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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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    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.