MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

## Euclidean Function at 0

In a few places where I have looked the Euclidean Function of a Euclidean Domain is only being defined for non-zero elements. I am teaching an undergraduate course and I am trying to make things simpler as possible. Is there any good reason why not to define it as $0$ at $0$?

-

You'll find your answer and much more in the little-known paper [1] which surveys all of the dozen known ways of axiomatizing Euclidean rings (including those of Nagata and Samuel), and explores in-depth all of their logical interrelations. It's a convenient reference to have at hand when you're comparing texts which use (seemingly) different definitions of Euclidean rings / domains.

[1] Euclidean Rings. A. G. Agargun, C. R. Fletcher
Tr. J. of Mathematics, 19, 1995, 291 - 299.
http://journals.tubitak.gov.tr/math/issues/mat-95-19-3/pp-291-299.pdf

-
 +1: this certainly answers the question. I found it somewhat unsatisfying though that -- so far as I could see -- Nagata's 1978 example of a Euclidean domain with a transfinite algorithm but no $\mathbb{Z}^+$-valued algorithm is listed in the bibliography but not discussed (or even cited!) in the text at all. Instead they give the example of $\mathbb{Z} \oplus \mathbb{Z}$, which is not a domain, so seems rather cheap. According to MathSciNet, before Nagata, Hiblot gave a similar example (first incorrectly, but corrected before Nagata's paper). – Pete L. Clark Jul 13 2010 at 22:14

I think this is only because you don't care dividing by zero. But you may as well define the value at zero to be 0, and any other value at least 1. See this expository paper by Keith Conrad for interesting remarks on euclidean domains and functions.

-