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The left-cancellation property by itself implies that if ab=0 then at least one of a,b must be zero, and this then implies the right-cancellation property. Of course, the same holds if left and right are interchanged.
The zero referred to here is 0=x-x, the unique ring element that is the difference of any ring element x from itself. It is easily shown that in any ring a0=0 and 0b=0, for any a,b. To see that left-cancellation implies ab=0 only if at least one of a,b is zero, just note that if a is not 0, but ab is 0, then ab=0=a0 so that b=0 by left cancellation. Hence it is impossible that ab=0 with both a and b nonzero. Now if ab=0 only if at least one of a,b is zero then, if b is not zero, xb=yb implies (x-y)b=0 so that x-y=0, or x=y and hence b can be right-cancelled. The second comment by JLo gives an excellent short proof that uses a little logic manipulation and avoids any detour through "ab=0 only if at least one of a,b is zero." |
