# Answer to Question #16738 in Algebra for sanches

Question #16738

Let a, b, c be such that ab + c = 1 in a ring R. If there exists x ∈ R such that a + cx ∈ U(R), show that there exists y ∈ R such that b + yc ∈ U(R).

Expert's answer

Write

Therefore, for

=

=

Adding the first and the last equation yields

*u*=*a*+*cx**∈**U(**R*). We claim that the element*y*: = (1*− bx*)*u*^{−}^{1}works, i.e.*v*: =*b*+ (1*− bx*)*u*^{−}^{1}*c**∈**U(**R*)*.*To see this, note that*vx*=*bx*+ (1*− bx*)*u*^{−}^{1}(*u − a*) = 1*−*(1*− bx*)*u*^{−}^{1}*a,**vx*(1*− ba*) = 1*− ba −*(1*− bx*)*u*^{−}^{1}*a*(1*− ba*) = 1*− ba −*(1*− bx*)*u*^{−}^{1}(1*− ab*)*a*= 1*−*[*b*+ (1*− bx*)*u*^{−}^{1}*c*]*a*= 1*− va.*Therefore, for

*w*: =*a*+*x*(1*− ba*)*,*we have*vw*= 1. We finish by showing that*wv*= 1 (for then*v**∈**U(**R*)). Note that*wb*=*ab*+*xb*(1*− ab*) =*ab*+*xbc,**w*(1*− bx*) =*a*+*x*(1*− ba*)*− abx − xbcx*=

*a*+ (1*− ab*)*x − xb*(*a*+*cx*)=

*a*+*cx − xbu,**w*(1*− bx*)*u*^{−}^{1}*c*=*c − xbc.*Adding the first and the last equation yields

*wv*=*ab*+*c*= 1, as desired.
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