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A Noetherian local ring is an integral domain with Krull dimension one, and the maximal ideal of R is principal, is called a discrete valuation ring (DVR). This is one out of many possibilities to define a discrete valuation ring (https://en.wikipedia.org/wiki/Discrete_valuation_ring).

But what do we get in addition, if we skip "dimension one"? The other way around, what are examples for noetherian local rings with maximal ideal principal, which are not DVR. What I can see is that fields are of this type, but I wasn't able to find other...

EDIT: In the comments a finite example is mentioned, that I've missed, but are there also infinite rings with this property? Also I want to keep integral domain as property if this is possible.

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    $\begingroup$ How about $\Bbb Z/p^n\Bbb Z$? $\endgroup$
    – Angina Seng
    Oct 23, 2017 at 15:33
  • $\begingroup$ Ok, thanks, I see, I missed something trivial, but are there also infinite rings? $\endgroup$
    – user302982
    Oct 23, 2017 at 15:41
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    $\begingroup$ Let $k$ be an infinite field and take $k[t]/(t^{n})$. $\endgroup$
    – Minseon Shin
    Oct 23, 2017 at 19:53
  • $\begingroup$ If $A$ is a Noetherian local integral domain whose maximal ideal $\mathfrak{m}$ is generated by a single nonzero element $\pi$, then $A$ is a DVR. The valuation $v$ can be described as $v(x) = \min\{n \ge 0 \;:\; x \in \mathfrak{m}^{n}\}$. $\endgroup$
    – Minseon Shin
    Oct 23, 2017 at 20:27
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    $\begingroup$ The conditions on $A$ in my previous comment imply in particular that $A$ has Krull dimension one. In general, if $A$ is a Noetherian local ring, then the Krull dimension of $A$ is less than or equal to the minimum number of generators of the maximal ideal, so if you assume that the maximal ideal can be generated by one element, then $A$ has dimension either one or zero. $\endgroup$
    – Minseon Shin
    Oct 23, 2017 at 20:37

2 Answers 2

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Let $A$ be a Noetherian local ring with maximal ideal $\mathfrak{m}$. Consider three conditions:

  1. $A$ is an integral domain;
  2. $\mathfrak{m}$ can be generated by one element;
  3. $A$ has Krull dimension one.

Conditions (1) + (2) imply that $A$ is a DVR, in particular (3). There are two examples of $A$ in the comments (namely $\mathbb{Z}/p^{n}\mathbb{Z}$ or $k[t]/(t^{n})$) that satisfy (2) but not (1) or (3). In general, the Krull dimension of $A$ is less than or equal to the minimum number of generators of $\mathfrak{m}$ (see for example Eisenbud, "Commutative Algebra with a View Toward Algebraic Geometry", Theorem 10.2) so (2) implies that the Krull dimension of $A$ is either zero or one.

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I think that $\mathbb Z/p^n \mathbb Z[[x]]$ works, being the power series ring over the example given in the comments.

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  • $\begingroup$ Thank you, I will try to understand it. $\endgroup$
    – user302982
    Oct 23, 2017 at 15:59
  • $\begingroup$ @sigmabe I'm not 100% on the first answer, so it would certainly be worth working out yourself, or waiting for further comments before accepting this one. $\endgroup$
    – Andres Mejia
    Oct 23, 2017 at 16:02

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