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Math Help - Nilpotent = Ideal

  1. #1
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    Nilpotent = Ideal

    Show that nilpotent elements of a commutative ring R constitute an ideal of R.

    Proof.

    So nilpotent element is an element a \in R such that a^n =0 for some n>0.

    Denote the set of all nilpotent elements of R by N.

    Pick  a \in N and r \in R

    We need to show that  ar \in N

    (ar)^n=a^nr^n=0

    That's it? Thanks.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tttcomrader View Post
    Show that nilpotent elements of a commutative ring R constitute an ideal of R.

    Proof.

    So nilpotent element is an element a \in R such that a^n =0 for some n>0.

    Denote the set of all nilpotent elements of R by N.

    Pick  a \in N and r \in R

    We need to show that  ar \in N

    (ar)^n=a^nr^n=0

    That's it? Thanks.
    looking good so far to me.

    another trivial thing you have to show though is that N is closed under addition, that is, x \pm y \in N for all x,y \in N
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  3. #3
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    Quote Originally Posted by tttcomrader View Post
    Show that nilpotent elements of a commutative ring R constitute an ideal of R.

    Proof.

    So nilpotent element is an element a \in R such that a^n =0 for some n>0.

    Denote the set of all nilpotent elements of R by N.

    Pick  a \in N and r \in R

    We need to show that  ar \in N

    (ar)^n=a^nr^n=0

    That's it? Thanks.
    Exactly. You need to show rN \subseteq N and that it exactly what you shown.
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