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It is well-known that there is examples of infinite (Krull) dimensional ring. For example, Kang and Park, in [Example, pages 111 and 112, A localization of a power series ring over a valuation domain , JPAA 140 (1999) 107-124], constructs an infinite-dimensional discrete valuation ring.

So, is there any example of a local ring $(R,M)$ such that $\dim(R)=\infty$ and $R/M$ is finite?

T. Ali
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2 Answers2

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Let $F$ be any finite field, and let $R$ be the localization of the polynomial ring $F[x_n:n\in\mathbb{N}]$ at the maximal ideal $\langle x_n:n\in\mathbb{N}\rangle$. $R$ is local, with unique maximal ideal $M=\langle x_n/1:n\in\mathbb{N}\rangle$, and $R\big/M\cong F$ is finite. But $$0<\langle x_1/1\rangle<\langle x_1/1,x_2/1\rangle<\dots$$ is an infinite strictly ascending chain of prime ideals of $R$, so that $\dim R=\infty$.

As an aside, I am not sure about your claim that the paper you cite constructs an "infinite-dimensional discrete valuation ring". By definition, a discrete valuation ring has Krull dimension $1$, at least for the usual definitions of "discrete valuation ring" and "Krull dimension"; see for example characterization 5 here. Are the authors of the paper you cite using different definitions?

Atticus Stonestrom
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  • The following is the comment stated before the example of Kang and Park in the mentioned paper: "For a finite-dimensional valuation domain V, V is discrete if and only if V is an SFT-ring [2, Lemma 2.7]. For an infinite-dimensional case a discrete valuation domain need not be an SFT-ring although the converse always holds. We illustrate this in the next example." – T. Ali Sep 11 '21 at 19:49
  • hi @T.Ali, thank you for the quote. do you know what their definition of a discrete valuation domain is? – Atticus Stonestrom Sep 11 '21 at 19:49
  • A DVR is by definition rank-one discrete valuation domain, and other authors used the following: a one-dimensional valuation domain with maximal principal ideal – T. Ali Sep 11 '21 at 20:05
  • So, from the previous comment discrete valuation ring is not necessary of rank one. – T. Ali Sep 11 '21 at 20:06
  • hi @T.Ali, yes, the second definition is what I was referring to, wherein a DVR by definition has dimension one. anyway, this is an aside to main question; does the example I have written above of a local infinite-dimensional ring with finite quotient make sense? :) – Atticus Stonestrom Sep 12 '21 at 02:21
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    Your example seems correct! – T. Ali Sep 12 '21 at 20:47
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According to Example 175 of [H.C. Hutchins, Examples of Commutative Rings, Polygonal Publishing House, (1981)], if we take $K_0$ a finite field then $R:=K_0+P$ is a local infinite-dimensional ring with maximal ideal $P$ and hence $R/P$ is finite.

T. Ali
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