If $P$ is a prime then $R/P$ is an integral domain.

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I know the same question has been already asked here. So, I am not asking for any proof rather to find out what's wrong with my proof.

So, this is what I did: Let, $a+p, b+p \in R/P$, since $P$ is a prime, $ab + p = 0+p, ab\in R \Rightarrow (a+p)(b+p) = 0+p \Rightarrow a+p = 0+p \vee b+p = 0+p.$ Thus, $R/P$ is an integral domain.

Please could anyone tell me in details what's wrong with my proof?

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When you say $(a+P)(b+P)=0+P$ you miss a lot of steps to arrive to $a+P=0+P$ or $b+P=0+P$.

You should say something like this: Let's suppose that $(a+P)(b+P)=0+P$ but $(a+P)(b+P)=ab+P$, so $ab+P=0+P\Rightarrow ab\in P\Rightarrow a\in P\vee b\in P$ (Why?) $\Rightarrow a+P=0\vee b+P=0$... And like that.