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 *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.

This shows that

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