Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $f:X\longrightarrow S$ be a morphism of schemes. What is the link between sheaf-sections of $O_X$ over an open set of $X$ and morphism-sections of $f$. Is there a kind of correspondence?

share|improve this question
    
Consider the case S = Spec k ... I suggest that you follow the advice of your username ;-) –  Kevin H. Lin Aug 31 '10 at 22:43
    
This is amusing, Kevin: in my answer (which I wrote before seeing your comment) I mention a case where $X$=Spec k... –  Georges Elencwajg Aug 31 '10 at 23:16
8  
Since $O_X$ has nothing to do with $S$, you're looking for the wrong dictionary. The correspondence is that sections of $O_X$ over an open $U$ correspond to sections of $\mathbf{A}^1_X \rightarrow X$ over $U$. In that respect, the two general concepts are related. To see that, I second Kevin's wise advice. Rather than ask the obvious follow-up with more general sheaves, I again refer back to your name. –  BCnrd Aug 31 '10 at 23:27
    
Let's take $f$ of relative dimension 1. In that case one sheaf-section correspond to one (positive) divisor, and, on the other hand, the schematic image of one morphism-section seems to come from one (positive) (relative? and usually more specific) divisor. (In other rel dim, a correspondence could involve one s-section and family of m-sections.) –  Workitout Sep 1 '10 at 10:48

5 Answers 5

Dear Workitout, of course I can't prove there is no link, but I'm rather pessimistic . Here is a fuzzy argument in support of my feeling.

The set of sections of $\mathcal O_X$ is never empty (after all it is a ring and so contains zero!) .But I would say that "in general" (in a non technical sense), $f:X\to S$ has an empty set of sections. For example if $X=Spec A$ is affine and $f:X\to S=Spec (\mathbb Z ) $ is the unique morphism, sections of $f$ correspond to ring morphisms $A\to \mathbb Z$, and my feeling is that there is no particular reason why they should exist: if $A$ is a field (say), certainly no ring morphism $A\to \mathbb Z$ exists .

share|improve this answer

The name "sections" of a sheaf comes from the old viewpoint of a sheaf as its espace étalé. That is, they are sections of the canonical map $\acute Et(\mathcal{O}_X)\to X$ (this is a continuous map, not a map of schemes).

share|improve this answer

In my answer here I indicated how sheaf sections are sections of maps. This doesn't involve the structure map, but clearly motivates the name, if that's what you were looking for.

A more detailed account is in "Sheaves in Geometry and Logic" (from p. 88) by MacLane/Moerdijk, and an even more detailed and intuition-emphasizing one in Goldblatt's "Topoi", online viewable under the link.

share|improve this answer

Sections to the morphism $X \to S$ are more or less $S$-rational points of $X$: if for example $S = Spec(R)$ and $X = Spec(R[x_1, \dots, x_n]/(f_1, \dots, f_m))$ with polynomials $f_1, \dots, f_m \in R[x_1, \dots, x_n]$, then sections to $X \to S$ correspond to points $(a_1, \dots, a_n) \in R^n$ with $f_i(a_1, \dots, a_n) = 0$ for all $i$.

On the other hand, the elements of $\mathcal{O}_X(U)$ can be seen as holomorphic functions on $U$.

So the one kind of sections can be seen as "points" of the geometric object, the others can be seen as "functions" on the geometric object.

share|improve this answer

At least in the case that $f:X\to S$ is a vector bundle the answer is given in Ex 5.18 Chapter 2 of "algebraic Geometry" by Hartshorne.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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