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Let $R$ be a non-Noetherian local commutative ring with identity such that it is of Krull-dimension zero. I am wondering if there are conditions which will force the maximal ideal to be nilpotent.

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  • $\begingroup$ What do you mean by "the only prime ideal"? There is no reason why there is only one prime ideal. $\endgroup$ – Martin Brandenburg Feb 13 '13 at 20:03
  • $\begingroup$ Martin is absolutely right: if $R$ has only one prime ideal, it is of dimension zero! (And then that prime ideal indeed consists of nilpotent elements but needn't be nilpotent) $\endgroup$ – Georges Elencwajg Feb 13 '13 at 20:23
  • $\begingroup$ Since your ring is local, did you perhaps mean to type "the only $maximal$ ideal"? I think this would make your question interesting. $\endgroup$ – user55407 Feb 13 '13 at 21:30
  • $\begingroup$ I must apologize. I was a bit confused by the definition of the dimension in a ring. Yes, it should be 0-dimensional ring. So we have a 0-dimensional local ring. Then the maximum ideal's elements are nilpotent. I was wondering if there are conditions that forces the ideal itself to be nilpotent. Thank you all for correcting me. $\endgroup$ – user62198 Feb 14 '13 at 1:25
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    $\begingroup$ Here is a class of examples: Let $R$ be a (non-noetheran) ring, $\mathfrak{m}$ be a maximal ideal and consider $R/\mathfrak{m}^n$ $\endgroup$ – Martin Brandenburg Feb 14 '13 at 1:54
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This is a broad question...

A necessary condition for the maximal ideal $\mathfrak m$ to be nilpotent is $R$ is separated for the $\mathfrak m$-adic topology: $$ \cap_{n\ge 1} \mathfrak m^n=0.$$ Now suppose moreover that $\mathfrak m/\mathfrak m^2$ is finite dimensional over $R/\mathfrak m$, then $\mathfrak m$ is nilpotent.

Proof: let $I\subseteq \mathfrak m$ be a finitely generated ideal which generates $\mathfrak m$ modulo $\mathfrak m^2$. Let $N>0$ be such that $I^N=0$ ($N$ exists because $I$ is finitely generated). For any $n>0$, we have $\mathfrak m=I+ \mathfrak m^n$ and $$\mathfrak m^{N}\subseteq I^N+\mathfrak m^n=\mathfrak m^n.$$ Therefore $\mathfrak m^{N}\subseteq \cap_n \mathfrak m^n=0$.

Of course in general, $ \mathfrak m/\mathfrak m^2$ is not finite dimensional. I don't know whether the above condition is reasonable for you.

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  • $\begingroup$ Can you give a nontrivial example where $\mathfrak{m}/\mathfrak{m}^2$ is finite, but $\mathfrak{m}$ is not? $\endgroup$ – Martin Brandenburg Feb 14 '13 at 16:23
  • $\begingroup$ @MartinBrandenburg: I don't know what do you call nontrivial. In general we can have $\mathfrak m/\mathfrak m^2=0$ and $\mathfrak m$ not of finite type (e.g. continuous functions in a neighborhood of $0\in\mathbb R$), but I don't have such an example in dimension $0$. $\endgroup$ – user18119 Feb 14 '13 at 16:44
  • $\begingroup$ Thank you very much QiL'8. $\endgroup$ – user62198 Feb 15 '13 at 16:55
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    $\begingroup$ @Martin: Dear Martin, Here is one example where $R$ is of dimension zero, $\mathfrak m = \mathfrak m^2$, but $\mathfrak m$ is not finitely generated: $R = k[X,X^{1/2},X^{1/3},...]/(X)$. Once you see this example, you can make many others. Regards, $\endgroup$ – Matt E Feb 15 '13 at 17:33
  • $\begingroup$ Thank you. By nontrivial I meant that $\mathfrak{m}/\mathfrak{m}^2$ has a given dimension $n>0$. $\endgroup$ – Martin Brandenburg Feb 15 '13 at 18:03

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