A question about thinking modules as free Abelian groups First of all, I am aware that my question is not so clear. I am trying to understand the structure of number fields and their ring of integers.
Let $K$ be a number field and $\mathcal{O}_K$ is its ring of integers . Then, we can say these:

$K$ is a vector space over $\mathbb{Q}$, it has some finite basis consisting of elements of $K$. Actually, there is more. $K$can be considered as a $\mathbb{Q}$ vector space with basis elements belonging to $\mathcal{O}_K$.
$\mathcal{O}_K$ as a vector space over $\mathbb{Z}$ such that it has a basis $\{\alpha_1, \dots,\alpha_n\}$ where each $\alpha_i$ belong to $\mathcal{O}_K$.

My problem is that I do not know when should I think $K$ or $\mathcal{O}_K$ as a free Abelian group like $\mathcal{O}_K = \alpha_1\mathbb{Z}\oplus\dots\oplus\alpha_n\mathbb{Z}$ or as a finitely generated module.
Can someone please explain all the connections between $K$ / $\mathcal{O}_K$ and modules/free Abelian groups?
 A: To "explain all the connections between $K / O_K$ and modules/free Abelian groups", one could simply say that $K\cong O_K \otimes \mathbf Q$, but immediately warn that doing so "kills" all the arithmetic of $O_K$. Since you must be aware of this, I guess that your question rather concerns the differences, in the context of linear algebra, between the structures of vector spaces and free modules (what can be done in a vector space which cannot be done in a free module).
Throughout, $V$ (resp. $M$) will denote implicitly a finitely generated vector space (resp. module) over a commutative field (resp. ring) $F$ (resp. $R$). The central notion in linear algebra is that of linear (in)dependence. This requires to get rid of the "torsion submodule" $tM$ of $M$, i.e. of the elements $m\in M$ such that $r.m =0$ for a certain non null $r\in R$ (just think of a finite abelian group). So the ring $R$ should be a $domain$ (no non null divisors of zero), but even if $R=\mathbf Z$, one should also avoid finite modules. In the usual definition, an $R$-module $M$ is free iff it admits a basis over $R$, i.e. $M\cong R^n$ as $R$-modules for a certain $n$, which is its (well defined) rank. An $F$-vector space is automatically free (why ?), but this is not true for a general $R$-module $M$, even if $R$ is a domain and $M$ has no torsion. The most well known smooth result is that, if the ring $R$ is principal, then $M$ is free iff $M$ has no torsion (apart from $0$ of course). This applies to $O_K$ over $\mathbf Z$ (the "absolute" case), but for a general finite extension $L/K$ of number fields (the "relative" case), it does not apply to $O_L$ over $O_K$. Besides, even in the absolute case, one cannot a priori complete a given linearly independent system of elements of $O_K$ to get a $\mathbf Z$-basis of $O_K$(same "why" as before). This is the difficult problem of finding an integral basis, which is not the same thing as a basis of $K$ consisting of integral elements.
As noticed above, another reason not to consider $O_K$ simply as a module lies in the arithmetic of $O_K$, which is a ring, not merely an additive group. For example, consider the multiplicative group $U_K$ of units (=invertible elements) of $O_K$. It is a $\mathbf Z$-module (in multiplicative notation), its torsion is the (finite) group $\mu_K$ of roots of unity of $K$, but the determination of the rank of the free $\mathbf Z$-module $U_K /\mu_K$ is not a trivial matter (Dirichlet's unit theorem). 
