Euclidean Valuation of Non-Unit is less than that of Product

Theorem
Let $\struct {D, +, \times}$ be a Euclidean domain whose zero is $0$, and unity is $1$.

Let the valuation function of $D$ be $\nu$.

Let $b, c \in D_{\ne 0}$.

Then:
 * If $c$ is not a unit of $D$ then $\map \nu b < \map \nu {b c}$

Proof
By principal ideal domain, $D$ is a principal ideal domain

Consider the principal ideal $U = \ideal b$ of $D$ generated by $b$.

As $\nu$ is a valuation function, we have:
 * $\forall x \in D, x \ne 0: \map \nu b \le \map \nu {b \times x}$

As $U$ is a principal ideal:
 * $\forall a \in \ideal b: \exists x \in D: a = x \times b$

and so:
 * $\forall a \in U: \map \nu b \le \map \nu a$

Let $\map \nu b = \map \nu {b c}$.

Then from above:
 * $\forall a \in U: \map \nu {b c} \le \map \nu a$

From the argument in Euclidean Domain is Principal Ideal Domain, we have that $U$ is also the ideal generated by $b c$.

Thus $a b$ is a divisor of $b$ itself, that is:
 * $\exists y \in D: b = b c y$

Thus:
 * $c y = 1$

and so $c$ is a unit of $D$.

So if $c$ is not a unit, then it must be the case that $\map \nu b < \map \nu {b c}$.