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Answer on Abstract Algebra Question 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 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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