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.

Question

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.

Answer 1

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 $b=0$.

Answer 2

In your proof, you say that $apr=aps\Rightarrow pr=ps$ "because $P$ is an integral domain."

Even if I do not like the fact of using domains with no unit, you can of course do it. But to say that $apr=aps$ implies $pr=ps$, you implicitly write $a(pr-ps)=0$ and assume that $a\not=0$ (that you did not write but implicitly you assume this) and want to deduce that $pr=ps$. This is true in an integral domain, which is your assumption on $P$, but the element $a$ does not belong to $P$. This is the problem in your proof.

For a correct (and simple) proof, see the answer of user26857.