Let $(R,\mathfrak m)$ be a Cohen-Macaulay local ring of dimension $1$ admitting a canonical ideal $\omega\subseteq R$. Then $\omega$ contains a non-zero divisor. Let $Q(R)$ be the total ring of fractions of $R$. Since $\omega$ contains a non-zero divisor, we have an Isomorphism $(\omega:_{Q(R)}\omega)\cong \text{Hom}_R(\omega, \omega)\cong R$. Hence we have $(\omega:_{Q(R)}\omega)=qR$ for some $q\in Q(R)$.
My question is: Is it actually true that $(\omega:_{Q(R)}\omega)=R$ ?