Almost-free \(E(R)\)-algebras and \(E(A,R)\)-modules (Q2759154)

Let \(R\) be a commutative ring and \(A\) an \(R\)-algebra with unit \(1\) common to \(R\) and \(A\). A left \(A\)-module \(M\) is called an \(E(A,R)\)-module if \(\Hom_R(A,M)=\Hom_A(A,M)\) and \(A\) is an \(E(R)\)-algebra if \(_AA\) is an \(E(A,R)\)-module. These concepts are generalizations of the well-established and important concepts of \(E\)-module and \(E\)-ring of Abelian group theory. Among other basic properties the authors prove that an \(E(R)\)-algebra is commutative (Theorem~3.3). This means that its \(R\)-endomorphism ring \(\text{End}_RA\cong A\) is commutative and therefore large \(E(R)\)-algebras are far from being free \(R\)-modules. Can a large \(E(R)\)-algebra \(A\) be almost free in the sense that every \(R\)-submodule of cardinality less than \(|A|\) is embedded in a free \(R\)-submodule of \(A\)? The answer to this question requires special set theoretic assumptions such as ``diamond'', ``weak diamond'' and ``half diamond''. A stronger version of ``almost free'' is considered, namely ``polynomial-almost-free'' which means that all subalgebras of \(A\) of smaller cardinality are contained in a polynomial ring over \(R\). Along the way there are interesting lemmas on polynomial rings with countably many commuting indeterminates.NEWLINENEWLINENEWLINEA representative result is Theorem~5.5. Assume ZFC + half diamond. Suppose that \(R\) is a countable torsion-free domain that is not a field. Then, for any regular non-weakly compact cardinal \(\kappa>\aleph_0\), there exists a polynomial-almost-free \(E(R)\)-algebra \(A\) of cardinality \(\kappa\). -- Definition~4.2 should read: \(M\) is \(\kappa\)-free if every submodule \(N\) of \(M\) of cardinality less than \(\kappa\) is contained in a free submodule of \(M\).