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Let $(R,\mathfrak m)$ be a Noetherian local domain of dimension at least $2$. Then, must there exist $x\in \mathfrak m \setminus \mathfrak m^2$ such that $xR$ is a prime ideal of $R$? What if we also assume $R$ is normal?

My thoughts: If $R$ is a UFD, then every height $1$ prime ideal is principal, and in that case the problem boils down to finding a height $1$ prime ideal not contained in $\mathfrak m^2$, can this be always done? Outside the UFD case, I have no idea.

Please help.

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    $\begingroup$ Why the downvote :( ? $\endgroup$
    – feder
    Commented Nov 18, 2022 at 10:29
  • $\begingroup$ You should at least look at Flenner's local Bertini theorem. $\endgroup$
    – Mohan
    Commented Nov 18, 2022 at 22:14
  • $\begingroup$ The UFD case is more elementary: Pick any $x \in \mathfrak{m} \setminus \mathfrak{m}^2$. Necessarily $x$ isn't a unit. $x$ has a prime factor $p$. Necessarily $p \in \mathfrak{m} \setminus \mathfrak{m}^2$. $\endgroup$ Commented Nov 23, 2022 at 0:55
  • $\begingroup$ +1 for this question $\endgroup$
    – TShiong
    Commented Feb 3, 2023 at 17:43

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