Exercise A0.V.2.17

Proposition

Let $R$ be a Euclidean domain that is not a field. Then there exists a non-zero non-unit element $c \in R$ such that dividing anything by $c$ has a remainder that is either a unit or $0.$

Proof

By 2.16 we know that units have minimal valuation. Let $v$ be the Euclidean valuation on $R$ and let

$$S = \{ r \in R \mid r \notin R^{\times}, r \neq 0 \}.$$

Since $R$ is not a field, it must have a non-zero non-unit, so $S$ is not empty. Let $c \in S$ be an element minimizing $v.$ Then, for any $a \in R,$ we can write $a = qc + r$ with either $r = 0$ or $v(r) < v(c).$ But only elements of $R$ with lower valuation than $c$ are units. Thus, $r$ is either $0$ or a unit, as required.

$\blacksquare$