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I am looking for an example of the following situation:

  • $R$ is a noetherian one-dimensional ring.
  • $x \in R$ is a non-nilpotent element contained in infinitely many primes of $R$.

Obviously, nilpotent elements behave in such a way and I was asking myself if other elements can behave in a similar fashion.

If there is a reference that no such elements exist, please tell me or post a proof if you like to.

Thanks a lot in advance!

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    $\begingroup$ What about $R= k \times \mathbb{Z}$, and $x=(1,0)$ (here $k$ is a field)? $\endgroup$ – Krish Sep 25 '17 at 10:55
  • $\begingroup$ @Krish Thank you for your example. Do you know another one with $R$ being an integral domain? $\endgroup$ – windsheaf Sep 25 '17 at 12:22
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Here is an example: Let k be an algebraically closed field, and $R = k[X,Y]/(X\cdot Y)$. For $a \in k\setminus \{0\}$ we have $$X = (-Xa^{-1}) \cdot (Y-a) \in \langle Y-a\rangle $$ so $X$ is contained in infinitely many prime ideals. Geometrically, $R$ corresponds to a union of two coordinate axes, and $X$ is a function which vanishes identically on one of them. As it is nonzero on the other, the element is not nilpotent.

EDIT: If $R$ is integral, this cannot happen: By Krull's principal ideal theorem, for any $x \in R \setminus\{0\}$, the quotient $R / \langle x \rangle$ is going to be noetherian $0$-dimensional and hence will only have finitely many prime ideals (i.e its Spectrum will have finitely many points). But the prime ideals of this quotient correspond exactly to the prime ideals of $R$ containing $x$.

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  • $\begingroup$ Thank you for your example. Do you know another one with $R$ being an integral domain? $\endgroup$ – windsheaf Sep 25 '17 at 12:22
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    $\begingroup$ No, your element $x$ will have to be a zero-divisor. I edited my answer to expand on this. $\endgroup$ – Johann Haas Sep 25 '17 at 12:47
  • $\begingroup$ I'm not sure of the meaning of the statement that the quotient "only consist of finitely many points," but I think it's clear that the quotient is semilocal, and thanks for that insight. $\endgroup$ – rschwieb Sep 25 '17 at 12:52
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    $\begingroup$ I think he wanted to say that $\operatorname{Spec}(R/xR)$ is finite which is what you said. $\endgroup$ – windsheaf Sep 25 '17 at 13:00
  • $\begingroup$ Indeed, thanks for the correction! Thinking about dimension carried me too far into the geometric side. I edited my answer accordingly. $\endgroup$ – Johann Haas Sep 25 '17 at 14:04

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