4

I am going to assume that in any case all schemes are Noetherian, separated and integral.

Can someone provide an example of a scheme that is normal but not locally factorial. I know that being locally factorial will imply normal, because UFDs are integrally closed.

I am just curious about the boundary between these two classes of objects.

Elle Najt
  • 20,740

1 Answers1

5

The simplest example is the coordinate ring $A=\mathbb C [X,Y,Z]/\langle Z^2-X^2-Y^2\rangle$ of the affine cone $V$ given by $Z^2=X^2+Y^2$ in $\mathbb A^3$.
A little down-to-earth calculation done in Shafarevich's Basic Algebraic Geometry ( Vol.1, Chapter II, 5.1, page 125) shows that $A$ is normal.
At a more abstract level we may deduce normality from the fact that $V$ is regular in codimension one and is a complete intersection.
That $A$ (or its localization at any point) is not a UFD is almost evident from the two factorizations $Y^2=(Z+X)(Z-X)$, although some details still need to be checked.

Georges Elencwajg
  • 150,790
  • How can you tell by looking at it that this ring is regular in codimension one? (Without first computing that it is normal.) – Elle Najt Nov 13 '15 at 13:36
  • 1
    The only singularity of the cone $V$ is the origin, which has codimension two in $V$. Hence the cone is regular in codimension one. – Georges Elencwajg Nov 13 '15 at 14:50
  • Thanks. So somehow the feeling is that if the singular locus has codimension at least two, then removing it gives a smooth variety containing all of the generic points of prime divisors of the original variety. It follows that the local rings of the generic points of prime divisors are regular since they are localizations of maximal ideals in the new smooth (quasiaffine) variety? – Elle Najt Nov 13 '15 at 16:08
  • 1
    Yes. It is indeed quite unintuitive to relate the classical point of view of properties of subvarieties with the more modern scheme-theoretic point of view expressing everything in terms of the local rings of the generic points of those subvarieties. Unfortunately, as far as I know, this correspondence is nowhere made explicit. – Georges Elencwajg Nov 13 '15 at 18:57