Problem with rings integral domain

[placid heron]

I wanted to make sure my proof was correct.
I proved that if R is a ring with unit element and 1 = 0, then R = {0}.
I then had show in any integral domain, 0 != 1.
I said that this was true because if 0 = 1, then R = {0} which only has 1 element and thus cannot be an integral domain.

This is the full proof.
Suppose R is and ring and 1 = 0.
Let a be an element in R such that a != 1 != 0.
Then 0 = a*1 = a.
So a = 0 = 1.
Hence R = {0}.

Now we show in any integral domain, 0 != 1.
Suppose for sake of contradiction 0 = 1.
Then if R is the integral domain R = {0} and thus contains 1 element and cannot be an integral domain.

[sage burrow]

Yes, if 1=0, it is the trivial ring

My definition of integral domain requires $1\neq0$, where is this problem from?

[dusky hare]

SelahW

[sage burrow]

I guess I should ask, what's your definition of integral domain

No nonzero zero divisors?
Or cancellation property of multiplication

@placid heron

[placid heron]

This is from Abstract Algebra a Concrete Introduction Robert H. Redfield. From the book A communutative ring R with unit element which has at least two elements but no zero divisors is called an integral domain. That is a communtative ring R with unitelement and at least two elements is an integral domain if it satifies the additional axiom.

DM If x,y $\in$ R satisfy xy = 0, then x = 0 or y = 0.