maximal ideal is prime (general case)

Theorem. In a ring (not necessarily commutativePlanetmathPlanetmathPlanetmath) with unity, any maximal idealMathworldPlanetmath is a prime idealMathworldPlanetmathPlanetmath.

Proof.  Let 𝔪 be a maximal ideal of such a ring R and suppose R has ideals 𝔞 and 𝔟 with 𝔞𝔟𝔪, but 𝔞𝔪. Since 𝔪 is maximal, we must have 𝔞+𝔪=R. Then,


Thus, either 𝔞𝔪 or 𝔟𝔪. This demonstrates that 𝔪 is prime.

Note that the condition that R has an identity elementMathworldPlanetmath is essential. For otherwise, we may take R to be a finite zero ringMathworldPlanetmath. 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