Answer to Question #138278 in Abstract Algebra for Raunaqe

Question

Show that d:Q[x]\{0}→NU{0}:d(f)=2^deg f is a Euclidean valuation on Q[x].

Answer 1

Let f(x),g(x)$\in Q[x]$

Then there exists q(x) and r(x)$\in Q[x]$ such that f(x)=g(x)q(x)+r(x) where either r(x)=0 or deg(r)<deg(g). (Since Q is a field, division algorithm holds in Q[x])

When r$\neq 0$ , deg(r)<deg(g) $\Rightarrow$ 2$^ {deg(r)}$ <2$^{deg(g)}$ $\Rightarrow d(r)<d(g)$

We also note d$(fg)=2^{deg(fg)}= 2^{deg(f)+deg(g)}=2^{deg(f)} 2^{deg(g)}=d(f)d(g)$

Hence d is an Euclidean evaluation.