There is a theorem in Neukirch's 'Algebraic number theory' which says
If $L | K$ is separable and $A$ is a principal ideal domain, then every finitely generated $B$-submodule $M \neq 0$ of $L$ is a free $A$-module of rank $[L:K]$
Where $K$ are the fractions of $A$ and $L$ is a finite extension of $K$ and $B$ are the integers of $A$ in $L$.
Recalling that $L = \{\frac{b}{a} \}_{b \in B, a \in A}$, the proof is given
Let $M \neq 0$ be a finitely generated $B$-submodule of $L$ and $(\alpha_i)_{1..n}$ a basis of $L|K$. Multiplying by an element of $A$, we can arrange for $a_i \in B$. Let $d =\text{disc}(\alpha_1,...,\alpha_n).$ By (2.9), $dB \subset A\alpha_1 + ... + A\alpha_n$, and in particular, $\text{rank}(B) \leq [L:K]$, and since a system of generators of the $A$-module $B$ is also a system of generators of the $K$-module $L$, $\text{rank}(B) = [L:K]$. Let $(\mu_i)_{1..r} \in M$ be a system of generators of the $B$-module $M$. There is an $a \neq 0 \in A$ such that $a \mu_i \in B$ so that $aM \subset B$. Then $$adM \subset dB \subset A\alpha_1 + ... + A\alpha_n = M_0$$ Since $M_0$ is a free $A$-module, so is $adM$, and hence also $M$.
Why $M$ is free because of $adM$ is free? Even if $M$ has torsion with order devising $ad$,$adM$ is free, but $M$ is not free..
P.S. Why is $M$ torsion free?