Lemma 48.20.2. In Situation 48.20.1 let $X$ be a scheme of finite type over $S$ and let $\mathcal{U}$ be a finite open covering of $X$ by schemes separated over $S$. If there exists a dualizing complex normalized relative to $\omega _ S^\bullet $ and $\mathcal{U}$, then it is unique up to unique isomorphism.
Proof. If $(K, \alpha _ i)$ and $(K', \alpha _ i')$ are two, then we consider $L = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(K, K')$. By Lemma 48.2.6 and its proof, this is an invertible object of $D(\mathcal{O}_ X)$. Using $\alpha _ i$ and $\alpha '_ i$ we obtain an isomorphism
\[ \alpha _ i^ t \otimes \alpha '_ i : L|_{U_ i} \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\omega _ i^\bullet , \omega _ i^\bullet ) = \mathcal{O}_{U_ i}[0] \]
This already implies that $L = H^0(L)[0]$ in $D(\mathcal{O}_ X)$. Moreover, $H^0(L)$ is an invertible sheaf with given trivializations on the opens $U_ i$ of $X$. Finally, the condition that $\alpha _ j|_{U_ i \cap U_ j} \circ \alpha _ i^{-1}|_{U_ i \cap U_ j}$ and $\alpha '_ j|_{U_ i \cap U_ j} \circ (\alpha '_ i)^{-1}|_{U_ i \cap U_ j}$ both give $\varphi _{ij}$ implies that the transition maps are $1$ and we get an isomorphism $H^0(L) = \mathcal{O}_ X$. $\square$
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Your email address will not be published.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)