ideals with maximal radicals are primary
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 .
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|
|Date of creation||2013-03-22 19:04:31|
|Last modified on||2013-03-22 19:04:31|
|Last modified by||joking (16130)|