Question #138278
Show that d:Q[x]\{0}→NU{0}:d(f)=2^deg f is a Euclidean valuation on Q[x].
1
Expert's answer
2020-10-14T14:41:13-0400

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

Then there exists q(x) and r(x)Q[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 r0\neq 0 , deg(r)<deg(g) \Rightarrow 2deg(r)^ {deg(r)} <2deg(g)^{deg(g)} d(r)<d(g)\Rightarrow d(r)<d(g)

We also note d(fg)=2deg(fg)=2deg(f)+deg(g)=2deg(f)2deg(g)=d(f)d(g)(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.


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