Show that \(I \normal R\) is prime iff \(R/I\) is an integral domain.
Solution
Let \(R\) be a commutative ring with identity and \(I\) an ideal of \(R\). Recall that \(I\) is prime if \(I\neq R\) and \(ab\in I\) implies \(a\in I\) or \(b\in I\), and that a commutative ring \(S\) is an integral domain if \(1_S\neq0_S\) and \(xy=0\) implies \(x=0\) or \(y=0\).
1. \(I\neq R\) if and only if \(1+I\neq 0+I\) in \(R/I\).
\(1+I=0+I\) if and only if \(1\in I\), if and only if \(I=R\).
2. For \(a,b\in R\), \(ab\in I\) if and only if \((a+I)(b+I)=0\) in \(R/I\), and \(a\in I\) if and only if \(a+I=0\).
\((a+I)(b+I)=ab+I\), and \(x+I=0+I\) if and only if \(x\in I\).
3. Q.E.D.
By step 1, the condition \(I\neq R\) is the condition \(1\neq0\) in \(R/I\). By step 2, the implication “\(ab\in I\Rightarrow a\in I\) or \(b\in I\)” for all \(a,b\in R\) is the implication “\(\bar a\bar b=0\Rightarrow\bar a=0\) or \(\bar b=0\)” for all \(\bar a,\bar b\in R/I\), since every element of \(R/I\) has the form \(a+I\). Hence \(I\) is prime if and only if \(R/I\) is an integral domain.