# Singular Loci in Noetherian normal schemes

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$$(\bar{B})\geq 2$$?

• Yes. Since $\mathcal{O}_x$ is normal, it is a field if $\dim(\mathcal{O}_x)=0$, and a discrete valuation ring if $\dim(\mathcal{O}_x)=1$, hence regular in both cases.
– abx
Oct 9, 2018 at 12:59
• @abx: I don't see why this answers the question. Oct 9, 2018 at 14:55
• @Laurent Moret-Bailly: OK, what I said gives $\operatorname{codim}(B)\geq 2$. Could it be that $\operatorname{codim}(\bar{B})<\operatorname{codim}(B)$?
– abx
Oct 9, 2018 at 17:06

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.
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 $$\operatorname{codim} \bar B = \operatorname{codim} B \ge 2$$.