follows that $S:\m=\m$. The fractional ideal $S:\m$ is isomorphic to $\m^\dag$ and so we obtain $\m\cong\m^\dag$.

\(3) $\implies$ (4): Applying the functor $(-)^\dag$ to the short exact sequence $0\to \m \to R \to k \to 0$, we see that the resulting exact sequence is $0 \to \omega \to \m^\dag \to \Ext^1_R(k,\omega)\cong k \to 0$. Replacing $\m^\dag$ by $\m$, using the assumption $\m\cong \m^\dag$, we get the desired exact sequence.

\(4) $\implies$ (3): Applying the functor $(-)^\dag$ to the short exact sequence $0\to \omega \to \m \to k \to 0$, we get an exact sequence $0 \to \m^\dag \to R \to \Ext^1_R(k,\omega)\cong k \to 0$. Then, the image of $\m^\dag$ in $R$ must be equal to $\m$ and hence one has an isomorphism $\m^\dag\cong \m$.

\(4) $\implies$ (5): The exact sequence $0 \to \omega \to \m \to k \to 0$ yields that there is an ideal $I\cong \omega$ such that the colength $\ell_R(\m/I)$ is one. The equality $\ell_R(\m/I)=2$ immediately follows from the above.

\(5) $\implies$ (4): Take an ideal $I\cong \omega$ such that $l(R/I)\le 2$. If $I=R$, then $R$ is Gorenstein and there is nothing to prove. So we may suppose that $I\subseteq \m$. If $I=\m$, then $R$ must be ， 