Let D be a domain. A non-negative integer valued function \(\displaystyle{N}:{D}-{\left\lbrace{0}\right\rbrace}\) is Euclidean , if given a, b in D there exist \(\displaystyle{q}{\quad\text{and}\quad}{r}\in{D}\) such that \(a=bq+r\), with either \(r=0\) or N(r) To show that f(x,y) going to highest degree in x is not a Euclidean function. (By symmetry) , the same argument shows that f(x,y) going to highest degree in y is not a Euclidean function.

\(\displaystyle{N}:{F}{\left[{x},{y}\right]}-{\left\lbrace{0}\right\rbrace}\rightarrow{\left\lbrace{0},{1},{2},{3},\ldots\right\rbrace}\)

\(N(f(x,y)) =\) highest degree wrt x

Claim: N is not Euclidean. Consider \(a=x+y, b=y\). If N were Euclidean, \(\displaystyle\exists{q}{\left({x},{y}\right)},{r}{\left({x},{y}\right)}\) with \(a=bq+r,r=0\) or N(r) Now \(\displaystyle{a}={b}{q}+{r}\Rightarrow{x}+{y}={p}{\left({x},{y}\right)}{y}+{r}{\left({x},{y}\right)}\)

Comparing degrees, we deduce

\(p(x,y)=1\ and\ r(x,y)=x.\)

But \(N(r)=1\), whereas \(N(b)=0\)

so, N(r) So, N is not a Euclidean norm

Proving that the total degree function is also not a Euclidean norm

\(\displaystyle{N}:{F}{\left[{x},{y}\right]}-{\left\lbrace{0}\right\rbrace}\rightarrow{\left\lbrace{0},{1},{2},{3},\ldots\right\rbrace}\)

\(N(f(x,y)) =\) total degree

Claim: N is not Euclidean. Consider \(\displaystyle{a}={x}+{y}^{{2}},{b}={x}\). If N were Euclidean, \(\displaystyle\exists{q}{\left({x},{y}\right)},{r}{\left({x},{y}\right)}\) with \(a=bq+r,r=0\) or N(r) Comparing degrees, we deduce

\(q(x,y)=1\) and \(r(x,y)=y^2\)

But \(N(r)=2\), whereas \(N(b)=2\)

So, N(r) So, N is not a Euclidean norm