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 x^iy^i appears with a nonzero coefficient) all fail o be norm making F[x,y] a Euclidean domain.

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 x^iy^i appears with a nonzero coefficient) all fail o be norm making F[x,y] a Euclidean domain.

Question
Abstract algebra
asked 2021-01-04
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.

Answers (1)

2021-01-05
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
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