# Answer to Question #17267 in Abstract Algebra for Tsit Lam

Question #17267

Show that rad R is a quasi-regular ideal which contains every quasi-regular left (resp. right) ideal of R. (In particular, rad R contains every nil left or right ideal of R.)

Expert's answer

To show that rad

Choose

(

shows that

*R is a left ideal*, it suffices to check that, if*a, b**∈*rad*R*, then*r*(*a*+*b*) is left quasi-regular for any*r**∈**R*. For any*c , d**∈**R*, note that*c ◦ d ◦ r*(*a*+*b*)=*c ◦*[*d*+*ra*+*rb − dr*(*a*+*b*)] =*c◦*[*d ◦ ra*+ (*r − dr*)*b*]*.*Choose

*d*such that*d ◦ ra*=0; then choose*c*such that*c ◦*(*r − dr*)*b*= 0. The above equation shows that*c ◦ d*is a left quasi-inverse of*r*(*a*+*b*), as desired. To show that rad*R is also a right ideal*,we must check that*a**∈*rad*R*and*s**∈**R*imply*as**∈*rad*R*. For any*r**∈**R*,*sra*is left quasiregular, so*r*(*as*) is also left quasi-regular. This shows that*as**∈*rad*R*.*Next, we show that*rad*R is a quasi-regular ideal*.It suffices to show that every element*a**∈*rad*R*is left quasi-regular. Since*a*^{2}*∈**Ra*is left quasi-regular, there exists*b**∈**R*such that*b ◦ a*^{2}= 0. But then(

*b ◦*(*−a*))*◦ a*=*b ◦*((*−a*)*◦ a*) =*b ◦ a*^{2}= 0shows that

*a*is left quasi-regular. By definition, rad*R*contains every quasi-regular left ideal. Let*I*be any quasi-regular*right*ideal. For*a**∈**I*,*aR*is right quasi-regular, therefore quasi-regular. We also see that*Ra*is (left) quasi-regular, so*I**⊆*rad*R*, as desired.Need a fast expert's response?

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