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

• Mar 2nd 2013, 06:22 AM
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?
• Mar 2nd 2013, 11:06 AM
Nehushtan
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. (Shake)

Where did you find this question? (Thinking)