Lemma 47.15.7. Let $A$ be a Noetherian ring. Let $f_1, \ldots , f_ n \in A$ generate the unit ideal. If $\omega _ A^\bullet $ is a complex of $A$-modules such that $(\omega _ A^\bullet )_{f_ i}$ is a dualizing complex for $A_{f_ i}$ for all $i$, then $\omega _ A^\bullet $ is a dualizing complex for $A$.

**Proof.**
Consider the double complex

The associated total complex is quasi-isomorphic to $\omega _ A^\bullet $ for example by Descent, Remark 35.3.10 or by Derived Categories of Schemes, Lemma 36.9.4. By assumption the complexes $(\omega _ A^\bullet )_{f_ i}$ have finite injective dimension as complexes of $A_{f_ i}$-modules. This implies that each of the complexes $(\omega _ A^\bullet )_{f_{i_0} \ldots f_{i_ p}}$, $p > 0$ has finite injective dimension over $A_{f_{i_0} \ldots f_{i_ p}}$, see Lemma 47.3.8. This in turn implies that each of the complexes $(\omega _ A^\bullet )_{f_{i_0} \ldots f_{i_ p}}$, $p > 0$ has finite injective dimension over $A$, see Lemma 47.3.2. Hence $\omega _ A^\bullet $ has finite injective dimension as a complex of $A$-modules (as it can be represented by a complex endowed with a finite filtration whose graded parts have finite injective dimension). Since $H^ n(\omega _ A^\bullet )_{f_ i}$ is a finite $A_{f_ i}$ module for each $i$ we see that $H^ i(\omega _ A^\bullet )$ is a finite $A$-module, see Algebra, Lemma 10.23.2. Finally, the (derived) base change of the map $A \to R\mathop{\mathrm{Hom}}\nolimits _ A(\omega _ A^\bullet , \omega _ A^\bullet )$ to $A_{f_ i}$ is the map $A_{f_ i} \to R\mathop{\mathrm{Hom}}\nolimits _ A((\omega _ A^\bullet )_{f_ i}, (\omega _ A^\bullet )_{f_ i})$ by More on Algebra, Lemma 15.99.2. Hence we deduce that $A \to R\mathop{\mathrm{Hom}}\nolimits _ A(\omega _ A^\bullet , \omega _ A^\bullet )$ is an isomorphism and the proof is complete. $\square$

## 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Your email address will not be published.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.

## Comments (0)

There are also: