Is there an easy example of valuation ring which is not noetherian？
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$\begingroup$ 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. $\endgroup$– Marc PalmCommented Nov 28, 2011 at 7:07
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7$\begingroup$ @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 $1$). $\endgroup$– Sándor KovácsCommented Nov 28, 2011 at 7:46
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4$\begingroup$ 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. $\endgroup$– Damian RösslerCommented Nov 28, 2011 at 9:52
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6 Answers
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The valuation ring of $\mathbb{C}_p$ is not noetherian.
answered Nov 28, 2011 at 9:34
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2$\begingroup$ 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. $\endgroup$ Commented Nov 17, 2014 at 23:29
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2$\begingroup$ ...and why is the valuation ring of $\mathbb C_p$ not Noetherian? $\endgroup$ Commented Nov 18, 2014 at 13:27
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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.
answered Nov 28, 2011 at 16:28
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1$\begingroup$ 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. $\endgroup$ Commented Nov 17, 2014 at 22:43
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1$\begingroup$ 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. $\endgroup$ Commented Nov 17, 2014 at 23:18
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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.
answered Nov 28, 2011 at 19:01
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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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$\begingroup$ 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 $p$. $\endgroup$– LeonardoCommented Nov 28, 2011 at 22:20
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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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$\begingroup$ 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 $\overline{\mathbb{F}_p}$. $\endgroup$ Commented Mar 14, 2013 at 14:52