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S Oct 26, 2017 at 10:41 history suggested user 1
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Oct 26, 2017 at 10:08 review Suggested edits
S Oct 26, 2017 at 10:41
Feb 7, 2015 at 3:48 answer added Jesse Elliott timeline score: 3
Feb 6, 2015 at 15:58 comment added dongrugose yes. $x$ is a non zero-divisor.
Feb 6, 2015 at 5:21 comment added Will Chen By regular you mean that $x$ is not a zero-divisor?
Feb 5, 2015 at 14:34 history edited dongrugose CC BY-SA 3.0
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Feb 5, 2015 at 14:18 comment added Francesco Polizzi Are you sure of this statement? Take $R = k[x,y]/(x^2)$. Then $R$ is Noetherian (quotient of a Noetherian ring) and $R/(x)=k[y]$ is a domain, hence $(x)$ is a prime ideal. Obvioulsy $\cap(x)^n=0$, however $R$ is not a domain because the element $x$ is nilpotent. Am I missing something?
Feb 5, 2015 at 13:41 history asked dongrugose CC BY-SA 3.0