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I'm reading the paper: SGA 7 II, Intersections sur les surfaces regulieres.

In Papge 6 , I cannot understand why there is sign $-1$ in the formula (1.10.4):

Let $S$ be a trait, for any $\mathcal O_S$-module $M$ concentrated at the closed point of $S$, then we have \begin{equation} (1.10.4)\quad\quad\quad\quad \chi(RHom(M,\mathcal O_S))=-1\times \chi(M) \end{equation}

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1 Answer 1

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For a module of finite length $M$ over a principal ideal domain $A$, we have $\mathrm{Hom}(M,A)=0$, and $\mathrm{Ext}^1(M,A)$ (the dual of $A$, see Bourbaki, Algebra VII, §4, No. 9) is non-canonically isomorphic to $M$. Therefore $\chi (\mathrm{RHom}(M,A))=-\mathrm{length}\,(\mathrm{Ext}^1(M,A))=-\chi (M)$.

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