Strongly comaximizable primes
\textit{W. Heinzer} and \textit{S. Weigand} [Proc. Am. Math. Soc. 107, No. 3, 577-586 (1989; Zbl 0693.13002)] showed that if \(P\) is a nonzero prime ideal in a Noetherian domain \(R\) and if either \(R\) is not Henselian or \(P\) is not maximal, then for any integer \(m \geq 1\), there exists a finitely generated integral extension domain \(T\) of \(R\) in which at least \(m\) primes lie over \(P\). -- \textit{R. Heitmann} and the author [Proc. Am. Math. Soc. 112, No. 3, 661-669 (1991; Zbl 0741.13004)] strengthened this by showing that if \(R\) is not Henselian, then in fact \(T\) can be chosen so that there is a set of \(m\) primes in \(T\) which lie over \(P\) and which are pairwise comaximal. However, in this last result, there may be other primes which lie over \(P\). A prime ideal \(P\) in \(R\) is called strongly comaximizable if for every integer \(m \geq 1\), there is a finitely generated integral extension domain \(T\) of \(R\) such that there are exactly \(m\) primes of \(T\) lying over \(P\) and that those \(m\) primes are pairwise comaximal. In the paper under review the author shows that if \(P\) is not contained in the Jacobson radical of \(R\), then \(P\) is strongly comaximizable. (If \(P\) is contained in the Jacobson radical of \(R\), then \(P\) may or may not be strongly comaximizable.) Moreover, if \(R\) is a nearly Noetherian domain, then the author considers the following problem: Question. Are the concepts of \(H\)-prime and not strongly comaximizable prime equivalent? The author shows that if \(R\) is integrally closed, the answer is yes and if \(R\) is quasi-local, then not strongly comaximizable primes are \(H\)- primes (but he does not know if the converse holds).
scientific article; zbMATH DE number 703028
scientific article