# Answer to Question #17275 in Algebra 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

Then rad

Consider

*V*=*R/*m, where mis a maximal left ideal of*R*containing*I*. Then*IV*= 0, so*I**⊆**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)*e*where_{i}k*k*is any division ring, and let*R*= End(*Vk*).Then*V*is a simple left*R*-module. However, ann*R*(*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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