Let R be a commutative ring, and let P be a prime ideal of R. Suppose that P has no nontrivial zero divisors in it. Show that R is an integral domain.
My proof:
Take $r,s,a \in R$ with $ar = as$, and $p \in P$. Then
\begin{align*}
par &= pas\\
a p r &= a p s && \text{R is commutative}
\end{align*}
and since $P$ is an ideal of $R$, then $pr, ps \in P$.
Since $P$ is an integral domain, then $a p r = a p s \implies p r = p s$, so $r = s$.
I'm not using the fact that $P$ is a prime ideal which leads me to believe that I messed up a step.
Best Answer
If $ab=0$, then from $0\in P$ we get $a\in P$ or $b\in P$. Since $P$ doesn't contain non-zero zero-divisors we get $a=0$ or $b=0$.