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单词 IdealsWithMaximalRadicalsArePrimary
释义

ideals with maximal radicals are primary


PropositionPlanetmathPlanetmathPlanetmath. Assume that R is a commutative ring and IR is an ideal, such that the radicalPlanetmathPlanetmathPlanetmathPlanetmath r(I) of I is a maximal idealMathworldPlanetmath. Then I is a primary idealMathworldPlanetmath.

Proof. We will show, that every zero divisorMathworldPlanetmath in R/I is nilpotentPlanetmathPlanetmath (please, see parent object for details).

First of all, recall that r(I) is an intersectionMathworldPlanetmath of all prime idealsMathworldPlanetmathPlanetmath containing I (please, see this entry (http://planetmath.org/ACharacterizationOfTheRadicalOfAnIdeal) for more details). Since r(I) is maximal, it follows that there is exactly one prime ideal P=r(I) such that IP. In particular the ring R/I has only one prime ideal (because there is one-to-one correspondence between prime ideals in R/I and prime ideals in R containing I). Thus, in R/I an ideal r(0) is prime.

Now assume that αR/I is a zero divisor. In particular α0+I and for some β0+IR/I we have

αβ=0+I.

But 0+Ir(0) and r(0) is prime. This shows, that either αr(0) or βr(0).

Obviously αr(0) (and βr(0)), because r(0) is the only maximal ideal in R/I (the ring R/I is local). Therefore elements not belonging to r(0) are invertiblePlanetmathPlanetmath, but α cannot be invertible, because it is a zero divisor.

On the other hand r(0)={x+IR/I|(x+I)n=0 for some n}. Therefore α is nilpotent and this completesPlanetmathPlanetmathPlanetmath the proof.

Corollary. Let p be a prime number and n. Then the ideal (pn) is primary.

Proof. Of course the ideal (p) is maximal and we have

r((pn))=r((p)n)=(p),

since for any prime ideal P (in arbitrary ring R) we have r(Pn)=P. The result follows from the proposition.

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更新时间:2025/5/4 13:22:31