
Prove integral domain.
Let R be a commutative ring with unity in which the cancellation law for multiplication holds. That is, if a, b, and c are elements of R, then a does not equal to 0 and ab = ac always imply b = c. Prove that R is an integral domain.
Here is my work:
since a can't be zero, then b = c is true. Since R is a commutative ring with unity, then eb = be = b and ec = ce = c which implies that b and c can't be zero. If not, then eb = be = 0 and ec = ce = 0. Therefore, R is an integral domain.
Am I right?

I don't think you have shown what is required. To show that this ring is an integral domain, we must show that it has no nonzero divisors, i.e.
or .
So, we assume that and that . But then . This is a contradiction, so either or must be zero.
Your definition of cancellation is a bit badly formed. You said
" if a, b, and c are elements of R, then a does not equal to 0 and ab = ac always imply b = c "
This is wrong: the fact that is not a consequence of the fact that , it is part of the definition of cancellation. Compare to the following:
" if and then ab=ac always implies b = c "
Do you see the difference?

I think you misread the question. "That is if a,b,c are elements of R then a is not equal to zero and ab=ac always implys b=c" is part of the question. It is right out of the text book. This is the given information. What we need to prove is R is an integral domain by proving the following:
1. D is a commutative ring
2. D has a unity e, and e is not = 0
3. D has no zero divisors

@natmov85
If you have a book about rings in your home which has the following statement in it:
"if a,b,c are in R then a is not zero",
then don't answer any of the questions in it, throw it away, and buy a new book. If you don't understand why then please read my original reply again, paying extra attention to the question at the end. If you still don't see why then write back and we'll discuss the consequences of the statement.
Furthermore, in the statement of the question we are allowed to assume that R is a commutative ring with unity, so that showing 1. and 2. is superfluous, and I believe I have shown 3. in my original reply.