Let \(R\) be a principal ideal domain that is not a field, and write \(F\) for its field of fractions. Prove that \(F\) is not a finitely generated \(R\)-module.
Solution
1. Finite generation of \(F\) would give a nonzero\(d\in R\) with \(dF\subseteq R\).
Suppose \(F\) is generated over \(R\) by finitely many fractions \(a_i/b_i\), where \(a_i,b_i\in R\) and \(b_i\ne0\). Since \(F\ne0\), the generating family may be taken nonempty. Set \(d=\prod_i b_i\), which is nonzero because \(R\) is a domain. Each \(d(a_i/b_i)=a_i\prod_{j\ne i}b_j\) belongs to \(R\). Multiplying any \(R\)-linear combination of the generators by \(d\) therefore gives an element of \(R\). Thus \(dF\subseteq R\).
2. This forces \(R\) to be a field, a contradiction.
Since \(d\) is a nonzero element of the field \(F\), multiplication by \(d\) is a bijection of \(F\) onto itself: \(x=d(x/d)\) for every \(x\in F\). Thus \(F=dF\subseteq R\). The reverse inclusion is the defining inclusion into the fraction field, so \(R=F\), contradicting the hypothesis that \(R\) is not a field. Hence \(F\) is not finitely generated. The argument uses only that \(R\) is a nonfield integral domain; principality is not needed.