Rings with distributive lattice of right ideals. (Q1092986)

Rings \(R\) without zero divisors are considered with a distributive lattice of right ideals of \(R\). This implies that (i) \(R\) is a right Ore-domain, and (ii) \(R\) has a right quotient field \(K\supseteq R\). 1. \textit{H. H. Brungs} proved [in J. Algebra 40, 392--400 (1976; Zbl 0346.16006)]: \(R\) is the intersection of right chain-rings \(R_ i\) \((i\in I)\), where all \(R_ i\) are subrings of \(K\). Here a right chain ring \(R_ i\) is a ring with the property that the right ideals of \(R_ i\) are linearly ordered. If \({\mathcal B}\) is the collection of all right chain-rings \(B\), such that \(R\subseteq B\subseteq K\), then \(R=\cap_{B\in {\mathcal B}}B\). 2. Let \(R\) be an algebraic extension of its center \(C(R)\); then (i) for each \(k\in K\), there is an element \(r\in R\) and \(0\neq c\in C(R)\), such that \(k=rc^{-1}\); (ii) for all \(0\neq k\in K\), and all \(B\in {\mathcal B}\), we have \(k\in B\) or \(k^{-1}\in B\). (iii) for all \(a,b,c\) in \(K\) we have \(aR\cap (bR+cR)=(aR\cap bR)+(aR\cap cR)\); (iv) for all \(k\in K\), \(\exists n\in\mathbb N\) with \(k^ nB=Bk^ n\) for all \(B\in\mathcal B\). 3. If \(R\) is an algebraic extension of \(C(R)\), then also the lattice of the left ideals of \(R\) is distributive.