Answer to Question #16669 in Algebra for Irvin
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.
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