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

The Wikipedia article on coherent sheaves makes the following claim (without any reference), which I had trouble proving or finding a reference for: on an algebraic variety X (or I guess possibly even on a locally noetherian scheme), the coherent sheaves can be defined as the smallest class of sheaves of $\mathcal{O}_X$-modules with the following two properties:

i) the sheaf $\mathcal{O}_X$ is itself coherent;

ii) if, in a short exact sequence of sheaves, two of the sheaves are coherent, then so is the third.

I'm skeptical, but I would still like to know if this is true. If so, does anyone know a reference?

share|improve this question
1  
Wouldn't the first Chern class of every sheaf in that class be zero, since $c_1$ is additive over short exact sequences? –  MartinG May 20 '10 at 14:25
    
Oh wow, I feel silly. You should post that as an answer so I can accept it. –  Dan Petersen May 20 '10 at 14:43
    
It isn't silly, after all it was claimed at Wikipedia, and you were skeptical.. –  MartinG May 20 '10 at 18:21
    
I think if you replace (ii) with the requirement that the kernel and cokernel of any morphism between coherent sheaves is coherent, then it works. –  Sándor Kovács Jun 19 '11 at 2:34

1 Answer 1

up vote 7 down vote accepted

This is false. Every sheaf in that class would have zero first Chern class, since $c_1$ is additive over short exact sequences.

share|improve this answer
1  
Perhaps the original Wikipedia article meant to have "$\mathcal{O}_{Z}$ is coherent for every subvariety $Z\subseteq X$" in place of (i). Then the fact that the smallest class of sheaves satisfying (i) and (ii) is the category of coherent sheaves is an application of the usual dévissage-type argument. –  Mike Roth May 26 '10 at 3:20

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.