When is the dual of an ideal a ring? (Q1972044)

Let \(R\) be a domain with quotient field \(K\). For a non-zero fractional ideal \(I\) of \(R\), the fractional ideal \(I^{-1}= (R:I)= \{x\in K \mid xI \subseteq R\}\) is called the inverse (or dual) of \(I\). -- The main results established by the authors are: (1) If \(I^{-1}\) is a ring, then \(P^{-1}\) is a ring for each minimal prime ideal \(P\) of \(I\); (2) If \(P\) is a prime ideal of \(R\), then: \(P^{-1}\) fails to be a ring \(\Leftrightarrow P\) has the form \((aR_R:b)\) and \(PR_P\) is principal; (3) The inverse of a radical ideal \(I\) is a ring \(\Leftrightarrow\) for each valuation overring \(V\) of \(R\) with \(IV \neq V\), we have \(I^{-1} \subseteq V_Q\), where \(Q\) is a prime of \(V\) which is minimal over \(IV\); (4) If \(I\) and \(J\) are radical ideals, then: \(I^{-1}\) and \(J^{-1}\) are rings \(\Leftrightarrow (I\cap J)^{-1}\) and \((I+J)^{-1}\) are rings; (5) If \(I\) is the irredundant intersection of prime ideals \(P_\alpha\), then: \(I^{-1}\) is a ring \(\Leftrightarrow\) each \(P_\alpha\) is a ring; (6) If \(I\) is an ideal in an integrally closed domain \(R\), then: \(I^{-1}\) is a ring \(\Leftrightarrow I^{-1}\subseteq V\) for each valuation overring \(V\) whose maximal ideal is minimal over \(IV\). Also the authors present examples tending to show that many of the established results are the best possible.