In Hartshorne, a scheme is regular in codimension one if the local ring at any (non-closed) point representing a codimension one subscheme is a regular local ring (of Krull dimension one).

For varieties, the most naive notion of being regular in codimension one (at least to me!) is just to say that the set of singular points is subvariety of codimension at least two.

Is my naive definition of "regular in codim one" equivalent to the definition in Hartshorne?

(I ask this because my naive definition is easy to verify: for instance, a surface with only ADE singularities is clearly regular in codimension one by my naive definition - there is no need to do any commutative algebra, which I'm terrible at. But I do need to know if this singular variety satisfies the condition in Hartshorne, because having DVRs in codimension one allows me to define Weil divisors.)

  • $\begingroup$ Typo: "Weyl" should be "Weil". I would change it but the software won't allow me to. $\endgroup$ – Nefertiti Feb 17 '17 at 11:01
  • $\begingroup$ Also note that the assumption 'regular in codimension one' is not needed to define the notion of a Weil Divisor. It is actually needed to associate to any rational function a Weil Divisor, so it is needed to define the notion of a principal Weil Divisor. $\endgroup$ – MooS Feb 17 '17 at 13:39
  • $\begingroup$ @MooS I don't think you need regular in codimension 1 to associate a Weil divisor to a rational function: Fulton in section 1.2 of Intersection Theory gets around this by defining order using length. $\endgroup$ – Takumi Murayama Feb 17 '17 at 17:34
  • $\begingroup$ Yes, you are of course correct. $\endgroup$ – MooS Feb 17 '17 at 20:00

The answer is yes.

The Hartshorne definition means that the generic point of any irreducible closed subset of codimension one is a regular point.

Since the local ring at the generic point can be obtained from the local ring at any other point by localizing (and localizing a regular local ring gives you a regular local ring again) this is equivalent to say that any irreducible closed subset of codimension one admits at least one regular point.

The latter is of course equivalent of saying that the singular locus (assuming we have already shown it is closed) has at least codimension two.

  • $\begingroup$ Thank you very much! That is really helpful. $\endgroup$ – Kenny Wong Feb 17 '17 at 11:10
  • $\begingroup$ Can I ask you a related question? Suppose my variety is singular, but regular in codimension one. I'm confused about how to construct the canonical divisor. How about this: I concentrate on the open subvariety of smooth points, and look at the canonical bundle on this open subvariety, which is locally free, hence a Cartier divisor, hence gives a Weil divisor on this open subvariety. Then I take the closure of this Weil divisor in the entire variety. Notice that the thing I've constructed, from the point of view of the whole variety, is a Weil divisor but NOT Cartier. Is this correct? $\endgroup$ – Kenny Wong Feb 17 '17 at 11:15
  • $\begingroup$ Yes. If I am not mistaken, it should be Cartier if the variety is Gorenstein. $\endgroup$ – MooS Feb 17 '17 at 11:19
  • $\begingroup$ How do I tell if a variety is Gorenstein? Are ADE surface singularities Gorenstein? $\endgroup$ – Kenny Wong Feb 17 '17 at 11:21
  • $\begingroup$ For instance complete intersections are Gorenstein. I am not an expert on singularity theory, so I cannot answer your last question. $\endgroup$ – MooS Feb 17 '17 at 11:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.