Lemma 47.15.2. Let $A$ be a Noetherian ring. Let $K, L \in D_{\textit{Coh}}(A)$ and assume $L$ has finite injective dimension. Then $R\mathop{\mathrm{Hom}}\nolimits _ A(K, L)$ is in $D_{\textit{Coh}}(A)$.
Proof. Pick an integer $n$ and consider the distinguished triangle
\[ \tau _{\leq n}K \to K \to \tau _{\geq n + 1}K \to \tau _{\leq n}K[1] \]
see Derived Categories, Remark 13.12.4. Since $L$ has finite injective dimension we see that $R\mathop{\mathrm{Hom}}\nolimits _ A(\tau _{\geq n + 1}K, L)$ has vanishing cohomology in degrees $\geq c - n$ for some constant $c$. Hence, given $i$, we see that $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(K, L) \to \mathop{\mathrm{Ext}}\nolimits ^ i_ A(\tau _{\leq n}K, L)$ is an isomorphism for some $n \gg - i$. By Derived Categories of Schemes, Lemma 36.11.5 applied to $\tau _{\leq n}K$ and $L$ we see conclude that $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(K, L)$ is a finite $A$-module for all $i$. Hence $R\mathop{\mathrm{Hom}}\nolimits _ A(K, L)$ is indeed an object of $D_{\textit{Coh}}(A)$. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: