maximal ideal is prime
Proof. Let be a maximal ideal of such a ring and let the ring product belong to but e.g. . The maximality of implies that . Thus there exists an element and an element such that . Now and belong to , whence
So we can say that along with , at least one of its factors (http://planetmath.org/Product) belongs to , and therefore is a prime ideal of .
|Title||maximal ideal is prime|
|Date of creation||2013-03-22 17:37:59|
|Last modified on||2013-03-22 17:37:59|
|Last modified by||pahio (2872)|