# Examples of nonstandard Euclidean functions on Euclidean domain

An integral domain $R$ is a Euclidean domain iff there exists a function $N: R\setminus\{0\} \rightarrow \mathbb{Z}$ such that

1. If $a,b\in R$, then there exists $q\in R$ such that either $a=qb$ or $N(a-qb)<N(b)$.
2. If $a,b \in R$, then $N(a)\leq N(ab)$.

I understand that condition 2 is unnecessary, as the existence of a function satisfying condition 1 implies the existence of a function satisfying both conditions.

To understand this better, I'd like to see examples of

1. A Euclidean function on some Euclidean domain satisfying condition 1 but not condition 2.
2. A Euclidean domain with two Euclidean functions $f$ and $g$ satisfying conditions 1 and 2, such that the orderings $f(x)\leq f(y)$ and $g(x) \leq g(y)$ are nonisomorphic.

Can someone give nice examples of these?

• The function $N$ should be to non-negative integers, but not to $\Bbb Z$, $a$ in the second condition and $b$ in both conditions should be non-zero, see, for instance, here. Jun 17, 2020 at 20:05

Fix the ring $$R=\mathbb{Z}_{(2)}.$$ So an element $$a \in R$$ can be uniquely written as $$a=\pm 2^{k_a} \cdot u_a, \;\; k_a\geq 0,\; u_a=\frac{x_a}{y_a}, \; x_a, y_a \text{ coprime odd positive integers}$$ This is a good testing place since the Euclidean algorithm for $$R$$ is very simple: for two nonzero elements $$a, b \in R,$$ we have either $$a \mid b$$ or $$b \mid a$$.
On $$R$$, define the norm $$N$$ as follows:
$$N(a)=2k_a+\delta_a,\;\;\; \delta_a=\begin{cases}1\; \text{ if }u_a=3 \\ 0\; \text{ else.} \end{cases}$$
Let us check that this Euclidean norm works. Firstly, it is a Euclidean norm, meaning that it satisfies the propety (1): Let us take $$a, b \in R$$. If $$b$$ divides $$a$$ there is nothing to prove (since the case $$a=qb$$ applies). So we may assume that $$a \mid b$$ and $$b \nmid a$$. Well, that means that $$k_a < k_b$$, and so it follows that $$N(a) no matter what $$\delta_a, \delta_b$$ are. Then we can set $$q=0$$, that is, $$a=0\cdot b+a,\;\; N(a-qb)=N(a) So the condition (1) is verified.
To see that the condition (2) fails, note that we have, for example, $$a=\frac{3}{1},\; b=\frac{1}{3}, N(a)=1>0=N(ab).$$