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 $\\mathfrak{a}\subseteq\\mathfrak{m}$ .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.