since $\m\cong \omega$. Now we deal with an assumption that $I\subsetneq \m$. The inequality $l(R/I)\le 2$ implies that the equality $l(\m/I)=1$. Thus the exact sequence $0\to I \to \m \to k \to 0$ is induced.

0

By assuming $S$ is complete, we can proof (2)$\implies$(3), using the theory of Auslander-Reiten sequence. We may also assume $S$ is not a ，. Now identify $R$ with the fractional ideal $\mathfrak{n}:\mathfrak{n}$. Then we see that $\mathfrak{n}=S:R\cong \Hom_S(R,S)$ and this module is a canonical module of $R$. Since $R$ has constant rank as an $S$-module, it is locally free on punctured spectrum of $S$. So one can take the Auslander-Reiten sequence $0 \to \tau(R) \to E_R \to R \to 0$ of the $S$-module $R$, see [@Yoshino Theorem 3.4]. The Auslander-Reiten transration $\tau(R)$ of $R$ over $S$ is isomorphic to $\Hom_S(\Omega_S \tr_S R,S)\cong \Omega_S R\cong \mathfrak{n}$ ([@Yoshino Proposition 3.11]). In particular, $E_R$ has constant rank $2$.

On the other hand, we have irreducible morphisms $\mathfrak{n} \to S$ and $\m \to R$, and their canonical duals $S\to R$ and $\mathfrak{n} \to \m^\dag$. Therefore, $S\oplus \m$ and $S\oplus \m^\dag$ are direct summands of $E_R$. Comparing the ranks of these modules, we have $S\oplus \m\cong E_R \cong S\oplus 