Maximal ideals are prime

Statement

Let $R$ be a commutative ring and let $M$ be a maximal ideal of $R$. Then $M$ is a prime ideal.

Remarks

The proof: $R/M$ is a field, so it is an integral domain, so $M$ is prime. The converse is false: the zero ideal of $\mathbb{Z}$ is prime but not maximal. In $\mathbb{Z}$, every nonzero prime ideal is maximal.

Explanations

L1 human

Every field is an integral domain. So when $R/M$ is a field, $R/M$ is an integral domain, and $M$ is prime. The zero ideal of the integers shows that a prime ideal need not be maximal.