$S$ and $R$ is not isomorphic to $S$, one has $R=\mathfrak{n}:(\mathfrak{n}:R)$. Therefore, $(\mathfrak{n}:\mathfrak{n})\subseteq(\mathfrak{n}:\mathfrak{n})R\subseteq R$.

Now we can explain the proof of the direction \[thA\] (1) $\implies$ (2) $\implies$ (3) $\iff$ (4) $\iff$ (5) of Theorem \[thA\].

\(1) $\implies$ (2): Assume $bg(R)\leq 1$. If $bg(R)=0$, then $R$ is Gorenstein, and there is nothing to prove. We may assume $bg(R)=1$. Then there is a Gorenstein local ring $(S,\mathfrak{n})$ and module-finite residually rational birational extension $S\subsetneq R$ with $\ell_S(R/S)=1$. In particular, $R$ is Cohen-Macaulay. By the previous lemma, we have $S\subsetneq \mathfrak{n}:\mathfrak{n}\subseteq R$. Therefore, it should follows that $\ell_S(R/\mathfrak{n}:\mathfrak{n})=0$, in other words, $R=\mathfrak{n}:\mathfrak{n}=\mathfrak{n}:S$.

\(2) $\Rightarrow$ (3): We may assume $S$ is not a ， (otherwise $R\cong S$ and hence $R$ is Gorenstein). Identify $R$ with $\mathfrak{n}:\mathfrak{n}$. By Lemma \[lem21\], one has $\ell_S(R/S)=1$. Hence we have that the colength $\ell_S(\m/\mathfrak{n})$ of the inclusion $\mathfrak{n}\subseteq \m$ is less than or equal to $1$. it is easy to check that $\m/\mathfrak{n}$ is an $R$-module and it has dimension one as a vector space over $R/\m$. Fix a preimage $t\in R$ of a basis $\overline{t}$ of $\m/\mathfrak{n}$. Then $\m=\mathfrak{n}+Rt$ and $\m^2=\mathfrak{n}^2+\m t\subseteq \mathfrak{n}$. This means $\m\subseteq S:\m$. We have another inclusion $$S:\m\subseteq S:\mathfrak{n}=R.$$ Since $Rt\not\subset S$, it 