Let $R$ be a commutative (non-trivially) graded ring. By a non-archimedean valuation I mean a map $v: R \to \Gamma \cup {0}$ such that for all $x,y \in R$, we have $v(x+y) \leq \max\{v(x),v(y)\}$, $v(xy)=v(x)v(y)$ and $v(0)=0,v(1)=1$. Here $\Gamma$ is a totally ordered commputative group and $0 \leq g$ for all $g \in \Gamma$.


Now, one says two valuations $v,w$ are equivalent for all $x,y \in R$ $$v(x) \geq v(y) \iff w(x) \geq w(y).$$ Now, my question is, can I check two valuations are equivalent by just looking at homogeneous elements, in other words if for all $a,b$ homogeneous elements in $R$ we have $$v(a) \geq v(b) \iff w(a) \geq w(b)$$ can we deduce that $v,w$ are equivalent?


It wouldn't be suprised if the answer is no, but I don't know much about this. If the answer is indeed no, can one add some extra assumtions to $R$ to make this the case, e.g free or something.


Thanks 



non-archimedean valuations on graded rings
Let $K$ be a field, and set $R=K[x]$ with the usual grading by degree. For each irreducible polynomial $p(x)\in R$, we get a valuation $v\_p$ on $R$ given by
$$
v\_p(f)=2^{-\operatorname{ord}\_p(f)},
$$
where
$\operatorname{ord}\_p(f)\in\mathbb{Z}\_{\geq 0}\cup\{\infty\}$ is the power to which $p(x)$ occurs in the prime factorization of $f$. Suppose $p,q\in R$ are non-associate irreducibles, neither of which is associate to $x$. Then $v\_{p}$ and $v\_{q}$ are not equivalent, since $v\_{p}(q)\geq v\_{p}(1)$ but $v\_{q}(q)<v\_{q}(1)$. However, for every non-zero homogeneous element $f\in R$, we have $v\_{p}(f)=v\_{q}(f)=1$.


