Theorem 48.34.2. Let $X$ be a nonempty separated scheme of finite type over a field $k$. Let $d = \dim (X)$. The following are equivalent

1. $H^ d(X, \mathcal{F}) = 0$ for all coherent $\mathcal{O}_ X$-modules $\mathcal{F}$ on $X$,

2. $H^ d(X, \mathcal{F}) = 0$ for all quasi-coherent $\mathcal{O}_ X$-modules $\mathcal{F}$ on $X$, and

3. no irreducible component $X' \subset X$ of dimension $d$ is proper over $k$.

Proof. Assume there exists an irreducible component $X' \subset X$ (which we view as an integral closed subscheme) which is proper and has dimension $d$. Let $\omega _{X'}$ be a dualizing module of $X'$ over $k$, see Lemma 48.27.1. Then $H^ d(X', \omega _{X'})$ is nonzero as it is dual to $H^0(X', \mathcal{O}_{X'})$ by the lemma. Hence we see that $H^ d(X, \omega _{X'}) = H^ d(X', \omega _{X'})$ is nonzero and we conclude that (1) does not hold. In this way we see that (1) implies (3).

Let us prove that (3) implies (1). Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module such that $H^ d(X, \mathcal{F})$ is nonzero. Choose a filtration

$0 = \mathcal{F}_0 \subset \mathcal{F}_1 \subset \ldots \subset \mathcal{F}_ m = \mathcal{F}$

as in Cohomology of Schemes, Lemma 30.12.3. We obtain exact sequences

$H^ d(X, \mathcal{F}_ i) \to H^ d(X, \mathcal{F}_{i + 1}) \to H^ d(X, \mathcal{F}_{i + 1}/\mathcal{F}_ i)$

Thus for some $i \in \{ 1, \ldots , m\}$ we find that $H^ d(X, \mathcal{F}_{i + 1}/\mathcal{F}_ i)$ is nonzero. By our choice of the filtration this means that there exists an integral closed subscheme $Z \subset X$ and a nonzero coherent sheaf of ideals $\mathcal{I} \subset \mathcal{O}_ Z$ such that $H^ d(Z, \mathcal{I})$ is nonzero. By Lemma 48.34.1 we conclude $\dim (Z) = d$ and $Z$ is proper over $k$ contradicting (3). Hence (3) implies (1).

Finally, let us show that (1) and (2) are equivalent for any Noetherian scheme $X$. Namely, (2) trivially implies (1). On the other hand, assume (1) and let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Then we can write $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits \mathcal{F}_ i$ as the filtered colimit of its coherent submodules, see Properties, Lemma 28.22.3. Then we have $H^ d(X, \mathcal{F}) = \mathop{\mathrm{colim}}\nolimits H^ d(X, \mathcal{F}_ i) = 0$ by Cohomology, Lemma 20.19.1. Thus (2) is true. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0G5F. Beware of the difference between the letter 'O' and the digit '0'.