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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?

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  • $\begingroup$ 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. $\endgroup$ Jun 17, 2020 at 20:05

1 Answer 1

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To answer the first question, I believe a trivial example to consider is the following.

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 $N(a)<N(b),$ no matter what $\delta_a, \delta_b$ are. Then we can set $q=0$, that is, $$a=0\cdot b+a,\;\; N(a-qb)=N(a)<N(b).$$ So the condition (1) is verified.

To see that the condition (2) fails, note that we have, for example, $$a=\frac{3}{1},\; b=\frac{1}{3}, N(a)=1>0=N(ab).$$

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