
    .........{( 0 c ) |.................}..............{( 0 0 ) |...........}

    values in between the {} are matrices.

    Show that:

    (a) R is a ring
    (b) I is an ideal of R
    (c) R/I ≈ F ⊕ F, where F is the field of real numbers
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Herstein's matrix proof

    Quote Originally Posted by ThatPinkSock View Post
    Herstein's abstract algebra: Chapter 4 section 3 number 19

    If R = {( a b ) | a, b, c real} and I = {( 0 b ) | b real }
    .........{( 0 c ) |.................}..............{( 0 0 ) |...........}

    values in between the {} are matrices.

    Show that:

    (a) R is a ring
    (b) I is an ideal of R
    (c) R/I ≈ F ⊕ F, where F is the field of real numbers
    What have you tried? I mean, there is no trickery here, it's just doing it. Are you stuck somewhere in particular, because I know you don't expect us to just do it for you.
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  3. #3
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    Re: Herstein's matrix proof

    Quote Originally Posted by Drexel28 View Post
    What have you tried? I mean, there is no trickery here, it's just doing it. Are you stuck somewhere in particular, because I know you don't expect us to just do it for you.


    I expected no such thing!

    (a) It suffices to show that R is a subring of M₂(R).

    Since the difference and products of upper triangular matrices are again upper triangular, this is indeed the case.

    For specifically, let
    A = (a b)
    ......(0 c), and

    B = (d e)
    ......(0 f) be in R.

    Then, A - B =
    (a-d b-e)
    (0 c-f) is also in R.

    AB =
    (ad ae+bf)
    (.0......cf...) is in R.



    b) For A = (0 a; 0 0), B = (0 b; 0 0) in I,
    A + B = (0 a+b; 0 0) is in R.

    For A = (0 a; 0 0) and I, and C = (b c; 0 d) in R:

    AC =
    (0 a)(b c)...(0 ac)
    (0 0)(0 d).=(0 0) is in R, and

    CA =
    (b c)(0 a)...(0 ab)
    (0 d)(0 0).=(0 0) is in R.

    Hence, I is an (two-sided) ideal of R.



    (c) Consider the map F : R → F ⊕ F defined by F((a b; 0 c)) = (a, c).
    It's easy to see that this is a surjective ring homomorphism,
    with ker F = {(0 b; 0 0) : b in F} = I.

    So, the result follows from the first isomorphism theorem.


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