Let $R$ be a noetherian commutative integrally closed domain whose field of fractions $K$ is a finite extension of the field of fractions $Q$ of $\Lambda = \mathbb{Z}_p[[T]]$. Let $L \subset R$ be a $\Lambda$-lattice of $K$. That is, $L$ is a $\Lambda$-module such that the $Q$-span of $L$ is all of $K$. I would like to show that $L$ necessarily contains a nonzero $R$-ideal.
This is what I have so far: We can define a subring of $R$ $$ R' = \{x \in R : xL \subseteq L\}. $$ Then $R'$ is also a $\Lambda$-lattice of $K$ and hence $R'$ is an order in $R$. Since $L$ is finitely generated, it follows that there is some nonzero $r \in R'$ such that $rL \subseteq R \cap L$. Thus, by replacing $L$ with $rL$ (which is still a lattice), we may assume $L \subseteq R'$. Then $L$ is an $R'$-ideal.
My hope is to somehow use the fact that $[R : R'] < \infty$ to show that $L$ must contain a (nonzero) $R$-ideal. The naive thing is to take a finite set $S$ of $R$ that generates $R$ over $R'$ and then consider $$ \bigcap_{s \in S} sL. $$ This is still a lattice (since $S$ is finite), but I think it will only be an $R$-ideal if $S$ happens to be multiplicatively closed. This also doesn't exploit the fact that $R'$ is actually an order in $R$, which I think I must use somehow. Ideas and suggestions are appreciated!