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# 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.
1
Expert's answer
2012-10-30T10:32:35-0400
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 &sube;rad R. Consider any a not fromrad R. Then e: = xa is not left quasi-regular for some x &isin; R. It is easy to check that I = {r &minus;re : r &isin; R} is a modular left ideal. Since e is not in I, we have I is less than R. I &sube;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 &sube; 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 &isin; R be such that r &equiv; re (mod m) for all r &isin; R. Write e = a + b where a &isin;rad R and b &isin;m. Then e &minus; a &isin;m so e &minus; ae &isin;m. Since rad R is left quasi-regular, there exists a&#039; &isin; R such that a&#039; + a &minus; a&#039;a = 0. But then
e = e &minus; (a&#039; + a &minus; a&#039;a)e =(e &minus; ae) &minus; a_(e &minus; ae) &isin;m,
a contradiction.

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