If $M$ is a $R$ Module which has a finite generating system and $T$ is a submodule of $M$, is there a submodule $F$ such that $M=F\oplus T$? 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 $M$? 
 A: 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
$$\underbrace{R/d_1R\times R/d_2R\times\dots\times R/d_rR}_{\text{torsion submodule}}\times R^{\text{rk}(F_0)-r}.$$
A: I presume that $R$ is always a PID here.

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?

No. Consider $R=M=\mathbb{Z}$ and $T=2\mathbb{Z}$ and note that $\mathbb{Z}$ is not decomposable as a group, let alone as a module (well, over $\mathbb{Z}$ it is the same thing).
In fact rings with this property (i.e. every submodule is a direct summand) are precisely semisimple rings and these are all well known by the Artin-Wedderburn theorem. As a consequence an integral domain is semisimple if and only if it is a field. And so we are back at vector spaces.

If it is not true for the general case, how can it be true if $T$ is the torsion submodule of $M$?

It is true. Since $M$ is finitely generated then so is $M/T$. Also $M/T$ is torsion free, which I leave as an exercise. Finally finitely generated torsion free modules over PID are free, in particular projective. Therefore
$$0\to T\to M\to M/T\to 0$$
splits which shows that $T$ is a direct summand of $M$.
