# Answer to Question #17369 in Algebra for Melvin Henriksen

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.

(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)

(3)

(4)

*⇒**(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*(1^{n}*− a*) = 0 implies that 1^{n}b*− a*^{n}b*∈**p. This shows that**R/*p is a field, so p is a maximal ideal.
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