Let F be a field and consider the ring of polynominals in two variables over F,F[x,y]. Prove that the functions sending a polyomial f(x,y) to its degr

Let D be a domain. A non-negative integer valued function ${\displaystyle N:D-\left\{0\right\}}$ is Euclidean , if given a, b in D there exist ${\displaystyle q\text{and}r\in D}$ 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.
${\displaystyle N:F[x,y]-\left\{0\right\}\to \{0,1,2,3,\dots \}}$
$N(f(x,y))=$ highest degree wrt x
Claim: N is not Euclidean. Consider $a=x+y,b=y$. If N were Euclidean, ${\displaystyle \mathrm{\exists }q(x,y),r(x,y)}$ with $a=bq+r,r=0$ or N(r) Now ${\displaystyle a=bq+r\Rightarrow x+y=p(x,y)y+r(x,y)}$
Comparing degrees, we deduce
$p(x,y)=1andr(x,y)=x.$
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
${\displaystyle N:F[x,y]-\left\{0\right\}\to \{0,1,2,3,\dots \}}$
$N(f(x,y))=$ total degree
Claim: N is not Euclidean. Consider ${\displaystyle a=x+y^2,b=x}$. If N were Euclidean, ${\displaystyle \mathrm{\exists }q(x,y),r(x,y)}$ 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