The Stacks project

Proof. By the Grothendieck's vanishing result in Cohomology, Proposition 20.20.7 we conclude that $\dim (U) \geq d$. Assume $\dim (U) = d$. Choose a compactification $U \to X$ such that $U$ is dense in $X$. (This is possible by More on Flatness, Theorem 38.33.8 and Lemma 38.32.2.) After replacing $X$ by its reduction we find that $X$ is a proper variety of dimension $d$ and we see that $U$ is proper if and only if $U = X$. Set $Z = X \setminus U$. We will show that $H^ d(U, \mathcal{F})$ is zero if $Z$ is nonempty.

Choose a coherent $\mathcal{O}_ X$-module $\mathcal{G}$ whose restriction to $U$ is $\mathcal{F}$, see Properties, Lemma 28.22.5. Let $\omega _ X^\bullet$ denote the dualizing complex of $X$ as in Section 48.27. Set $\omega _ U^\bullet = \omega _ X^\bullet |_ U$. Then $H^ d(U, \mathcal{F})$ is dual to

by Lemma 48.33.1. By Lemma 48.27.1 we see that the cohomology sheaves of $\omega _ X^\bullet$ vanish in degrees $< -d$ and $H^{-d}(\omega _ X^\bullet ) = \omega _ X$ is a coherent $\mathcal{O}_ X$-module which is $(S_2)$ and whose support is $X$. In particular, $\omega _ X$ is torsion free, see Divisors, Lemma 31.11.10. Thus we see that the cohomology sheaf

is torsion free, see Divisors, Lemma 31.11.12. Consequently this sheaf has no nonzero sections vanishing on any nonempty open of $X$ (those would be torsion sections). Thus it follows from Lemma 48.33.3 that $H^{-d}_ c(U, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ U}(\mathcal{F}, \omega _ U^\bullet ))$ is zero, and hence $H^ d(U, \mathcal{F})$ is zero as desired. $\square$

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

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

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. $\square$