Let \(R\) be a PID and let \((p)\) be a prime ideal of \(R\). Prove that \(R/(p)\) is a field.
The statement needs \(p\neq0\): the zero ideal of a PID is prime, and \(R/(0)\cong R\) is a field only when \(R\) is; for example \((0)\subset\ZZ\) is prime and \(\ZZ\) is not a field.
Solution
Let \((p)\) be a nonzero prime ideal in the PID \(R\). Since every nonzero prime ideal in a PID is maximal, \((p)\) is maximal.
1. Every \(x+(p)\ne (p)\) in \(R/(p)\) is invertible.
Here \(x\notin(p)\). Because \((p)\) is maximal and \((p)\subsetneq (p,x)\), one has \[ (p,x)=R. \] Hence there exist \(a,b\in R\) such that \[ ap+bx=1. \] Reducing modulo \((p)\) gives \[ (b+(p))(x+(p))=1+(p). \] Thus every nonzero class in \(R/(p)\) has a multiplicative inverse.
2. Therefore \(R/(p)\) is a field.
A commutative ring with identity is a field exactly when every nonzero element is invertible. This holds by 1.