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For valuation rings I know examples which are Noetherian.

I know there are good standard non Noetherian Valuation Rings. Can anybody please give some examples of rings of this kind?

I am very eager to know. Thanks.

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5 Answers 5

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Consider the tower of domains

$$ K[x]\subset K[x^{1/2}]\subset \cdots \subset K[x^{1/2^k}]\subset\cdots $$

where $K$ is a field and $x$ is transcendental over $K$. Every ring in the chain is a polynomial ring in one variable over $K$. Thus the localizations $O_k:=K[x^{1/2^k}]_{P_k}$, where $P_k$ is the prime ideal generated by $x^{1/2^k}$ are discrete valuation rings. Since $P_{k+1}\cap K[x^{1/2^k}]=P_k$ one has $O_k\subset O_{k+1}$ and $M_{k+1}\cap O_k =M_k$ for the maximal ideals $M_k$ of the rings $O_k$.

Now $O:=\bigcup\limits_k O_k$ is a non-noetherian valuation ring of the field $K(x^{1/2^k} : k\in\mathbb{N})$. The value group of an associated valuation is order-isomorphic to the subgroup $\{z/2^k : z\in\mathbb{Z}, k\in\mathbb{N}\}\subset\mathbb{Q}$. Hence this example yields a non-noetherian valuation ring of Krull dimension $1$.

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Valuation rings that have dimension $\geq 2$ are not Noetherian. The dimension of a valuation ring is equal to the rank of its value group.

To get a simple example of a valuation ring that has dimension $2$, take $R = k[x,y]$, where $k$ is a field. Define the standard valuation $v: k(x,y) \rightarrow \mathbb{Z}^2$ with $v(x) = (1,0) \leq v(y) = (0,1)$, and take the value of a polynomial as the minimal values among those of its monomials. The value group is $\mathbb{Z}^2$, which has rank $2$. So the valuation ring is not Noetherian. This example is "standard" in the sense that it is encountered more often. However, Hagen's example is more interesting.

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    $\begingroup$ Can you please provide some reference to these statements- "Valuation rings that have dimension ≥2 are not Noetherian. The dimension of a valuation ring is equal to the rank of its value group. " $\endgroup$
    – Babai
    May 22, 2017 at 8:43
  • $\begingroup$ Sorry but it seems that $(R,v)$ is not a valuation ring. By definition, under the valuation $v$ the valuation ring should be $\{f/g: v(f)\ge v(g)\}$ which contains much more than $R$, for example all $(y+...)/(x^n+...)$. But fortunately the dimension argument is correct because in the new ring $(y) = \{f/g: v(f/g)\ge (0,1)\}$ is a prime ideal different from the maximal ideal $(x)$. This is essentially because the lexicographical order on $\mathbb{Z}^2$ is not Archimedean, i.e., the sum of $(1,0)$'s is always smaller than $(0,1)$. $\endgroup$ Mar 2 at 19:08
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In order to obtain a non Noetherian valuation ring, take $\mathbb{Z}^2$ with the lexicographic order. Define the valuation $v:k(x,y)^* \to \mathbb{Z}^2$ as follows: for any $a \in k^*$ and $0 \le n,m \in \mathbb{Z}$ set $v(ax^ny^m)=(n,m)$. For a polynomial $\: f=\sum f_i \in k[x,y]^*$ set $v(f)= \inf \{v(f_0),...,v(f_d)\} $ where the $f_i$ are distinct monomials. Finally for a rational function $f \in k(x,y)^*$ there are $ g,h \in k[x,y]$ such that $f= \frac{g}{h}$ set $v(f)= v(g)-v(h)$. The corresponding valuation ring $R_v= \{f \:|\: v(f) \ge 0\}\cup \{0\}$ contains $k[x,y]$, but it also contains $xy^{-1}$ since $(0,0) < (1,-1)$. In fact $xy^n \in R_v$ for any $n \in \mathbb{Z}$. It follows that $R_v=k[x,y,x/y,x/y^2,x/y^3...]_{(y)}$.

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    $\begingroup$ Just want to say that if you localise in (y) as in the very last line, then x will be invertible but it has valuation -1 so something seems wrong. $\endgroup$
    – neptun
    Jan 15, 2017 at 17:52
  • $\begingroup$ @neptun You are right. The valuation of x is positive so x is also in the maximal ideal. I need to invert everything that is not divisible by x nor by y, so maybe need to localize at (x,y). I'll think about a bit more before editing. $\endgroup$ Jan 16, 2017 at 10:05
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    $\begingroup$ For others who were confused by this, $x$ is an element of the ideal $(y)$ since $x = y\cdot \frac{x}{y}$, so localising at $(y)$ doesn't invert $x$. The description of $R_v$ as given is correct. $\endgroup$
    – BHT
    Dec 9, 2021 at 4:51
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This was bumped to the front page for some reason, so I apologize for resurrecting this. But I think that there is an exceedingly natural example. In fact, it comes up all the time in 'nature'. Namely, consider $\mathbb{Q}_p$ with the standard valuation $v_p$. Then, there is a unique extension of this valuation to $\overline{\mathbb{Q}_p}$. The value group is $\mathbb{Q}$, and so if $\mathcal{O}$ is its valuation ring (it's just the integral closure $\overline{\mathbb{Z}_p}$ of $\mathbb{Z}_p$ in $\mathbb{Q}_p$), then $\mathcal{O}$ is a non-Noetherian valuation ring.

Other examples which come up are $\mathcal{O}_{\mathbb{C}_p}$, the valuation ring of the $p$-adic complex numbers.

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Let $(K, \lvert\cdot\rvert)$ be a complete algebraically closed field with a non trivial absolute value. Let $R$ be its valuation ring and $\mathfrak{m}$ the maximal ideal of $R$. Since every element of $K$ has a square root in $K$, therefore $\mathfrak{m}=\mathfrak{m}^2$. By Nakayama, $R$ cannot be noetherian. Such fields $K$ exist of course. Start with any $K$ with a non-trivial absolute value. Complete it, take the algebraic closure of the completion, and complete that. So for example the valuation ring of $\mathbb{C}_p$, with $p$ a prime number, would be such an example.

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  • $\begingroup$ Actually, in my example, I don't see why we need completion. Algebraically closed $K$ is enough. $\endgroup$ Oct 2, 2019 at 8:40

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