# Answer on Algebra Question for Irvin

Question #16669

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.

Expert's answer

Define

This shows that

*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.Need a fast expert's response?

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