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I want to find a ring $R$ which satisfies

$R$ is not Noether and $\operatorname{Spec}R$ is Hausdorff.

I found the latter condition is equivalent to R's Krull dimension is $0$. So, I just need to find an example of Non-noether and $0$ dimensional ring. Are there any good examples? Thank you advance.

user26857
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Pont
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1 Answers1

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If one takes $R=K[x_1,x_2,\ldots]/(x_1^2,x_2^2,\ldots)$ to be the polynomial ring over a field in infinitely many variables modulo the ideal $(x_1^2,x_2^2,\ldots)$, then $R$ is not noetherian and $\operatorname{Spec}R$ is a single point, in particular Hausforff.

user26857
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Thanks.
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  • Thank you! Why specR is a single point? – Pont Apr 09 '20 at 17:59
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    @川村はるか Because this is clearly a local ring. The nilradical $(x_1,\ldots)$ is clearly maximal. – rschwieb Apr 09 '20 at 20:55
  • Why specR is a local ring? If so, I can say specR is a singleton because R is PID. – Pont Apr 11 '20 at 09:00
  • I think specR is local if R is algebraically closed. But if not so, I doubt specR is local. – Pont Apr 11 '20 at 14:57
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    The prime ideals of $R$ are the prime ideals of the polynomial ring containing the ideal $(x_1^2,x_2^2,...)$. But then, using the prime property, each such prime contains also the ideal $(x_1,x_2,...)$ which is maximal, hence we have equality, and $Spec(R)$ ist just the image of $(x_1,x_2,...)$ under the quotient map. – Thanks. Apr 11 '20 at 21:26