TITLE: If $R$ is a integral domain and $S$ is a subring of $R$ then is $S$ an integral domain automatically?
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: Rings (should) always have an identity; otherwise call them pseudo-rings or rngs or non-unital rings. Subrings are injective ring homomorphisms, and of course ring homomorphisms preserve (by definition) the whole structure, including the unit. This is no special convention - it follows from general principles of general algebra.