**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.
$\square$

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