# Prove if a ring has a unity, then it is unique

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.

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This might be not work, since I don't remember all the definitions off-hand... But if a and b are both unities in some ring R, then ab = a and ab = b. Hence a = b.

Muzza said:
This might be not work, since I don't remember all the definitions off-hand... But if a and b are both unities in some ring R, then ab = a and ab = b. Hence a = b.
I believe that's what we are trying to prove. If a and b are both unity's in a ring, then a = b.

Apparently my deduction wasn't explicit enough, or you looked past one of the sentences in my post.

Since a was a unity in R, ax = x for all x in R. In particular, it must hold if we set x = b. Hence ab = b.

But b was also a unity in R, so that xb = x for all x in R. In particular, it must hold if we set x = a, so that ab = a.

So we have proved that ab = a and ab = b. By transitivity, we must have that a = b.

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Muzza said:
Apparently my deduction wasn't explicit enough, or you looked past one of the sentences in my post.

Since a was a unity in R, ax = x for all x in R. In particular, it must hold if we set x = b. Hence ab = b.

But b was also a unity in R, so that xb = x for all x in R. In particular, it must hold if we set x = a, so that ab = a.

So we have proved that ab = a and ab = b. By transitivity, we must have that a = b.
Your right. I didn't even read the sentence ab = a and ab = b. My bad, sorry for the confusion and tanks very much for the help.