Quotients in a non-discrete valuation ring

Question

Let $R$ be a valuation ring, $\mathfrak{m}$ the maximal ideal of $R$. Let $k$ be the residue field, and $K$ the field of fractions of $R$. Assume that the valuation on $K$ is such that $v(K)=\mathbb{Q}$. For every element $\lambda \in \mathbb{Q}$ we have

$$ \mathfrak{m}_{\lambda}:=\left\{ x\in K\:\middle|\:v(x)\geq \lambda \right\} \\ \mathfrak{m}^+_{\lambda}:=\left\{ x\in K\:\middle|\:v(x)> \lambda \right\}$$

Now the quotient $V_{\lambda}=\mathfrak{m}_{\lambda}/\mathfrak{m}^+_{\lambda}$ is $k$-vector space, and I would like to know its dimension.

I know that it's 1 in a discrete valuation case, but clearly $v$ isn't discrete.

Answer 1

Let $x_0\in K$ such that $v(x_0)=\lambda$. Then $\{\bar x_0\}$ is a $k$-basis of $V_{\lambda}$.

If $\bar x\in V_{\lambda}$, $\bar x\ne\bar 0$, then $v(x)=\lambda$. We get $v(xx_0^{-1})=0$, so $xx_0^{-1}\in R$. Therefore there is $a\in R$ such that $x=ax_0$, and then it follows that $\bar x=\hat a\bar x_0\in k\bar x_0$.

If $\hat a\bar x_0=\bar 0$ then $v(ax_0)>\lambda$, that is, $v(a)>0$. Thus $a\in\mathfrak m$, that is, $\hat a=\hat 0$.