Commutative orders revisited. (Q485397)

The authors give a new description of commutative orders (semigroups having a semigroup of quotients). Let \(S\) be a commutative semigroup and \(\mathcal S(S)\) be its subsemigroup of square-cancellable elements (meaning that \(a^2x=a^2y\) always implies \(ax=ay\)). The authors prove that \(S\) is an order in a semigroup \(Q\) (that is, \(Q\) is a semigroup of quotients of \(S\)) if and only if \(S\) has a set \(\{C_e,I_e:e\in E\}\) of subsets such that \(\mathcal S(S)\) is a semilattice \(E\) of subsemigroups \(C_e\), \(e\in E\), \(S=\bigcup_{e\in E}I_e\) and four more conditions are satisfied. It is also proved (among other results) that if \(S\) is a commutative order, \(\rho\) the smallest semilattice congruence on \(\mathcal S(S)\) and \(R\) a semigroup of quotients of \(\mathcal S(S)\) inducing \(\rho\), then every semigroup of quotients of \(S\) is a homomorphic image of \(R\otimes_{\mathcal S(S)}S\). The authors introduce the notion of generalized quotient semigroups: \(Q\) is called a \textit{generalized quotient semigroup} of \(S\) and \(S\) is called a \textit{generalized order in} \(Q\) if every square-cancellable element of \(S\) lies in a subgroup of \(Q\) and \(Q=\langle S\cup\{a^\#:a\in\mathcal S(S)\}\rangle\) where \(a^\#\) is the inverse of \(a\) in a subgroup of \(Q\). Let \(S\) be a commutative semigroup and let \(\rho\) be a semilattice congruence on \(\mathcal S(S)\) with associated pre-order \(\leq\). Then \(S\) is a generalized order in a semigroup \(Q\) inducing \(\leq\) if and only if 1) For all \(a,b\in\mathcal S(S)\) and \(c\in S\), if \(a=bc\), then \(a\leq b\) and 2) For all \(b,c\in S\) and \(a\in\mathcal S(S)\) with \(b\,\overline\leq\,au\), \(c\,\overline\leq\,av\) for some \(u,v\in S^1\), if \(ab=ac\), then \(b=c\) (\(\overline\leq\) denotes the smallest compatible pre-order on \(S\) containing \(\leq\)).