# When can we choose non-zero-divisor $x\in \mathfrak m$ in a reduced local ring $(R,\mathfrak m)$ such that $R/xR$ is also reduced?

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.

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

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.