Let $ R$ be an integral domain and $ \bar R$ denote its integral closure in the fraction field (i.e. normalization). If $ p\in R$ is a prime element in $ R$ , then does $ p$ remain prime in $ \bar R$ also ?
By a prime element in an integral domain $ R$ , I mean a non-zero non unit $ p\in R$ such that $ p |ab $ for some $ a,b \in R$ implies $ p|a$ or $ p|b$ i.e. if $ pR$ is a prime ideal in $ R$ . I can see that $ p$ still remains a non-unit in $ \bar R$ , but I’m unable to say anything about the ideal $ p\bar R$ .
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