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 non discrete valuation ring [closed] 2 answers
Is there an easy example of valuation ring which is not noetherian？
This question already has an answer here:
Is there an easy example of valuation ring which is not noetherian？ 

marked as duplicate by YCor, S. Carnahan♦ Nov 18 '14 at 2:45This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question. 


The valuation ring of $\mathbb{C}_p$ is not noetherian. 


For a very explicit example, consider the ring $$k[x, y, x/y, x/y^2, x/y^3, \dots]$$ localized at the origin (ie, localize at the maximal ideal $\langle x, y, x/y,x/y^2, \ldots \rangle$. This has value group $\mathbb{Z} \oplus \mathbb{Z}$ with lexicographic ordering (in other words, the $x$value always is more important than the $y$value). It's easy to see it's not Noetherian but it does have finite Krull dimension, equal to $2$. You can obtain this example geometrically, and explicitly, by repeated blowings up of the origin. See Hartshorne Chapter II, Exercise 4.12. 


Another good example inside the field of rational functions in two variables goes like this: Choose an irrational positive number $\alpha$, and look at all rational functions $R(x,y)=\frac{P(x,y)}{Q(x,y)}$ such that when $R(x,x^\alpha)$ is written out as a formal linear combination of powers of $x$ there are no negative powers occurring. 


Take a finite prime of the algebraic closure A of Q and complete the ring of integers of A with respect to this prime. 


Consider the subring $A$ of $\Bbb Q_p(X)$ consisting of rational functions defined at $X=0$ and such that $f(0)\in\Bbb Z_p$. In other words, let $B$ denote the localization of the ring $\Bbb Q_p[X]$ at the maximal ideal $(X)$ and set $A=\Bbb Z_p+XB$. It is a twodimensional valuation ring which is therefore not noetherian (cf. Damian Rössler's comment above). 


Construction of valuation domains of Krull dimension $>1$: Let $O\neq K:=\mathrm{Frac}(O)$ be a valuation domain. Consider the natural map $h:O\rightarrow k$, where $k$ is the residue field of $O$. Let $\overline{O}$ be a valuation domain of $k$. Then $O^\prime:=h^{1}(\overline{O})\subseteq O$ is a valuation domain of $K$ with the following properties:
In particular: $O^\prime$ is never noetherian. 

