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Math Help - To prove {ab-ba | a,b in R} is an ideal of the ring R

  1. #1
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    To prove {ab-ba | a,b in R} is an ideal of the ring R

    R is a ring. Prove that J = {ab-ba | a,b in R} is an ideal of R.

    To prove this, I tried proving the following things:-
    1. J is nonempty. I could do this.
    2. J is a subset of R. I could do this.
    3. J is closed to negatives. I could do this.
    4. J is closed to addition. Don't know how to do it.
    5. J absorbs the products in R. I could do this.

    I tried a lot but could not do #4. Could you help me? Please let me see the proof of J is closed to addition.
    Alternatively, please let me know the proof of J is closed to subtraction because #3 and #4 can be clubbed into "J is closed to subtraction"

    Any help?
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  2. #2
    Junior Member Nehushtan's Avatar
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    Re: Commutator ideal

    I did a Web search and came across the notion of a “commutator ideal”; this is smallest ideal of the ring R containing all elements of the form ab-ba for a,b\in R. This suggests that the set of all ab-ba itself is not generally an ideal.

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