Matrix invertible extensions over commutative rings. Part III: Hermite rings

We reobtain and often refine prior criteria due to Kaplansky, McGovern, Roitman, Shchedryk, Wiegand, and Zabavsky–Bilavska and obtain new criteria for a Hermite ring to be an \textsl{EDR}. We mention three criteria: (1) a Hermite ring $R$ is an \textsl{EDR} iff for all pairs $(a,c)\in R^2$, the product homomorphism $U(R\mathrm{/}Rac)\times U(R\mathrm{/}Rc(1-a))\to U(R\mathrm{/}Rc)$ between groups of units is surjective; (2) a reduced Hermite ring $R$ is an \textsl{EDR} iff it is a pre-Schreier ring and for each $a\in R$, every zero determinant unimodular $2\times 2$ matrix with entries in $R\mathrm{/}Ra$ lifts to a zero determinant matrix with entries in $R$; (3) a Bézout domain $R$ is an \textsl{EDD} iff for all triples $(a,b,c)\in R^3$ there exists a unimodular pair $(e,f)\in R^2$ such that $(a,e)$ and $(be+af,1-a-bc)$ are unimodular pairs. We use these criteria to show that each Bézout ring $R$ that is an $(SU{)}_2$ ring (as introduced by Lorenzini) such that for each nonzero $a\in R$ there exists no nontrivial self-dual projective $R\mathrm{/}Ra$-module of rank $1$ generated by $2$ elements (e.g., all its elements are squares), is an \textsl{EDR}.