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Given a prime ideal $P$ in a Dedekind domain $R$ can you always find an element $r \in R$ such that $P$ divides the ideal $(r)$ but $P^{2}$ does not?

Rupert
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  • See https://math.stackexchange.com/questions/1643534/property-of-a-noetherian-ring-how-come-p-setminus-p2-is-non-empty-p-is – Watson Dec 15 '16 at 16:01

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Hint: That's the same as proving that $P^2\subsetneq P$. That $P^2\subseteq P$ is true is easy. What property of Dedekind domains might you use to prove that $P^2\neq P$?

Thomas Andrews
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