Example 2

Let A be a principal ideal domain that is not a field, and let K be its field of fractions. As a K-module, K is without radical. Let us prove that the radical of K, viewed as an A-module, is equal to K; equivalently, we must prove that every A-linear mapping $f$ from K to a simple A-module S is zero. By VII, §4, No. 8, p. 25, we may assume that S is equal to $A/(\pi )$, where $\pi$ is an irreducible element of A. We have $f(x) =f(\pi \frac{x}{\pi})=\pi f(\frac{x}{\pi})= 0$ for every $x\in K$ because $\pi S = 0$.