Lemma 48.2.3. Let $A$ be a Noetherian ring and let $X = \mathop{\mathrm{Spec}}(A)$. Let $K, L$ be objects of $D(A)$. If $K \in D_{\textit{Coh}}(A)$ and $L$ has finite injective dimension, then

in $D(\mathcal{O}_ X)$.

Lemma 48.2.3. Let $A$ be a Noetherian ring and let $X = \mathop{\mathrm{Spec}}(A)$. Let $K, L$ be objects of $D(A)$. If $K \in D_{\textit{Coh}}(A)$ and $L$ has finite injective dimension, then

\[ R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\widetilde{K}, \widetilde{L}) = \widetilde{R\mathop{\mathrm{Hom}}\nolimits _ A(K, L)} \]

in $D(\mathcal{O}_ X)$.

**Proof.**
We may assume that $L$ is given by a finite complex $I^\bullet $ of injective $A$-modules. By induction on the length of $I^\bullet $ and compatibility of the constructions with distinguished triangles, we reduce to the case that $L = I[0]$ where $I$ is an injective $A$-module. In this case, Derived Categories of Schemes, Lemma 36.10.8, tells us that the $n$th cohomology sheaf of $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\widetilde{K}, \widetilde{L})$ is the sheaf associated to the presheaf

\[ D(f) \longmapsto \mathop{\mathrm{Ext}}\nolimits ^ n_{A_ f}(K \otimes _ A A_ f, I \otimes _ A A_ f) \]

Since $A$ is Noetherian, the $A_ f$-module $I \otimes _ A A_ f$ is injective (Dualizing Complexes, Lemma 47.3.8). Hence we see that

\begin{align*} \mathop{\mathrm{Ext}}\nolimits ^ n_{A_ f}(K \otimes _ A A_ f, I \otimes _ A A_ f) & = \mathop{\mathrm{Hom}}\nolimits _{A_ f}(H^{-n}(K \otimes _ A A_ f), I \otimes _ A A_ f) \\ & = \mathop{\mathrm{Hom}}\nolimits _{A_ f}(H^{-n}(K) \otimes _ A A_ f, I \otimes _ A A_ f) \\ & = \mathop{\mathrm{Hom}}\nolimits _ A(H^{-n}(K), I) \otimes _ A A_ f \end{align*}

The last equality because $H^{-n}(K)$ is a finite $A$-module, see Algebra, Lemma 10.10.2. This proves that the canonical map

\[ \widetilde{R\mathop{\mathrm{Hom}}\nolimits _ A(K, L)} \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\widetilde{K}, \widetilde{L}) \]

is a quasi-isomorphism in this case and the proof is done. $\square$

Your email address will not be published.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.

## Comments (0)

There are also: