# Answer to Question #17270 in Abstract Algebra for sanches

Question #17270

Show that rad R is the intersection of all modular maximal left (resp. right) ideals of R.

Expert's answer

Let

a contradiction.

*J*be the intersection of all modular maximal left ideals of*R*. (If there are no modular maximal left ideals, we define*J*=*R*.)*First we prove J**⊆*rad*R*. Consider any*a*not fromrad*R*. Then*e*: =*xa*is not left quasi-regular for some*x**∈**R*. It is easy to check that*I*=*{r −re*:*r**∈**R}*is a modular left ideal. Since*e*is not in*I*, we have*I*is less than*R*.*I**⊆*m for some modular maximal left ideal m, with*e*not fromm.In particular,*e*=*xa*is not in*J*, so*a*is not in*J*.*We now finish by proving that*rad*R**⊆**J*. Assume the contrary. Then rad*R*is not contained in m for some modular maximal left ideal m. In particular, rad*R*+m =*R*. Let*e**∈**R*be such that*r ≡ re*(mod m) for all*r**∈**R*. Write*e*=*a*+*b*where*a**∈*rad*R*and*b**∈*m. Then*e − a**∈*m so*e − ae**∈*m. Since rad*R*is left quasi-regular, there exists*a'**∈**R*such that*a'*+*a − a'a*= 0. But then*e*=*e −*(*a'*+*a − a'a*)*e*=(*e − ae*)*− a_*(*e − ae*)*∈*m*,*a contradiction.

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