Here is the problem that I have:
Let $R$ be an integral domain and $S$ be a subring of $R$ containing the one of $R$. Prove that $S$ is also an integral domain.
Here is my answer: Suppose for a contradiction that $S$ is not an integral domain then there exists $x,y \in S$ s.t $x,y \neq 0$ and $x \cdot y=0$ but since $S$ is a subset of $R$ then $x,y \in R$ and so $R$ is not an integral domain i.e. a contradiction. So $S$ is an integral domain. $~\square$
My problem is why does the question stipulate that $S$ must contain the one from $R$. Why is this necessary?
Best Answer
Actually, you don't even have to stipulate that $1_S=1_R$ as long as you do stipulate that S does have a nonzero identity: it can then be proven that R and S share identity.
This is because any nonzero idempotent element of a ring without nonzero zero divisors is automatically the identity of the ring.
However, requiring a subring to share the identity of the containing ring is a standard thing to do, and sidesteps having to think about the path I just outlined. integral domains "without identity" are not usually called integral domains.