TITLE: Question about the Euclidean ring definition
QUESTION [8 upvotes]: I recently came across the following definition for a euclidean ring:
There exists a function $g:R\to\Bbb N_0$ with the following properties:
1.) $\forall x,y \in R$ with $ y \neq 0$ there exist $q,r \in R$ with $x = qy + r$ where $r=0$ or $g(r) < g(y)$.
2.) $\forall x,y\in R\setminus\{0\}$ we have $g(xy)\ge g(x)$
What I don't understand, is the point of the second requirement. I have seen definitions where this is not a requirement. What does 2.) achieve? 
Thanks very much

REPLY [5 votes]: The second condition can be helpful to describe the values of the Euclidean function. For instance, if $R$ is a Euclidean ring with such a function $g$, you can easily deduce 
$$g(0)<g(u)=g(v)<g(a)$$
for all units $u,v\in R^*$ and $a\in R\backslash (\{0\}\cup R^*)$ or that $g(R)$ has an upper bound if and only if $R$ is a field.
Both definitions are actually the same. Each ring, that permits a function satisfying the first condition, also has a function satisfying both conditions.
To see this, let $g$ be a function satisfying 1.) and define $g^*:R\to\mathbb{N}_0$ with $g^*(0)=g(0)$ and 
$$g^*(a)=\min\{g(b) \mid b\in (a)\backslash\{0\}\}$$
for any $a\in R\backslash\{0\}$. You can check that $g^*$ satisfies 1.). For $x,y\in R\backslash\{0\}$ we have $(xy)\subset (x)$, so $g^*(xy)\geq g^*(x)$.
Check this paper of Pierre Samuel for more details.