F}),R) {\longrightarrow}H_0({\bf F}){\longrightarrow}Ext^{n-1}_R(H_1({\bf F}),R) {\longrightarrow}0$$

\[relativemodule\] In the above situation we put $$M_S:=Ext^n_R(H_0({\bf F}),R)$$

As $H_0({\bf F})=R/J_S$, we see as before from the exact sequence that there exists an ideal $I_S \subset R$ such that $M_S=I_S/J_S$. Completely analoguous to \[isoextij\] we have

\[isoextrelativ\] $$Ext^n_R(M_S,R)=M_S$$

We omit the proof, that is identical to that of \[isoextij\].

However, we want to understand this duality in terms of a [*family of pairings*]{}, parametrised by $S$. We can not just apply the local duality theorem, but rather we would have to use the duality statement for the morphism $Spec(R) {\longrightarrow}Spec(S)$.\
To treat this in an elementary way, we express higher $Ext$-groups $Ext_R^i(M,R)$ as groups $Hom_{\overline{R}}(M,\overline{R})$ of homomorphisms, where $\overline{R}$ arises from $R$ by dividing out suitable elements. In this way we can reduce to the case of a finite ring extension and use the duality for a finite map, a change-of-rings isomorphism. First we recall

\[exthom\] Let $R$ be a ring and let $M$, $N$ be two non-trivial finite $R$-modules. If $\textup{ann}(M)+\textup{ann}(N)=R$, $Ext_R^i(M,N)$ is zero for all $i\in{{\mathbb N}}$. Otherwise $$d=\textup{depth}(\textup{ann}(M),N)$$ is the smalles number $i$ with $Ext^i_R(M,N)$ not equal to zero. If $t_1,\ldots,t_d$ is a maximal regular $N$-sequence in $\textup{ann}(M)$ and we define 