# Krull dimension of the smooth locus

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?

$$\DeclareMathOperator\Spec{Spec}$$

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 Flenner-Trivedi 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.

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