maximal ideal is prime (general case)
Proof. Let be a maximal ideal of such a ring and suppose has ideals and with , but . Since is maximal, we must have . Then,
Thus, either or . This demonstrates that is prime.
Note that the condition that has an identity element is essential. For otherwise, we may take to be a finite zero ring. Such rings contain no proper prime ideals. As long as the number of elements of is not prime, will have a non-zero maximal ideal.
|Title||maximal ideal is prime (general case)|
|Date of creation||2013-03-22 17:38:02|
|Last modified on||2013-03-22 17:38:02|
|Last modified by||mclase (549)|