Let $R$ be a commutative ring. Let $P$ be a prime ideal of $R$ such that $P$ has no (non-zero) zero divisors. Prove that $R$ has no (non-zero) zero divisors.
My try:
Let $ab=0$ where $a,b\in R$ .Let us assume that $a\neq 0$ then we prove that $b=0$. Assume that $b \neq 0$.
Now $ab=0\in P$ then either $a\in P $ or $b\in P$ .Now both $a,b$ can't be in $P$ as then $P$ would have zero divisors.
Suppose that $a\notin P$ then $b\in P$.
But how to arrive at a contradiction from here?
On
If you mean by zero divisors in $P$ that for both non zero $x \in P$ and $y \in R$ s.t $xy=0$ then, From your try let $b \in P$ then consider $<b> \subseteq P$ then $ab=0$ in $P$ contradiction.
Otherwise if you say that there are no non-zero $x,y \in P$ s.t. $xy=0$ then the statement is false: take $\Bbb Z_{15}$, and $P=<3>$. Then $P$ is a prime ideal which does not contain $x,y\ne0$ such that $xy=0$, but $\Bbb Z_{15}$ has zero divisors.
On
Your argument is good, but you miss a premise: suppose $a$ is a zero divisor in $R$; then $a\ne0$ and there is $b\ne0$ such that $ab=0$.
Since $0\in P$, either $a\in P$ or $b\in P$, because $P$ is a prime ideal.
Now $a\notin P$, because it's a zero divisor and none belongs to $P$. So $b\in P$; but $b$ is also a zero divisor: contradiction.
Claim: Ring $R$ is an integral domain.
Proof: 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 $a=0$
One more "similar" kind of exercise from Atiyah Macdonald (If you need hint to solve this please let me know):