# Extension of a valuation on a function field

Let $K$ be a field, and $K(x)$ the field of rational functions over $K$. Consider the degree valuation $v$ on $K(x)$, That is $v\left(\frac{f(x)}{g(x)}\right)=\deg(g)-\deg(f)$. So for every $f(x)\in K[x]$, $v(f(x))\leq 0$.
Now let $F$ be a finite extension of $K(x)$, and $R$ the integral closure of $K[x]$ in $F$. Let $w$ be some extension of $v$ to a valuation on $F$. My question is, if for every $a\in R$, it happens that $w(a)\leq 0$.
I checked it for some private cases, like $F=K(x^{1/n})$, and $w$ some extension of $v$, and it turned out to be true. But I can't manage to prove it for the general case.
P.S. I also asked this question on Math.SE, but I have a feeling it won't get an answer there.

• $K(x^{1/n})$ won't give you a counterexample because there is only one extension of $v$. I suggest you try $L=K\left(\sqrt{x(x-1)}\right)$ and $a=x-\sqrt{x(x-1)}$. Commented Feb 10, 2016 at 11:00
• I don't think this will give a counter example, since if we take $w$ to be an extension of $v$ to $L$, then $-2=w(x(x-1))=w(\sqrt{x(x-1)}^2)=2w(\sqrt{x(x-1)}$. Hence $w(\sqrt{x(x-1)})=-1$ Commented Feb 10, 2016 at 13:44
• @ Laurent Moret-Bailly: Can you please explain more how this is a counter example? I can see how for every extension $w$, $w(a)>-1$ ,but what extension $w$ should we choose such that $w(a)>0$?. Commented Feb 11, 2016 at 9:43
• @giladude: Sorry, my example was wrong. Take $L=K\left(\sqrt{x^2-1}\right)$ and $a=x-\sqrt{x^2-1}$. If $b:=x+\sqrt{x^2-1}$ is the conjugate of $a$, then $ab=1$, hence $w(a)+w(b)=0$. But we cannot have $w(a)=w(b)=0$ since $a+b=2x$. So either $w(a)>0$ or $w(b)>0$. In fact you can check that there are two extensions $w^\pm$, with $w^\pm(a)=\pm1$. Commented Feb 11, 2016 at 21:17