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Tag 0A9V

Chapter 35: Derived Categories of Schemes > Section 35.15: An example generator

Lemma 35.15.3. Let $A$ be a ring. Let $X = \mathbf{P}^n_A$. Then $$ E = \mathcal{O}_X \oplus \mathcal{O}_X(-1) \oplus \ldots \oplus \mathcal{O}_X(-n) $$ is a generator (Derived Categories, Definition 13.33.2) of $D_\textit{QCoh}(X)$.

Proof. Let $K \in D_\textit{QCoh}(\mathcal{O}_X)$. Assume $\mathop{\rm Hom}\nolimits(E, K[p]) = 0$ for all $p \in \mathbf{Z}$. We have to show that $K = 0$. By Derived Categories, Lemma 13.33.3 we see that $\mathop{\rm Hom}\nolimits(E', K[p])$ is zero for all $E' \in \langle E \rangle$ and $p \in \mathbf{Z}$. By Lemma 35.15.2 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 $$ \mathop{\rm Hom}\nolimits(\mathcal{O}_X(-m), K[p]) = H^p(X, K \otimes_{\mathcal{O}_X}^\mathbf{L} \mathcal{O}_X(m)) = H^p(X, K \otimes_{\mathcal{O}_X}^\mathbf{L} \mathcal{O}_X(1)^{\otimes m}) $$ we conclude that $K = 0$ by Lemma 35.15.1. (This also uses that $\mathcal{O}_X(1)$ is an ample invertible sheaf on $X$ which follows from Properties, Lemma 27.26.12.) $\square$

    The code snippet corresponding to this tag is a part of the file perfect.tex and is located in lines 3266–3276 (see updates for more information).

    \begin{lemma}
    \label{lemma-generator-P1}
    Let $A$ be a ring. Let $X = \mathbf{P}^n_A$. Then
    $$
    E =
    \mathcal{O}_X \oplus \mathcal{O}_X(-1) \oplus \ldots \oplus \mathcal{O}_X(-n)
    $$
    is a generator
    (Derived Categories, Definition \ref{derived-definition-generators})
    of $D_\QCoh(X)$.
    \end{lemma}
    
    \begin{proof}
    Let $K \in D_\QCoh(\mathcal{O}_X)$. Assume
    $\Hom(E, K[p]) = 0$ for all $p \in \mathbf{Z}$.
    We have to show that $K = 0$.
    By Derived Categories, Lemma
    \ref{derived-lemma-right-orthogonal}
    we see that $\Hom(E', K[p])$ is zero for all $E' \in \langle E \rangle$
    and $p \in \mathbf{Z}$.
    By Lemma \ref{lemma-construct-the-next-one}
    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
    $$
    \Hom(\mathcal{O}_X(-m), K[p]) =
    H^p(X, K \otimes_{\mathcal{O}_X}^\mathbf{L} \mathcal{O}_X(m)) =
    H^p(X, K \otimes_{\mathcal{O}_X}^\mathbf{L} \mathcal{O}_X(1)^{\otimes m})
    $$
    we conclude that $K = 0$ by Lemma \ref{lemma-nonzero-some-cohomology}.
    (This also uses that $\mathcal{O}_X(1)$ is an ample
    invertible sheaf on $X$ which follows from
    Properties, Lemma \ref{properties-lemma-open-in-proj-ample}.)
    \end{proof}

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