Let $R$ be a commutative ring with $1 \neq 0$. Show that if $I$ is a prime ideal, then $R/I$ is an integral domain.
I have a proof for "Let $R$ be a commutative ring. Show that if $I$ is a prime ideal, then $R/I$ is an integral domain." Do you have to change anything when $1 \neq 0$?
Best Answer
Thm. Let $P$ be an ideal of a commutative ring $R$ with identity $1$. Then $P$ is a prime ideal of $R$ if and only if $R/P$ is an integral domain.
Proof.
Suppose that $P$ is a prime ideal of a commutative ring $R$ with $1$. Then $P$≠$R$ implies $1+P≠0+P$. As you stated $1≠0$. Hence $R/P$ is a commutative ring $R$ with identity. Assume that $(a+P)(b+P)=0+P$. Then $ab+P=0+P$ and $ab∈P$. By the definition of a prime ideal P we get $a∈P$ or $b∈P$. That is, $a+P=0+P$ or $b+P=0+P$. Thus $R/P$ is an integral domain.
Conversely, if $R/P$ is an integral domain, then $1+P≠0+P$ and $R/P$ is a commutative ring $R$ which has no zero divisors. Hence $P≠R$. Assume $ab∈P$. Then $ab+P=0+P$ and $(a+P)(b+P)=0+P$. Since $R/P$ is an integral domain, we get $a+P=0+P$ or $b+P=0+P$. So $a∈P$ or $b∈P$. Thus $P$ is a prime ideal.