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

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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.