Existence of $x\in \mathfrak m \setminus \mathfrak m^2$ such that $xR$ is a prime ideal

Asked Nov 16 '22 at 10:59

Active Feb 04 '23 at 02:07

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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.

asked Nov 16 '22 at 10:59

feder

1

Why the downvote :( ? – feder Nov 18 '22 at 10:29
You should at least look at Flenner's local Bertini theorem. – Mohan Nov 18 '22 at 22:14
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 $p \in \mathfrak{m} \setminus \mathfrak{m}^2$. – Aryaman Maithani Nov 23 '22 at 00:55
+1 for this question – TShiong Feb 03 '23 at 17:43

Existence of $x\in \mathfrak m \setminus \mathfrak m^2$ such that $xR$ is a prime ideal

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