The Stacks project

Lemma 59.62.1. Let $S$ be a scheme. Let $\mathcal{F}$ and $\mathcal{G}$ be finite locally free sheaves of $\mathcal{O}_ S$-modules of positive rank. If there exists an isomorphism $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ S}(\mathcal{F}, \mathcal{F}) \cong \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ S}(\mathcal{G}, \mathcal{G})$ of $\mathcal{O}_ S$-algebras, then there exists an invertible sheaf $\mathcal{L}$ on $S$ such that $\mathcal{F} \otimes _{\mathcal{O}_ S} \mathcal{L} \cong \mathcal{G}$ and such that this isomorphism induces the given isomorphism of endomorphism algebras.

Proof. Fix an isomorphism $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ S}(\mathcal{F}, \mathcal{F}) \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ S}(\mathcal{G}, \mathcal{G})$. Consider the sheaf $\mathcal{L} \subset \mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{F}, \mathcal{G})$ generated as an $\mathcal{O}_ S$-module by the local isomorphisms $\varphi : \mathcal{F} \to \mathcal{G}$ such that conjugation by $\varphi $ is the given isomorphism of endomorphism algebras. A local calculation (reducing to the case that $\mathcal{F}$ and $\mathcal{G}$ are finite free and $S$ is affine) shows that $\mathcal{L}$ is invertible. Another local calculation shows that the evaluation map

\[ \mathcal{F} \otimes _{\mathcal{O}_ S} \mathcal{L} \longrightarrow \mathcal{G} \]

is an isomorphism. $\square$

Comments (0)

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Your email address will not be published.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 0A2K. Beware of the difference between the letter 'O' and the digit '0'.