K[x] is a Euclidean domain

Statement

Let $K$ be a field. Then the polynomial ring $K[x]$ is a Euclidean domain, with the degree as Euclidean function. That is, for all $f, g \in K[x]$ with $g \neq 0$, there are $q, r \in K[x]$ with $f = q g + r$ and $r = 0$ or $\deg r < \deg g$.

Remarks

The ring $K[x]$ is an integral domain because $K$ is one. The existence of $q$ and $r$ is the division algorithm over a field. A consequence is that $K[x]$ is a principal ideal domain and a unique factorization domain. The ring $\mathbb{Z}[x]$ is not a Euclidean domain, because it is not a principal ideal domain.

Explanations

L1 human

Over a field, you can divide any polynomial by any nonzero polynomial and get a remainder of smaller degree. This is the same situation as division with remainder for integers. So all the theory of the Euclidean algorithm works for polynomials over a field.