of $R$). If $R$ has $\depth R \ge 1$, then every $\m$-primary ideal is a fractional ideal of $R$. For a fractional ideal $I$ and $J$, we can naturally identify $\Hom_R(I,J)$ with the set $J:I=\{a\in Q(R)\mid aI\subseteq J\}$. In this way, the endomorphism ring $\End_R(\m)$ of $\m$ is identified with the $R$-subalgebra $\m:\m$ of $Q(R)$. Here we note that the extension $R\subseteq \m:\m$ is module-finite and hence $\m:\m$ is commutative semilocal ring.

We give the following well-known lemma in order to use in the proof of Theorem \[thA\].

\[lem21\] Let $(R,\m)$ be a Cohen-Macaulay local ring of dimension one. Assume $R$ is not ，. Then

-   $R\subsetneq\m:\m=R:\m$

-   $\ell_R(\m:\m/R)$ is equal to the Cohen-Macaulay type $r(R)$ of $R$.

-   There exists a short exact sequence $$0 \to \m^{\oplus t(R)} \to R^{\oplus r(R)+1} \to \m:\m \to 0.$$ In particular, $\mu_R(\m:\m)=r(R)+1$ and $\syz_R(\m:\m)=\m^{\oplus r(R)}$.

\[lem22\] Let $(S,\mathfrak{n}) \subsetneq (R,\m)$ be a module-finite birational extension of one-dimensional local rings. Assume $R$ is reflexive as an $S$-module. Then we have birational extensions $S\subsetneq \mathfrak{n}:\mathfrak{n} \subseteq R$.

Note that $S$ is not a ，, and so $S$ is properly contained in $\mathfrak{n}:\mathfrak{n}$ (Lemma \[lem21\]). By the assumption, $R=S:(S:R)$. Since $R$ has constant rank one over 