R/M is a field iff M is maximal

Statement

Let $R$ be a commutative ring and let $M$ be an ideal of $R$. Then the quotient ring $R/M$ is a field if and only if $M$ is a maximal ideal.

Remarks

The proof uses the correspondence theorem: the ideals of $R/M$ are the images of the ideals of $R$ that contain $M$. So $M$ is maximal if and only if $R/M$ is nonzero and has only the two ideals $\{0\}$ and $R/M$. A nonzero commutative ring with only these two ideals is a field, by the converse for fields. For example, $\mathbb{Z}/p\mathbb{Z}$ is a field because $p\mathbb{Z}$ is maximal for $p$ prime.

Explanations

L1 human

The quotient by an ideal is a field exactly when the ideal is as large as possible. In the integers, $(5)$ is maximal, and $\mathbb{Z}/5\mathbb{Z}$ is a field. The ideal $(6)$ is not maximal, and $\mathbb{Z}/6\mathbb{Z}$ is not a field.