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

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?

• I don't see you conclude that $a+P = 0+P$ or $b+P = 0+P$ from $(a+P)(b+P) = 0 + P$ – Callus - Reinstate Monica Apr 27 '15 at 0:21
• Yes, I did. Please see the line before conclusion. – Jellyfish Apr 27 '15 at 0:23
• You should use that $P$ is a prime ideal. I don't see you really using that $ab\in P \Rightarrow (a\in P\vee b\in P)$ – LeviathanTheEsper Apr 27 '15 at 0:29
• I don't understand your last comment. For example: $xy = 0$ means that $x = 0$ or $y = 0$. Doesn't it? I applied the same concept here. – Jellyfish Apr 27 '15 at 0:35
• I mean, if you say $xy=0\Rightarrow x=0\vee y=0$ you are supposing $R/P$ doesn't have zero divisors and that is which you have to prove. Not every ring has the property of $ab=0\Rightarrow a=0\vee b=0$, see $\mathbb{Z}_4$ for example, in which $2*2=0$. ------------------------------------------------------------------------- Edit: I've deleted my last comment so I put what it said: "When you say $(a+P)(b+P)=0+P$ you miss a lot of steps to arrive to $a+P=0+P$ or $b+P=0+P$.". – LeviathanTheEsper Apr 27 '15 at 0:37

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.