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Background

Definition 1: Let $R$ be an integral domain. A mapping $\theta:R\to\Bbb{N}$ is called Dedekind-Hasse norm on $R$ if

$(i)$ $\theta(rs)=\theta(r)\theta(s)$ for all $r,s\in R$ ($\theta$ is a norm).

$(ii)$ $\theta >0$ for all $r\in R\setminus \{0\}$ ($\theta$ is positive).

$(iii)$ for $r,s\in R$, either $s$ divides $r$ or there are $x,y\in R$ such that $0<\theta(rx-sy)<\theta(s)$.

$\quad$ Note that with $s=1$, this is a Euclidean norm.

Definition 2 An integral domain $R$ is a Euclidean domain if there is a function $\delta$ from the nonzero elements of $R$ to the nonnegative integers with these properties;

(i) If $a$ and $b$ are nonzero elements of $R$, then $\delta(a)\leq \delta(ab)$.

(ii) If $a,b\in R$ and $b\neq 0_R$, then there exist $q,r\in R$ such that $a=bq+r$ and either $r=0_R$ or $\delta(r) < \delta(b)$.

Question

I want to ask the motivation behind Definition 1, the dedekind hasse norm. I understand that the Euclidean norm for Definition 2 has to do with conditions for allowing for euclidean algorithm in integral domains while for the case of Dedekind Hasse Norm, it more have to do with integral domain allowing for GCD? (I am guessing h ere since $rx-sy$ reminds me of gcd). Also, can someone exhibit an example of integral domain that satisfy and a non-example of an integral domain that violates conditions of the Dedekind Hasse Norm. I know the Dedekind hasse norm is applied showing that $\Bbb{Z}[\frac{1+\sqrt{-19}}{2}]$ is not a euclidean domain, or that if an integral domain is a PID iff it satisfies the conditions for the Dedekind Hasse Norm. But I am looking for examples of explicit integral domain like in undergraduate algebra exercises showing whether a particular integral domain is not a Dedekind Hasse norm or an integral domain with a particular $\theta$ function satisfies the Dedekind Hasse norm.

Thank you in advance.

Seth
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  • Dedekind=Hasse can be viewed as a sort of generalization of the Euclidean algorithm to PIDs, see here. – Bill Dubuque May 31 '24 at 00:24
  • @BillDubuque in your answer in that post, a Bezout domain, how is that different from a GCD domain? Dedekind Hasse norm has really nothing to do with GCD, I mean we are talking about existence of a linear combination of two integers. Another thing, you list some reference I think in one of my earlier posts or some other ones about some journal articles about definitions of associates in integral domains. I think you mentioned soemthing along the line that they are deceptively complicated. I can't find those posts. I am wondering if you can provide them again. Thank you. – Seth May 31 '24 at 00:30
  • GCD domains needn't be Bezout, e..g. see the Beware here. Dedekind-Hasse is a Euclidean-like characteriation of PIDs ($\Rightarrow$ GCD). See here on "associate" definitions in non-domains. $\ \ $ – Bill Dubuque May 31 '24 at 01:07
  • @BillDubuque ah okay okay, thank you for clarifying that. I have seen both Bezout and GCD domain in algebra textbooks. I have not taken a closer look at them but at first glance, i would have thought without aasking on here, I would have mistaken for the two to mean the same thing. – Seth May 31 '24 at 02:42

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