ideals with maximal radicals are primary
Proposition. Assume that is a commutative ring and is an ideal, such that the radical of is a maximal ideal. Then is a primary ideal.
Proof. We will show, that every zero divisor in is nilpotent (please, see parent object for details).
First of all, recall that is an intersection of all prime ideals containing (please, see this entry (http://planetmath.org/ACharacterizationOfTheRadicalOfAnIdeal) for more details). Since is maximal, it follows that there is exactly one prime ideal such that . In particular the ring has only one prime ideal (because there is one-to-one correspondence between prime ideals in and prime ideals in containing ). Thus, in an ideal is prime.
Now assume that is a zero divisor. In particular and for some we have
But and is prime. This shows, that either or .
Obviously (and ), because is the only maximal ideal in (the ring is local). Therefore elements not belonging to are invertible, but cannot be invertible, because it is a zero divisor.
On the other hand . Therefore is nilpotent and this completes the proof.
Corollary. Let be a prime number and . Then the ideal is primary.
Proof. Of course the ideal is maximal and we have
since for any prime ideal (in arbitrary ring ) we have . 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 |