Is there a Noetherian integral domain that is regular in codimension 1 but not integrally closed?

Question

Regular in codimension 1 means that for all height 1 prime ideals $p$, $R_p$ is a regular local ring.

The motivation behind this question is that Hartshorne's Algebraic Geometry in chapter 2.6 defines (*) as : Noetherian, integral separated scheme that is regular in codimension 1. "Normal" is missing. While Vakil's Foundations of Algebraic Geometry, chapter 15.4 uses a normality hypothesis extensively. I want to know why there is a difference.

My attempt is: since it's regular in codimension 1 but not regular, the dimension should be at least 2. I can think of the example $k[x,y,z]/(y^2-x^3-z)$ as a potential example . But I don't know how to prove that it's integrally closed and regular in codimension 1.

Serre’s normality criterion says that normality is equivalent to $(R_1)$ and $(S_2)$, where $(R_1)$ is “regular in codimension $1$” and $(S_2)$ means that “any localization of $R$ at a prime ideal of height $h \geq 2$ has depth at least $2$”. — Aphelli, Jul 24 '24 at 09:29

score 2 · Accepted Answer · answered Jul 24 '24 at 12:14

2

Take for example the subring of $k[x,y]$ formed by all polynomials $f(x,y)$ such that $f(0,0) = f(1,0)$.

answered Jul 24 '24 at 12:14

Sasha

20,727

Is it obvious that this example is Noetherian? – walkar Jul 24 '24 at 17:25
@walkar the proof in the answer over at one of OP's previous questions shows that this is true. There are also quicker, higher-tech options to see this. – KReiser Jul 24 '24 at 18:31
Thank you. How do we show it is not integrally closed? $(x - 1/2)$ is not in the ring but its square is in the ring. But I don't know why $x-1/2 \in Quot(R)$. – David Lui Jul 25 '24 at 00:17
Because both $y$ and $y(x - 1/2)$ are in the ring. – Sasha Jul 25 '24 at 06:55
@KReiser I see, thanks! Maybe it's not appropriate to ask this here in the comments, but what higher-tech options do you have in mind? – walkar Jul 30 '24 at 17:53

Is there a Noetherian integral domain that is regular in codimension 1 but not integrally closed?

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