In Chapter III,$\S 4$ of Milne's Etale cohomology a correspondence between twisted forms and Cech cohomology cocycles is described.

Fix some Grothendieck topology, say, etale, and let $A$ be a sheaf of algebras over a scheme $X$. A sheaf of algebras $A'$ is called a twisted form of $A$ if there exists a cover $\mathscr{U}=(U_i \to X)$ such that $A \times_X U_i \cong A' \times_X U_i$. Milne then writes that to such an isomorphism one can associate a cocycle in $\check{H}^1(\mathscr{U}, \underline{Aut}(A))$ where $\underline{Aut}(A)$ is the sheaf associated to the presheaf of groups $Aut(U)=Aut_U(A \times_X U)$.

Why does one need to sheafify $Aut$? Isn't it true that $Aut$ is a sheaf already? It is clearly separated, because $A$ is separated, and I don't see why the second sheaf condition would fail, and I couldn't find a counterexample.


1 Answer 1


You are absolutely right, it is a sheaf, you can glue local automorphisms.

  • $\begingroup$ Thank you for confirming my suspicions. I was left confused after reading this passage from Milne's book. $\endgroup$ Nov 11, 2011 at 21:53

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