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) MaximalIdealIsPrimegeneralCase 2013-03-22 17:38:02 2013-03-22 17:38:02 mclase (549) mclase (549) 8 mclase (549) Theorem msc 16D25 MaximalIdealIsPrime