maximal ideal is prime (general case)

Theorem. In a ring (not necessarily commutative) with unity, any maximal ideal is a prime ideal.

Proof. Let $\mathfrak{m}$ be a maximal ideal of such a ring $R$ and suppose $R$ has ideals $\mathfrak{a}$ and $\mathfrak{b}$ with $\mathfrak{a}\mathfrak{b}\subseteq\mathfrak{m}$ , but $\mathfrak{a}\nsubseteq\mathfrak{m}$ . Since $\mathfrak{m}$ is maximal, we must have $\mathfrak{a}+\mathfrak{m}=R$ . Then,

$\mathfrak{b}=R\mathfrak{b}=(\mathfrak{a}+\mathfrak{m})\mathfrak{b}=\mathfrak{a% }\mathfrak{b}+\mathfrak{m}\mathfrak{b}\subseteq\mathfrak{m}+\mathfrak{m}=% \mathfrak{m}.$

Thus, either $\mathfrak{a}\subseteq\mathfrak{m}$ or $\mathfrak{b}\subseteq\mathfrak{m}$ . This demonstrates that $\mathfrak{m}$ is prime.

Note that the condition that $R$ has an identity element is essential. For otherwise, we may take $R$ to be a finite zero ring. Such rings contain no proper prime ideals. As long as the number of elements of $R$ is not prime, $R$ will have a non-zero maximal ideal.

Title	maximal ideal is prime (general case)
Canonical name	MaximalIdealIsPrimegeneralCase
Date of creation	2013-03-22 17:38:02
Last modified on	2013-03-22 17:38:02
Owner	mclase (549)
Last modified by	mclase (549)
Numerical id	8
Author	mclase (549)
Entry type	Theorem
Classification	msc 16D25
Related topic	MaximalIdealIsPrime