K[x]/(f) is a field iff f is irreducible

Statement

Let $K$ be a field and let $f \in K[x]$. Then the quotient ring $K[x]/(f)$ is a field if and only if $f$ is irreducible over $K$.

Remarks

The proof: $K[x]$ is a principal ideal domain. So $(f)$ is maximal if and only if $f$ is irreducible, by the theorem on PIDs. Then $R/M$ is a field iff $M$ is maximal. For example, $\mathbb{R}[x]/(x^2 + 1) \cong \mathbb{C}$ and $\mathbb{Q}[x]/(x^2 - 2) \cong \mathbb{Q}(\sqrt{2})$. The field $\mathbb{F}_2[x]/(x^2 + x + 1)$ has $4$ elements. If $f$ is irreducible of degree $n$, then $K[x]/(f)$ is an extension of $K$ of degree $n$. The class of $x$ in $K[x]/(f)$ is a root of $f$.

Explanations

L1 human

You can build new fields from polynomials. Take an irreducible polynomial $f$ and declare $f(x) = 0$. The result is a field in which $f$ has a root, as the complex numbers come from $x^2 + 1 = 0$.