ideals with maximal radicals are primary
Proposition. Assume that R is a commutative ring and I⊆R is an ideal, such that the radical
r(I) of I is a maximal ideal
. Then I is a primary ideal
.
Proof. We will show, that every zero divisor in R/I is nilpotent
(please, see parent object for details).
First of all, recall that r(I) is an intersection of all prime ideals
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 I⊆P. 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 α∈R/I is a zero divisor. In particular α≠0+I and for some β≠0+I∈R/I we have
αβ=0+I. |
But 0+I∈r(0) and r(0) is prime. This shows, that either α∈r(0) or β∈r(0).
Obviously α∈r(0) (and β∈r(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 invertible, but α cannot be invertible, because it is a zero divisor.
On the other hand r(0)={x+I∈R/I|(x+I)n=0 for some n∈ℕ}. Therefore α is nilpotent and this completes the proof. □
Corollary. Let p∈ℕ be a prime number and n∈ℕ. Then the ideal (pn)⊆ℤ 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. □
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 |