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Let $R$ be a Noetherian ring. Let $(x)$ be a prime ideal such that $\bigcap_n (x)^n=0$. Then $R$ is a domain.

Is this a known result? I heard its known as the Davis lemma. Can anyone give a reference?

EDIT: $x$ should be regular. if the height of the prime is not positive, examples like the excellent one given by Francesco will crop up.

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  • $\begingroup$ 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? $\endgroup$
    – Francesco Polizzi
    Commented Feb 5, 2015 at 14:18
  • $\begingroup$ By regular you mean that $x$ is not a zero-divisor? $\endgroup$
    – Will Chen
    Commented Feb 6, 2015 at 5:21
  • $\begingroup$ yes. $x$ is a non zero-divisor. $\endgroup$
    – dongrugose
    Commented Feb 6, 2015 at 15:58

1 Answer 1

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No reference, but the proof is trivial. If $ab = 0$, then $x^{2n} | ab$ (for any $n$), so since $x$ is prime and a nonzerodivisor, $x^n | a$ or $x^n | b$, and since this is true for all $n$, either $a$ or $b$ lies in $\bigcap_n (x^n)$ and hence equals $0$. There's no need to assume $R$ is Noetherian.

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