---
tag: A1QC3J125A
kind: proof
name: Proof that maximal ideals are prime
msc: [13A15]
proves: 05WAF5T0GZ
uses: [0GYGHKX834, 09J6KGT65G, 06HC6HK0Z8, 0WD3SE7S1X, 0QA8CHH8MG, 09S3PFBEPM, 0AVXBEC05W]
sources:
  - url: https://proofwiki.org/wiki/Maximal_Ideal_of_Commutative_and_Unitary_Ring_is_Prime_Ideal
---

## Proof

1. Let $R$ be a [[commutative ring|0GYGHKX834]] and let $M$ be a [[maximal ideal|09J6KGT65G]] of $R$.
2. Since $M$ is maximal, the quotient ring [[$R/M$ is a field|0WD3SE7S1X]].
3. Every [[field|0QA8CHH8MG]] $K$ is an [[integral domain|09S3PFBEPM]]. It is commutative with $1 \neq 0$. If $x y = 0$ in $K$ and $x \neq 0$, then $y = (x^{-1} x) y = x^{-1} (x y) = x^{-1} \cdot 0 = 0$.
4. So $R/M$ is an integral domain.
5. Since $R/M$ is an integral domain, [[$M$ is a prime ideal|0AVXBEC05W]].

## Idea

Maximal ideals are exactly those with a field as quotient, and prime ideals are exactly those with an integral domain as quotient. Since every field is an integral domain, every maximal ideal is prime.