Let $v$ be a valuation of DVR of character 0, If $v(p)＞0$, then for all $q≠p$, $v(q)＝0$

Question

Let $v$ be a valuation of DVR of character 0. Let $p$ be a prime.

I would like to prove

If $v(p)＞0$, then for all $q≠p$, $v(q)＝0$.

If $v$ is usual $p$-adic order, this is trivial. But I want to check this in general valuation of DVR.

I may not need assumption valuation of DVR, may be arbitrary valuation of valuation ring.

Answer 1

OP's question is poorly worded but the point is to assume that $ap+qb=1$ for some $a,b,p,q$ in a DVR.

If $v(p)>0$ then $v(q)=0$ because otherwise we'd have $v(1)>0$.