I have asked this question in Math Stack exchange but got no answer. Maybe it fits Mathoverflow better.

All rings below are assumed to be Noetherian.

Let $E$ be an etale algebra over $\mathbb{Q}$. In other words, $E$ is a finite sum of number fields. Let $L$ be a lattice in $E$, and $R$ the order associated to $L$. More explicitly, $$R=\{ e\in E\mid eL\subset L\}.$$ Then is it true that $L$ is an invertible $R$-module?

I know this is true in several cases. For example, if $E$ is quadratic imaginary field (Serge Lang, Elliptic functions, Theorem 9, chaper 8), or if $R$ is the maximal order in $E$ (Then $R$ is a product of Dedkind Domains). On a somewhat related note, if we take an eichler order $R$ in a quaternion algebra $B$, and a lattice $L$ in $B$ whose associated left order coincides with $R$, then $L_p$ is in fact a principal $R_p$-module for all prime $p$ (here subscript $p$ means tensoring with $\mathbb{Z}_p$).

More generally, I am wondering if the following is true. Let $R$ be a 1-dimensional Cohen-Macaulay ring with total fractional ring $K(R)$. Let $I\subset K(R)$ be a fractional ideal (i.e., a finitely generated $R$-module) containing a unit of $E$ (and thus containing a nonzero-divisor of $R$). Is it true that the condition $\{ e\in K(R)\mid e I\subset I\}=R$ implies that $I$ is an invertible $R$-module? Note that since $I$ contains a nonzero-divisor, $\{ e\in K(R)\mid e I\subset I\}=End_R(I)$.

I have check the ring $k[x,y]/(x^2)$ and the answer is positive (In fact, for this ring all fractional ideals satisfying the assumption must be principal).

Of course the dimension=1 assumption is crucial. Otherwise, one may take a UFD of dimension $\geq 2$ (UFD of dim=1 are PIDs), and a proper ideal $I=(x,y)$ generated by two primes. Then $I^{-1}:=\{e\in K(R)\mid e I \subseteq R\}=R$. But clearly $I^{-1}$ contains the associated order. So the associate order of $I$ must be $R$ as well. Now it is easy to produce examples such that $I$ is not invertible.