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Let $(R,m)$ be a Gorenstein local ring, $I\subset R$ a perfect ideal of grade $g$ and $S = R/I$. Prove that $S$ is Gorenstein iff $\operatorname{Tor}_g^R(S,S)=S$.

This question is Exercise 3.3.25(c) in the book of Winfried Bruns and Jürgen Herzog, Cohen-Macaulay Rings, Cambridge University Press, 1998.

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Suppose $S$ is Gorenstein. Then $\operatorname{Tor}^g _R(S,S)=\operatorname{Hom}_S(\omega_S, S)=\operatorname{Hom}_S(\omega_S, \omega_S)= S$.
Conversely, suppose $\operatorname{Tor}^g_R(S, S) = S$. Then $\omega^*_S=\operatorname{Hom}_S(\omega_S, S)=\operatorname{Tor}^g_R(S,S)=S$. Now in order to prove $S$ is Gorenstein you need to prove $r(S)=1$. This we do: by usual reductions in the section of canonical modules you can assume $\dim S=0$ and so $\omega_S=E(k)$. Let $r=r(S)=μ(E)=μ$. Then $$r=\dim_k \hom(k,\hom(E,S))=\dim_k \hom(k^μ,S)=\dim_kk^{r^2}=r^2,$$ so $r=1$.

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