We know that in vector spaces such a complementary substructure exists. But in this proof we took advantage of the fact that every vector space has a basis, ie. the subvectorspace has a basis and this basis then can be extended to a full basis of the whole vector space and then the basis vectors which are only in the extension create a basis for the desired complement.
In an $R$-module $M$, where $R$ is a principal ideal domain, we cannot use this argument with a basis, is it still possible to find a submodule $F$ such that every element of $M$ can be expressed as $x=f+t$ with $f\in F$ and $t\in T$ and that every such description is unique? If it is not true for the general case, how can it be true if $T$ is the torsion submodule of $M$?
It works for the torsion module primarily because a finitely generated module $M$ over a P.I.D. $R$, which is a noetherian ring, has a finite presentation, i.e. there exists an exact sequence $$F_1\longrightarrow F_0\longrightarrow M\longrightarrow 0,$$ where $F_0$ and $F_1$ are finitely generated free modules. The map $F_1\longrightarrow F_0$ is represented by a matrix, and in suitable bases for $F_0$ and $F_1$, this matrix has a Smith normal form: $$\begin{pmatrix} d_1&0&0&\dots&0&\dots&0\\ 0&d_2&0&\dots&0&\dots&0\ \\ 0&0&d_3&\dots&0 &\dots&0\\ \vdots&&&\ddots&&&\vdots \\ 0&0&0&&d_r&\dots&0\\ 0&0&0&\dots&0&\dots&0 \\ \vdots&&&&&\ddots&\vdots\\ 0&0&0&\dots&0&\dots&0 \end{pmatrix}$$ where $\;d_i\mid d_{i+1}$ for each $1\le i<r$. There results that the module $M$ is isomorphic to $$\underbrace{R/d_1R\times R/d_2R\times\dots\times R/d_rR}_{\text{torsion submodule}}\times R^{\text{rk}(F_0)-r}.$$