If a subring $B$ of an integral domain $A$ contains 1, then $B$ is an integral domain.

This is an exercise out of Pinter's Abstract Algebra. What I did was just noted that since $B$ is a subring of $A$, it is closed under subtraction and multiplication. Therefore, if $\exists x, y \in B$ such that $xy = 0$, $x \neq 0$ and $y \neq 0$ then this also implies that $A$ is not an integral domain, since $x, y \in A$ as well. However, this did not use the assumption that $1 \in B$, so I'm wondering if I missed something.

1 Answer

Some authors use a definition of a ring that does not include a multiplicative identity, but typically an integral domain is required to have a multiplicative identity. Therefore $A,$ being an integral domain, has a $1$, but $B$ a general subring need not include that $1$. Once $1\in B$, then you can conclude that $B$ is an integral domain.