Consider a commutative ring, \(R\), and a maximal ideal \(I\), what can you say about the structure of \(R/I\)? What if \(I\) were prime?
Solution
Let \(R\) be commutative with \(1\).
If \(I\) is maximal, then the ideals of \(R/I\) correspond to ideals of \(R\) containing \(I\). Maximality leaves only \[ (0)\quad\text{and}\quad R/I. \] Thus every nonzero class in \(R/I\) generates the whole ring, hence is invertible. Therefore \(R/I\) is a field.
If \(I\) is prime, then \[ (a+I)(b+I)=0 \] means \(ab\in I\). Primality gives \(a\in I\) or \(b\in I\), so \[ a+I=0\quad\text{or}\quad b+I=0. \] Hence \(R/I\) has no zero divisors and is an integral domain.
Since every field is an integral domain, every maximal ideal is prime.