Can inequivalent Krull valuations induce the same topology? I was working with (Krull) valuations and I have a question I could not answer.
Let me recall you some notions first.
Let $K$ be a field. Given a totally ordered abelian group $\Gamma$, we say $v: K^\times \to \Gamma$ is a valuation on $K$ if it is a surjective group homomorphism and satisfies
$$v(x+y) \geq \min\{v(x),v(y)\}$$
for every $x,y \in K^\times$ such that $x+y \neq 0$. $\Gamma$ is called the value group of the valuation.
For every $x \in K$ and $\gamma \in \Gamma$, consider the ball
$$B(x,\gamma):=\{y \in K - \{x\} \ | \ v(x-y) > \gamma\} \cup \{x\}.$$
There is a unique topology on $K$ such that for every $x \in K$, $\{B(x,\gamma) \ | \ \gamma \in \Gamma\}$ is a local base for $x$.
Let $w: K^\times \to \Delta$ another valuation on $K$. We say $v$ and $w$ are equivalent if there exists an isomorphism $\phi: \Gamma \to \Delta$ of ordered groups such that $\phi \circ v = w$.
It is easy to see that equivalent valuations induce the same topology on $K$. My question is: is the converse also true?
I know the converse is true if the value groups are ordered subgroups of $\mathbb{R}$. Ordered abelian groups $G$ that can be embedded in $\mathbb{R}$ are characterised by the property that for every $a,b \in G$, $a > 0$, there exists $n \geq 1$ such that $na \geq b$. As far as I know, this property turns to be key to prove the result. This is why I doubt the converse is true in general. Can anyone think about a counterexample if that is the case, or a proof if not?
Thanks in advance!
 A: I now give a full answer to my question based on @fyx1123581347 's comment above.
Let $K$ be a field. Recall that there is a bijective correspondence between equivalence classes of valuations on $K$ and valuation rings of $K$. Therefore we reinterpret the problem in terms of valuation rings.
Let $\mathcal{O}$ be a valuation ring of $K$ with maximal ideal $\mathfrak{m}$. For every $x \in K$ and $a \in K^\times$, define the ball
$$B(x,a):=\Big\{y \in K \ \Big| \ \frac{x-y}{a} \in \mathfrak{m} \Big\}.$$
The topology on $K$ induced by the equivalence class of valuations associated to $\mathcal{O}$ is the unique topology on $K$ such that for every $x \in K$, $\{B(x,a) \ | \ a \in K^\times \}$ is a local base for $x$.
Besides, there is an order-reversing one-to-one correspondence between $\text{Spec} \ \mathcal{O}$ and the intermediate subrings between $\mathcal{O}$ and $K$. Note that the latter are automatically valuation rings of $K$. The correspondence is given by $\mathfrak{p} \mapsto \mathcal{O}_\mathfrak{p}$, and its inverse sends an intermediate subring to its maximal ideal.

Proposition. If $\mathfrak{p}$ is a nonzero prime ideal of $\mathcal{O}$, then $\mathcal{O}$ and $\mathcal{O}_\mathfrak{p}$ induce the same topology on $K$.

Proof. If $x \in K$ and $a \in K^\times$, denote by $B_\mathfrak{p}(x,a)$ the corresponding ball for $\mathcal{O}_\mathfrak{p}$. Let $p$ be a nonzero element of $\mathfrak{p}$. Then, for every $x \in K$ and $a \in K^\times$,
$$B(x,ap) \subseteq B_\mathfrak{p}(x,a),$$
$$B_\mathfrak{p}(x,a) \subseteq B(x,a).$$
$\Box$
In conclusion, if we pick a prime ideal $\mathfrak{p}$ of $\mathcal{O}$ different from $\{0\}$ and $\mathfrak{m}$, then $\mathcal{O}_\mathfrak{p}$ is a valuation ring of $K$ different from $\mathcal{O}$ but inducing the same topology as $\mathcal{O}$. Note that such a prime ideal exists if and only if $\dim \mathcal{O} > 1$; i.e, if the value group of the valuation cannot be embedded into $\mathbb{R}$.
