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 Euclidean norm: For any natural number $n$, the set $\{0\}\cup \{a\in R\mid N(a)<n\}$ is an additive subgroup of $R$.
• $N$ is a multiplicatively monotone Euclidean norm:

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 $a=bq+r$ and $r=0$ or $N(r)<N(b)$.

Proof

Given: Integral domain $R$, Euclidean norm $N:R\setminus \{0\}\to \mathbb {N} _{0}$ 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 $a=bq+r$ 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:

$a=bq_{1}+r_{1}=bq_{2}+r_{2}$.

1. We have $b(q_{2}-q_{1})=r_{1}-r_{2}$: This follows by manipulating the equation $bq_{1}+r_{1}=bq_{2}+r_{2}$.
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 $\{0\}\cup \{r\in R\mid N(r)<N(b)\}$. Since $N$ is filtrative, this set is a subgroup, so $r_{1}-r_{2}$ also belongs to this set.
3. 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 $N(b(q_{2}-q_{1}))\geq N(b)$.
4. $(q_{1},r_{1})=(q_{2},r_{2})$: 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, $(q_{1},r_{1})=(q_{2},r_{2})$.