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Answer on Algebra Question for sanches

Question #17275
Defining rad' R to be the intersection of all maximal ideals of R, show that rad R ⊆ rad'R, and give an example to show that this may be a strict inclusion.
Expert's answer
Let V = R/m, where mis a maximal left ideal of R containing I. Then IV = 0, soI ⊆ ann(V ). Since ann(V ) is an ideal of R, themaximality of I implies that I = ann(V ).
Then rad R ⊆ I. Therefore, rad R ⊆ rad' R.
Consider V =(direct sum)eikwhere k is any division ring, and let R = End(Vk).Then V is a simple left R-module. However, annR(V )= 0 is not a maximal ideal in R. R is von Neumann regular,so rad R = 0. On the other hand, the only maximal ideal of R is I= {f ∈ R : dim f(V ) < ∞}so we have here rad' = I contains but not equal rad R =0.

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