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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?

Thanks in advance!

EDIT: changed the requirement on the valuation due to comment

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  • $\begingroup$ 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/… $\endgroup$
    – dsh
    Dec 21, 2023 at 5:15
  • $\begingroup$ @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? $\endgroup$
    – Gauss
    Dec 21, 2023 at 10:18
  • $\begingroup$ 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. $\endgroup$
    – dsh
    Dec 21, 2023 at 11:14
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    $\begingroup$ 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 $\mathbb{Q}$. $\endgroup$
    – Hagen Knaf
    Dec 27, 2023 at 17:12
  • $\begingroup$ @HagenKnaf Thank you for reference and correction. $\endgroup$
    – dsh
    Dec 28, 2023 at 6:38

1 Answer 1

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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.

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  • $\begingroup$ 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... $\endgroup$
    – Gauss
    Dec 28, 2023 at 0:01
  • $\begingroup$ 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. $\endgroup$
    – Hagen Knaf
    Dec 28, 2023 at 2:12
  • $\begingroup$ In that case, I thank you for the two books you provided and thank you again for your answer! $\endgroup$
    – Gauss
    Dec 28, 2023 at 12:10

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