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Is there an easy example of valuation ring which is not noetherian?

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marked as duplicate by YCor, S. Carnahan Nov 18 '14 at 2:45

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Proposition 2, page 7 in Serre: Local fields says that a commutative ring is a discrete valuation ring iff it is local and noetherian and that its maximal ideal is finitely generated. Hence you noetherian ring must be nondiscrete. – Marc Palm Nov 28 '11 at 7:07
@pm: if you quote a theorem, it is a reasonable expectation that you quote it correctly. Someone might think that Serre made a mistake! The criterion you describe holds for any noetherian local ring. Serre actually says (correctly) that the maximal ideal is generated by a single non-nilpotent element (in other words its a regular local noetherian ring of dimension $1$). – Sándor Kovács Nov 28 '11 at 7:46
A valuation ring is a discrete valuation ring if and only if it is noetherian. See Matsumura, "Commutative rings", Th. 11.1, p. 78. So that provides you with a slew of examples. – Damian Rössler Nov 28 '11 at 9:52

6 Answers 6

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

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May I comment that this answer might be difficult to parse for anybody who doesn't know what $\mathbb C_p$ is. A couple of explanatory words might be useful. – André Henriques Nov 17 '14 at 23:29
Yes.… – Martin Brandenburg Nov 18 '14 at 6:41
...and why is the valuation ring of $\mathbb C_p$ not Noetherian? – André Henriques Nov 18 '14 at 13:27

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.

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Actually, that's not a valuation ring, since it isn't local (e.g. $(x)$ is a maximal ideal, but so is $(x-1,y)$). However, $R_{(x)}$ is a valuation ring that has the properties you claim, where $R$ is the ring you gave. – Neil Epstein Nov 17 '14 at 22:43
Hi Neil, you are right of course. I was thinking to localize at the maximal ideal generated by all the monomials. I'll fix this. – Karl Schwede Nov 17 '14 at 23:18

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.

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

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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 two-dimensional valuation ring which is therefore not noetherian (cf. Damian Rössler's comment above).

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You can also replace $\Bbb Q_p$ by $\Bbb Q$ and $\Bbb Z_p$ by $\Bbb Z_{(p)}$, the localization of $\Bbb Z$ at the prime $p$. – Leonardo Nov 28 '11 at 22:20

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:

  • $\mathrm{Spec}(O)\subset\mathrm{Spec} (O^\prime)$,
  • $O=O^\prime_M$, where $M$ is the maximal ideal of $O$,
  • $O^\prime/M\cong\overline{O}$.

In particular: $O^\prime$ is never noetherian.

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Nice generalisation of Leonardo's answer. But you have to assume that in $k$ there is some proper $\overline{\mathcal{O}}$, i.e. $k$ does not embed into some $\overline{\mathbb{F}_p}$. – Torsten Schoeneberg Mar 14 '13 at 14:52

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