Question #17369

For a commutative ring R, show that the following are equivalent:
(1) R has Krull dimension 0.
(2) rad R is nil and R/rad R is von Neumann regular.
(3) For any a ∈ R, the descending chain Ra ⊇ Ra2 ⊇ . . . stabilizes.
(4) For any a ∈ R, there exists n ≥ 1 such that an is regular.

Expert's answer

(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 *= *Ra*^{2}.Let p be any prime ideal of *R*. Since *R *is reduced, so is *R*p.But p*R*p is the only prime ideal of *R*p, so we have p*R*p = 0.Therefore, *R*p is a field. In particular, *R*p*a *= *R*p*a*^{2}.It follows that (*Ra/Ra*^{2})p = 0 for every prime ideal p *⊂** R*; hence *Ra *= *Ra*^{2}, asdesired.

(2)*⇒** *(3). Let *a **∈** R*. By (2 ),*a *= *a*^{2}*b **∈** R/*rad *R *for some *b **∈** R*, so (*a − a*^{2}*b*)*n *=0 for some *n ≥ *1. Expanding the LHS and transposing, we get *a*^{n}*∈** Ra*^{n}^{+1}, and hence *Ra*^{n} = *Ra*^{n}^{+1}= *· · · *.

(3)*⇒** *(4) is clear.

(4)*⇒** *(1). Let p be any prime ideal, and *a is not in *p. By (4), *a*^{n}= *a*^{2n}*b *for some *b **∈** R *and some *n ≥ *1. Then *a*^{n}(1*− a*^{n}b) = 0 implies that 1 *− a*^{n}b *∈** *p. This shows that *R/*p is a field, so p is a maximal ideal.

(2)

(3)

(4)

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