Question #17270

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

Expert's answer

Let *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.

a contradiction.

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