Valuation of polynomials

Mathematics Asked by Muselive on November 27, 2020

My paper defines a valuation on a Ring $R$ to be a map $v:R-{0} rightarrow A$ where $A$ is an ordered abelian group. This map has the following properties;


2.$v(a+b)geq min{v(a),v(b)}$
(note this is missing the usual conditon wich grants equality if $v(a) neq v(b)$)

My question is the following: Let $w:R[x]-0_{R[x]} rightarrow A$ by;
$$w(sum_{i=0}^nr_ix^i)=min_{0leq i leq n}v(r_i)$$

and I’m to show this is a valuation. I can easily show property 2 but I cannot show property 1. I’m starting to doubt if it is possible; I wrote out some examples and I don’t see how I can get anything other than a lower bound for a $w(fg)$. A hint or even confirmation that this is solvable would be very appreciated.

One Answer

If $R$ is an integral domain, your formula for $w$ does yield a valuation. Here is a hint toward proving condition 1 holds. We will use the observation in my comment that $v(a+b)=min{v(a),v(b)}$ if $v(a)ne v(b)$.

Let $f=sum_i r_ix^i$ and $g=sum_j s_jx^j$ and suppose $w(f)=v(r_p)$, $w(g)=v(s_q)$, where $p,q$ are as small as possible. In $fg$, the coefficient $t_{p+q}$ of $x^{p+q}$ is a sum containing the product $r_ps_q$ and other products $r_is_j$ where either $i<p$ or $j<q$. Show $v(r_is_j)>v(r_ps_q)$ for those terms, and so $v(t_{p+q})=v(r_ps_q)=v(r_p)+v(s_q)$. Show that no coefficient of another $t^k$ can be smaller than $v(r_p)+v(s_q)$. Conclude that $w(fg)=v(r_p)+v(s_q)=w(f)+w(g)$.

Correct answer by Allen Bell on November 27, 2020

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