Filtrative and multiplicatively monotone Euclidean implies uniquely Euclidean

Statement

Suppose $R$ is an integral domain with a Euclidean norm $N$ satisfying the following two conditions:

$N$ is a filtrative norm, so is a filtrative Euclidean norm: For any natural number $n$ , the set ${0} \cup {a \in R ∣ N (a) < n}$ is an additive subgroup of $R$ .
$N$ is a multiplicatively monotone norm, so is a multiplicatively monotone Euclidean norm: If $a b \neq 0$ , we have $N (a b) \geq max {N (a), N (b)}$ .

Then, $N$ is a uniquely Euclidean norm: for any $a, b \in R$ with $b \neq 0$ , there exists a unique pair $(q, r)$ satisfying $a = b q + r$ and $r = 0$ or $N (r) < N (b)$ .

Proof

Given: Integral domain $R$ , Euclidean norm $N : R ∖ {0} \to N_{0}$ that is filtrative and multiplicatively monotone.

To prove: $N$ is uniquely Euclidean: for any $a, b \in R$ with $b \neq 0$ , there exists a unique pair $(q, r)$ for which $a = b q + r$ and $r = 0$ or $N (r) < N (b)$ .

Proof: Suppose there are two solution pairs: $(q_{1}, r_{1})$ and $(q_{2}, r_{2})$ . Then, we have:

$a = b q_{1} + r_{1} = b q_{2} + r_{2}$ .

We have $b (q_{2} - q_{1}) = r_{1} - r_{2}$ : This follows by manipulating the equation $b q_{1} + r_{1} = b q_{2} + r_{2}$ .
Either $r_{1} = r_{2}$ or $N (r_{1} - r_{2}) < N (b)$ : By the definition of Euclidean norm, both $r_{1}$ and $r_{2}$ belong to the set ${0} \cup {r \in R ∣ N (r) < N (b)}$ . Since $N$ is filtrative, this set is a subgroup, so $r_{1} - r_{2}$ also belongs to this set.
Either $q_{1} = q_{2}$ or $N (b (q_{2} - q_{1})) \geq N (b)$ : If $q_{1} \neq q_{2}$ , then $q_{2} - q_{1} \neq 0$ . Since $R$ is an integral domain, the product $b (q_{2} - q_{1})$ is nonzero, so by the fact that $N$ is multiplicatively monotone, we get $N (b (q_{2} - q_{1})) \geq N (b)$ .
$(q_{1}, r_{1}) = (q_{2}, r_{2})$ : Combining steps (1), (2) and (3), we obtain that we cannot have both $r_{1} \neq r_{2}$ and $q_{1} \neq q_{2}$ . Thus, either $q_{1} = q_{2}$ or $r_{1} = r_{2}$ . But $q_{1} = q_{2}$ yields, by step (1), that $r_{1} = r_{2}$ . Similarly $r_{1} = r_{2}$ , along with step (1) and the fact that $R$ is an integral domain, yields $q_{1} = q_{2}$ . Hence, $(q_{1}, r_{1}) = (q_{2}, r_{2})$ .