# Answer to Question #24887 in Abstract Algebra for john.george.milnor

Question #24887

For any ring R and any ordinal α, define Nα(R) as follows. For α = 1, N1(R) is a nil subideal of Nil*R which contains all nilpotent one-sided ideals of R. If α is the successor of an ordinal β, define

Nα(R) = {r ∈ R : r + Nβ(R) ∈ N1 (R/Nβ(R))}. If α is a limit ordinal, define Nα(R) = (Union over β<α) Nβ(R).

Show that Nil*R = Nα(R) for any ordinal α with Card α > Card R.

Nα(R) = {r ∈ R : r + Nβ(R) ∈ N1 (R/Nβ(R))}. If α is a limit ordinal, define Nα(R) = (Union over β<α) Nβ(R).

Show that Nil*R = Nα(R) for any ordinal α with Card α > Card R.

Expert's answer

The

*N**α*(*R*)’s form an ascending chainof ideals in Nil**R*. Write*P*(*R*) =*N**α*(*R*) where*α*is an ordinal with Card*α**>*Card*R*. Then, for any ordinal*α**'*with Card*α**' >*Card*R*, we have*P*(*R*) =*N**α**'*(*R*). Since*P*(*R*)*⊆**Nil***R*, it suffices to show that Nil**R**⊆**P*(*R*). Now*R/P*(*R*) has nononzero nilpotent ideals, so it is a semiprime ring. This means that*P*(*R*)is a semiprime ideal. Hence Nil**R**⊆**P*(*R*) since Nil**R*is the smallest semiprime ideal of*R*.Need a fast expert's response?

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