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Math Help - Proof Help! Modules and left ideals

  1. #1
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    Proof Help! Modules and left ideals

    Here is the question:
    Show that if L is a left ideal of the ring M2(F), F any field, then L consists of all left multiples of some element A in M2(F).

    Show that this A can be chosen to be idempotent, such that A^2=A.

    Show that the foregoing assertions hold not just for M2(F) but also for Mn(F), any n.

    Any help would be great!
    Last edited by dabien; October 26th 2009 at 04:44 PM. Reason: M2(F)
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  2. #2
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    Quote Originally Posted by dabien View Post
    Here is the question:

    Show that if L is a left ideal of the ring M2(F), F any field, then L consists of all left multiples of some element A in F.

    Show that this A can be chosen to be idempotent, such that A^2=A.

    Show that the foregoing assertions hold not just for M2(F) but also for Mn(F), any n.

    Any help would be great!
    that's a false claim! fix the mistake you made in writing your question and then we'll help you.
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  3. #3
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    maximum rank in L

    Hi, what do you mean by it is not accurate?
    I was thinking about looking at the maximum rank in L,
    but I still need help trying to figure this out.

    Anything would be great!
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  4. #4
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    help!

    Ok I caught the error. But I am still very lost!
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  5. #5
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    there's a quick way to solve your problem if you know a little bit about the "ring structure" of R=M_n(F). what we need here is to know that R is "semsimple" and thus every left ideal of R is a

    direct summand of R. so if L is a left ideal of R, then R=L \oplus K, for some left ideal K of R. therefore 1=e+f, for some e \in L, \ f \in K. thus for any a \in L we have a+0=a \cdot 1=ae + af,

    and ae \in L, \ af \in K. thus ae=a (and hence af = 0, but we'll not use this one). this proves both e^2=e and L=Re.
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