(1) ⇒ (2). By (1), rad R is the intersection of all prime ideals, so itis nil. For the rest, we may assume that rad R = 0. For any a ∈ R, it suffices to show that Ra = Ra2.Let p be any prime ideal of R. Since R is reduced, so is Rp.But pRp is the only prime ideal of Rp, so we have pRp = 0.Therefore, Rp is a field. In particular, Rpa = Rpa2.It follows that (Ra/Ra2)p = 0 for every prime ideal p ⊂ R; hence Ra = Ra2, asdesired.
(2) ⇒ (3). Let a ∈ R. By (2 ),a = a2b ∈ R/rad R for some b ∈ R, so (a − a2b)n =0 for some n ≥ 1. Expanding the LHS and transposing, we get an∈ Ran+1, and hence Ran = Ran+1= · · · .
(3) ⇒ (4) is clear.
(4) ⇒ (1). Let p be any prime ideal, and a is not in p. By (4), an= a2nb for some b ∈ R and some n ≥ 1. Then an(1− anb) = 0 implies that 1 − anb ∈ p. This shows that R/p is a field, so p is a maximal ideal.
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