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Let me provide some background before I begin (although I feel as though it's hardly needed):

Let $R$ be an integral domain. I call a function $d:R\setminus \{0\}\to\mathbb{N}\cup\{0\}$ a Euclidean function if for every $a,b\in R$, $a\ne0$, there exists $q,r\in R$ with $b=aq+r$ and either $r=0$ or $d(r).
I say that a Euclidean function satisfies the $d$-inequality if $x\mid y$ implies $d(x)\leqslant d(y)$.

It is often assumed that Euclidean functions for example, in the context of Euclidean domains, always satisfy the $d$-inequality since given a Euclidean function $d$ the function $\displaystyle \widetilde{d}(x)=\min_{y\ne 0}d(xy)$ is a Euclidean function satisfying the $d$-inequality. Thus, Euclidean domains (i.e. rings for which there exists a Euclidean function on) are precisely the same as the rings that admit Euclidean functions satisfying the $d$-inequality.

Now, while, as I said above, the study of such rings (where the only key is the existence of such functions) is no different, practically it's much nicer to have Euclidean functions satisfying the $d$-inequality since they enjoy such benefits as $a\in R^\times$ if and only if $d(a)=d(1)$.

The strange thing is, most naturally occurring Euclidean functions satisfy the $d$-inequality (e.g. the degree function on $F[x]$, field norms, etc.) And, for the life of me, I can't think of a non-contrived example of a Euclidean function that does not satisfy the $d$-inequality. So, what are some? Moreover, there will undoubtedly be some trivial, common one that I've overlooked, then I would still love to hear more obscure ones that arise naturally in more advanced contexts.

Thanks for your time!

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    I cannot think of a non-trivial one, but here is a trivial example: Let $k$ be a field with at least three elements. Then define the norm $d: k - \{0\} \longrightarrow \mathbb{N} \cup \{0\}$ by $d(1) = 1$ and $d(x)=0$ if $x \neq 1$. Then this is a norm, because for any $a,b \in k$ such that $b \neq 0$ we have $a = ab^{-1}b$, and $1|x$ for all nonzero $x$, but $d(1) > d(x)$. However, as this is in some sense an artificial example, this does not answer your question.2011-10-14
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    Rankeya, thank you, I appreciate your response. This is precisely the example I had come up with--so of course we know now that there is no implication that every Euclidean function satisfies the $d$-inequality, but this example isn't "natural"--not that you ever said it was.2011-10-14
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    Yes, precisely why I did not post this as an answer, but rather chose to just put it as a comment.2011-10-14

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