6
$\begingroup$

My question is as follows:

Given a field $k$, is it always possible to find a valuation ring of $k(x, y)$ of dimension $2$?

1 Answers 1

3

If by $k(x,y)$ you mean the rational function field in two variables over $k$, then the answer is yes. Consider the map $v:k[x,y]\rightarrow\mathbb{Z}\times\mathbb{Z}$ defined as follows: express a polynomial $f\neq 0$ in the form $f=x^ay^bg$, where $g$ is neither divisible by $x$ nor by $y$, and set $v(f):=(a,b)$. Since $v$ is multiplicative it can be extended to the fraction field $k(x,y)$ of $k[x,y]$. Order the group $\mathbb{Z}\times\mathbb{Z}$ lexikographically: $(a,b)<(c,d):\Leftrightarrow (a. Then $v$ is a valuation having a valuation ring of Krull dimension $2$.

Remark: all valuation rings of $k(x,y)$ containing $k$ can be obtained using the construction just described but replacing $x,y$ with other prime polynomials.

  • 0
    @a local property: Ask the same question for a Z valuation. Only one proper convex subgroup, but two prime ideals in a DVR.2011-04-13