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}



A problem in intersection theorem
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 $\chi (\mathrm{RHom}(M,A))=-\mathrm{length}\,(\mathrm{Ext}^1(M,A))=-\chi (M)$.


