Question

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.

Answer 1

The N&alpha;(R)&rsquo;s form an ascending chainof ideals in Nil*R. Write P(R) = N&alpha;(R) where &alpha; is an ordinal with Card &alpha; &gt; Card R. Then, for any ordinal &alpha;' with Card &alpha;' &gt; Card R, we have P(R) = N&alpha;' (R). Since P(R) &sube; Nil*R, it suffices to show that Nil*R &sube; 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 &sube; P(R) since Nil*R is the smallest semiprime ideal of R.