1

Let $X$ be an integral scheme of finite type over field $k$, which corresponds to an irreducible variety $V$ over $k$.

I want to show that $X$ is regular in codimension one (every local ring at any point of scheme $X$ of dimension one is a regular local ring) if and only if singular points of $V$ have codimension at least two.

This question has been asked in this post, but I am stuck at the "only if" part.

Let $Y$ be an irreducible component of the singular locus of $V$, and let $\zeta$ be its generic point in $X$. Assume that $Y$ has codimension one, then $\mathcal{O}_{X,\zeta}$ has dimension one and hence is a regular local ring by assumption. In order to get a contradiction, I want to show that there is a closed point of $Y$ (corresponds to a point in the subvariety) that is not singular.

My question is:

How to implies that there is a closed point of $Y$ such that the local ring at that closed point is a regular local ring?

Eric
  • 497
  • 1
    So what you actually want to show is the following: let $X$ be integral of finite type over a field and $x \in X$ such that $O_{X,x}$ is regular. Then there is a specialization $y \in X$ of $x$ such that $O_{X,y}$ is regular and $y$ is a closed point. Note that it’s much weaker than the openness of the regular locus of $X$, which holds in this case, see https://stacks.math.columbia.edu/tag/07R2. I’ll try to think of a more ad hoc argument. – Aphelli Feb 16 '23 at 07:11
  • I think you are confused by the fact that $X$ can be viewed either as a scheme or a variety. Thinking of $X$ as a scheme, one knows that the set ${\rm Sing}(X)$ of points of $X$ (closed or not), whose local ring is not regular, is closed. So your point $\zeta$ is not in ${\rm Sing}(X)$. So the closure $Y$ of $\zeta$ meets the complement ${\rm Reg}(X)$ of ${\rm Sing}(X)$. But now $Y\cap{\rm Reg}(X)$ is a closed subscheme of an open subscheme of $X$, and hence is of finite type over $k$. Hence it has closed points. Such points are also closed in $Y$, because they are the points... – Damian Rössler Feb 17 '23 at 14:41
  • ... whose residue fields are finite extensions of $k$. – Damian Rössler Feb 17 '23 at 14:41
  • @DamianRössler Thanks for your reply! Your proof is right but I think the set $Sing(X)$ is closed is not a trivial fact. In fact $Sing(X)$ is not closed in general, see https://math.stackexchange.com/questions/1542078/is-the-set-of-regular-points-in-a-scheme-open-in-general – Eric Feb 17 '23 at 16:38
  • Yes you are right but it is when X is of finite type over a field. See eg https://stacks.math.columbia.edu/tag/07R2 – Damian Rössler Feb 17 '23 at 22:45

0 Answers0