The Stacks project

Proof. Recall that as $X$ has an ample invertible sheaf, it is quasi-compact and separated (Properties, Definition 28.26.1 and Lemma 28.26.7). Thus we may apply Proposition 36.7.5 and represent $K$ by a complex $\mathcal{F}^\bullet$ of quasi-coherent modules. Pick any $p$ such that $\mathcal{H}^ p = \mathop{\mathrm{Ker}}(\mathcal{F}^ p \to \mathcal{F}^{p + 1})/ \mathop{\mathrm{Im}}(\mathcal{F}^{p - 1} \to \mathcal{F}^ p)$ is nonzero. Choose a point $x \in X$ such that the stalk $\mathcal{H}^ p_ x$ is nonzero. Choose an $n \geq 0$ and $s \in \Gamma (X, \mathcal{L}^{\otimes n})$ such that $X_ s$ is an affine open neighbourhood of $x$. Choose $\tau \in \mathcal{H}^ p(X_ s)$ which maps to a nonzero element of the stalk $\mathcal{H}^ p_ x$; this is possible as $\mathcal{H}^ p$ is quasi-coherent and $X_ s$ is affine. Since taking sections over $X_ s$ is an exact functor on quasi-coherent modules, we can find a section $\tau ' \in \mathcal{F}^ p(X_ s)$ mapping to zero in $\mathcal{F}^{p + 1}(X_ s)$ and mapping to $\tau$ in $\mathcal{H}^ p(X_ s)$. By Properties, Lemma 28.17.2 there exists an $m$ such that $\tau ' \otimes s^{\otimes m}$ is the image of a section $\tau '' \in \Gamma (X, \mathcal{F}^ p \otimes \mathcal{L}^{\otimes mn})$. Applying the same lemma once more, we find $l \geq 0$ such that $\tau '' \otimes s^{\otimes l}$ maps to zero in $\mathcal{F}^{p + 1} \otimes \mathcal{L}^{\otimes (m + l)n}$. Then $\tau ''$ gives a nonzero class in $H^ p(X, K \otimes ^\mathbf {L}_{\mathcal{O}_ X} \mathcal{L}^{(m + l)n})$ as desired. $\square$

Proof. Recall that $\mathbf{P}^ n_ A$ is $\text{Proj}(A[X_0, \ldots , X_ n])$, see Constructions, Definition 27.13.2. Consider the Koszul complex

over $S = A[X_0, \ldots , X_ n]$ on $X_0, \ldots , X_ n$. Since $X_0, \ldots , X_ n$ is clearly a regular sequence in the polynomial ring $S$, we see that (More on Algebra, Lemma 15.30.2) that the Koszul complex $K_\bullet$ is exact, except in degree $0$ where the cohomology is $S/(X_0, \ldots , X_ n)$. Note that $K_\bullet$ becomes a complex of graded modules if we put the generators of $K_ i$ in degree $+i$. In other words an exact complex

Applying the exact functor $\tilde{\ }$ functor of Constructions, Lemma 27.8.4 and using that the last term is in the kernel of this functor, we obtain the exact complex

Twisting by the invertible sheaves $\mathcal{O}_ X(n + a)$ we get the exact complexes of the lemma. $\square$

Proof. Let $K \in D_\mathit{QCoh}(\mathcal{O}_ X)$. Assume $\mathop{\mathrm{Hom}}\nolimits (E, K[p]) = 0$ for all $p \in \mathbf{Z}$. We have to show that $K = 0$. By Derived Categories, Lemma 13.36.4 we see that $\mathop{\mathrm{Hom}}\nolimits (E', K[p])$ is zero for all $E' \in \langle E \rangle$ and $p \in \mathbf{Z}$. By Lemma 36.16.2 applied with $a = -n - 1$ we see that $\mathcal{O}_ X(-n - 1) \in \langle E \rangle$ because it is quasi-isomorphic to a finite complex whose terms are finite direct sums of summands of $E$. Repeating the argument with $a = -n - 2$ we see that $\mathcal{O}_ X(-n - 2) \in \langle E \rangle$. Arguing by induction we find that $\mathcal{O}_ X(-m) \in \langle E \rangle$ for all $m \geq 0$. Since

we conclude that $K = 0$ by Lemma 36.16.1. (This also uses that $\mathcal{O}_ X(1)$ is an ample invertible sheaf on $X$ which follows from Properties, Lemma 28.26.12.) $\square$