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Let $R$ be an integral domain and $P$ a prime ideal. Let $x$ be an element such that $xP^{m-1}=P^m$ for some $m>0$.

Is $P$ generated by $x$?

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    There's no reason to expect this if $P$ isn't invertible (that is if there doesn't exist an ideal $P^{-1}$ such that $P^{-1} P$ is principal).2011-08-29

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Let $R=\mathbb{Z}[\sqrt{-3}]$ and let $P=(2,1+\sqrt{-3})$. Then $2P=P^2$ but $P$ is not generated by $2$.

An example where $R$ is the coordinate ring of a variety would be $R=k[x,y]/(y^2-x^3)$, and $P=(x,y)$, where we have $P^2=xP$, but $P$ is not generated by $x$.

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    That was fast! Thanks a lot!2011-08-29
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    No problem, glad to help! You may want to hold off on accepting my answer, I am sure that someone more knowledgeable than I will come along with a more general answer about why this can occur.2011-08-29
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    Do you have a counterexample when $R$ is the coordinate ring of an affine variety btw?2011-08-29
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    If I had to guess, this should be possible when $R$ is not normal, so maybe a coordinate ring like $R=k[x,y]/(y^2-x^3)$ would do the trick - I'm not sure though.2011-08-29
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    @Zev: I think that works; if $P=(x,y)$, then $P^2=(x^2,y^2,xy) = (x^2,x^3,xy) = (x^2,xy)$, and $xP = (x^2,xy)=P^2$. But $P\neq (x)$.2011-08-29
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    @Arturo: Aha, thanks! I'll put this example in my answer. My vague recollection is that when the coordinate ring of a variety isn't normal, this indicates that there is a singularity, and the local ring at a singularity won't be a DVR, so there won't be a uniformizer (or something like that?).2011-08-29
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    This is great. So what if $R$ is normal? Is it true in that case?2011-08-29
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If the domain is not integrally closed then counterexamples may be obtained via non-integral witnesses $\rm\:w = a/b\:,\:$ which yield $\rm\: b\:(a,b)^{n-1}= (a,b)^n\:,\:$ but $\rm\:(a,b) \ne (b)\:.\:$ Zev's example is a special case. See my posts in this May 22, 2009 sci.math thread for much more.