Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've got a question concerning sheaves created by global sections. Serre's vanishing theorem says:

Let $X$ be a projective scheme over a noetherian ring $A$ with a very ample invertible sheaf $\mathcal{O}_X(1)$ on $X$ over $\text{Spec}\ A$. Let $\mathcal{F}$ be a coherent sheaf. Then $H^i(X,F(n))=0$ for $i\ge 1$ and sufficiently large $n$.

There is another theorem by Serre saying that:

If $X$ is a projective scheme over a noetherian ring $A$, let $\mathcal{O}_X (1)$ denote a very ample invertible sheaf on $X$. Then there is a $d_0$ for each coherent sheaf $\mathcal{F}$ on $X$ so that $F(d)$ is generated by its global sections, whenever $d ≥ d_0$ .

Are these theorems related in the way that one of them follows from the other one? Especially: Is it true that for every sheaf created by global sections the higher cohomology vanishes?



share|cite|improve this question
up vote 4 down vote accepted

No, it is not true that the higher cohomology vanishes for any sheaf generated by its global sections. Consider the structure sheaf itself: for a smooth genus $g$ curve $C$, for example, $H^1(C, \mathcal{O}_C) = g$.

share|cite|improve this answer
Nice answer on the second question...should have known this myself... Any ideas on the first question? – Betti Meyer Nov 6 '12 at 12:21
QiL's answer provides the relevent reference, except for a typo: it should be Hartshorne, III, Prop. 5.3. (I can only comment here...) – Rhys Nov 7 '12 at 9:44
Thanks @Rhys for pointing out the typo! – user18119 Nov 7 '12 at 22:26

The first result implies the second one. You can find a proof in Hartshorne, III, Prop. 5.3 ((ii) implies (i)). Also the second result characterizes the ample sheaves (see the same reference).

share|cite|improve this answer

Your Answer


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.