Let \(R\)be a Bézout domain with quotient field \(K\) and group of divisibility \(G(R)=K^{\ast }/U(R)\). Then \(G(R)\) is lattice-ordered, i.e., an \(l\)-group. Conversely, the Krull-Kaplansky-Jaffard-Ohm Theorem says that every \(l\)-group arises in this way. More precisely, if \(G\)is an \(l\)-group and \(k\) is any field, the map \(v:k[G]\rightarrow G\cup \{\infty \}\) given by \(v(\Sigma \alpha _{g}g)=\inf \{g\in G\mid \alpha _{g}\neq 0\}\) extends to a semivaluation on the quotient field \(k(G)\) of \(k[G]\) and \(R_{v}=\{\alpha \in k(G)\mid v(\alpha )\geq 0\}\) is a Bezout domain with \(G(R_{v})\cong G\). Now since every overring (a ring between \(R\) and its quotient field) of a Bezout domain \(R\) is a localization, the overrings of \(R\) are in a one-to-one correspondence with the factor \(l\)-groups of \(G(R)\).
M. Anderson [in: Ordered algebraic structures, Ordered algebraic structures, Proc. Caribb. Math. Found. Conf., Curaçao/Neth. Antilles 1988, Math. Appl., D. Reidel Publ. Co. 55, 3–9 (1989; Zbl 0721.06018)] conjectured that every \(l\)-embedding \(G(R)\rightarrow H\)comes from an extension \(R\hookrightarrow S\) of Bézout domains where \(S\) is no longer required to be an overring of \(R\). The paper under review proves this conjecture. A connection with the spectral spaces of Hochster is also given.