An integral domain $R$ is said to be Euclidean if it admits some Euclidean norm: i.e., a function $N: R \rightarrow \mathbb{N} = \mathbb{Z}^{\geq 0}$ such that: for all $x, y \in R$ with $N(y) > 0$, either $y$ divides $x$ or there exists $q \in R$ such that $N(x-qy) < N(y)$. A well-known "descent" argument shows that any Euclidean domain is a PID. In fact, the argument that a Euclidean domain is necessarily a UFD is a little more direct and elementary than the argument that shows that a PID is a UFD (because, in the latter case, one needs some kind of ideal-theoretic argument to show the existence of factorizations into irreducible elements). Because of this, Euclidean domains are a familiar staple of undergraduate algebra.

A lot of texts seem to emphasize the fact that a PID need not be a Euclidean domain. In order to show this, one has to show not only that some particular norm (and often there is a preferred norm in sight, see below) is not Euclidean, but that there is no Euclidean norm whatsoever. In general this is a very delicate question: for instance, the proof of the most standard example -- that the ring of integers of $\mathbb{Q}(\sqrt{-19})$ is a PID but does not admit *any* Euclidean norm -- is already rather intricate.

My question is this:

Given a ring $R$ that we already know is a PID, why do we care whether or not it admits some Euclidean norm?

Note that in contrast, many domains admit natural norms. A class of domains which I have been thinking about recently are the infinite domains satisfying (FN): the quotient by every nonzero ideal is finite. In this case, the map $0 \mapsto 0$, $x \in R \setminus \{0\} \mapsto \# R/(x)$ is a multiplicative norm, which I call **canonical**. For instance, the usual absolute value on $\mathbb{Z}$ is the canonical norm, as is the norm on any ring of integers in a number field that you meet in an algebraic number theory course.

I have recently realized that I care quite a bit about whether certain specific norms on integral domains are Euclidean. (This has come up in my work on quadratic forms and the Davenport-Cassels theorem.) There is some very natural algebra and discrete geometry here. But why do I care if some crazy Euclidean norm exists?

Here are three reasons that one might care about this:

If a domain admits an "effective" Euclidean norm, one can give effective algorithms for linear algebra over that ring, whereas the structure theory of modules over an arbitrary PID is not

*a priori*algorithmic in nature.(in algebraic K-theory): If $R$ is Euclidean, $\operatorname{SK}_1(R) = 0$, but there exists a PID with nonvanishing $\operatorname{SK}_1$. (Thanks to Charles Rezk for giving the precise result based on my vague allusion to it.)

In algebraic number theory, there has been a lot of work towards proving the conjecture that if $K$ is a number field which is

*not*$\mathbb{Q}(\sqrt{D})$ for $D = -19, -43, -67, -163$, then the ring $\mathbb{Z}_K$ of integers of $K$ is a PID iff it is Euclidean (for some crazy norm). In particular, disproving this would disprove the generalized Riemann hypothesis.

Comments on 1: There is something to this, but I somehow doubt that it's such a big deal. For instance, the ring of integers of $\mathbb{Q}(\sqrt{-19})$ is not Euclidean, but I'm pretty sure that there are algorithms for modules over it. In particular, it seems to me that for algorithmic purposes, having a Dedekind-Hasse norm is just as good as a Euclidean norm, and every PID has a Dedekind-Hasse norm. In fact, for every PID which satisfies (FN), the canonical norm is a Dedekind-Hasse norm. (See p. 27 of http://math.uga.edu/~pete/factorization2010.pdf for this.)

Comments on 3: if I knew more about this result, I might appreciate it better. It does seem to involve some interesting geometry of numbers. But this convinces me why I should be interested in the special case of rings of integers in number fields, which, as a number theorist, I am already convinced are more worthy of scrutiny from every possible angle than an arbitrary domain.

If there are other good reasons to care, I'd certainly like to know.

$SK_1$ of an interesting PID, JPAA 1981); the PID with $SK_1(A)\neq 0$ is $A=S^{-1}\mathbb{Z}[T]$, where $S=\{T^m-1\,,\,m\geq 1\}\cup \{T\}$. Nifty. – Charles Rezk Sep 15 '10 at 5:22isEuclidean, ..." Hopefully this should say "is not Euclidean". – Ricky Demer Sep 15 '10 at 5:31