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Let $(R,\mathfrak m)$ be a local Cohen-Macaulay reduced ring of dimension at least $2$. Then, can we find a non-zero-divisor $x\in \mathfrak m$ such that $R/xR$ is again a reduced ring?

If needed, I am willing to assume $R$ is excellent https://en.m.wikipedia.org/wiki/Excellent_ring.

Note: $\dim R\ge 2$ is needed, otherwise $R$ must be regular. Indeed, if $\dim R=1$ and $R/xR$ is reduced for some non-zero-divisor $x$, then $R/xR$ is also Artinian. So $R/xR$ is a field, hence $\mathfrak m=xR$, so $R$ is regular.

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    $\begingroup$ If your ring is equicharacteristic, this follows from Bertini-type theorems, cf. the book of Jouanolou. $\endgroup$ Commented Oct 10, 2022 at 12:06

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Since $R$ is Cohen-Macaulay $R/xR$ is unmixed. In particular, it suffices to show that $R/xR$ is generically reduced. But now the Flenner-Trivedi local Bertini (as explained in the work of Trivedi, https://www.tandfonline.com/doi/abs/10.1080/00927879408824878 ) implies that we can pick an $x \in \mathfrak{m}$ such that $x \notin P^{(2)} = P^2 R_P \cap R$ for any prime $P \subseteq \mathfrak{m}$. Furthermore we can ensure that $x$ avoids any height one prime $P$ of $R$ such that $R_P$ is not regular (there are only finitely many). In particular, at each height one prime $P$ lying over $(x)$, we have that $R_P$ is a DVR and $x$ is a uniformizer since its not in $P^2R_P$. Thus $R_P/xR_P$ is a field and in particular reduced. That should do it.

Probably the Cohen-Macaulay hypothesis can be weakened substantially too.

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