Let x, y be elements in a ring R such that Rx = Ry. Show that there exists a right R-module isomorphism f : xR → yR such that f(x) = y.
Define f(xr) = yr for any r ∈ R. Note that if xr = xr', then x(r − r') = 0 implies that y(r − r') ∈ Rx(r − r') = 0. This shows that yr = yr', so f is well-defined. Clearly, f is an R-module epimorphism from xR to yR, with f(x) = y. Finally, if xr ∈ker(f), then yr = 0 and we have xr ∈(Ry)r = 0. Therefore, f is the isomorphism we want.