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Math Help - ideal,nil,nilpotent ideal in prime ring

  1. #1
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    Red face ideal,nil,nilpotent ideal in prime ring

    Let R be non-commutative prime ring
    let t \in R such that {t}^{2}=0
    i found statement:
    {(tR)}^{ 3}=0 , thus tR is nonzero nil right ideal satisfying {z}^{ 3}=0 for all z \in tR, then R has nonzero nilpotent ideal.
    my question is:
    1. tR is nonzero nil right ideal, how can i proof tR is right ideal on R?(proof about closure in substraction) and any guarantee that tR is nonzero?
    2. R has nonzero nilpotent ideal. how can i say that? (my mentor say there is lemma proof this in book "I.N. Herstein, Topics in Ring Theory, univ. chicago press,1969" i can't find this book, pretty rare i guess.
    thanks
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  2. #2
    MHF Contributor

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    Quote Originally Posted by Shurelia View Post
    Let R be non-commutative prime ring
    let t \in R such that {t}^{2}=0
    i found statement:
    {(tR)}^{ 3}=0 , thus tR is nonzero nil right ideal satisfying {z}^{ 3}=0 for all z \in tR, then R has nonzero nilpotent ideal.
    my question is:
    1. tR is nonzero nil right ideal, how can i proof tR is right ideal on R?(proof about closure in substraction) and any guarantee that tR is nonzero?
    2. R has nonzero nilpotent ideal. how can i say that? (my mentor say there is lemma proof this in book "I.N. Herstein, Topics in Ring Theory, univ. chicago press,1969" i can't find this book, pretty rare i guess.
    thanks
    your question is a mess!

    are you assuming that (tR)^3=0? i'm saying this because t^2=0 does not imply that (tR)^3=0, even in prime rings. if you're assuming (tR)^3=0, why do you need the condition t^2=0? are you trying to prove that a prime ring has no non-zero nilpotent ideal? to prove this, you only need R to be semiprime.

    anyway, tR is obviously a right ideal. i don't know why you can't prove this! it is non-zero becasue 1 \in R and so 0 \neq t \in tR. i guess you forgot to mention in your question that 1 \in R and t \neq 0.

    for the second part of your question, you don't need Herstein! suppose that I is a right ideal of a ring R and I^n=0 for some integer n \geq 2. then RI is clearly an ideal of R and

    (RI)^n=R(IR)^{n-1}I \subseteq RI^{n-1}I = RI^n=0.

    thus (RI)^n=0 and so RI is a nilpotent ideal of R. note that if I \neq 0, then  RI \neq 0 because I \subseteq RI.
    Last edited by NonCommAlg; May 24th 2011 at 09:07 AM.
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