On \(\phi\)-sharp rings (Q2808593)

Let \(R\) be a commutative ring with unity, \(\mathrm{Nil}(R)\) its nilradical and \(Z(R)\) its subset of zerodivisors. The class of rings \(R\) for which \(\mathrm{Nil}(R)\) is a prime ideal comparable to each ideal of \(R\) is denoted by \(\mathcal H\). A ring \(R\) is called a \textbf{\(\phi\)-sharp} (resp. sharp) ring if for any ideals \(A\), \(B\), \(I\) which are not contained in \(\mathrm{Nil}(R)\) (resp. \(Z(R)\)) satisfying \(AB\subset I\) there exist two ideals \(A'\), \(B'\) such that \(A\subseteq A'\), \(B\subseteq B'\) and \(A'B'=I\). Even if \(R\notin\mathcal H\) it is obvious that each \(\phi\)-sharp ring \(R\) is sharp and the converse holds if \(Z(R)=\mathrm{Nil}(R)\). But it is easy to give an example of a sharp ring \(R\) belonging to \(\mathcal H\) which is not \(\phi\)-sharp: for instance \(R\) the trivial ring extension of \(D\) by \(E\), where \(D\) a local domain which is not sharp and \(E\) a torsion divisible \(D\)-module. Let \(R\in\mathcal H\), \(T(R)\) its quotient ring and \(\phi:T(R)\rightarrow R_{\mathrm{Nil }R)}\) the natural map. The authors call \(R\) a \(\phi\)-Dedekind ring if \(\phi(I)\) is an invertible ideal of \(\phi(R)\) for any ideal \(I\) which is not contained in \(\mathrm{Nil}(R)\). They prove that each \(\phi\)-Dedekind ring \(R\) is \(\phi\)-sharp. The converse holds if \(R\) satisfies some additional conditions. Other classes of rings are studied. But this paper contains no interesting result showing that the study of these classes of rings is useful. And there is no example.