Suppose R is a commutative ring with identity. Such that for all a in R there exists a unique b, such that aba=a. How will we show that R is an Integral domain.
The property you mention is more or less Von Neumann regularity. In fact you say that a ring R is Von Neumann regular if for any element x in R there exists an element y such that xyx=x. For a commutative ring R, this is equivalent to say that for all x there is a unique y such that xyx=x and yxy=y. In your question there is probably something missing because if you take a=0, then aba=a for all b in R, so you cannot have uniqueness (unless R is trivial).
Anyway, any commutative reduced ring with Krull dimension 0 is Von Neumann regular, so your ring does not need to be a domain (e.g. take a product of fields). On the other hand, if it is an integral domain, then it is a field.
@Simone Yeah this is a property of regular rings. But how will we show that this R is an integral domain? I want to show that R is an integral domain. I hope you got my point
Yes... maybe I was not clear enough. Take two primes p, q and take the finite fields Z/pZ and Z/qZ. Then, the product R=Z/pZ x Z/qZ is not a domain, (e.g. the element (1,0) is a 0-divisor) but it is Von Neumann regular.
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