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.

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*.

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