Let \(R\) be a commutative ring with \(1\ne0\). Suppose that every proper principal ideal of \(R\) is prime. Prove that \(R\) is a field.
The source states the hypothesis for every principal ideal. A prime ideal is proper, so \((1)=R\) is not prime and that hypothesis is never satisfied; the problem states it for proper principal ideals. The hypothesis \(1\ne0\) excludes the zero ring, which has no proper ideals and is not a field.
Solution
1. The ring \(R\) is an integral domain.
Since \(1\ne0\), the principal ideal \((0)\) is proper and hence prime by hypothesis. If \(ab=0\), then \(ab\in(0)\), so primality gives \(a\in(0)\) or \(b\in(0)\). Thus \(R\) has no zero divisors and is an integral domain.
2. Every nonzero element of \(R\) is a unit.
Let \(a\ne0\). Suppose that \(a\) is not a unit. Then \((a^2)\) is proper: if \(1=a^2b\) for some \(b\in R\), the element \(ab\) would be an inverse of \(a\). By hypothesis \((a^2)\) is therefore prime. Since \(a\cdot a\in(a^2)\), primality implies \(a\in(a^2)\). Write \(a=a^2c\). Then \(a(1-ac)=0\), and step 1 together with \(a\ne0\) gives \(1=ac\). This makes \(a\) a unit, a contradiction. Hence every nonzero element is invertible, and the nonzero commutative unital ring \(R\) is a field.