1.
let n be an integer greater than 1. In a ring in which x^n=x for all x, show that ab=0 implies
ba=0.
Solution
For part a), note that ba=(ba)^n=(ba)(ba)(ba)...(ba) n-times, but since multiplication is
associative, this can be re-written as b[(ab)(ab)(ab)...(ab)]a with n-1 times of the term (ab) inside,
but our hypothesis is that ab=0, so this whole product is equal to 0 (since 0 times anything is 0)
For part b) we can write the product ab in two different ways, first ab=(ab)^2=abab and secondly
ab=a^2b^2=aabb. Thus abab=aabb, and now you can cancel the a's on the left and the b's on the
right to obtain ba=ab