Singular Loci in Noetherian normal schemes

Question

Let $X$ be a Noetherian normal scheme. Denote by $B$ the singular locus of $X$. By normality, points in $B$ have codimension strictly greater 1. If $X$ is of finite type over field (or some J-2 ring), then it is known that $B$ is closed.

I am interested in the general situation without the J-2/finite type hypothesis. Specifically, I want to know the following:

Let $X$ be a Noetherian normal scheme and $\bar{B}$ denote the closure of the singular locus $B$. Then is codim$(\bar{B})\geq 2$?

Answer 1

Counterexamples surely exist. For instance, let $(R, \mathfrak{m})$ be a Noetherian normal local domain of dimension $3$ whose regular locus is not open; an explicit such $R$ can be found in Example 2.11 of Nishimura's "A few examples of local rings I." By normality, the singular locus $B$ of $\mathrm{Spec}(R)$ intersects the punctured spectrum $U_R := \mathrm{Spec}(R) \setminus \{\mathfrak{m}\}$ in a collection of closed points, an infinite collection because $B$ is not closed. The closure of $B$ cannot be of codimension $\ge 2$: if it were, its (closed in $U_R$) intersection with $U_R$ would consist of finitely many points, but this fails already for $B$ itself.

Answer 2

This is only a partial answer: The singular locus of an excellent ring $R$ is closed. So if you are willing to assume that $X$ is excellent, then $\operatorname{codim} \bar B = \operatorname{codim} B \ge 2$.