Let $A$ be a domain, $K=\operatorname{frac}A$ and $v_i, i=1,2$ are non-archimedean (real) valuations on $K$ s.t. $v_i(A)\geq 0$. We know that $\mathfrak p_i=\{x\in A:v_i(x)>0\}$ are primes of $A$. I wonder if $\mathfrak p_1=\mathfrak p_2$ implies $v_1\sim v_2$.
(It seems true for the ring of integers in algebraic number field, as we have $v_i\sim v_{\mathfrak p_i}$)
No, $$\Bbb{Q}[ x(x-1),x(x-1)^2]$$ has two discrete valuations above $\mathfrak{p}=(x(x-1),x(x-1)^2)$,
the order of the zero at $0$ and the order of the zero at $1$.
If $\mathfrak{p}$ is invertible in $A_\mathfrak{p}$ then yes though this is not necessary (try with $\Bbb{Q}[x^2,x^3]$)