I recently came across a remark about Cartier divisors in a textbook on algebraic geometry. I'm not sure how to interpret the remark. I've attached the previous paragraph as well for context. enter image description here

The phrasing I am confused about is "thus the function $f$ may be finite over a point". What exactly is meant by this, and why does it follow from the previous sentence about $f$ being integral? Does this mean finite as a morphism of schemes? A morphism to where though, the spectrum of the residue field?

For reference, this is from page 139 of Mumford & Oda. A draft electronic copy is available here (on Mumford's web page) in which the remark appears on page 109.

  • $\begingroup$ If I understand correctly, "function" in this context means "rational function", i.e. it might have singularities. "Function is a morphism on U" means that it has no singularities on the given open set U (this is somewhat similar to distinction between birational map and birational morphism). In principle, you can think of these "morphisms" as morphisms to the affine line over integers (stacks.math.columbia.edu/tag/01JH). $\endgroup$ – cardinal Mar 29 at 12:32
  • $\begingroup$ So f is an element of the ring of rational functions. To say that f is integral over O, the subring of functions without singularities, means that it satisfies a polynomial equation with coefficients in O. This is equivalent to O[f] being a finite O-module (the distinction between integral and finite extensions arises only when your extension is infinitely generated stacks.math.columbia.edu/tag/02JJ). To say that f is finite over a point means that its image in the stalk of the point satisfies a polynomial equation over the stalk of the structure sheaf... $\endgroup$ – cardinal Mar 29 at 12:43
  • $\begingroup$ ...so "f is integral over O" implies "f is finite over a point". If you take the affine irreducible complex curve $y^2=x^3$, then in the stalk of the singular point we have a rational function $x/y$ which satisfies a polynomial equation $(y/x)^2-x=0$ but it is not a function without singularities itself. $\endgroup$ – cardinal Mar 29 at 12:43
  • $\begingroup$ Note that the sheaf of rational functions on an arbitrary scheme is a pretty confusing object. See the expository note "Misconceptions about $K_X$" for some clarification. $\endgroup$ – cardinal Mar 29 at 12:45

Your Answer

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

Browse other questions tagged or ask your own question.