Is the Euclidean valuation on a Euclidean domain unique?

Question

I have just started studying Euclidean domain which is defined as follows :

An integral domain $R$ is called a Euclidean domain if there exists a function $\delta : R \setminus {\{0\}} \longrightarrow \mathbb {N_{0}}$ which satisfies the following properties.

(1) $\delta (a) \leq \delta (ab)$, for all $a,b \in R \setminus {\{0\}}$.

(2) For any $a,b \in R$ with $b \neq 0$ there exist $q,r \in R$ such that $a = bq + r$, where either $r = 0$ or $\delta (r) < \delta (b)$.

Here $q$ is called quotient and $r$ is called remainder.

The function $\delta$ is called a Euclidean norm function (or Euclidean valuation) on $R$.

Now my question is :

Is such Euclidean valuation on a Euclidean domain unique?

I am in a fix.Please tell me.

Thank you in advance.

Answer 1

No. For instance, if $\delta$ is a Euclidean valuation on $R$, then so is the function $\delta'$ defined by $\delta'(a)=\delta(a)+1$.

Answer 2

Another example (to show non-uniqueness): On the ring of integers, define $\delta (x) = x^2$.

Answer 3

I have another example too :D

I suppose that you must be aware that in $K[X]$ (here $K$ is a field) the degree of a polynomial $\deg$ is used to defined an euclidean valuation. Just take $\delta(p(x))=\deg(p(x))$, where $p(x)\in K[x]$. But it turn outs that we can define $\delta(p(x))=2^{\deg(p(x))}$ and this is an euclidean valuation too.

Last example is given in N. Jacobson's book "Basic Algebra I", exactly after definition 2.5. The advantages of this euclidean valuation is that we can define it even for the zero polynomial. Just take the convention $\deg(0)=-\infty$ and $2^{-\infty}=0$. Also, this euclidean valuation is multiplicative, i.e., $\delta(ab)=\delta(a)\delta(b)$.