# The residue field of valuations are finite extension

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 $$\mathbb{Z}_{\geq 0}$$.

• Isn't $[k_\mathit{v} : k]$ the definition of the degree of $m_\mathit{v}$ (which is finite) for the function field case? – Sqyuli Feb 12 at 20:55
• Yes, I guess you are right. That would be the case in which $K$ has transcendence degree 1 and the valuation is discrete. – Walter Simon Feb 12 at 21:01

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 $$n-1$$.
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.