Let $R$ be a normal complete local domain of dimension $n \geq 4$. Does there exist a prime ideal $\mathfrak{p}$ of height $\dim(R)  1$ such that $R_{\mathfrak{p}}$ is a regular local ring? In general, can the dimension of the smooth locus be as low as possible or can the codimension of the singular locus be exactly $2$ in excellent schemes?
1 Answer
$\DeclareMathOperator\Spec{Spec}$
Maybe I'm misreading this, but I don't see why you need dimension $\geq 4$, normal, domain, etc.
EDIT: I originally wrote this requiring R1 but I don't think we need R1 (regular in codimension 1). Instead, we should just require R0, regular at the minimal primes (ie, $R_Q$ is a field if $Q$ is a minimal prime). The proof is modified to reflect this.
Theorem. If $(R,m)$ is excellent and regular in codimension 0 and $\dim R > 0$ then there exists a prime $P$ such that $\dim R_P = \dim R  1$ and such that $R_P$ is regular.
Let's prove it.
Let $J$ denote a radical ideal such that $\dim V(J) \leq \dim R  1$ and such that the singular locus of $R$ is contained in $V(J)$ (we can do this since $R$ is excellent). Finally set $U = \Spec R \setminus V(J)$ (note $U$ is nonsingular).
If $J = m$, then the statement is obvious as any height one prime will do the job. In particular, we may suppose $\dim R/J > 0$.
We proceed by induction on dimension of $R$. If $R$ is dimension 1, then since $R$ is local and R0, we can take $J = m$ and we are done with the base case.
Note $V(J)$ has codimension $\geq 1$ and say minimal associated primes $Q_1, ... Q_t$ of $V(J)$ of heights $\geq 1$. By FlennerTrivedi local Bertini, there exists $x \in m$ such that $x$ is not in any of the $Q_i$, and not in any minimal prime of $R$, and such that $x \notin P^{(2)}$ for any $P \in U$.
We set $R' = R/xR$, $J' = (J + xR)/xR$, and $U' = U \cap V(x)$.
The first statement of the local Bertini theorem guarantees that $\dim( R/J ) = \dim (R' / J') + 1$ (we also have $\dim(R) = \dim(R') + 1$). Next, for any $P \in U$, $R_{P}$ is a regular ring, the second statement in the local Bertini means that $(R/x)_P$ is also regular. Hence $R/x$ is R0 too (its regular locus contains $U \cap V(x)$).
Thus, by the induction hypothesis, there exists $P' = (P + xR)/(xR) \in U' \subseteq \Spec R/x$ of height $\dim(R/x)  1 = \dim R  2$ such that $(R/x)_{P'}$ is regular. Hence $R_P$ is also regular by the above ($P \in U$). But then $\dim R_P = \dim R'_{P'} + 1$ and so $P$ has height $\dim R  1$ as desired.

$\begingroup$ Thanks! Indeed dimension $\geq 4$ and being normal are not necessary, as is the complete local domain assumption. $\endgroup$ Jun 4 at 17:42