Prove if a ring has a unity, then it is unique: Here is what I have so far: Proof: Assume there exists a ring R that contains two distinct unity's, call a and b, where a != b. By the definition of a unity, we get ax = xa = x and bx = xb = x for all x != 0 in R. So, ax = xa = bx = xb = x. If the ring is an integral domain, we get a = b because there are no zero divisors. Problem occurs for the case of an integral domain. Thanks for help.