7
$\begingroup$

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.

$\endgroup$
15
$\begingroup$

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 $1$.

$\endgroup$
12
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '17 at 8:43
8
$\begingroup$

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 $R_v=k[x,y,x/y,x/y^2,x/y^3...]_{(y)}$.

$\endgroup$
  • 1
    $\begingroup$ 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 '17 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. $\endgroup$ – Uri Brezner Jan 16 '17 at 10:05
5
$\begingroup$

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.

$\endgroup$

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.