 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 Dolly Robinson 2021-01-04 Answered

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 degree in x, its degree in y, and its total degree (i.e, the highest $$i+j$$ where $$\displaystyle{x}^{{i}}{y}^{{i}}$$ appears with a nonzero coefficient) all fail o be norm making F[x,y] a Euclidean domain.

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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