## Every subring of a field which contains 1 is an integral domain

Prove that any subring of a field which contains the identity is an integral domain.

Let $R \subseteq F$ be a subring of a field. (We need not yet assume that $1 \in R$). Suppose $x,y \in R$ with $xy = 0$. since $x,y \in F$ and the zero element in $R$ is the same as that in $F$, either $x = 0$ or $y = 0$. Thus $R$ has no zero divisors. If $R$ also contains 1, then $R$ is an integral domain.

In general though there are fields which have only the trivial subrings, and thus any subring containing 1 *is* a field. (For instance $\mathbb{Z}/(p)$.)