## 36.7 The coherator for Noetherian schemes

In the case of Noetherian schemes we can use the following lemma.

Lemma 36.7.1. Let $X$ be a Noetherian scheme. Let $\mathcal{J}$ be an injective object of $\mathit{QCoh}(\mathcal{O}_ X)$. Then $\mathcal{J}$ is a flasque sheaf of $\mathcal{O}_ X$-modules.

Proof. Let $U \subset X$ be an open subset and let $s \in \mathcal{J}(U)$ be a section. Let $\mathcal{I} \subset X$ be the quasi-coherent sheaf of ideals defining the reduced induced scheme structure on $X \setminus U$ (see Schemes, Definition 26.12.5). By Cohomology of Schemes, Lemma 30.10.4 the section $s$ corresponds to a map $\sigma : \mathcal{I}^ n \to \mathcal{J}$ for some $n$. As $\mathcal{J}$ is an injective object of $\mathit{QCoh}(\mathcal{O}_ X)$ we can extend $\sigma$ to a map $\tilde s : \mathcal{O}_ X \to \mathcal{J}$. Then $\tilde s$ corresponds to a global section of $\mathcal{J}$ restricting to $s$. $\square$

Lemma 36.7.2. Let $f : X \to Y$ be a morphism of Noetherian schemes. Then $f_*$ on quasi-coherent sheaves has a right derived extension $\Phi : D(\mathit{QCoh}(\mathcal{O}_ X)) \to D(\mathit{QCoh}(\mathcal{O}_ Y))$ such that the diagram

$\xymatrix{ D(\mathit{QCoh}(\mathcal{O}_ X)) \ar[d]_{\Phi } \ar[r] & D_\mathit{QCoh}(\mathcal{O}_ X) \ar[d]^{Rf_*} \\ D(\mathit{QCoh}(\mathcal{O}_ Y)) \ar[r] & D_\mathit{QCoh}(\mathcal{O}_ Y) }$

commutes.

Proof. Since $X$ and $Y$ are Noetherian schemes the morphism is quasi-compact and quasi-separated (see Properties, Lemma 28.5.4 and Schemes, Remark 26.21.18). Thus $f_*$ preserve quasi-coherence, see Schemes, Lemma 26.24.1. Next, let $K$ be an object of $D(\mathit{QCoh}(\mathcal{O}_ X))$. Since $\mathit{QCoh}(\mathcal{O}_ X)$ is a Grothendieck abelian category (Properties, Proposition 28.23.4), we can represent $K$ by a K-injective complex $\mathcal{I}^\bullet$ such that each $\mathcal{I}^ n$ is an injective object of $\mathit{QCoh}(\mathcal{O}_ X)$, see Injectives, Theorem 19.12.6. Thus we see that the functor $\Phi$ is defined by setting

$\Phi (K) = f_*\mathcal{I}^\bullet$

where the right hand side is viewed as an object of $D(\mathit{QCoh}(\mathcal{O}_ Y))$. To finish the proof of the lemma it suffices to show that the canonical map

$f_*\mathcal{I}^\bullet \longrightarrow Rf_*\mathcal{I}^\bullet$

is an isomorphism in $D(\mathcal{O}_ Y)$. To see this it suffices to prove the map induces an isomorphism on cohomology sheaves. Pick any $m \in \mathbf{Z}$. Let $N = N(X, Y, f)$ be as in Lemma 36.4.1. Consider the short exact sequence

$0 \to \sigma _{\geq m - N - 1}\mathcal{I}^\bullet \to \mathcal{I}^\bullet \to \sigma _{\leq m - N - 2}\mathcal{I}^\bullet \to 0$

of complexes of quasi-coherent sheaves on $X$. By Lemma 36.4.1 we see that the cohomology sheaves of $Rf_*\sigma _{\leq m - N - 2}\mathcal{I}^\bullet$ are zero in degrees $\geq m - 1$. Thus we see that $R^ mf_*\mathcal{I}^\bullet$ is isomorphic to $R^ mf_*\sigma _{\geq m - N - 1}\mathcal{I}^\bullet$. In other words, we may assume that $\mathcal{I}^\bullet$ is a bounded below complex of injective objects of $\mathit{QCoh}(\mathcal{O}_ X)$. This follows from Leray's acyclicity lemma (Derived Categories, Lemma 13.16.7) via Cohomology, Lemma 20.12.5 and Lemma 36.7.1. $\square$

Proposition 36.7.3. Let $X$ be a Noetherian scheme. Then the functor (36.3.0.1)

$D(\mathit{QCoh}(\mathcal{O}_ X)) \longrightarrow D_\mathit{QCoh}(\mathcal{O}_ X)$

is an equivalence with quasi-inverse given by $RQ_ X$.

Proof. This follows from Lemma 36.6.5 and Lemma 36.7.2. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

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.

All contributions are licensed under the GNU Free Documentation License.

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 09T1. Beware of the difference between the letter 'O' and the digit '0'.