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 as simple as possible. Is there any good reason why not to define it as $0$ at $0$?
3 Answers
$\begingroup$
$\endgroup$
1
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$\begingroup$ +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). $\endgroup$ Commented Jul 13, 2010 at 22:14
$\begingroup$
$\endgroup$
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.
$\begingroup$
$\endgroup$
Old thread, but nobody wrote the simple practical reason, so I'll do it now. In $K[T]$, we want to use $\deg$ as a Euclidean function, so $\deg(c) = 0$ for $c \in K - \{0\}$. But $\deg(0)$ is best thought of as $-\infty$, which breaks the usual definitions of "Euclidean function". It's common to sidestep this by simply not requiring the Euclidean function to be defined at 0 at all, which is fine since you never need it anyway.