The residue field of valuations are finite extension

Question

Let $K\mid k$ be a finitely generated extension (maybe of transcendence degree bigger than one) and consider a rank one valuation of $K$ over $k$, that is, a function $$v:K^\times\rightarrow \mathbb{R}_{\geq 0}$$ such that $v(xy)=v(x)+v(y)$, $v(x+y)\geq v(x)+v(y)$ and $v(c)=0$ for all $c\in k$.

Let $\mathcal{O}_v=\{x\in K\mid v(x)\geq 0\}$ be the valuation ring of $v$, $m_v=\{x\in K\mid v(x)> 0\}$ its unique maximal ideal and $k_v=\mathcal{O}_v/m_v$. Notice that we have an inclusion $k\hookrightarrow k_v$.

My question are

1. Is the extension $k_v\mid k$ algebraic?
2. Is it finitely generated?

Also I am interested in the same questions in case of discrete valuations, say changing $\mathbb{R}_{\geq 0}$ by $\mathbb{Z}_{\geq 0}$.

Answer 1

Now I can answer this.

In general it is neither algebraic or finitely generated even if the valuation is discrete.

To give an example in which is not algebraic consider a Weil divisor $E$ over a variety $X$ of dimension $n$. It induces a valuation $\text{ord}_E:K(X)\rightarrow \mathbb{Z}$ by looking at the order of vanishing at its generic point $\eta$. The residue field of this valuation is $\mathcal{O}_{X,\eta}/\mathfrak{m}$ and this is isomorphic to the function field of $E$. As $E$ has codimension $1$ this function field has transcendence degree $n-1$.

Also there are discrete valuations in which the residue field is not finitely generated, see example 9 here.

The important result here is the Abhanyankar inequality that says that for any valuation in this context we have $$\text{tr.}\deg(k_v\mid k)+\text{rat.rk}(\Gamma_v)\leq \text{tr.deg}(K\mid k)$$ and equality implies that the extension $k_v\mid k$ is finitely generated.