Divisor group of a integral scheme of dimension one

divisors-algebraic-geometryschemes

In Hartshorne's book, for an scheme $X$ (noetherian integral separated and regular in codimension one) defined $Div(X)$ as the free abelian group generated by the prime divisors (closed integral subscheme of codimension one).

My doubt is if the definition in the Silverman`s book are the same for a algebraic curve.
Silverman defined $Div(X)$ as the free abelian group generated by the points of $X$.

In definition of Hartshorne not include the generic point true?

Do these groups differ?

Thanks you all.

Best Answer

If $X$ is an algebraic curve then $X$ is a Noetherian, integral, separated scheme of finite type over a field (AKA a variety) and it's of dimension 1 (curves are $1$-dimensional).

So if $X$ is $1$ dimensional, then a codimension one subvariety of $X$ is a dimension $0$ variety which is just a (closed) point.

And, in general if $X$ is a:

  • curve, then divisors are points
  • surface, then divisors are curves
  • 3-fold, then divisors are surfaces

Etc. We want divisors to be closed subschemes if that's what you're asking. I don't know exactly what Hartshorne says but it should just be a generalization of Silverman's definition to schemes higher dimension and not-necessarily over a field.