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 $D$.
$\blacksquare$
Sources
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $6$: Polynomials and Euclidean Rings: $\S 30$. Unique Factorization: Theorem $58 \ \text{(ii)}$