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Can someone please give me an example of a Noetherian normal local domain of dimension two such that there exists a prime ideal $P$ of height one having the property $P^{(n)}$ is not a principal ideal for any $n \geq 1$. Here $P^{(n)}$ is the symbolic $n$-power.

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  • $\begingroup$ Is $P^{(n)}$ the same as $P^n$? $\endgroup$
    – YCor
    Nov 20, 2021 at 12:03
  • $\begingroup$ @YCor Are you suggesting that the symbolic powers coincide with ordinary powers in this case? $\endgroup$
    – Z. M
    Nov 20, 2021 at 17:10
  • $\begingroup$ @Z.M I'm suggesting nothing, I'm actually asking about terminology. So you seem to answer my question: $P^{(n)}$ is the symbolic $n$-power. $\endgroup$
    – YCor
    Nov 20, 2021 at 17:33
  • $\begingroup$ The condition is equivalent to the class group being not torsion. Such examples exist though I can not recall an easy one. I will try to find one when I have a little more time. $\endgroup$
    – Mohan
    Nov 21, 2021 at 22:56
  • $\begingroup$ @YCor Yes, $P^{(n)}$ is the symbolic n-power. $\endgroup$ Nov 22, 2021 at 9:17

1 Answer 1

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Take $E$ an elliptic curve $zy^2 - x(x-z)(x-tz)$ say over $\mathbb{C}$ and choose a point $Q$ of infinite order (or so for instance the divisor $Q - O$ has infinite order in the divisor class group, here $O$ is the point at infinity).

It follows that in the graded ring of dimension $2$, $$\mathbb{C}[x,y,z]/(zy^2 - x(x-z)(x-tz))$$ that the homogeneous ideal corresponding to $Q$, call it $P$, has the property that $P^{(n)}$ is never principal. Indeed, if $P^{(n)} = (f)$, then $(f)$ is homogeneous and the corresponding divisor $\mathrm{Div}_E(f/z^{\deg f})$ is linearly equivalent to 0. This contradicts the infinite order of $P$.

One should point out that for any rational double point, the divisor class group is finite by a result of Lipman (if and only if under some hypotheses), so one has to leave the setting of rational singularities. The cone over an elliptic curve is the probably the simplest singularity that is not rational.

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    $\begingroup$ Karl, did you mean $zy^2-x(x-z)(x-tz)$? $\endgroup$ Nov 24, 2021 at 4:12
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    $\begingroup$ Yes, thanks, I've fixed it! $\endgroup$ Nov 24, 2021 at 19:34

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