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 ( 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


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


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

Title ideals with maximal radicals are primary
Canonical name IdealsWithMaximalRadicalsArePrimary
Date of creation 2013-03-22 19:04:31
Last modified on 2013-03-22 19:04:31
Owner joking (16130)
Last modified by joking (16130)
Numerical id 4
Author joking (16130)
Entry type Theorem
Classification msc 13C99