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

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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 $b c$.


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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 $\map \nu b < \map \nu {b c}$.

$\blacksquare$


Sources

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