$$\overline{R}=R/(t_1,\ldots,t_d), \overline{N}=N/(t_1,\ldots,t_d)N,$$ then there is an isomorphism $$Ext^d_R(M,N) \cong Hom_{\overline{R}}(M,\overline{N}).$$

This is well-known. The first part of the statement is essentially [@brunsherzog], 1.2.10 and the rest follows by induction from the long exact Ext-sequence, obtained by dividing out an element.

\[pairingfamily\] An isomorphism $\phi: R \stackrel{\cong}{{\longrightarrow}} \omega_{R/S}$ defines a bilinear pairinng $$B_{\phi}: M_S \times M_S {\longrightarrow}S$$ This pairing is non-degenerate in the sense that the adjoint map $$M_S \cong Hom_S(M_S,S)$$ is an isomorphism of $S$-modules.

From \[isoextrelativ\] there is is an isomorphism $M_S \cong Ext^n_R(M_S,R)$. We take a maximal regular sequence $t_1,\ldots,t_d$ in the annihilator of $M_S$. We divide out these elements and obtain a factor ring $\overline{R}$ of $R$ and applying \[exthom\] we get an isomorphism $Ext^n_R(M_S,R)= Hom_{\overline{R}}(M_S,\overline{R})$. As $M_S$ is a finite $S$-module, the ring $\overline{R}$ is finite over $S$. From the isomorphism $\phi: R \stackrel{\cong}{{\longrightarrow}} \omega_{R/S}$ we obtain by dividing out $t_1,\ldots,t_d$ an isomorphism $\overline{R} \cong \omega_{\overline{R}/S}$. Duality for the finite map $S {\longrightarrow}\overline{S}$ tells us $\omega_{\overline{R}/S}=Hom_S(\overline{R},S)$. From the change-of-rings isomorphism we get $$Hom_{\overline{R}}\left(M_S,Hom_S\left(\overline{R},S\right)\right) \cong Hom_{S}(M_S,S).$$ Combining these isomorphisms we obtain $$Ext_R^n(M_S,R)\cong Hom_{\overline{R}}(M_S,\overline{R})\cong Hom_S(M_S,S)$$ Hence, in total we obtain a natural isomorphism $M_S\cong Hom_S\left(M_S,S\right)$, which can be seen as a family of non-degenerate pairings $B_{\phi}: M_S\times M_S{\longrightarrow}S.$

