Pullbacks of Prüfer rings
Let \(R\) be a commutative ring with identity, and let \(Q(R)\) be the total quotient ring of \(R\). An ideal is \textit{regular} if it contains an element that is not a zero-divisor. There are five extensions of Prüfer domain notion to commutative rings with zero-divisors: [1] \(R\) is \textit{semi-hereditary} if every finitely generated ideal is projective. [2] \(R\) has \textit{weak global dimension \(\leq 1\)} if every finitely generated ideal is flat. [3] \(R\) is an \textit{arithmetical ring} if the lattice formed by its ideals is distributive. [4] \(R\) is a \textit{Gaussian ring} if \(c(fg) = c(f)c(g)\) for every \(f, g \in R[X]\), where \(c(h)\) is the content of the polynomial \(h \in R[X]\). [5] \(R\) is a \textit{Prüfer ring} if every finitely generated regular ideal is invertible. It is known that the implications \((1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (4) \Rightarrow (5)\) hold [\textit{S. Glaz}, in: Arithmetical properties of commutative rings and monoids. Boca Raton, FL: Chapman \& Hall/CRC. Lecture Notes in Pure and Applied Mathematics 241, 272--281 (2005; Zbl 1107.13023)]. Let \(R \subseteq T\) be an extension of commutative rings with identity, \(C = \{x \in T| xT \subseteq R\}\) (so \(C\) is a common ideal of \(R\) and \(T\)), \(A = R/C\) and \(B = T/C\). Suppose that \(C\) is a regular ideal; hence \(T\) is an overring of \(R\), i.e., \(R \subseteq T \subseteq Q(R)\). It is well known that if \(T\) is an integral domain and \(C\) is a maximal ideal of \(T\), then \(R\) is a Prüfer domain if and only if \(A\) and \(T\) are Prüfer domains and \(B\) is the quotient field of \(A\) [\textit{M. Fontana}, Ann. Mat. Pura Appl., IV. Ser. 123, 331--355 (1980; Zbl 0443.13001)]. In this paper, the author shows (i) If \(R\) is a Prüfer ring, then \(A\) and \(T\) are Prüfer rings and \(B_P\) is an overring of \(A_P\) for each prime (maximal) ideal \(P\) of \(R\). Conversely, for each prime (maximal) ideal \(P\) of \(R\), if \(A_P\) and \(T_P\) are Prüfer rings and \(B_P\) is an overring of \(A_P\), then \(R\) is a Prüfer ring. (ii) For \(n = 1, 2, 3, 4\), \(R\) has Prüfer condition \((n)\) if and only if \(T\) has Prüfer condition \((n)\), \(A_P\) is a Prüfer ring and \(B_P\) is an overring of \(A_P\) for each prime (maximal) ideal \(P\) of \(R\).
scientific article; zbMATH DE number 5365179