A proof about prime ideals Assume $R$ is commutative. Prove that if $P$ is a prime ideal of $R$ and $P$ contains no zero-divisors then $R$ is an integral domain.
Proof: let $ab \in P$ where $ab \not= 0$. that means $a \in P$ or $b \in P$. Which implies that $a+P=P$ or $b+P=P$ the only way that $a+P=P$. If $a =0$ similarly $b+P=P$ of $b=0$. Which is a contradiction thus $ab$ is an integral domain. I do not know if this is a right approach. 
 A: Try to show that: if there exists a zero-divisor, then there is a zero-divisor  in $P$.
This then shows that if there is no zero-divisor in $P$, then there is none at all and you are done. 
To do this assume: $ab=0$ with non-zero $a,b$, that is assume a zero-divisor. Now, $0 \in P$ so $a \in P$ or $b \in P$ and you are essentially done. 
A: Assuming $\mathfrak p$ is prime in $R$ and contains no nonzero zero divisors, take a minimal prime $\mathfrak q$ under $\mathfrak p$. Since $\mathfrak q$ consists entirely of zero divisors but contains no nonzero zero divisors, it must be the zero ideal. Since we've shown the zero ideal is prime, $R$ is an integral domain. 
Of course, it remains to show minimal primes consist of zero divisors. You can find some discussion here. 
A proof is sketched there: In $R_\mathfrak q$,  we know that $\mathfrak q R_\mathfrak q$ is a maximal ideal, and that the primes of $R_\mathfrak q$ are those under $\mathfrak q$. Since $\mathfrak q $ is minimal, it follows the only prime ideal of this ring is $\mathfrak q R_\mathfrak q$, so the radical of this ring is $\mathfrak q R_\mathfrak q$. This means that for every $q\in \mathfrak q$ there is a smallest $n$ such that $q^n/1=0$, i.e. there is $s\notin \mathfrak q $ such that $sq^n=0$. Since $n$ was chosen smallest, $sq^{n-1}$ is not zero, and hence $q$ is a zero divisor.  
A: Suppose $a,b\in R$ with $ab=0$, we must show that $a=0$ or $b=0$. Certainly the images $\overline a,\overline b$ of $a,b$ in $A/P$ satisfy $\overline a\overline b=0$, and since $R/P$ is an integral domain (by the definition of prime ideal) one of $\overline a,\overline b$ is zero, that is $a\in P$ or $b\in P$. Assuming by symmetry the former is the case, then either $b=0$ (in which case we are done), or else $b\neq0$ and the fact that $P$ does not contain zero divisors (of $R$) imply that $a=0$ (and we are done as well).
I think this is more or less the proof you wanted to give, but your sentences get confused from "the only way" on; I simply cannot decipher them.
