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

Suppose that $X$ is a reduced rigid space and $\scr{F}$ is a coherent sheaf on $X$. For a section $f\in {\scr F}(X)$, the zero locus of $f$ is the set of points $x\in X$ at which $f$ vanishes in the fiber ${\scr F}(x) = {\scr F}\otimes_{{\scr O}_X}k(x)$.

If ${\scr F}$ is locally free, then the zero locus of a section is an analytic set (by which I mean the zero locus of a coherent ideal sheaf). In general, this is quite false (consider a sky-scraper sheaf at a point).

Here a probably too general question:

Which sets are the zero loci of such sections?

Let's call ${\scr F}$ torsion-free if it is without torsion by non-zero-divisors in the structure sheaf. Equivalently, if the natural map ${\scr F}\to {\scr F}\otimes_{{\scr O}_X} {\scr M}_X$ is injective, where ${\scr M}_X$ is the sheaf of meromorphic functions on $X$.

What if ${\scr F}$ is torsion-free? Are the zero loci analytic in this generality?

Though I've phrased the problem in the context of rigid spaces, there are obvious analogues for schemes and complex analytic spaces. I'd welcome comments in any of these contexts.

share|improve this question
6  
Even if F is torsion-free, the zero locus of a section need not be closed. Example: X=plane, F=ideal sheaf of the origin, f=linear function (vanishing at the origin). I doubt you can say more than constructibility. –  t3suji Jan 24 '11 at 20:07
    
Neat example, t3suji ! –  Georges Elencwajg Jan 24 '11 at 20:57
    
Nice example indeed, t3suji! I guess the local go-to example of a non-fat but torsion-free module would have been a good place to start :) –  Ramsey Jan 24 '11 at 21:32

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.