maximal ideal is prime

Proof.  Let π”ͺ be a maximal ideal of such a ring R and let the ring product r⁒s belong to π”ͺ but e.g.  rβˆ‰π”ͺ. The maximality of π”ͺ implies that  π”ͺ+(r)=R=(1).  Thus there exists an element  m∈π”ͺ  and an element  x∈R  such that  m+x⁒r=1.  Now m and r⁒s belong to π”ͺ, whence


So we can say that along with r⁒s, at least one of its factors ( belongs to π”ͺ, and therefore π”ͺ is a prime ideal of R.

Title maximal ideal is prime
Canonical name MaximalIdealIsPrime
Date of creation 2013-03-22 17:37:59
Last modified on 2013-03-22 17:37:59
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 8
Author pahio (2872)
Entry type Theorem
Classification msc 16D25
Classification msc 13A15
Related topic SumOfIdeals
Related topic MaximumIdealIsPrimeGeneralCase
Related topic CriterionForMaximalIdeal