For all $f(x)∈R[x]$ and $G(x)∈R[x]$, there exists $q(x)∈R[x]$ such that $f(x)＝G(x)q(x)+P(x)$ such that $\deg P(x)＜\deg G(x)$.

Asked Jan 19 '22 at 15:08

Active Jan 19 '22 at 18:20

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Let $R$ be a local ring, $K＝\mathrm{Frac}(R)$. Let $g(x)∈R[x]$ be polynomial whose highest coefficient is a unit.

How can I take $q(x)∈R[x]$ (not from $K[x]$) which satisfies the following

For all $f(x)∈R[x]$ and $G(x)∈R[x]$, there exists $q(x)∈R[x]$ such that $f(x)＝G(x)q(x) +P(x)$ such that $\deg P(x)＜\deg G(x)$.

We can take $q(x)∈K[x]$ because $K[x]$ is euclidean, but how can we take from $R[x]$ ?

Background: This is from Neukirch 'Algebraic number theory', p.130. I don't know how to take 7th line from the bottom, I don't understand why $q(x)∈O[x]$. So I asked this question.

edited Jan 19 '22 at 18:20

user26857

52,094

asked Jan 19 '22 at 15:08

Pont

5,909

Does this answer your question? – mathcounterexamples.net Jan 19 '22 at 15:21
What is Frac(R)? As far as I can see R is not assumed a domain. – user26857 Jan 19 '22 at 18:22
Use induction on deg (F). – DanielWainfleet Jan 19 '22 at 18:40
Basically the same proof as the usual Euclidean algorithm, noting that we only need the leading coefficient of $g(x)$ to be nonzero. As pointed out you don't necessarily have Frac$(R)$, but that is not a problem. – Kushi Jan 19 '22 at 18:41
See the linked dupe - the division algorithm works for any polynomial with unit lead coef $,u,$ over any commutative ring. Or we can reduce to the monic case, i.e. divide $f$ by the monic $,u^{-1} g,$ so $,f = q(u^{-1}g) + r = (qu^{-1}) g + r\ \ $ – Bill Dubuque Jan 30 '22 at 19:54

For all $f(x)∈R[x]$ and $G(x)∈R[x]$, there exists $q(x)∈R[x]$ such that $f(x)＝G(x)q(x)+P(x)$ such that $\deg P(x)＜\deg G(x)$.

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