Element is Unit iff its Euclidean Valuation equals that of 1

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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 $a \in D$.

Then:

$a$ is a unit of $D$ if and only if $\map \nu a = \map \nu 1$


Proof

For $a \in D$ we have that:

$\map \nu 1 \le \map \nu {1 a} = \map \nu a$

by definition of Euclidean valuation.


Let $a$ be a unit of $D$.

Then:

$\exists b \in D: a b = 1$

Then:

$\map \nu a \le \map \nu {a b} = \map \nu 1$

and so:

$\map \nu a = \map \nu 1$

$\Box$


Let $\map \nu a = \map \nu 1$.

We can write this as:

$\map \nu {1 a} = \map \nu 1$

and it follows from Euclidean Valuation of Non-Unit is less than that of Product that $a$ is a unit of $D$.

$\blacksquare$


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