\(R\) is a commutative ring with 1. Prove that if \(I\) is a maximal ideal in \(R\), then \(R/I\) is a field. Prove that if \(R\) is a PID, then every nonzero prime ideal in \(R\) is maximal. Conclude that if \(R\) is a PID and \(p\in R\) is prime, then \(R/(p)\) is a field.
Solution
1. If \(I\) is maximal in a commutative unital ring \(R\),then \(R/I\) is a field.
A maximal ideal is proper, so \(1+I\ne0+I\). For a nonzero coset \(a+I\), one has \(a\notin I\). The ideal \(I+(a)\) strictly contains \(I\), and hence equals \(R\). Write \(1=i+ba\) with \(i\in I\) and \(b\in R\). Passing to the quotient gives \((b+I)(a+I)=1+I\). Thus every nonzero element of the nonzero commutative quotient is invertible.
2. Every nonzero prime ideal \(P\) of a PID is maximal.
Write \(P=(a)\) with \(a\ne0\). Let \(J\) be an ideal containing \(P\), and write \(J=(b)\). The containment gives \(a=bc\) for some \(c\in R\). Since \(bc\in P\) and \(P\) is prime, either \(b\in P\) or \(c\in P\). In the first case \(J=(b)\subseteq P\), so \(J=P\). In the second case write \(c=ad\). Then \(a=bad\), and cancellation of the nonzero \(a\) in the domain gives \(1=bd\). Thus \(b\) is a unit and \(J=R\). There is no ideal strictly between the proper ideal \(P\) and \(R\), proving maximality.
3. If \(p\) is a prime element of a PID, then \(R/(p)\) is a field.
By definition a prime element is nonzero and not a unit, and \(p\mid ab\) implies \(p\mid a\) or \(p\mid b\). These assertions say respectively that \((p)\) is nonzero, proper, and a prime ideal. Step 2 makes it maximal, and step 1 makes its quotient a field.