My question may be nestled in the sense that there might be confusion about several notions, please let me know if it is the case.
In "The Arithmetic of Hyperbolic 3-manifolds" by Maclachlan and Reid, the proof of the following Lemma is left as an Exercise:
Let $L$ be a complete Lattice in $V$ and $M$ an $R$-submodule of $V$. Then $M$ is a complete Lattice $\iff$ there exists $a \in R$ such that $aL \subset M \subset a^{-1}L$
Note: $R$ is a Dedekind Domain whose Field of Fraction $k$ is either a number field or a $\mathscr{P}$-adic Field. An $R$-Lattice is a finitely generated $R$-module contained in a vector space $V$ of finite dimension over $k$. Completeness of an $R$-Lattice $L$ is defined through $L \otimes_R k \cong V$.
My attempt:
$\implies$ since both are complete, we have $L \otimes_R k \cong M \otimes_R k$ as vector spaces. Therefore, for $l_i,m_i$ bases of $L,M$, we have that $l_i \otimes_R 1$ and $m_i \otimes_R 1$ are bases of their respective spaces and we can can $\alpha_j^i \in k$ such that: $$ l_i \otimes_R 1 = \sum_j \alpha_j^i \ m_j \otimes_R 1 $$ Since $k$ is the field of fraction, we can express each $\alpha_j^i = n_j^i/d_j^i$ with $n_j^i, d_j^i$ both in $R$.
At this point I get stuck. I figured I could use a method similar to the proof of the gaussian lemma (roughly speaking factoring out the denominators).
But I am falling short.
Have you tried a 2 by 2 example? The entries $\alpha^i_j$ form an invertible change of basis matrix. Mutiply both sides of the equation you wrote by the common denominator of all entries of the matrix, denoted by $a$, we have $$a \cdot l_i\otimes_R 1 = \sum_j (a \cdot \alpha^i_j ) m_j\otimes_R 1,$$ which implies $a\cdot L\subset M$. Invert your equation by the inverse of the matrix of $\alpha^i_j$, you will have $M \subset a^{-1}L$.