AnyidealisasubspaceofA,soadescendingchainofidealscorrespondstoadescendingchainofsubspaces,whichmustbefinitebydimensionalconsiderations.Weconcludei).9DiscretevaluationringsandDedekinddomains3.Exercise.Avaluationring(otherthanafield)isNoetherianifandonlyifitisadiscretevaluationring.Solution.LetRbeaNoetherianvaluationringthatisnotafield.ThendimR≥1.FromProposition5.18,Risalocalringandisintegrallyclosed,sobyProposition9.2,itisenoughtoshowthatdimR=1.LetpbeaprimeidealofRandmitsmaximalideal.Supposethatp6=m.Thenchoosea∈pandb∈m\p.SinceRisavaluationring,eitherb/a∈R,inwhichcaseb∈p(contradiction),ora/b∈R,inwhichcasea/b∈pbecausepisprimeandb∈/p.Similarly,wemusthavea/bn∈pforalln≥1.Butthen(a/b)⊂(a/b2)⊂···⊂(a/bn)⊂···givesaninfiniteincreasingchain,whichcontradictsthatRisNoetherian.Weconcludethenthatp=m,sodimR=1.
全文预览
