# Valuation on $K(x, y)$ similar to $\mathfrak{p}$-adic valuation

Let $$K$$ be a field. Since $$K[x]$$ is a Dedekind domain, constructing valuations for $$K(x)$$ is easy - we can just take any prime ideal of $$K[x]$$ and consider the $$\mathfrak{p}$$-adic valuation.

So consider the prime ideal $$\mathfrak{p} = \langle 1 - x \rangle$$. It yields a valuation of $$K(x)$$ and, since it's $$\mathfrak{p}$$-adic, it has a very nice extension property: in the field $$K(\sqrt{x})$$, $$\mathfrak{p} = \langle 1 + \sqrt{x} \rangle\langle 1 - \sqrt{x} \rangle$$, where these are prime ideals in the ring of algebraic integers of the extension. Then, I can obtain a valuation such that $$v(1 + \sqrt{x}) = 1$$ and $$v(1 - \sqrt{x}) = 0$$.

Now, my question is: can this be generalized? More specifically:

Can I construct a valuation on $$K(\sqrt{x_1}, ..., \sqrt{x_n})$$ such that $$v(1 - \sqrt{x_i}) \neq v(1 + \sqrt{x_i})$$ for all $$i$$?

Even for $$n = 2$$, I don't know how to proceed. My biggest problem is that $$K[x, y]$$ isn't a Dedekind domain, so $$\mathfrak{p}$$-adic valuations don't really work to construct them directly, while the only transcendental extension I know is the minimum valuation, and so doesn't provide the nice property I want of differentiating the two root signs...

Can this be done? If so, how? Could you provide any references?

EDIT: changed the requirement on the valuation due to comment

• If you consider valuation with values in ordered group of real numbers under addition then you cannot construct required valuation, because such valuations are of height 1 necessarily. You need valuation of height > 1 to satisfy requirements if the number of variables in polynomial ring is > 1 (Krull dimension is > 1). Reference on valuations is Bourbaki, Commutative Algebra, ch VI. math.stackexchange.com/questions/2404312/… math.stackexchange.com/questions/462574/…
– dsh
Dec 21, 2023 at 5:15
• @dsh When you say it can’t have values on the additive group of “real numbers” do you mean integers instead? Also, in the case when I consider values on any ordered group, how can I do it? Dec 21, 2023 at 10:18
• In this case, real-valued valuations maybe automatically equivalent to integer-valued, but according to reference, height 1 valuation rings take values in real numbers.
– dsh
Dec 21, 2023 at 11:14
• It is not true that a valuation satisfying the requirements is automatically of rank greater than $1$. Using the method described on the pages 102/103 of Zariski-Samuel, Commutative Algebra 2, one can construct a valuation satisfying the requirements but having value group equal to $\mathbb{Q}$. Dec 27, 2023 at 17:12
• @HagenKnaf Thank you for reference and correction.
– dsh
Dec 28, 2023 at 6:38

A solution for the case $$n=2$$: For convenience I write $$x$$ and $$y$$ for the indeterminates. We work in the algebraic extension $$K(\sqrt{x},\sqrt{y})$$ of the rational function field $$K(x,y)$$. The rings $$K[\sqrt{x}]$$ and $$K[\sqrt{y}]$$ are polynomial rings in the indeterminates $$u:=\sqrt{x}$$ and $$v:=\sqrt{y}$$ and $$u,v$$ are algebraically independent over $$K$$. Hence $$K[u,v]$$ is a polynomial ring in two indeterminates.

As you already mentioned there exist discrete valuations $$v_1$$ and $$v_2$$ on $$K(u)$$ and $$K(v)$$ such that $$v_1(1-u)=0=v_2(1-v)$$ and $$v_1(1+u)\neq 0$$, $$v_2(1+v)\neq 0$$. The valuation rings of these valuations satisfy $$M_{v_1}\cap K[u]=(1-u)K[u]$$ and $$M_{v_2}\cap K[v]=(1-v)K[v]$$ and these equations uniquely determine those valuations among all valuations of $$K(u)$$ resp. $$K(v)$$. Here $$M_{v_i}$$ denote the maximal ideals of the valuation rings.

The ideal $$p:=(1-u)K[u,v]+(1-v)K[u,v]$$ is maximal hence prime and satisfies $$p\cap K[u]=(1-u)K[u]$$ and $$p\cap K[v]=(1-v)K[v]$$. By the extension theorem ("Chevalley's Theorem") there exists a valuation $$w$$ of $$K(u,v)$$ such that $$K[u,v]\subset O_w$$ and $$M_w\cap K[u,v]=p$$. Consequently $$w$$ extends both valuations $$v_1$$ and $$v_2$$ to $$K(u,v)$$ and therefore satisfies the requirements.

A concrete $$w$$ can be given as follows: we have $$K(u,v)=K(1-u,1-v)$$. Hence every element $$f\in K(u,v)$$ can be expressed as

$$f=(1-u)^a\cdot (1-v)^b\cdot\frac{z}{n}$$

with $$a,b\in\mathbb{Z}$$, $$z,n\in K[1-u,1-v]$$ both neither divisible by $$1-u$$ nor by $$1-v$$. Since $$K[1-u,1-v]$$ is a factorial ring, this representation is unique. The map

$$w: K(u,v)\setminus 0\rightarrow\mathbb{Z}^2, f\mapsto (a,b)$$

is a valuation provided one orders the group $$\mathbb{Z}^2$$ lexicographically.

• Thank you, that seems to work! Do you have a reference for these results you cited, and the construction you provided so I can study further? I know Ribbenboim's book on valuations, but I couldn't find anything to the ends of this question... Dec 28, 2023 at 0:01
• Not really: I learned valuation theory partially from the books Valuation Theory by Otto Endler and Volume 2 of Zariski-Samuel, but most of it from lectures given by Peter Roquette and Franz-Viktor Kuhlmann at the University of Heidelberg. Dec 28, 2023 at 2:12
• In that case, I thank you for the two books you provided and thank you again for your answer! Dec 28, 2023 at 12:10